Communication method and apparatus

Abstract

The present application provides a communication method and apparatus for reducing the cost and complexity of baseband chips for processing low-frequency and high-frequency signals. The present application is applied to a wireless local area network system supporting a WLAN protocol, for example, supporting 802.11 series protocols such as IEEE 802.11be next-generation Wi-Fi protocol, for example, Wi-Fi 8. In the present application, the amount of subcarriers corresponding to a first channel in a first frequency band is greater than or equal to the amount of subcarriers corresponding to a second channel in a second frequency band, the ratio of the bandwidth of the first channel to the minimum channel bandwidth supported in the first frequency band is equal to the ratio of the bandwidth of the second channel to the minimum channel bandwidth supported in the second frequency band, and the minimum channel bandwidth supported in the first frequency band is greater than the minimum channel bandwidth supported in the second frequency band.

Description

[Technical field]

[0001] CROSS-REFERENCE TO RELATED APPLICATIONS This application claims priority to Chinese Patent Application No. 202210369283.1, entitled "COMMUNICATION METHOD AND APPARATUS," filed with the China State Intellectual Property Office on April 8, 2022, the entire contents of which are incorporated herein by reference.

[0002] The present application relates to the field of communication technologies, and in particular to communication methods and devices. [Background technology]

[0003] With the development of wireless local area networks (WLANs), wireless communication has become more and more popular. The standards (i.e., 802.11 protocol suite) established by the Institute of Electrical and Electronics Engineers (IEEE) for WLANs are constantly evolving. For example, protocol standards for low frequency bands include 802.11a / g, 802.11n, 802.11ac, 802.11ax, and 802.11be, and the supported frequency bands include one or more of 2.4GHz / 5GHz / 6GHz, and protocol standards for high frequency bands include 802.11aj / ay, and the supported frequency bands include one or more of 45GHz / 60GHz.

[0004] The frame structures of the PHY protocol data units (PPDUs) corresponding to the above different protocols are different. In particular, the frame structure of the PPDU corresponding to the protocol for the low frequency band is significantly different from the frame structure of the PPDU corresponding to the protocol for the high frequency band. Therefore, the method for processing the low frequency signal is also different from the method for processing the high frequency signal. In this case, the hardware for processing the signal (e.g., the baseband chip of the dedicated circuit) is also significantly different.

[0005] In view of this, baseband chips are sometimes designed for high-frequency signals and low-frequency signals, respectively, but the cost is high. It is complicated to design a converged baseband chip that can be used to process both low-frequency signals and high-frequency signals. Summary of the Invention [Means for solving the problem]

[0006] The present application provides a communication method and apparatus for reducing the cost and complexity of baseband chips for processing low and high frequency signals.

[0007] According to a first aspect, the present application provides a communication method. The method may be implemented by a transmitting device, which may be an access point, a station, a chip, or a circuit. The method includes generating a first orthogonal frequency division multiplexing (OFDM) signal and sending the first OFDM signal on a first channel in a first frequency band. An amount of subcarriers corresponding to the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band, a lowest frequency of the first frequency band is greater than a highest frequency of the second frequency band, and the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band. M The first basic channel includes 2 Mnumber of second fundamental channels, the bandwidth of the first fundamental channel being a minimum channel bandwidth supported in the first frequency band, the bandwidth of the second fundamental channel being a minimum channel bandwidth supported in the second frequency band, the bandwidth of the first fundamental channel being greater than the bandwidth of the second fundamental channel, and M being an integer greater than or equal to 0.

[0008] In this embodiment of the present application, the ratio of the bandwidth of the first channel to the minimum channel bandwidth supported in the first frequency band is equal to the ratio of the bandwidth of the second channel to the minimum channel bandwidth supported in the second frequency band, for example, both ratios are 2 M and the amount of subcarriers corresponding to the first channel is greater than or equal to the amount of subcarriers corresponding to the second channel in the second frequency band. In the present application, it can be seen that the correspondence between the bandwidth and the amount of subcarriers in the first frequency band is similar to the correspondence between the bandwidth and the amount of subcarriers in the second frequency band, so that the first frequency band and the second frequency band can share signal processing devices, for example, IDFT devices or DFT devices are reused. Thus, the device cost is low. The amount of subcarriers can also be understood as inverse discrete fourier transform (IDFT) size / discrete fourier transform (DFT) size.

[0009] According to a second aspect, the present application provides a communication method. The method may be implemented by a receiving device, which may be an access point, a station, a chip, or a circuit. The method includes receiving a first OFDM signal on a first channel in a first frequency band and processing a first OFDM symbol. An amount of subcarriers corresponding to the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band, a lowest frequency of the first frequency band is greater than a highest frequency of the second frequency band, and the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band. M The first basic channel includes 2 M number of second fundamental channels, the bandwidth of the first fundamental channel being a minimum channel bandwidth supported in the first frequency band, the bandwidth of the second fundamental channel being a minimum channel bandwidth supported in the second frequency band, the bandwidth of the first fundamental channel being greater than the bandwidth of the second fundamental channel, and M being an integer greater than or equal to 0.

[0010] In this embodiment of the present application, the correspondence between the bandwidth and the amount of subcarriers (or IDFT size / DFT size) in the second frequency band is reused in the first frequency band, so that the first frequency band and the second frequency band can share signal processing devices, for example, the IDFT device or the DFT device is reused. The device cost is low.

[0011] According to a third aspect, the present application further provides a communication device. The device may be a transmitting device or a chip in a transmitting device, and the transmitting device may be an access point or a station. The communication device has a function of performing any method provided in the first aspect. The communication device may be implemented by hardware, or by hardware executing corresponding software. The hardware or software includes one or more units or modules corresponding to the above functions.

[0012] In a possible design, the communication device includes a processor, the processor configured to support the communication device in performing the corresponding functions of the distance measurement initiator in the above method. The communication device may further include a memory. The memory may be coupled to the processor, the memory storing necessary program instructions and data for the communication device. Optionally, the communication device further includes an interface circuit. The interface circuit is configured to support communication between the communication device and a device such as a receiving device.

[0013] For example, a communications device may have the capability to implement the method provided in the first aspect, and the processor may be configured to generate a first OFDM signal. The interface circuit may be configured to send the first OFDM signal on a first channel in a first frequency band, where an amount of subcarriers corresponding to the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band, where a lowest frequency of the first frequency band is greater than a highest frequency of the second frequency band, and where the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band. M The first basic channel includes 2 M number of second fundamental channels, the bandwidth of the first fundamental channel being a minimum channel bandwidth supported in the first frequency band, the bandwidth of the second fundamental channel being a minimum channel bandwidth supported in the second frequency band, the bandwidth of the first fundamental channel being greater than the bandwidth of the second fundamental channel, and M being an integer greater than or equal to 0.

[0014] In a possible design, the communication device includes corresponding functional modules, each configured to perform a step in the above method. The functions may be implemented by hardware or by hardware executing corresponding software. The hardware or software includes one or more modules corresponding to the above functions.

[0015] In a possible design, the structure of the communication device includes a processing unit (or a processing module) and a transceiver unit (or a transceiver module). These units may implement corresponding functions in the above method examples. For details, please refer to the description in the method provided in the first aspect. Details will not be described again in this specification.

[0016] For example, a communications device may have the capability to implement the method provided in the first aspect, and the processing unit may be configured to generate a first OFDM signal. The transceiver unit may be configured to send the first OFDM signal on a first channel in a first frequency band, where an amount of subcarriers corresponding to the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band, where a lowest frequency of the first frequency band is greater than a highest frequency of the second frequency band, and where the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band. M The first basic channel includes 2 M number of second fundamental channels, the bandwidth of the first fundamental channel being a minimum channel bandwidth supported in the first frequency band, the bandwidth of the second fundamental channel being a minimum channel bandwidth supported in the second frequency band, the bandwidth of the first fundamental channel being greater than the bandwidth of the second fundamental channel, and M being an integer greater than or equal to 0.

[0017] According to a fourth aspect, the present application further provides a communication device. The device may be a receiving device or a chip in a receiving device, and the receiving device may be an access point or a station. The communication device has a function of performing any method provided in the second aspect. The communication device may be implemented by hardware, or by hardware executing corresponding software. The hardware or software includes one or more units or modules corresponding to the above functions.

[0018] In a possible design, the communication device includes a processor, the processor configured to support the communication device in performing the corresponding functions of the terminal device in the above method. The communication device may further include a memory. The memory may be coupled to the processor, the memory storing necessary program instructions and data for the communication device. Optionally, the communication device further includes an interface circuit. The interface circuit is configured to support communication between the communication device and a device such as a transmitting device.

[0019] For example, a communications device may have the capability to implement the method provided in the second aspect, and the interface circuit may be configured to receive a first OFDM signal on a first channel in a first frequency band, where an amount of subcarriers corresponding to the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band, where a lowest frequency of the first frequency band is greater than a highest frequency of the second frequency band, and where the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band. M The first basic channel includes 2 M number of second fundamental channels, a bandwidth of the first fundamental channel being a minimum channel bandwidth supported in the first frequency band, a bandwidth of the second fundamental channel being a minimum channel bandwidth supported in the second frequency band, the bandwidth of the first fundamental channel being greater than the bandwidth of the second fundamental channel, and M being an integer greater than or equal to 0. The processor may be configured to process the first OFDM symbol.

[0020] In a possible design, the communication device includes corresponding functional modules, each configured to perform a step in the above method. The functions may be implemented by hardware or by hardware executing corresponding software. The hardware or software includes one or more modules corresponding to the above functions.

[0021] In a possible design, the structure of the communication device includes a processing unit (or a processing module) and a transceiver unit (or a transceiver module). These units may implement corresponding functions in the above method examples. For details, please refer to the description in the method provided in the second aspect. Details will not be described again in this specification.

[0022] For example, a communications device may have the capability to implement the method provided in the second aspect, and the transceiver module may be configured to receive a first OFDM signal on a first channel in a first frequency band, where an amount of subcarriers corresponding to the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band, a lowest frequency of the first frequency band is greater than a highest frequency of the second frequency band, and the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band. M The first basic channel includes 2 M number of second fundamental channels, a bandwidth of the first fundamental channel being a minimum channel bandwidth supported in the first frequency band, a bandwidth of the second fundamental channel being a minimum channel bandwidth supported in the second frequency band, the bandwidth of the first fundamental channel being greater than the bandwidth of the second fundamental channel, and M being an integer greater than or equal to 0. The processor may be configured to process the first OFDM symbol.

[0023] Regarding the first to fourth aspects:

[0024] In a possible design, the subcarrier spacing of the first channel is greater than or equal to the threshold. In this manner, the effect of the frequency offset on the signal transmission on the first channel may be reduced, and the signal transmission performance may be improved.

[0025] In a possible design, the bandwidth of the first fundamental channel is N times the bandwidth of the second fundamental channel, where N is an integer greater than 1. In this manner, the signal in the first frequency band and the signal in the second frequency band can be transmitted by reusing the same crystal oscillator. This can reduce hardware complexity and reduce hardware overhead.

[0026] In a possible design, the bandwidth of the first fundamental channel is 320 MHz. In this manner, the spectrum resources in the first frequency band can be used as much as possible through finer channel division, which reduces resource waste and improves resource utilization.

[0027] In a possible design, the amount of subcarriers corresponding to the first channel may be: If the bandwidth of the first channel is 320 MHz, the first channel corresponds to 64 subcarriers; When the bandwidth of the first channel is 640 MHz, the first channel corresponds to 128 subcarriers; If the bandwidth of the first channel is 1280 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 2560 MHz, the first channel supports 512 subcarriers; or If the bandwidth of the first channel is 5120 MHz, the first channel can accommodate 1024 subcarriers. Satisfy one of the following:

[0028] In a possible design, the bandwidth of the first fundamental channel is 640 MHz.

[0029] In a possible design, the amount of subcarriers corresponding to the first channel may be: If the bandwidth of the first channel is 640 MHz, the first channel corresponds to 64 subcarriers; If the bandwidth of the first channel is 1280 MHz, the first channel corresponds to 128 subcarriers; If the bandwidth of the first channel is 2560 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 5120 MHz, the first channel supports 512 subcarriers; or If the bandwidth of the first channel is 10240 MHz, the first channel corresponds to 1024 subcarriers. Satisfy one of the following:

[0030] In a possible design, the bandwidth of the first fundamental channel is 80 MHz.

[0031] In a possible design, the amount of subcarriers corresponding to the first channel may be: If the bandwidth of the first channel is 80 MHz, the first channel corresponds to 64 subcarriers; If the bandwidth of the first channel is 160 MHz, the first channel corresponds to 128 subcarriers; If the bandwidth of the first channel is 320 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 640 MHz, the first channel supports 512 subcarriers; or If the bandwidth of the first channel is 1280 MHz, the first channel corresponds to 1024 subcarriers. Satisfy one of the following:

[0032] In a possible design, the bandwidth of the first fundamental channel is 160 MHz.

[0033] In a possible design, the amount of subcarriers corresponding to the first channel may be: If the bandwidth of the first channel is 160 MHz, the first channel corresponds to 64 subcarriers; If the bandwidth of the first channel is 320 MHz, the first channel corresponds to 128 subcarriers; If the bandwidth of the first channel is 640 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 1280 MHz, the first channel supports 512 subcarriers; or If the bandwidth of the first channel is 2560 MHz, the first channel can accommodate 1024 subcarriers. Satisfy one of the following:

[0034] In a possible design, the bandwidth of the first fundamental channel is obtained by performing R separate bisections on a frequency band obtained by dividing the first frequency band based on a minimum channel bandwidth supported in the millimeter-wave band, where R is an integer greater than or equal to 0. In this manner, signals in the first frequency band, signals in the second frequency band, and signals in the millimeter-wave band may be transmitted by reusing the same crystal oscillator.

[0035] In a possible design, the bandwidth of the first fundamental channel is 270 MHz. With finer granularity of channel division, the spectrum resources in the first frequency band can be used as much as possible. This reduces resource waste and improves resource utilization.

[0036] In a possible design, the amount of subcarriers corresponding to the first channel may be: If the bandwidth of the first channel is 270 MHz, the first channel corresponds to 64 subcarriers; If the bandwidth of the first channel is 540 MHz, the first channel corresponds to 128 subcarriers; If the bandwidth of the first channel is 1080 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 2160 MHz, the first channel supports 512 subcarriers; or If the bandwidth of the first channel is 4320 MHz, the first channel can accommodate 1024 subcarriers. Satisfy one of the following:

[0037] In a possible design, the bandwidth of the first fundamental channel is 540 MHz. With finer granularity of channel division, the spectrum resources in the first frequency band can be used as much as possible. This reduces resource waste and improves resource utilization.

[0038] In a possible design, the amount of subcarriers corresponding to the first channel may be: If the bandwidth of the first channel is 540 MHz, the first channel corresponds to 64 subcarriers; If the bandwidth of the first channel is 1080 MHz, the first channel corresponds to 128 subcarriers; If the bandwidth of the first channel is 2160 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 4320 MHz, the first channel supports 512 subcarriers; or If the bandwidth of the first channel is 8640 MHz, the first channel can accommodate 1024 subcarriers. Satisfy one of the following:

[0039] In a possible design, the bandwidth of the first fundamental channel is obtained by performing K bisections on the first subband, where the bandwidth of the first subband is smaller than a minimum channel bandwidth supported in the millimeter-wave band, and K is an integer greater than or equal to 0. A portion of the bandwidth is selected from the minimum channel bandwidth supported in the millimeter-wave band, such that inter-channel interference can be further reduced.

[0040] In a possible design, the bandwidth of the first fundamental channel is 250 MHz. By using finer-grained channel division, the spectrum resources in the first frequency band can be used as much as possible. This reduces resource waste and improves resource utilization.

[0041] In a possible design, the amount of subcarriers corresponding to the first channel may be: If the bandwidth of the first channel is 250 MHz, the first channel corresponds to 64 subcarriers; If the bandwidth of the first channel is 500 MHz, the first channel corresponds to 128 subcarriers; If the bandwidth of the first channel is 1000 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 2000 MHz, the first channel supports 512 subcarriers; or If the bandwidth of the first channel is 4000 MHz, the first channel can accommodate 1024 subcarriers. Satisfy one of the following:

[0042] In a possible design, the bandwidth of the first fundamental channel is 500 MHz. By using finer-grained channel division, the spectrum resources in the first frequency band can be used as much as possible. This reduces resource waste and improves resource utilization.

[0043] In a possible design, the amount of subcarriers corresponding to the first channel may be: If the bandwidth of the first channel is 500 MHz, the first channel corresponds to 64 subcarriers; If the bandwidth of the first channel is 1000 MHz, the first channel corresponds to 128 subcarriers; If the bandwidth of the first channel is 2000 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 4000 MHz, the first channel supports 512 subcarriers; or If the bandwidth of the first channel is 8000 MHz, the first channel can accommodate 1024 subcarriers. Satisfy one of the following:

[0044] In possible designs, the first frequency band is a 60 GHz frequency band, or the first frequency band is a 45 GHz frequency band, and the second frequency band is a 2.45 GHz frequency band, or the second frequency band is a 6 GHz frequency band.

[0045] According to a fifth aspect, a communication device is provided. The communication device includes a processor and an interface circuit. The interface circuit is configured to receive a signal from a communication device other than the communication device and transmit the signal to the processor or transmit a signal from the processor to the communication device other than the communication device. The processor is configured to implement the method of the first aspect and any one of the possible designs of the first aspect by logic circuits or by executing code instructions.

[0046] According to a sixth aspect, a communication device is provided. The communication device includes a processor and an interface circuit. The interface circuit is configured to receive a signal from a communication device other than the communication device and transmit the signal to the processor or transmit a signal from the processor to the communication device other than the communication device. The processor is configured to implement the method of the second aspect and any one of the possible designs of the second aspect by logic circuits or by executing code instructions.

[0047] According to a seventh aspect, a computer-readable storage medium is provided. The computer-readable storage medium stores a computer program or instruction. When the computer program or instruction is executed by a processor, the method of the first aspect or the second aspect and any one of the possible designs of the first aspect or the second aspect is performed.

[0048] According to an eighth aspect, there is provided a computer program product storing instructions which, when executed by a processor, perform the method of the first aspect or the second aspect and any one of possible designs of the first aspect or the second aspect.

[0049] According to a ninth aspect, a chip system is provided. The chip system includes a processor and may further include a memory configured to implement the method in the first aspect and possible designs of the first aspect. The chip system may include a chip, or may include a chip and another discrete device.

[0050] According to a tenth aspect, a chip system is provided. The chip system includes a processor and may further include a memory configured to implement the method in the second aspect and possible designs of the second aspect. The chip system may include a chip, or may include a chip and another discrete device.

[0051] According to an eleventh aspect, there is provided a communication system, the communication system including an apparatus according to the first aspect (eg, a transmitting device) and / or an apparatus according to the second aspect (eg, a receiving device). [Brief description of the drawings]

[0052] [Figure 1] FIG. 1 is a diagram of the architecture of a communication system according to an embodiment of the present application. [Diagram 2] FIG. 2 is a diagram of the structure of a DFT-s-OFDM transmitter according to an embodiment of the present application. [Diagram 3]1 is a schematic flowchart of a communication method according to an embodiment of the present application; [Figure 4] FIG. 2 is a diagram of channel splitting according to an embodiment of the present application. [Diagram 5] FIG. 2 is a diagram of channel splitting according to an embodiment of the present application. [Figure 6] FIG. 2 is a diagram of channel splitting according to an embodiment of the present application. [Figure 7] FIG. 2 is a diagram of channel splitting according to an embodiment of the present application. [Figure 8] FIG. 2 is a diagram of channel splitting according to an embodiment of the present application. [Figure 9] FIG. 2 is a diagram of channel splitting according to an embodiment of the present application. [Figure 10] FIG. 2 is a diagram of channel splitting according to an embodiment of the present application. [Figure 11] FIG. 2 is a diagram of channel splitting according to an embodiment of the present application. [Figure 12] FIG. 2 is a diagram of the structure of a communication device according to an embodiment of the present application; [Figure 13] FIG. 2 is a diagram of the structure of a communication device according to an embodiment of the present application; DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENTS

[0053] Hereinafter, embodiments of the present application will be described in detail with reference to the accompanying drawings.

[0054] With the development of wireless local area networks (WLANs), wireless communication has become more and more popular. The standards (i.e., 802.11 protocol suite) established by the Institute of Electrical and Electronics Engineers (IEEE) for WLANs are constantly evolving. For example, protocol standards for low frequency bands include 802.11a / g, 802.11n, 802.11ac, 802.11ax, and 802.11be, and the supported frequency bands include one or more of 2.4GHz / 5GHz / 6GHz, and protocol standards for high frequency bands include 802.11aj / ay, and the supported frequency bands include one or more of 45GHz / 60GHz.

[0055] Based on the principle of forward compatibility, the next generation WLAN technology supports both high and low frequency bands and is compatible with WLAN technology (such as 802.11n / 802.11ac / 802.11ax / 802.11be) on the low frequency band. However, the frame structures of the physical layer protocol data units (PHY protocol data units, PPDUs) corresponding to the above different protocols are different. In particular, the frame structure of the PPDU corresponding to the protocol for the low frequency band is significantly different from the frame structure of the PPDU corresponding to the protocol for the high frequency band. Therefore, the method for processing the low frequency signal is also different from the method for processing the high frequency signal. In this case, the hardware for processing the signal (e.g., the baseband chip of the dedicated circuit) is also significantly different. Regarding this issue, the baseband chip may be designed for high frequency signals and low frequency signals respectively, but the cost is high. It is complicated to design a converged baseband chip that can be used to process both low frequency signals and high frequency signals.

[0056] In view of this, the embodiments of the present application provide a communication method and apparatus for reducing the cost and complexity of baseband chips for processing low-frequency signals and high-frequency signals. The method and the apparatus are based on the same concept. The method and the apparatus have similar principles for solving problems. Therefore, the implementation forms of the apparatus and the method should refer to each other. No repeated description is provided.

[0057] The technical solutions provided in the present application are applicable to WLAN systems, for example, IEEE 802.11 system standards, for example, 802.11a / b / g standard, 802.11n standard, 802.11ac standard, 802.11ax standard, or a next-generation standard of 802.11ax standard, for example, 802.11be standard, wireless-fidelity (Wi-Fi) Wi-Fi 7, or extremely high throughput (EHT), in another example, a next-generation standard of 802.11be, Wi-Fi 8, or a next-generation standard of Wi-Fi 8.

[0058] Although the embodiments of the present application are mainly described by using an example of a deployed WLAN network, particularly a network to which the IEEE 802.11 system standard is applied, those skilled in the art will readily understand that various aspects in the present application can be extended to other networks using various standards or protocols, such as Bluetooth, high performance radio LAN (HIPERLAN) (a wireless standard similar to the IEEE 802.11 standard and used mainly in Europe), wide area network (WAN), personal area network (PAN), or other networks known or developed in the future. Thus, various aspects provided in the present application are applicable to any suitable wireless network, regardless of the coverage area and wireless access protocol used.

[0059] Alternatively, the embodiments of the present application are applicable to wireless local area network systems, such as internet of things (IoT) or vehicle to everything (V2X) networks. Certainly, the embodiments of the present application are further applicable to other possible communication systems, such as narrow band internet of things (NB-IoT) systems, long term evolution (LTE) systems, LTE frequency division duplex (FDD) systems, LTE time division duplex (TDD) systems, or worldwide interoperability for microwave access (WiMAX) communication systems, 5th generation (5G) communication systems, hybrid architectures of LTE and 5G, 5G NR systems, 6th generation (6G) communication systems, new communication systems emerging in future communication developments, etc. The above communication systems applicable to the present application are merely illustrative examples, and the communication systems applicable to the present application are not limited thereto. This is explained throughout this specification and the details will not be explained again below.

[0060] This application may be applied to a communication system. The communication system may include one or more access point (AP) stations and one or more non-access point stations (non-AP STAs). For ease of explanation, in this specification, the access point stations are referred to as access points (APs) and the non-access point stations are referred to as stations (STAs). In FIG. 1, the explanation is given by using an example in which the network structure includes one AP and two stations (STA102 and STA103). WLAN communication is implemented between the AP and the STAs, and the STAs may be in fixed locations or may be mobile. The amount of APs and STAs included in the communication system is not limited in this application. The communication method provided in this application is applicable to data communication between an access point and one or more stations (e.g., data communication between AP101 and STA102 and between AP101 and STA103), and is also applicable to data communication between APs and data communication between STAs (e.g., data communication between STA102 and STA103).

[0061] AP is also called access point, hotspot, etc., and is configured to provide wireless access service, allow another wireless device to access, and provide data access. In other words, AP may be an access point used by a terminal device (such as a mobile phone) to access a wired (or wireless) network, and is mainly deployed at home, in a building, and in a park. The typical coverage radius is tens of meters to hundreds of meters. Certainly, access points can alternatively be deployed outdoors. AP is equivalent to a bridge that connects wired and wireless networks. AP is mainly used to connect wireless network clients together and then connect the wireless network to Ethernet. Specifically, an access point may be a terminal (such as a mobile phone) or a network device (such as a router) equipped with a Wi-Fi chip. An access point may be a device that supports the 802.11be standard. The access point may alternatively be a device that supports multiple wireless local area network standards of the 802.11 family, such as 802.11ax, 802.11ac, 802.11n, 802.11g, 802.11b, 802.11a, and 802.11be next generation standards. The access point in this application may be a high efficient (HE) AP or an EHT AP, or may be an access point applicable to future generation Wi-Fi standards.Alternatively, the AP may be a base station, an evolved NodeB (eNodeB), a transmission reception point (TRP), a next generation NodeB (gNB) in a 5G communication system, or a base station in a future communication system, or may be a module or unit that completes some functions of a base station, for example, a central unit (CU), a distributed unit (DU), a router, a switch, or a network bridge, etc. The specific technology and the specific device form used by the AP are not limited in this application.

[0062] The STA is a communication device connected to a wireless network, and may be, for example, a wireless communication chip, a wireless sensor, or a wireless communication terminal. The station may also be referred to as a terminal, a user equipment (UE), a mobile station, a mobile terminal, or the like. For example, the station may be a mobile phone, a tablet computer, a set-top box, a smart television, a smart wearable device, an in-vehicle communication device, or a computer supporting Wi-Fi communication functions. Optionally, the station may support the 802.11be standard. The station may alternatively support multiple WLAN standards of the 802.11 family, such as 802.11ax, 802.11ac, 802.11n, 802.11g, 802.11b, 802.11a, 802.11be, Wi-Fi 7, Wi-Fi 8, or a next-generation standard of Wi-Fi 8.

[0063] The access point in this application may be a HE STA or an EHT STA, or may be a STA applicable to future generation Wi-Fi standards.

[0064] For example, the access points and stations may each be devices used in the Internet of Vehicles, Internet of Things nodes, sensors in the Internet of Things, etc., smart cameras, smart remote controls, and smart water or electricity meters in a smart home, sensors in a smart city, etc.

[0065] The AP and the STA may communicate with each other on the 2.4 gigahertz (GHz) frequency band, may communicate with each other on the 5 GHz or 6 GHz frequency band, or may communicate with each other on the 45 GHz or 60 GHz frequency band. For example, the 45 GHz frequency band may include 42.3-47.0 GHz and 47.2-48.4 GHz, the 60 GHz frequency band may include 57-66 GHz frequency band, or the 60 GHz frequency band may include 56.16-73.44 GHz. It should be understood that with the development of communication, the 45 GHz frequency band and the 60 GHz frequency band may each further include another frequency band. This is not particularly limited in this specification.

[0066] It should be noted that the AP and the STA may also communicate with each other on different frequency bands, which is not limited in the embodiments of this application.

[0067] In a possible implementation, the communication device may be a Discrete Fourier Transform Orthogonal Frequency Division Multiplexing (DFT-s-OFDM) transmitter. FIG. 2 is a diagram of the structure of a DFT-s-OFDM transmitter according to the present application. The DFT-s-OFDM transmitter includes a Mapping module. Optionally, the DFT-s-OFDM may further include an IDFT module, a parallel / serial (P / S) converter, a radio frequency (RF) module, and an antenna. The IDFT module may be configured to perform IDFT processing on the signal to be sent, and the mapping module may be configured to map the signal output by the IDFT module to a subcarrier. It may be understood that the communication device may alternatively be a DFT-s-OFDM receiver. For the structure of the DFT-s-OFDM receiver, please refer to the above figure illustrating the structure of the DFT-s-OFDM transmitter. The signal processing process performed by a DFT-s-OFDM receiver is the inverse of the signal processing performed by a DFT-s-OFDM transmitter.

[0068] In the embodiments of the present application, "at least one" means one or more, and "multiple" means two or more. The term "and / or" represents a relational relationship between related objects and indicates that three relations may exist. For example, A and / or B may indicate the following cases: only A is present, both A and B are present, and only B is present, where A and B may be singular or plural. The character " / " usually indicates an "or" relationship between related objects. "At least one of the following members" or similar expressions means any combination of these members, including any combination of a single member or multiple members. For example, at least one member of a, b, or c may indicate a, b, c, ab, ac, bc, or abc, where a, b, and c may be singular or plural.

[0069] In addition, unless otherwise specified, ordinals such as "first" and "second" in the embodiments of the present application are intended to distinguish between multiple objects, and are not intended to limit the size, content, order, time sequence, priority, or importance of the multiple objects. For example, a first basic channel and a second basic channel are only used to distinguish between different basic channels, and do not indicate different bandwidth values, priorities, importance, etc., of the two basic channels.

[0070] The network architecture and service scenarios described in the embodiments of the present application are intended to more clearly explain the technical solutions in the embodiments of the present application, and do not constitute limitations on the technical solutions provided in the embodiments of the present application. Those skilled in the art will understand that with the evolution of network architecture and the emergence of new service scenarios, the technical solutions provided in the embodiments of the present application can also be applied to similar technical problems.

[0071] Please note that in the following description, an example in which the first device is a transmitting device and the second device is a receiving device is used for description. The first device may be an AP, and the second device is an STA, in other words, the AP sends a signal to the STA, or the first device is an STA, and the second device is an AP, in other words, the STA sends a signal to the AP. In the following description, only the first device and the second device are used as the performers for description. Optionally, the operation of the first device may be alternatively implemented by a processor, chip, or functional module in the first device, and the operation of the second device may be alternatively implemented by a processor, chip, or functional module in the second device. This is not limited in the present application.

[0072] In an embodiment of the present application, the bandwidth of the channel may be in units of Hertz, the details of which will not be described again below.

[0073] The method provided in the present application may be applied to a first frequency band. The following further relates to a second frequency band, and the lowest frequency of the first frequency band is higher than the highest frequency of the second frequency band. For example, the first frequency band is a 60 GHz frequency band, or the first frequency band is a 45 GHz frequency band, etc., and the second frequency band is a 2.45 GHz frequency band, or the second frequency band is a 5 GHz / 6 GHz frequency band, etc. To facilitate the understanding of the present solution, the present application is described by using an example in which the 45 GHz frequency band includes 42.3-47.0 GHz and 47.2-48.4 GHz, the 60 GHz frequency band includes 57-66 GHz, and the 60 GHz frequency band includes 56.16-73.44 GHz. The method for another frequency band is similar to the method for the 45 GHz frequency band and the 60 GHz frequency band, and the details will not be described again herein.

[0074] In the embodiments of the present application, the amount of subcarriers corresponding to a channel may also be understood as the IDFT size / DFT size corresponding to a channel, and the two are alternatively described. The IDFT size / DFT size may also be understood as the amount of sampling points used during IDFT / DFT processing or the size of the filter used during IDFT / DFT processing.

[0075] In the embodiment of the present application, the millimeter wave band may be understood as a frequency band whose wavelength is at the millimeter level, for example, a frequency band whose wavelength is from 1 to 10 millimeters. For example, the millimeter wave band may be a 30-300 GHz frequency band, etc. It should be understood that the 30-300 GHz frequency band is just one example of a millimeter wave band for illustration, or may be another frequency band, which is not particularly limited in this specification.

[0076] EMBODIMENT 1

[0077] 3 is a schematic flow chart of a communication method according to the present application. The method includes the following steps:

[0078] S301: A first device generates a first OFDM signal.

[0079] In a possible implementation, the first device may perform an IDFT based on a first size on the signal to be sent to obtain the first OFDM signal. Correspondingly, the second device may perform a DFT based on the first size on the received first OFDM signal to convert the received first OFDM signal into a signal in the frequency domain. The first size may be an amount of subcarriers corresponding to a first channel in a first frequency band.

[0080] In another possible implementation, the first device generates a single carrier (SC) signal, and correspondingly, the second device performs a DFT based on a first size on the received SC signal to convert the received SC signal into a signal in the frequency domain for processing.

[0081] The amount of subcarriers corresponding to the first channel is greater than or equal to the amount of subcarriers corresponding to the second channel in the second frequency band, the lowest frequency of the first frequency band is greater than the highest frequency of the second frequency band, and the first channel is M The first basic channel includes 2 M number of second fundamental channels, the bandwidth of the first fundamental channel being a minimum channel bandwidth supported in the first frequency band, the bandwidth of the second fundamental channel being a minimum channel bandwidth supported in the second frequency band, the bandwidth of the first fundamental channel being greater than the bandwidth of the second fundamental channel, and M being an integer greater than or equal to 0.

[0082] Optionally, the subcarrier spacing of the first channel is greater than or equal to a threshold (e.g., 2 MHz, 2.5 MHz, or 5 MHz). In this manner, the effect of frequency offset on signal transmission on the first channel may be reduced, and signal transmission performance may be improved.

[0083] Please note that the "first fundamental channel" is merely an exemplary name and represents a channel corresponding to the minimum channel bandwidth supported in the first frequency band. For example, if the bandwidth supported in the first frequency band is 320 MHz, 640 MHz, 1280 MHz, or 2560 MHz, the minimum channel bandwidth supported in the first frequency band is 320 MHz. Correspondingly, a channel having a bandwidth of 320 MHz in the first frequency band may be understood as the first fundamental channel in this application. The understanding of the second fundamental channel is similar to that of the first fundamental channel. For example, if the bandwidth supported in the second fundamental channel is 20 MHz, 40 MHz, 80 MHz, 160 MHz, or 320 MHz, the minimum channel bandwidth supported in the second frequency band is 20 MHz. Correspondingly, a channel having a bandwidth of 20 MHz in the second frequency band may be understood as the second fundamental channel in this application.

[0084] The relevant description of the first channel will be described in detail below.

[0085] S302: A first device sends a first OFDM signal on a first channel in a first frequency band, and a second device correspondingly receives a first OFDM symbol on the first channel in the first frequency band.

[0086] S303: A second device processes the first OFDM symbol.

[0087] It should be noted that S303 may be an optional step.

[0088] In the following, the amount of subcarriers of the first channel is described by using an example for the bandwidth of the first channel.

[0089] Example 1: The bandwidth of the first fundamental channel is N times the bandwidth of the second fundamental channel, where N is an integer greater than 1. For example, the bandwidth of the second fundamental channel is 20 MHz, in other words, the minimum channel bandwidth supported in the second frequency band is 20 MHz. The bandwidth of the first fundamental channel is (20*N) MHz, for example, 40 MHz, 80 MHz, 160 MHz, 320 MHz, or 640 MHz.

[0090] Example 1 is explained below by using an example.

[0091] In Example 1, the minimum channel bandwidth supported in the first frequency band is 320 MHz, in other words, the bandwidth of the first fundamental channel is 320 MHz. The first frequency band may support one or more bandwidths of 320 MHz, 640 MHz, 1280 MHz, 2560 MHz, and 5120 MHz.

[0092] Based on this example, the amount of subcarriers in the first channel may satisfy any one of the following:

[0093] If the bandwidth of the first channel is 320 MHz, the first channel corresponds to 64 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 20 MHz bandwidth.

[0094] For example, the second signal may be a signal in a second frequency band, for example, 802.11ax / be legacy preamble signaling, such as L-SIG (legacy signaling or non-HT SIGNAL) signaling, or a data field in 802.11n / ac. The following understanding of the second signal is the same as that described in this specification, and the details will not be described again below.

[0095] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 64. For example, the first channel may correspond to 128 subcarriers, 256 subcarriers, and so on.

[0096] If the bandwidth of the first channel is 640 MHz, the first channel corresponds to 128 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 40 MHz bandwidth.

[0097] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 128. For example, the first channel may correspond to 256 subcarriers, 512 subcarriers, and so on.

[0098] If the bandwidth of the first channel is 1280 MHz, the first channel corresponds to 256 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 80 MHz bandwidth.

[0099] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, and so on.

[0100] If the bandwidth of the first channel is 2560 MHz, the first channel corresponds to 512 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 160 MHz bandwidth.

[0101] Optionally, the first channel corresponds to 512 subcarriers. For example, the first channel may alternatively correspond to 1024 subcarriers, etc.

[0102] If the bandwidth of the first channel is 5120 MHz, the first channel corresponds to 1024 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 320 MHz bandwidth.

[0103] For example, the correspondence between each bandwidth and the amount of subcarriers (or IDFT size / DFT size) supported in the first frequency band may be shown in Table 1.

[0104] [Table 1]

[0105] In Example 2, the minimum channel bandwidth supported in the first frequency band is 640 MHz, in other words, the bandwidth of the first fundamental channel is 640 MHz. The first frequency band may support one or more bandwidths of 640 MHz, 1280 MHz, 2560 MHz, 5120 MHz, and 10240 MHz.

[0106] Based on this example, the amount of subcarriers in the first channel may satisfy any one of the following:

[0107] If the bandwidth of the first channel is 640 MHz, the first channel corresponds to 64 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal (i.e., the signal in the second frequency band) in the 20 MHz bandwidth.

[0108] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 64. For example, the first channel may correspond to 128 subcarriers, 256 subcarriers, 512 subcarriers, etc.

[0109] If the bandwidth of the first channel is 1280 MHz, the first channel corresponds to 128 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal (i.e., the signal in the second frequency band) in the 40 MHz bandwidth.

[0110] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 128. For example, the first channel may correspond to 256 subcarriers, 512 subcarriers, 1024 subcarriers, etc.

[0111] If the bandwidth of the first channel is 2560 MHz, the first channel corresponds to 256 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal (i.e., the signal in the second frequency band) in the 80 MHz bandwidth.

[0112] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, and so on.

[0113] If the bandwidth of the first channel is 5120 MHz, the first channel corresponds to 512 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal (i.e., the signal in the second frequency band) in the 160 MHz bandwidth.

[0114] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 512. For example, the first channel may correspond to 1024 subcarriers, etc.

[0115] If the bandwidth of the first channel is 10240 MHz, the first channel corresponds to 1024 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal (i.e., the signal in the second frequency band) in the 320 MHz bandwidth.

[0116] For example, the correspondence between each bandwidth and the amount of subcarriers (or IDFT size / DFT size) supported in the first frequency band may be shown in Table 2.

[0117] [Table 2]

[0118] In Example 3, the minimum channel bandwidth supported in the first frequency band is 40 MHz, in other words, the bandwidth of the first fundamental channel is 40 MHz. The first frequency band may support one or more bandwidths of 40 MHz, 80 MHz, 160 MHz, 320 MHz, and 640 MHz.

[0119] Based on this example, the amount of subcarriers in the first channel may satisfy any one of the following:

[0120] If the bandwidth of the first channel is 40 MHz, the first channel corresponds to 64 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 20 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 64. For example, the first channel may correspond to 128 subcarriers, 256 subcarriers, etc.

[0121] If the bandwidth of the first channel is 80 MHz, the first channel corresponds to 128 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 40 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 128. For example, the first channel may correspond to 256 subcarriers, 512 subcarriers, etc.

[0122] When the bandwidth of the first channel is 160 MHz, the first channel corresponds to 256 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 80 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, etc.

[0123] If the bandwidth of the first channel is 320 MHz, the first channel corresponds to 512 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 160 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 512. For example, the first channel may correspond to 1024 subcarriers, etc.

[0124] If the bandwidth of the first channel is 640 MHz, the first channel corresponds to 1024 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 320 MHz bandwidth.

[0125] For example, the correspondence between each bandwidth and the amount of subcarriers (or IDFT size / DFT size) supported in the first frequency band may be shown in Table 3.

[0126] [Table 3]

[0127] In Example 4, the minimum channel bandwidth supported in the first frequency band is 80 MHz, in other words, the bandwidth of the first fundamental channel is 80 MHz. The first frequency band may support one or more bandwidths of 80 MHz, 160 MHz, 320 MHz, 640 MHz, and 1280 MHz.

[0128] Based on this example, the amount of subcarriers in the first channel may satisfy any one of the following:

[0129] If the bandwidth of the first channel is 80 MHz, the first channel corresponds to 64 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 20 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 64. For example, the first channel may correspond to 128 subcarriers, 256 subcarriers, etc.

[0130] If the bandwidth of the first channel is 160 MHz, the first channel corresponds to 128 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 40 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 128. For example, the first channel may correspond to 256 subcarriers, 512 subcarriers, etc.

[0131] When the bandwidth of the first channel is 320 MHz, the first channel corresponds to 256 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 80 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, etc.

[0132] When the bandwidth of the first channel is 640 MHz, the first channel corresponds to 512 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 160 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 512. For example, the first channel may correspond to 1024 subcarriers, etc.

[0133] If the bandwidth of the first channel is 1280 MHz, the first channel corresponds to 1024 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 320 MHz bandwidth.

[0134] For example, the correspondence between each bandwidth and the amount of subcarriers (or IDFT size / DFT size) supported in the first frequency band may be shown in Table 4.

[0135] [Table 4]

[0136] In Example 5, the minimum channel bandwidth supported in the first frequency band is 160 MHz, in other words, the bandwidth of the first fundamental channel is 160 MHz. The first frequency band may support one or more bandwidths of 160 MHz, 320 MHz, 640 MHz, 1280 MHz, and 2560 MHz.

[0137] Based on this example, the amount of subcarriers in the first channel may satisfy any one of the following:

[0138] If the bandwidth of the first channel is 160 MHz, the first channel corresponds to 64 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal of 802.11ax / be in a 20 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 64. For example, the first channel may correspond to 128 subcarriers, 256 subcarriers, etc.

[0139] If the bandwidth of the first channel is 320 MHz, the first channel corresponds to 128 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 40 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 128. For example, the first channel may correspond to 256 subcarriers, 512 subcarriers, etc.

[0140] When the bandwidth of the first channel is 640 MHz, the first channel corresponds to 256 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 80 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, etc.

[0141] If the bandwidth of the first channel is 1280 MHz, the first channel corresponds to 512 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 160 MHz bandwidth. Optionally, the amount of subcarriers corresponding to the first channel may be greater than 512. For example, the first channel may correspond to 1024 subcarriers, etc.

[0142] If the bandwidth of the first channel is 2560 MHz, the first channel corresponds to 1024 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 320 MHz bandwidth.

[0143] For example, the correspondence between each bandwidth and the amount of subcarriers (or IDFT size / DFT size) supported in the first frequency band may be shown in Table 5.

[0144] [Table 5]

[0145] Example 2: The bandwidth of the first fundamental channel is obtained by performing R bisections separately on a frequency band obtained by dividing the first frequency band based on the minimum channel bandwidth supported in the millimeter wave band of the IEEE 802.11ad / ay standard, where R is an integer greater than or equal to 0. For example, it is assumed that the minimum channel bandwidth supported in the millimeter wave band is 2160 MHz. In this case, the bandwidth of the first fundamental channel may be 2160 MHz, 1080 MHz, 540 MHz, 270 MHz, etc.

[0146] In Example 6, the minimum channel bandwidth supported in the first frequency band is 270 MHz, in other words, the bandwidth of the first fundamental channel is 270 MHz. The first frequency band may support one or more bandwidths of 270 MHz, 540 MHz, 1080 MHz, 2160 MHz, and 4320 MHz.

[0147] Based on this example, the amount of subcarriers in the first channel may satisfy any one of the following:

[0148] If the bandwidth of the first channel is 270 MHz, the first channel corresponds to 64 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 20 MHz bandwidth.

[0149] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 64. For example, the first channel may correspond to 128 subcarriers, 256 subcarriers, 512 subcarriers, etc.

[0150] If the bandwidth of the first channel is 540 MHz, the first channel corresponds to 128 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 40 MHz bandwidth.

[0151] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 128. For example, the first channel may correspond to 256 subcarriers, 512 subcarriers, 1024 subcarriers, etc.

[0152] If the bandwidth of the first channel is 1080 MHz, the first channel corresponds to 256 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 80 MHz bandwidth.

[0153] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, and so on.

[0154] If the bandwidth of the first channel is 2160 MHz, the first channel corresponds to 512 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 160 MHz bandwidth.

[0155] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 512. For example, the first channel may correspond to 1024 subcarriers, etc.

[0156] If the bandwidth of the first channel is 4320 MHz, the first channel corresponds to 1024 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 3200 MHz bandwidth.

[0157] For example, the correspondence between each bandwidth and the amount of subcarriers (or IDFT size / DFT size) supported in the first frequency band may be shown in Table 6.

[0158] [Table 6]

[0159] In Example 7, the minimum channel bandwidth supported in the first frequency band is 540 MHz, in other words, the bandwidth of the first fundamental channel is 540 MHz. The first frequency band may support one or more bandwidths of 540 MHz, 1080 MHz, 2160 MHz, 4320 MHz, and 8640 MHz.

[0160] Based on this example, the amount of subcarriers in the first channel may satisfy any one of the following:

[0161] If the bandwidth of the first channel is 540 MHz, the first channel corresponds to 64 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 20 MHz bandwidth.

[0162] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 64. For example, the first channel may correspond to 128 subcarriers, 256 subcarriers, 512 subcarriers, etc.

[0163] If the bandwidth of the first channel is 1080 MHz, the first channel corresponds to 128 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 40 MHz bandwidth.

[0164] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 128. For example, the first channel may correspond to 256 subcarriers, 512 subcarriers, 1024 subcarriers, etc.

[0165] If the bandwidth of the first channel is 2160 MHz, the first channel corresponds to 256 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 80 MHz bandwidth.

[0166] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, and so on.

[0167] If the bandwidth of the first channel is 4320 MHz, the first channel corresponds to 512 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 160 MHz bandwidth.

[0168] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 512. For example, the first channel may correspond to 1024 subcarriers, etc.

[0169] If the bandwidth of the first channel is 8640 MHz, the first channel corresponds to 1024 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 320 MHz bandwidth.

[0170] For example, the correspondence between each bandwidth and the amount of subcarriers (or IDFT size / DFT size) supported in the first frequency band may be shown in Table 7.

[0171] [Table 7]

[0172] Example 3: The bandwidth of the first fundamental channel is obtained by performing K bisections on the first subband, where the bandwidth of the first subband is smaller than the minimum channel bandwidth supported in the millimeter wave band, and K is an integer greater than or equal to 0. For example, it is assumed that the minimum channel bandwidth supported in the millimeter wave band is 2160 MHz. In this case, the first subband may be 1000 MHz, 1800 MHz, 1940 MHz, 2000 MHz, 2100 MHz, etc. In a possible implementation, the bandwidth of the first subband may be an integer multiple of 20 MHz.

[0173] For ease of explanation, an example in which the first subband is 2000 MHz is used for explanation. The bandwidth of the first fundamental channel may be 2000 MHz, 1000 MHz, 500 MHz, 250 MHz, etc.

[0174] In Example 8, the minimum channel bandwidth supported in the first frequency band is 250 MHz, in other words, the bandwidth of the first fundamental channel is 250 MHz. The first frequency band may support one or more bandwidths of 250 MHz, 500 MHz, 1000 MHz, 2000 MHz, and 4000 MHz.

[0175] Based on this example, the amount of subcarriers in the first channel may satisfy any one of the following:

[0176] If the bandwidth of the first channel is 250 MHz, the first channel corresponds to 64 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 20 MHz bandwidth.

[0177] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 64. For example, the first channel may correspond to 128 subcarriers, 256 subcarriers, 512 subcarriers, etc.

[0178] If the bandwidth of the first channel is 500 MHz, the first channel corresponds to 128 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 40 MHz bandwidth.

[0179] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 128. For example, the first channel may correspond to 256 subcarriers, 512 subcarriers, 1024 subcarriers, etc.

[0180] If the bandwidth of the first channel is 1000 MHz, the first channel corresponds to 256 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 80 MHz bandwidth.

[0181] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, and so on.

[0182] If the bandwidth of the first channel is 2000 MHz, the first channel corresponds to 512 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 160 MHz bandwidth.

[0183] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 512. For example, the first channel may correspond to 1024 subcarriers, etc.

[0184] If the bandwidth of the first channel is 4000 MHz, the first channel corresponds to 1024 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 320 MHz bandwidth.

[0185] For example, the correspondence between each bandwidth and the amount of subcarriers (or IDFT size / DFT size) supported in the first frequency band may be shown in Table 8.

[0186] [Table 8]

[0187] In Example 9, the minimum channel bandwidth supported in the first frequency band is 250 MHz, in other words, the bandwidth of the first fundamental channel is 250 MHz. The first frequency band may support one or more bandwidths of 500 MHz, 1000 MHz, 2000 MHz, and 4000 MHz.

[0188] Based on this example, the amount of subcarriers in the first channel may satisfy any one of the following:

[0189] If the bandwidth of the first channel is 500 MHz, the first channel corresponds to 64 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 20 MHz bandwidth.

[0190] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 64. For example, the first channel may correspond to 128 subcarriers, 256 subcarriers, 512 subcarriers, etc.

[0191] If the bandwidth of the first channel is 1000 MHz, the first channel corresponds to 128 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 40 MHz bandwidth.

[0192] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 128. For example, the first channel may correspond to 256 subcarriers, 512 subcarriers, 1024 subcarriers, etc.

[0193] If the bandwidth of the first channel is 2000 MHz, the first channel corresponds to 256 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 80 MHz bandwidth.

[0194] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, and so on.

[0195] If the bandwidth of the first channel is 4000 MHz, the first channel corresponds to 512 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 160 MHz bandwidth.

[0196] Optionally, the amount of subcarriers corresponding to the first channel may be greater than 512. For example, the first channel may correspond to 1024 subcarriers, etc.

[0197] If the bandwidth of the first channel is 8000 MHz, the first channel corresponds to 1024 subcarriers. This implementation can be achieved by a straightforward extension of the subcarrier spacing of the second signal in the 320 MHz bandwidth.

[0198] For example, the correspondence between each bandwidth and the amount of subcarriers (or IDFT size / DFT size) supported in the first frequency band may be shown in Table 9.

[0199] [Table 9]

[0200] In this embodiment of the present application, the ratio of the bandwidth of the first channel to the minimum channel bandwidth supported in the first frequency band is equal to the ratio of the bandwidth of the second channel to the minimum channel bandwidth supported in the second frequency band, for example, both ratios are 2 M and the amount of subcarriers corresponding to the first channel is greater than or equal to the amount of subcarriers corresponding to the second channel in the second frequency band. The correspondence between the bandwidth and the IDFT size / DFT size (or the amount of subcarriers) in the high frequency band (e.g., the first frequency band) is similar to the correspondence between the bandwidth and the amount of subcarriers in the low frequency band (e.g., the second frequency band), so that the high frequency band and the low frequency band can share signal processing devices, for example, it can be seen that the IDFT device or the DFT device is reused. Therefore, the device cost is low.

[0201] The first embodiment describes the amount of subcarriers (or IDFT size / DFT size) corresponding to each bandwidth supported in the first frequency band. The following describes a channel division method for the first frequency band as an example. For details, please refer to the second embodiment, the third embodiment, and the fourth embodiment.

[0202] The channel division method provided in the present application may be implemented in combination with embodiment 1. Specifically, the channel division method provided in embodiment 2 may be implemented in combination with example 1 of embodiment 1, or it can be understood that when the first frequency band is divided into channels in the manner of embodiment 2, the correspondence between each channel bandwidth and the amount of subcarriers refers to example 1 of embodiment 1. The channel division method provided in embodiment 3 may be implemented in combination with example 2 of embodiment 1, or it can be understood that when the first frequency band is divided into channels in the manner of embodiment 3, the correspondence between each channel bandwidth and the amount of subcarriers refers to example 2 of embodiment 1. The channel division method provided in embodiment 4 may be implemented in combination with example 3 of embodiment 1, or it can be understood that when the first frequency band is divided into channels in the manner of embodiment 4, the correspondence between each channel bandwidth and the amount of subcarriers refers to example 3 of embodiment 1.

[0203] EMBODIMENT 2

[0204] It is assumed that the minimum bandwidth of the second frequency band (for example, the low frequency minimum bandwidth defined by 802.11be) is 20 MHz. In the channel division method provided in the embodiment 2 of the present application, the bandwidth of the first frequency band can be divided based on an integer multiple of 20 MHz, for example, 40 MHz, 80 MHz, 160 MHz, 320 MHz, or 640 MHz. In the embodiment 2 of the present application, the channel bandwidth of the high frequency band (i.e., the first frequency band) is an integer multiple of the minimum channel bandwidth of the low frequency band (i.e., the second frequency band), so that the signal in the high frequency band and the signal in the low frequency band can be transmitted by reusing the same crystal oscillator. This can reduce hardware complexity and reduce hardware overhead.

[0205] In the following, the channel division method for the first frequency band is described by using an example for a specific frequency band.

[0206] Example 4: The minimum channel bandwidth supported in the first frequency band is 320 MHz, in other words, the bandwidth of the first fundamental channel is 320 MHz. The bandwidths supported in the first frequency band may include one or more of the following bandwidths: 320 MHz, 640 MHz, 1280 MHz, 2560 MHz, and 5120 MHz.

[0207] Optionally, for the amount of subcarriers (or IDFT size / DFT size) corresponding to each channel bandwidth in the channel division manner of Example 4, refer to the relevant description of Example 1 in Example 1 of Embodiment 1. Details are not described again in this specification.

[0208] For example, the first frequency band is a 57-66 GHz frequency band. The first frequency band may be divided into 28 320 MHz channels, 14 640 MHz channels, 7 1280 MHz channels, 3 2560 MHz channels, or 1 5120 MHz channel, as shown in FIG.

[0209] A channel division scheme for the 57-66 GHz frequency band is described herein, in which a 20 MHz bandwidth is left in each of the front and rear portions of the first frequency band, as shown in FIG.

[0210] 320MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.180 + (n-1) × 0.320, 1 ≤ n ≤ 28, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.120 + (n-1) × 0.320, 1 ≤ n ≤ 28, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.340 + (n-1) × 0.320, where 1 ≤ n ≤ 28. C (n) is the center frequency, and f L (n) indicates the lowest frequency, and f H (n) indicates the highest frequency.C (n), f L (n), and f H The meaning of (n) is the same as that in this specification and will not be explained one by one.

[0211] 640MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.340 + (n-1) × 0.640, 1 ≤ n ≤ 14, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 0.640, 1 ≤ n ≤ 14, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.660 + (n-1) × 0.640, where 1≦n≦14.

[0212] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.660 + (n-1) × 1.280, 1 ≤ n ≤ 7, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 1.280, 1 ≤ n ≤ 7, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.300 + (n-1) × 1.280, where 1≦n≦7.

[0213] 2560MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 58.300 + (n-1) × 2.560, 1 ≤ n ≤ 3, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 2.560, 1 ≤ n ≤ 3, and the formula for calculating the maximum frequency is f H (n) [GHz] = 59.580 + (n-1) × 2.560, 1≦n≦3.

[0214] 5120MHz channel division: The formula for calculating the center frequency is f C (n)[GHz]=59.580+(n-1)×5.120, n=1, and the formula for calculating the minimum frequency is f L(n)[GHz]=57.020+(n-1)×5.120, n=1, and the formula for calculating the maximum frequency is f H (n) [GHz] = 62.140 + (n-1) × 5.120, n = 1.

[0215] In the channel division manner provided in this application, n is the sequence number or index number of the channel. The details are not described again below.

[0216] For example, the first frequency band includes the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band. The first frequency band may be divided into seventeen 320 MHz channels, eight 640 MHz channels, three 1280 MHz channels, or one 2560 MHz channel.

[0217] A channel division scheme for the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band is described herein, in which the remaining bandwidth in both the front and rear portions of the 42.3-47.0 GHz frequency band is 110 MHz, and the remaining bandwidth in both the front and rear portions of the 47.2-48.4 GHz frequency band is 120 MHz.

[0218] 320MHz channel division: The 42.3-47.0GHz frequency band contains 14 320MHz channels, and the 47.2-48.4GHz frequency band contains 3 320MHz channels. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.570 + (n-1) × 0.320, 1 ≤ n ≤ 14, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.410 + (n-1) × 0.320, 1 ≤ n ≤ 14, and the formula for calculating the maximum frequency is f H (n) [GHz] = 42.730 + (n-1) × 0.320, 1 ≤ n ≤ 14. For the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C(n) [GHz] = 47.480 + (n-15) × 0.320, 15 ≤ n ≤ 17, and the formula for calculating the minimum frequency is f L (n) [GHz] = 47.320 + (n-15) × 0.320, 15 ≤ n ≤ 17, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.640 + (n-15) × 0.320, where 15≦n≦17.

[0219] 640MHz Channel Division: The 42.3-47.0GHz frequency band contains seven 640MHz channels, and the 47.2-48.4GHz frequency band contains one 640MHz channel. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.730 + (n-1) × 0.640, 1 ≤ n ≤ 7, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.410 + (n-1) × 0.640, 1 ≤ n ≤ 7, and the formula for calculating the maximum frequency is f H (n) [GHz] = 43.050 + (n-1) × 0.640, 1 ≤ n ≤ 7, for the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz]=47.640+(n-8)×0.640, n=8, and the formula for calculating the minimum frequency is f L (n)[GHz]=47.320+(n-8)×0.640, n=8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.960 + (n-8) × 0.640, n = 8.

[0220] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 43.050 + (n-1) × 1.280, where 1 ≤ n ≤ 3, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.410 + (n-1) × 1.280, 1 ≤ n ≤ 3, and the formula for calculating the maximum frequency is f H (n) [GHz] = 43.690 + (n-1) × 1.280, where 1≦n≦3.

[0221] 2560MHz channel division: The formula for calculating the center frequency is f C (n)[GHz]=43.690+(n-1)×2.560, n=1, and the formula for calculating the minimum frequency is f L (n)[GHz]=42.410+(n-1)×2.560, n=1, and the formula for calculating the maximum frequency is f H (n) [GHz] = 44.970 + (n-1) × 2.560, n = 1.

[0222] For example, the first frequency band is the 56.16 to 73.44 GHz frequency band. The first frequency band may be divided into 54 320 MHz channels, 27 640 MHz channels, 13 1280 MHz channels, 6 2560 MHz channels, or 3 5120 MHz channels.

[0223] A channel division scheme for the 56.16-73.44 GHz frequency band is described herein, in which there is no remaining bandwidth in the front and rear portions of the first frequency band.

[0224] 320MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.320 + (n-1) × 0.320, 1 ≤ n ≤ 54, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 0.320, 1 ≤ n ≤ 54, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.480 + (n-1) × 0.320, where 1≦n≦54.

[0225] 640MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.480 + (n-1) × 0.640, 1 ≤ n ≤ 27, and the formula for calculating the minimum frequency is f L(n)[GHz]=56.160+(n-1)×0.640, 1≦n≦27, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.800 + (n-1) × 0.640, where 1≦n≦27.

[0226] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.800 + (n-1) × 1.280, 1 ≤ n ≤ 13, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 1.280, 1 ≤ n ≤ 13, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.440 + (n-1) × 1.280, where 1≦n≦13.

[0227] 2560MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.440 + (n-1) × 2.560, 1 ≤ n ≤ 6, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 2.560, 1 ≤ n ≤ 6, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.720 + (n-1) × 2.560, where 1≦n≦6.

[0228] 5120MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 58.720 + (n-1) × 5.120, 1 ≤ n ≤ 3, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 5.120, 1 ≤ n ≤ 3, and the formula for calculating the maximum frequency is f H (n) [GHz] = 61.280 + (n-1) × 5.120, 1≦n≦3.

[0229] Example 5: The minimum channel bandwidth supported in the first frequency band is 80 MHz, in other words, the bandwidth of the first fundamental channel is 80 MHz. The bandwidths supported in the first frequency band may include one or more of the following bandwidths: 80 MHz, 160 MHz, 320 MHz, 640 MHz, and 1280 MHz.

[0230] Optionally, for the amount of subcarriers (or IDFT size / DFT size) corresponding to each channel bandwidth in the channel division manner of Example 5, refer to the relevant description of Example 4 in Example 1 of Embodiment 1. Details are not described again in this specification.

[0231] For example, the first frequency band is the 57-66 GHz frequency band. The first frequency band may be divided into 112 80 MHz channels, 56 160 MHz channels, 28 320 MHz channels, 14 640 MHz channels, or 7 1280 MHz channels. In this division manner, a 20 MHz bandwidth is left in each of the front and rear portions of the first frequency band.

[0232] 80MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.060 + (n-1) × 0.080, 1 ≤ n ≤ 112, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 0.080, 1 ≤ n ≤ 112, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.100 + (n-1) × 0.080, 1≦n≦112.

[0233] 160MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.100 + (n-1) × 0.160, 1 ≤ n ≤ 56, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 0.160, 1 ≤ n ≤ 56, and the formula for calculating the maximum frequency is f H(n) [GHz] = 57.180 + (n-1) × 0.160, where 1≦n≦56.

[0234] 320MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.180 + (n-1) × 0.320, 1 ≤ n ≤ 28, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 0.320, 1 ≤ n ≤ 28, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.340 + (n-1) × 0.320, where 1≦n≦28.

[0235] 640MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.340 + (n-1) × 0.640, 1 ≤ n ≤ 14, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 0.640, 1 ≤ n ≤ 14, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.660 + (n-1) × 0.640, where 1≦n≦14.

[0236] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.660 + (n-1) × 1.280, 1 ≤ n ≤ 7, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 1.280, 1 ≤ n ≤ 7, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.300 + (n-1) × 1.280, where 1≦n≦7.

[0237] For example, the first frequency band includes the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band. The first frequency band may be divided into seventy-three 80 MHz channels, thirty-six 160 MHz channels, seventeen 320 MHz channels, eight 640 MHz channels, or three 1280 MHz channels.

[0238] A channel division scheme for the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band is described herein, in which the remaining bandwidth in both the front and rear parts of the 42.3-47.0 GHz frequency band is 30 MHz, and there is no remaining bandwidth in the front and rear parts of the 47.2-48.4 GHz frequency band.

[0239] 80MHz Channel Division: The 42.3-47.0GHz frequency band contains 58 80MHz channels, and the 47.2-48.4GHz frequency band contains 15 80MHz channels. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.370 + (n-1) × 0.080, 1 ≤ n ≤ 58, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.330 + (n-1) × 0.080, 1 ≤ n ≤ 58, and the formula for calculating the maximum frequency is f H (n) [GHz] = 42.410 + (n-1) × 0.080, 1 ≤ n ≤ 58. For the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 47.240 + (n-58) × 0.160, 58 ≤ n ≤ 73, and the formula for calculating the minimum frequency is f L (n) [GHz] = 47.200 + (n-58) × 0.080, 58 ≤ n ≤ 73, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.280 + (n-58) × 0.080, where 58≦n≦73.

[0240] 160MHz Channel Division: The 42.3-47.0GHz frequency band contains 29 160MHz channels, and the 47.2-48.4GHz frequency band contains 7 160MHz channels. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C(n) [GHz] = 42.410 + (n-1) × 0.160, 1 ≤ n ≤ 29, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.330 + (n-1) × 0.160, 1 ≤ n ≤ 29, and the formula for calculating the maximum frequency is f H (n) [GHz] = 42.490 + (n-1) × 0.160, 1 ≤ n ≤ 29. For the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 47.280 + (n-30) × 0.160, 30 ≤ n ≤ 36, and the formula for calculating the minimum frequency is f L (n)[GHz]=47.200+(n-30)×0.160, 30≦n≦36, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.360 + (n - 30) × 0.160, where 30 ≤ n ≤ 36.

[0241] 320MHz channel division: The 42.3-47.0GHz frequency band contains 14 320MHz channels, and the 47.2-48.4GHz frequency band contains 3 320MHz channels. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.490 + (n-1) × 0.320, 1 ≤ n ≤ 14, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.330 + (n-1) × 0.320, 1 ≤ n ≤ 14, and the formula for calculating the maximum frequency is f H (n) [GHz] = 42.650 + (n-1) × 0.320, 1 ≤ n ≤ 14. For the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 47.360 + (n-15) × 0.320, 15 ≤ n ≤ 17, and the formula for calculating the minimum frequency is f L (n)[GHz]=47.200+(n-15)×0.320, 15≦n≦17, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.520 + (n-15) × 0.320, where 15≦n≦17.

[0242] 640MHz Channel Division: The 42.3-47.0GHz frequency band contains seven 640MHz channels, and the 47.2-48.4GHz frequency band contains one 640MHz channel. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.650 + (n-1) × 0.640, 1 ≤ n ≤ 7, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.330 + (n-1) × 0.640, 1 ≤ n ≤ 7, and the formula for calculating the maximum frequency is f H (n) [GHz] = 42.970 + (n-1) × 0.640, 1 ≤ n ≤ 7, for the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz]=47.520+(n-8)×0.640, n=8, and the formula for calculating the minimum frequency is f L (n)[GHz]=47.200+(n-8)×0.640, n=8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.840 + (n-8) × 0.640, n = 8.

[0243] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 42.970 + (n-1) × 1.280, 1 ≤ n ≤ 3, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.330 + (n-1) × 1.280, 1 ≤ n ≤ 3, and the formula for calculating the maximum frequency is f H (n) [GHz] = 43.610 + (n-1) × 1.280, where 1≦n≦3.

[0244] For example, the first frequency band is the 56.16 to 73.44 GHz frequency band. The first frequency band may be divided into 216 80 MHz channels, 108 160 MHz channels, 54 320 MHz channels, 27 640 MHz channels, or 13 1280 MHz channels.

[0245] A channel division scheme for the 56.16-73.44 GHz frequency band is described herein, in which there is no remaining bandwidth in the front and rear portions of the first frequency band.

[0246] 80MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.200 + (n-1) × 0.080, 1 ≤ n ≤ 216, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 0.080, 1 ≤ n ≤ 216, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.240 + (n-1) × 0.080, where 1≦n≦216.

[0247] 160MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.240 + (n-1) × 0.160, 1 ≤ n ≤ 108, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 0.160, 1 ≤ n ≤ 108, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.320 + (n-1) × 0.160, 1≦n≦108.

[0248] 320MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.320 + (n-1) × 0.320, 1 ≤ n ≤ 54, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 0.320, 1 ≤ n ≤ 54, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.480 + (n-1) × 0.320, where 1≦n≦54.

[0249] 640MHz channel division: The formula for calculating the center frequency is f C(n) [GHz] = 56.480 + (n-1) × 0.640, 1 ≤ n ≤ 27, and the formula for calculating the minimum frequency is f L (n)[GHz]=56.160+(n-1)×0.640, 1≦n≦27, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.800 + (n-1) × 0.640, where 1≦n≦27.

[0250] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.800 + (n-1) × 1.280, 1 ≤ n ≤ 13, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 1.280, 1 ≤ n ≤ 13, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.440 + (n-1) × 1.280, where 1≦n≦13.

[0251] Example 6: The minimum channel bandwidth supported in the first frequency band is 160 MHz, in other words, the bandwidth of the first fundamental channel is 160 MHz. The bandwidths supported in the first frequency band may include one or more of the following bandwidths: 160 MHz, 320 MHz, 640 MHz, 1280 MHz, and 2560 MHz.

[0252] Optionally, for the amount of subcarriers (or IDFT size / DFT size) corresponding to each channel bandwidth in the channel division manner of Example 6, refer to the relevant description of Example 5 in Example 1 of Embodiment 1. Details are not described again in this specification.

[0253] For example, the first frequency band is the 57-66 GHz frequency band. The first frequency band may be divided into 56 160 MHz channels, 28 320 MHz channels, 14 640 MHz channels, 7 1280 MHz channels, or 3 2560 MHz channels. In this division manner, a 20 MHz bandwidth is left in each of the front and rear portions of the first frequency band.

[0254] 160MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.100 + (n-1) × 0.160, 1 ≤ n ≤ 56, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 0.160, 1 ≤ n ≤ 56, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.180 + (n-1) × 0.160, where 1≦n≦56.

[0255] 320MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.180 + (n-1) × 0.320, 1 ≤ n ≤ 28, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 0.320, 1 ≤ n ≤ 28, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.340 + (n-1) × 0.320, where 1≦n≦28.

[0256] 640MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.340 + (n-1) × 0.640, 1 ≤ n ≤ 14, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 0.640, 1 ≤ n ≤ 14, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.660 + (n-1) × 0.640, where 1≦n≦14.

[0257] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.660 + (n-1) × 1.280, 1 ≤ n ≤ 7, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 1.280, 1 ≤ n ≤ 7, and the formula for calculating the maximum frequency is f H(n) [GHz] = 58.300 + (n-1) × 1.280, where 1≦n≦7.

[0258] 2560MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 58.300 + (n-1) × 2.560, 1 ≤ n ≤ 3, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 2.560, 1 ≤ n ≤ 3, and the formula for calculating the maximum frequency is f H (n) [GHz] = 59.580 + (n-1) × 2.560, 1≦n≦3.

[0259] For example, the first frequency band includes the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band. The first frequency band may be divided into thirty-five 160 MHz channels, seventeen 320 MHz channels, eight 640 MHz channels, three 1280 MHz channels, or one 2560 MHz channel.

[0260] A channel division scheme for the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band is described herein, in which the remaining bandwidth in both the front and rear portions of the 42.3-47.0 GHz frequency band is 110 MHz, and the remaining bandwidth in both the front and rear portions of the 47.2-48.4 GHz frequency band is 40 MHz.

[0261] 160MHz channel division: The 42.3-47.0GHz frequency band contains 28 160MHz channels, and the 47.2-48.4GHz frequency band contains 7 160MHz channels. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.490 + (n-1) × 0.160, 1 ≤ n ≤ 28, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.410 + (n-1) × 0.160, 1 ≤ n ≤ 28, and the formula for calculating the maximum frequency is fH (n) [GHz] = 42.570 + (n-1) × 0.160, 1 ≤ n ≤ 28. For the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 47.320 + (n-29) × 0.160, 29 ≤ n ≤ 35, and the formula for calculating the minimum frequency is f L (n) [GHz] = 47.240 + (n-29) × 0.160, 29 ≤ n ≤ 35, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.400 + (n - 29) × 0.160, where 29 ≤ n ≤ 35.

[0262] 320MHz channel division: The 42.3-47.0GHz frequency band contains 14 320MHz channels, and the 47.2-48.4GHz frequency band contains 3 320MHz channels. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.570 + (n-1) × 0.320, 1 ≤ n ≤ 14, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.410 + (n-1) × 0.320, 1 ≤ n ≤ 14, and the formula for calculating the maximum frequency is f H (n) [GHz] = 42.730 + (n-1) × 0.320, 1 ≤ n ≤ 14. For the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 47.400 + (n-15) × 0.320, 15 ≤ n ≤ 17, and the formula for calculating the minimum frequency is f L (n)[GHz]=47.240+(n-15)×0.320, 15≦n≦17, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.560 + (n-15) × 0.320, where 15≦n≦17.

[0263] 640MHz Channel Division: The 42.3-47.0GHz frequency band contains seven 640MHz channels, and the 47.2-48.4GHz frequency band contains one 640MHz channel. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.370 + (n-1) × 0.640, 1 ≤ n ≤ 7, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.410 + (n-1) × 0.640, 1 ≤ n ≤ 7, and the formula for calculating the maximum frequency is f H (n) [GHz] = 43.050 + (n-1) × 0.640, 1 ≤ n ≤ 7, for the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz]=47.560+(n-8)×0.640, n=8, and the formula for calculating the minimum frequency is f L (n)[GHz]=47.240+(n-8)×0.640, n=8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.880 + (n-8) × 0.640, n = 8.

[0264] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 43.050 + (n-1) × 1.280, where 1 ≤ n ≤ 3, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.410 + (n-1) × 1.280, 1 ≤ n ≤ 3, and the formula for calculating the maximum frequency is f H (n) [GHz] = 43.690 + (n-1) × 1.280, where 1≦n≦3.

[0265] 2560MHz channel division: The formula for calculating the center frequency is f C (n)[GHz]=43.690+(n-1)×2.560, n=1, and the formula for calculating the minimum frequency is f L (n)[GHz]=42.410+(n-1)×2.560, n=1, and the formula for calculating the maximum frequency is f H(n) [GHz] = 44.970 + (n-1) × 2.560, n = 1.

[0266] For example, the first frequency band is the 56.16 to 73.44 GHz frequency band. The first frequency band may be divided into 108 160 MHz channels, 54 320 MHz channels, 27 640 MHz channels, 13 1280 MHz channels, or 6 2560 MHz channels.

[0267] A channel division scheme for the 56.16-73.44 GHz frequency band is described herein, in which there is no remaining bandwidth in the front and rear portions of the first frequency band.

[0268] 160MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.240 + (n-1) × 0.160, 1 ≤ n ≤ 108, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 0.160, 1 ≤ n ≤ 108, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.320 + (n-1) × 0.160, 1≦n≦108.

[0269] 320MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.320 + (n-1) × 0.320, 1 ≤ n ≤ 54, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 0.320, 1 ≤ n ≤ 54, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.480 + (n-1) × 0.320, where 1≦n≦54.

[0270] 640MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.480 + (n-1) × 0.640, 1 ≤ n ≤ 27, and the formula for calculating the minimum frequency is f L(n)[GHz]=56.160+(n-1)×0.640, 1≦n≦27, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.800 + (n-1) × 0.640, where 1≦n≦27.

[0271] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.800 + (n-1) × 1.280, 1 ≤ n ≤ 13, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 1.280, 1 ≤ n ≤ 13, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.440 + (n-1) × 1.280, where 1≦n≦13.

[0272] 2560MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.440 + (n-1) × 2.560, 1 ≤ n ≤ 6, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 2.560, 1 ≤ n ≤ 6, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.720 + (n-1) × 2.560, where 1≦n≦6.

[0273] Example 7: The minimum channel bandwidth supported in the first frequency band is 640 MHz, in other words, the bandwidth of the first fundamental channel is 640 MHz. The bandwidths supported in the first frequency band may include one or more of the following bandwidths: 640 MHz, 1280 MHz, 2560 MHz, and 5120 MHz.

[0274] Optionally, for the amount of subcarriers (or IDFT size / DFT size) corresponding to each channel bandwidth in the channel division manner of Example d, refer to the relevant description of Example 2 in Example 1 of Embodiment 1. Details are not described again in this specification.

[0275] For example, the first frequency band is a 57-66 GHz frequency band. The first frequency band may be divided into fourteen 640 MHz channels, seven 1280 MHz channels, three 2560 MHz channels, or one 5120 MHz channel, as shown in Figure 6. In this division manner, a 20 MHz bandwidth is left in each of the front and rear portions of the first frequency band, as shown in Figure 7.

[0276] A channel division scheme for the 57-66 GHz frequency band is described herein.

[0277] 640MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.340 + (n-1) × 0.640, 1 ≤ n ≤ 14, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.020 + (n-1) × 0.640, 1 ≤ n ≤ 14, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.660 + (n-1) × 0.640, where 1≦n≦14.

[0278] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.660 + (n-1) × 1.280, 1 ≤ n ≤ 7, and the formula for calculating the low frequency is f L (n) [GHz] = 57.020 + (n-1) × 1.280, where 1 ≤ n ≤ 7, and the formula for calculating the high frequency is f H (n) [GHz] = 58.300 + (n-1) × 1.280, where 1≦n≦7.

[0279] 2560MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 58.300 + (n-1) × 2.560, where 1 ≤ n ≤ 3, and the formula for calculating the low frequency is f L (n) [GHz] = 57.020 + (n-1) × 2.560, where 1 ≤ n ≤ 3, and the formula for calculating the high frequency is f H(n) [GHz] = 59.580 + (n-1) × 2.560, 1≦n≦3.

[0280] 5120MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 59.580 + (n-1) × 5.120, n = 1, and the formula for calculating the low frequency is f L (n) [GHz] = 57.020 + (n-1) × 5.120, n = 1, and the formula for calculating the high frequency is f H (n) [GHz] = 62.140 + (n-1) × 5.120, n = 1.

[0281] For example, the first frequency band includes the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band. The first frequency band may be divided into eight 640 MHz channels, three 1280 MHz channels, or one 2560 MHz channel.

[0282] A channel division scheme for the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band is described herein, in which the remaining bandwidth in both the front and rear portions of the 42.3-47.0 GHz frequency band is 110 MHz, and the remaining bandwidth in both the front and rear portions of the 47.2-48.4 GHz frequency band is 120 MHz.

[0283] 640MHz Channel Division: The 42.3-47.0GHz frequency band contains seven 640MHz channels, and the 47.2-48.4GHz frequency band contains one 640MHz channel. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.730 + (n-1) × 0.640, 1 ≤ n ≤ 7, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.410 + (n-1) × 0.640, 1 ≤ n ≤ 7, and the formula for calculating the maximum frequency is f H(n) [GHz] = 43.050 + (n-1) × 0.640, 1 ≤ n ≤ 7, for the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz]=47.640+(n-8)×0.640, n=8, and the formula for calculating the minimum frequency is f L (n)[GHz]=47.320+(n-8)×0.640, n=8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.960 + (n-8) × 0.640, n = 8.

[0284] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 43.050 + (n-1) × 1.280, where 1 ≤ n ≤ 3, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.410 + (n-1) × 1.280, 1 ≤ n ≤ 3, and the formula for calculating the maximum frequency is f H (n) [GHz] = 43.690 + (n-1) × 1.280, where 1≦n≦3.

[0285] 2560MHz channel division: The formula for calculating the center frequency is f C (n)[GHz]=43.690+(n-1)×2.560, n=1, and the formula for calculating the minimum frequency is f L (n)[GHz]=42.410+(n-1)×2.560, n=1, and the formula for calculating the maximum frequency is f H (n) [GHz] = 44.970 + (n-1) × 2.560, n = 1.

[0286] For example, the first frequency band is the 56.16 to 73.44 GHz frequency band. The first frequency band may be divided into twenty-seven 640 MHz channels, thirteen 1280 MHz channels, six 2560 MHz channels, three 5120 MHz channels, or one 10240 MHz channel.

[0287] A channel division scheme for the 56.16-73.44 GHz frequency band is described herein, in which there is no remaining bandwidth in the front and rear portions of the first frequency band.

[0288] 640MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.480 + (n-1) × 0.640, 1 ≤ n ≤ 27, and the formula for calculating the minimum frequency is f L (n)[GHz]=56.160+(n-1)×0.640, 1≦n≦27, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.800 + (n-1) × 0.640, where 1≦n≦27.

[0289] 1280MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.800 + (n-1) × 1.280, 1 ≤ n ≤ 13, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 1.280, 1 ≤ n ≤ 13, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.440 + (n-1) × 1.280, where 1≦n≦13.

[0290] 2560MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.440 + (n-1) × 2.560, 1 ≤ n ≤ 6, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 2.560, 1 ≤ n ≤ 6, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.720 + (n-1) × 2.560, where 1≦n≦6.

[0291] 5120MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 58.720 + (n-1) × 5.120, 1 ≤ n ≤ 3, and the formula for calculating the minimum frequency is fL (n) [GHz] = 56.160 + (n-1) × 5.120, 1 ≤ n ≤ 3, and the formula for calculating the maximum frequency is f H (n) [GHz] = 61.280 + (n-1) × 5.120, 1≦n≦3.

[0292] 10240MHz channel division: The center frequency is 61.280, the lowest frequency is 56.160, and the highest frequency is 66.400.

[0293] In the channel division manner provided in this application, n is the sequence number or index number of the channel. The details are not described again below.

[0294] It should be understood that the division patterns of Examples 4 to 7 are merely illustrative examples, and the locations of the center frequency, the lowest frequency, and the highest frequency of the channel are not particularly limited. Any channel division pattern may be used, provided that the bandwidth is divided based on an integer multiple of 20 MHz and the subcarrier spacing is appropriate.

[0295] In the channel division manner provided in the embodiment 2 of the present application, the spectrum resources in the first frequency band can be used as much as possible by channel division with smaller granularity. This reduces resource waste and improves resource utilization. In addition, the first frequency band is divided into channels based on an integer multiple of the minimum channel bandwidth of the second frequency band, so that the signals in the first frequency band and the signals in the second frequency band can be transmitted by reusing the same crystal oscillator. In addition, the correspondence between the bandwidth and the IDFT size / DFT size (or the amount of subcarriers) in the second frequency band is reused in the first frequency band, so that the first frequency band and the second frequency band can share a signal processing device, for example, the IDFT device or the DFT device is reused. Therefore, according to the above embodiment 2, the hardware complexity can be reduced, the hardware device overhead can be reduced, and the device cost can be reduced.

[0296] EMBODIMENT 3

[0297] In the channel division method provided in the embodiment 3 of the present application, the bandwidth of the first frequency band can be divided according to the minimum channel bandwidth of the millimeter wave band. In other words, the channel boundary of the embodiment 3 coincides with the channel boundary of the millimeter wave band in the first frequency band. In the embodiment 3, to obtain a channel in the first frequency band, R bisections are performed on the minimum channel bandwidth of the millimeter wave band, or the minimum channel bandwidth and performance of T millimeter wave bands are aggregated to obtain a channel in the first frequency band, where R is an integer greater than or equal to 0, and T is an integer greater than or equal to 0.

[0298] It is assumed that the minimum channel bandwidth of the millimeter wave band is 2160MHz. In the channel division method provided in embodiment 3 of the present application, the bandwidth of the first frequency band may be divided based on 2160MHz, in other words, the channel boundary of embodiment 3 coincides with the channel boundary of the millimeter wave band in the first frequency band. In embodiment 3, to obtain a channel in the first frequency band, R bisections are performed on the 2160MHz bandwidth, or T 2160MHz channels are aggregated to obtain a channel in the first frequency band, where R is an integer greater than or equal to 0, and T is an integer greater than or equal to 0.

[0299] It is assumed that the minimum channel bandwidth of the millimeter wave band is 540 MHz. In the channel division method provided in embodiment 3 of the present application, the bandwidth of the first frequency band may be divided based on 540 MHz, in other words, the channel boundary of embodiment 3 coincides with the channel boundary of the millimeter wave band in the first frequency band. In embodiment 3, to obtain a channel in the first frequency band, R bisections are performed on the 540 MHz bandwidth, or T 540 MHz channels are aggregated to obtain a channel in the first frequency band, where R is an integer greater than or equal to 0, and T is an integer greater than or equal to 0.

[0300] Therefore, the first frequency band can support the minimum channel bandwidth of the millimeter wave band so as to fit the device of the millimeter wave band. In the third embodiment, the bandwidth supported in the first frequency band may be an integer multiple of the minimum channel bandwidth of the second frequency band, or may not be an integer multiple of the minimum channel bandwidth of the second frequency band. Optionally, when the bandwidth of the channel in the first frequency band is not an integer multiple of the minimum channel bandwidth of the second frequency band, the channel carrier frequency may be implemented by using a fractional-N phase-locked loop, in other words, the multiple sampling clocks are implemented by using an external crystal oscillator and by internal fractional frequency multiplication.

[0301] According to the solution provided in embodiment 3, the millimeter wave band device using the low frequency band (i.e., the second frequency band) can be compatible, so that the signal in the high frequency band, the signal in the low frequency band, and the signal in the millimeter wave band can be transmitted by reusing the same crystal oscillator. This can reduce the hardware complexity and reduce the hardware overhead.

[0302] In the following, the channel division method for the first frequency band is described by using an example for a specific frequency band.

[0303] Example 8: The minimum channel bandwidth supported in the first frequency band is 270 MHz, in other words, the bandwidth of the first fundamental channel is 270 MHz. The bandwidths supported in the first frequency band may include one or more of the following bandwidths: 270 MHz, 540 MHz, 1080 MHz, 2160 MHz, and 4320 MHz.

[0304] Optionally, for the amount of subcarriers (or IDFT size / DFT size) corresponding to each channel bandwidth in the channel division manner of Example 8, refer to the relevant description of Example 6 in Example 2 of Embodiment 1. Details are not described again in this specification.

[0305] For example, the first frequency band is a 57-66 GHz frequency band. The first frequency band may be divided into 32 270 MHz channels, 16 540 MHz channels, 8 1080 MHz channels, 4 2160 MHz channels, or 2 4320 MHz channels, as shown in FIG.

[0306] A channel division scheme for the 57-66 GHz frequency band is described herein, in which the remaining bandwidths in the front and rear parts of the first frequency band are 240 MHz and 120 MHz, respectively, as shown in FIG.

[0307] 270MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.375 + (n-1) × 0.270, 1 ≤ n ≤ 32, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.240 + (n-1) × 0.270, 1 ≤ n ≤ 32, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.510 + (n-1) × 0.270, where 1≦n≦32.

[0308] 540MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.510 + (n-1) × 0.540, 1 ≤ n ≤ 16, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.240 + (n-1) × 0.540, 1 ≤ n ≤ 16, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.780 + (n-1) × 0.540, where 1≦n≦16.

[0309] 1080MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.780 + (n-1) × 1.080, 1 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.240 + (n-1) × 1.080, 1 ≤ n ≤ 8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.320 + (n-1) × 1.080, where 1≦n≦8.

[0310] 2160MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 58.320 + (n-1) × 2.160, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.240 + (n-1) × 2.160, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 59.400 + (n-1) × 2.160, where 1≦n≦4.

[0311] 4320MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 59.400 + (n-1) × 4.320, 1 ≤ n ≤ 2, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.240 + (n-1) × 4.320, 1 ≤ n ≤ 2, and the formula for calculating the maximum frequency is f H(n) [GHz] = 61.560 + (n-1) × 4.320, where 1≦n≦2.

[0312] For example, the first frequency band includes the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band. The first frequency band may be divided into twenty 270 MHz channels, ten 540 MHz channels, five 1080 MHz channels, two 2160 MHz channels, or one 4320 MHz channel. Optionally, in this channel division manner, the channel division boundaries coincide with the channel boundaries of the millimeter wave bands in the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band.

[0313] A channel division scheme for the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band is described herein, in which the remaining bandwidths in the front and rear parts of the 42.3-47.0 GHz frequency band are 90 MHz and 290 MHz, respectively, and the remaining bandwidths in the front and rear parts of the 47.2-48.4 GHz frequency band are 50 MHz and 70 MHz, respectively.

[0314] 270MHz channel division: The 42.3-47.0GHz frequency band contains 16 270MHz channels, and the 47.2-48.4GHz frequency band contains 4 270MHz channels. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.525 + (n-1) × 0.270, 1 ≤ n ≤ 16, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.390 + (n-1) × 0.270, 1 ≤ n ≤ 16, and the formula for calculating the maximum frequency is f H (n) [GHz] = 42.660 + (n-1) × 0.270, 1 ≤ n ≤ 16. For the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C(n) [GHz] = 47.385 + (n-9) × 0.270, 17 ≤ n ≤ 20, and the formula for calculating the minimum frequency is f L (n) [GHz] = 47.250 + (n-9) × 0.270, 17 ≤ n ≤ 20, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.520 + (n-9) × 0.270, where 17≦n≦20.

[0315] 540MHz channel division: The 42.3-47.0GHz frequency band contains eight 540MHz channels, and the 47.2-48.4GHz frequency band contains two 540MHz channels. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.660 + (n-1) × 0.540, 1 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n)[GHz]=42.390+(n-1)×0.540, 1≦n≦8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 42.930 + (n-1) × 0.540, 1 ≤ n ≤ 8, for the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 47.520 + (n-9) × 0.540, 9 ≤ n ≤ 10, and the formula for calculating the minimum frequency is f L (n) [GHz] = 47.250 + (n-9) × 0.540, 9 ≤ n ≤ 10, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.790 + (n-9) × 0.540, where 9≦n≦10.

[0316] 1080MHz Channel Division: The 42.3-47.0GHz frequency band contains four 1080MHz channels, and the 47.2-48.4GHz frequency band contains one 1080MHz channel. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.930 + (n-1) × 1.080, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L(n) [GHz] = 42.390 + (n-1) × 1.080, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 43.470 + (n-1) × 1.080, 1 ≤ n ≤ 4, for the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz]=47.790+(n-5)×1.080, n=5, and the formula for calculating the minimum frequency is f L (n)[GHz]=47.250+(n-5)×1.080, n=5, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.330 + (n-5) × 1.080, n = 5.

[0317] 2160MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 43.470 + (n-1) × 2.160, 1 ≤ n ≤ 2, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.390 + (n-1) × 2.160, 1 ≤ n ≤ 2, and the formula for calculating the maximum frequency is f H (n) [GHz] = 44.550 + (n-1) × 2.160, where 1≦n≦2.

[0318] 4320MHz channel division: The formula for calculating the center frequency is f C (n)[GHz]=44.550+(n-1)×4.320, n=1, and the formula for calculating the minimum frequency is f L (n)[GHz]=42.390+(n-1)×4.320, n=1, and the formula for calculating the maximum frequency is f H (n) [GHz] = 46.710 + (n-1) × 4.320, n = 1.

[0319] For example, the first frequency band is the 56.16 to 73.44 GHz frequency band. The first frequency band may be divided into 64 270 MHz channels, 32 540 MHz channels, 16 1080 MHz channels, 8 2160 MHz channels, or 4 4320 MHz channels.

[0320] A channel division scheme for the 56.16-73.44 GHz frequency band is described herein, in which there is no remaining bandwidth in the front and rear portions of the first frequency band.

[0321] 270MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.295 + (n-1) × 0.270, 1 ≤ n ≤ 64, and the formula for calculating the minimum frequency is f L (n)[GHz]=56.160+(n-1)×0.270, 1≦n≦64, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.430 + (n-1) × 0.270, 1≦n≦64.

[0322] 540MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.430 + (n-1) × 0.540, 1 ≤ n ≤ 32, and the formula for calculating the minimum frequency is f L (n)[GHz]=56.160+(n-1)×0.540, 1≦n≦32, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.700 + (n-1) × 0.540, where 1≦n≦32.

[0323] 1080MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.700 + (n-1) × 1.080, 1 ≤ n ≤ 16, and the formula for calculating the minimum frequency is f L (n)[GHz]=56.160+(n-1)×1.080, 1≦n≦16, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.240 + (n-1) × 1.080, where 1≦n≦16.

[0324] 2160MHz channel division: The formula for calculating the center frequency is f C(n) [GHz] = 57.240 + (n-1) × 2.160, 1 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n)[GHz]=56.160+(n-1)×2.160, 1≦n≦8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.320 + (n-1) × 2.160, 1≦n≦8.

[0325] 4320MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 58.320 + (n-1) × 4.320, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 4.320, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 60.480 + (n-1) × 4.320, where 1≦n≦4.

[0326] Example 9: The minimum channel bandwidth supported in the first frequency band is 540 MHz, in other words, the bandwidth of the first fundamental channel is 540 MHz. The bandwidths supported in the first frequency band may include one or more of the following bandwidths: 540 MHz, 1080 MHz, 2160 MHz, and 4320 MHz.

[0327] Optionally, for the amount of subcarriers (or IDFT size / DFT size) corresponding to each channel bandwidth in the channel division manner of Example 9, refer to the relevant description of Example 7 in Example 2 of Embodiment 1. Details are not described again in this specification.

[0328] For example, the first frequency band is a 57-66 GHz frequency band, and may be divided into sixteen 540 MHz channels, eight 1080 MHz channels, four 2160 MHz channels, two 4320 MHz channels, or one 8640 MHz channel.

[0329] A channel division scheme for the 57-66 GHz frequency band is described herein, in which the remaining bandwidth in the front and rear parts of the first frequency band is 240 MHz and 120 MHz, respectively.

[0330] 540MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.510 + (n-1) × 0.540, 1 ≤ n ≤ 16, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.240 + (n-1) × 0.540, 1 ≤ n ≤ 16, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.780 + (n-1) × 0.540, where 1≦n≦16.

[0331] 1080MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.780 + (n-1) × 1.080, 1 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.240 + (n-1) × 1.080, 1 ≤ n ≤ 8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.320 + (n-1) × 1.080, where 1≦n≦8.

[0332] 2160MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 58.320 + (n-1) × 2.160, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.240 + (n-1) × 2.160, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 59.400 + (n-1) × 2.160, where 1≦n≦4.

[0333] 4320MHz channel division: The formula for calculating the center frequency is f C(n) [GHz] = 59.400 + (n-1) × 4.320, 1 ≤ n ≤ 2, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.240 + (n-1) × 4.320, 1 ≤ n ≤ 2, and the formula for calculating the maximum frequency is f H (n) [GHz] = 61.560 + (n-1) × 4.320, where 1≦n≦2.

[0334] 8640MHz channel division: The formula for calculating the center frequency is f C (n)[GHz]=61.560+(n-1)×8640, n=1, and the formula for calculating the minimum frequency is f L (n)[GHz]=57.240+(n-1)×8640, n=1, and the formula for calculating the maximum frequency is f H (n) [GHz] = 65.880 + (n-1) × 8640, n = 1.

[0335] For example, the first frequency band includes the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band. The first frequency band may be divided into ten 540 MHz channels, five 1080 MHz channels, two 2160 MHz channels, or one 4320 MHz channel. Optionally, in this channel division manner, the channel division boundaries coincide with the channel boundaries of the millimeter wave bands in the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band.

[0336] A channel division scheme for the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band is described herein, in which the remaining bandwidths in the front and rear parts of the 42.3-47.0 GHz frequency band are 90 MHz and 290 MHz, respectively, and the remaining bandwidths in the front and rear parts of the 47.2-48.4 GHz frequency band are 50 MHz and 70 MHz, respectively.

[0337] 540MHz channel division: The 42.3-47.0GHz frequency band contains eight 540MHz channels, and the 47.2-48.4GHz frequency band contains two 540MHz channels. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.660 + (n-1) × 0.540, 1 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n)[GHz]=42.390+(n-1)×0.540, 1≦n≦8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 42.930 + (n-1) × 0.540, 1 ≤ n ≤ 8, for the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 47.520 + (n-9) × 0.540, 9 ≤ n ≤ 10, and the formula for calculating the minimum frequency is f L (n) [GHz] = 47.250 + (n-9) × 0.540, 9 ≤ n ≤ 10, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.790 + (n-9) × 0.540, where 9≦n≦10.

[0338] 1080MHz Channel Division: The 42.3-47.0GHz frequency band contains four 1080MHz channels, and the 47.2-48.4GHz frequency band contains one 1080MHz channel. For the 42.3-47.0GHz frequency band, the formula for calculating the center frequency is f C (n) [GHz] = 42.930 + (n-1) × 1.080, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.390 + (n-1) × 1.080, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 43.470 + (n-1) × 1.080, 1 ≤ n ≤ 4, for the 47.2 to 48.4 GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz]=47.790+(n-5)×1.080, n=5, and the formula for calculating the minimum frequency is f L(n)[GHz]=47.250+(n-5)×1.080, n=5, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.330 + (n-5) × 1.080, n = 5.

[0339] 2160MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 43.470 + (n-1) × 2.160, 1 ≤ n ≤ 2, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.390 + (n-1) × 2.160, 1 ≤ n ≤ 2, and the formula for calculating the maximum frequency is f H (n) [GHz] = 44.550 + (n-1) × 2.160, where 1≦n≦2.

[0340] 4320MHz channel division: The formula for calculating the center frequency is f C (n)[GHz]=44.550+(n-1)×4.320, n=1, and the formula for calculating the minimum frequency is f L (n)[GHz]=42.390+(n-1)×4.320, n=1, and the formula for calculating the maximum frequency is f H (n) [GHz] = 46.710 + (n-1) × 4.320, n = 1.

[0341] For example, the first frequency band is the 56.16 to 73.44 GHz frequency band. The first frequency band may be divided into 32 540 MHz channels, 16 1080 MHz channels, 8 2160 MHz channels, 4 4320 MHz channels, or 2 8640 MHz channels.

[0342] A channel division scheme for the 56.16-73.44 GHz frequency band is described herein, in which there is no remaining bandwidth in the front and rear portions of the first frequency band.

[0343] 540MHz channel division: The formula for calculating the center frequency is f C(n) [GHz] = 56.430 + (n-1) × 0.540, 1 ≤ n ≤ 32, and the formula for calculating the minimum frequency is f L (n)[GHz]=56.160+(n-1)×0.540, 1≦n≦32, and the formula for calculating the maximum frequency is f H (n) [GHz] = 56.700 + (n-1) × 0.540, where 1≦n≦32.

[0344] 1080MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 56.700 + (n-1) × 1.080, 1 ≤ n ≤ 16, and the formula for calculating the minimum frequency is f L (n)[GHz]=56.160+(n-1)×1.080, 1≦n≦16, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.240 + (n-1) × 1.080, where 1≦n≦16.

[0345] 2160MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 57.240 + (n-1) × 2.160, 1 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n)[GHz]=56.160+(n-1)×2.160, 1≦n≦8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.320 + (n-1) × 2.160, 1≦n≦8.

[0346] 4320MHz channel division: The formula for calculating the center frequency is f C (n) [GHz] = 58.320 + (n-1) × 4.320, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.160 + (n-1) × 4.320, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 60.480 + (n-1) × 4.320, where 1≦n≦4.

[0347] 8640MHz channel division: The formula for calculating the center frequency is f C (n)[GHz]=60.480+(n-1)×8640, n=1,2, and the formula for calculating the minimum frequency is f L (n)[GHz]=56.160+(n-1)×8640, n=1,2, and the formula for calculating the maximum frequency is f H (n) [GHz] = 64.800 + (n-1) × 8640, n = 1, 2.

[0348] It should be understood that the division patterns of Examples 8 and 9 are merely illustrative examples, and the locations of the center frequency, the lowest frequency, and the highest frequency of the channel are not particularly limited. Any channel division pattern may be used, provided that the bandwidth is divided based on the minimum channel bandwidth of the millimeter wave band, and the subcarrier spacing is appropriate.

[0349] In the channel division manner provided in the embodiment 2 of the present application, the spectrum resource in the first frequency band can be used as much as possible by the channel division with smaller granularity. This reduces resource waste and improves resource utilization. In addition, according to the embodiment 3, the signal in the first frequency band, the signal in the second frequency band, and the signal in the millimeter wave band can be transmitted by reusing the same crystal oscillator. In addition, the correspondence between the bandwidth and the IDFT size / DFT size (or the amount of subcarriers) in the second frequency band is reused in the first frequency band, so that the first frequency band and the second frequency band can share the signal processing device, for example, the IDFT device or the DFT device is reused. Therefore, according to the above embodiment 3, the hardware complexity can be reduced, the hardware device overhead can be reduced, and the device cost can be reduced.

[0350] EMBODIMENT 4

[0351] In the fourth embodiment of the present application, the bandwidth is also divided according to the minimum channel bandwidth of the millimeter wave band. In other words, the channel boundaries of the fourth embodiment coincide with the channel boundaries of the millimeter wave band in the first frequency band.

[0352] It is assumed that the minimum channel bandwidth of the millimeter wave band is 2160 MHz. In the fourth embodiment of the present application, the bandwidth is divided based on 2160 MHz, in other words, the channel boundary of the fourth embodiment coincides with the channel boundary of the millimeter wave band in the first frequency band.

[0353] It is assumed that the minimum channel bandwidth of the millimeter wave band is 540 MHz. In the fourth embodiment of the present application, the bandwidth is divided based on 540 MHz, in other words, the channel boundary of the fourth embodiment coincides with the channel boundary of the millimeter wave band in the first frequency band.

[0354] The difference between embodiment 4 and embodiment 3 includes, for example, that in embodiment 3, bisection or aggregation is performed based on the minimum channel bandwidth of the millimeter wave band, while in embodiment 4, bisection or aggregation is performed based on the first subband, the bandwidth of the first subband is smaller than the minimum channel bandwidth of the millimeter wave band, the first subband is selected from the minimum channel bandwidth supported in the millimeter wave band protocol IEEE 802.11ad / ay or 802.11aj, and the center frequencies of the millimeter wave band and the first subband overlap. The center frequencies of the millimeter wave band and the first subband overlap can also be understood as the center frequencies of the millimeter wave band and the first subband being the same. For example, the minimum channel bandwidth of the millimeter wave band is 540MHz, and the bandwidth of the first subband is 360MHz, 480MHz, 500MHz, 520MHz, etc. The minimum channel bandwidth of the millimeter wave band is 2160 MHz, and the bandwidth of the first sub-band is 1000 MHz, 1800 MHz, 1940 MHz, 2000 MHz, 2100 MHz, etc. In a possible implementation, the bandwidth of the first sub-band may be an integer multiple of 20 MHz.

[0355] For ease of understanding, the following uses an example in which the minimum channel bandwidth of the millimeter wave band is 2160 MHz and the first sub-band is 2000 MHz for description. In embodiment 4, the bandwidth supported in the first frequency band may be an integer multiple of the minimum channel bandwidth of the second frequency band, or may not be an integer multiple of the minimum channel bandwidth of the second frequency band. Optionally, when the bandwidth of the channel in the first frequency band is not an integer multiple of the minimum channel bandwidth of the second frequency band, the channel carrier frequency may be implemented by using a fractional-N phase-locked loop, in other words, the multiple sampling clocks are implemented by using an external crystal oscillator and by internal fractional frequency multiplication.

[0356] According to the solution provided in embodiment 4, the millimeter wave band device using the low frequency band (i.e., the second frequency band) can be compatible, so that the signal in the high frequency band, the signal in the low frequency band, and the signal in the millimeter wave band can be transmitted by reusing the same crystal oscillator. This can reduce the hardware complexity and reduce the hardware overhead.

[0357] In the following, the channel division method for the first frequency band is described by using an example for a specific frequency band.

[0358] Example 10: The minimum channel bandwidth supported in the first frequency band is 250 MHz, in other words, the bandwidth of the first basic channel is 250 MHz. The bandwidth supported in the first frequency band may include one or more of the following bandwidths: 250 MHz, 500 MHz, 1000 MHz, 2000 MHz, and 4000 MHz. In this example, the center frequency of 2000 MHz overlaps with the center frequency of 2160 MHz obtained by dividing based on the millimeter wave band. The 1000 MHz channel is obtained by dividing the 2000 MHz channel by two times, and the 500 MHz channel is obtained by dividing the 2000 MHz channel by three times. The 4000 MHz channel is obtained by aggregating two 2000 MHz channels. It should be understood that the two 2000 MHz channels included in the 4000 MHz bandwidth are not contiguous.

[0359] Optionally, for the amount of subcarriers (or IDFT size / DFT size) corresponding to each channel bandwidth in the channel division manner of Example 10, refer to the relevant description of Example 8 in Example 3 of Embodiment 1. Details are not described again in this specification.

[0360] For example, the first frequency band is a 57-66 GHz frequency band. The first frequency band may be divided into 32 250 MHz channels, 16 500 MHz channels, 8 1000 MHz channels, 4 2000 MHz channels, or 2 4000 MHz channels, as shown in FIG.

[0361] A channel division scheme for the 57-66 GHz frequency band is described herein, in which the remaining bandwidth in the front and rear parts of the first frequency band is 320 MHz and 200 MHz, respectively.

[0362] 250MHz Channel Division: The 57-66GHz frequency band is divided based on the millimeter wave band to obtain four 2160MHz frequency bands, which are consecutively indicated as the first 2160MHz frequency band, the second 2160MHz frequency band, the third 2160MHz frequency band, and the fourth 2160MHz frequency band. 2000MHz is selected from each of the above 2160MHz frequency bands to obtain four 2000MHz frequency bands, which are consecutively indicated as the first 2000MHz frequency band, the second 2000MHz frequency band, the third 2000MHz frequency band, and the fourth 2000MHz frequency band. The center frequency of the 2000MHz frequency band selected from each 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. For example, the center frequency of the first 2160MHz frequency band overlaps with or is the same as the center frequency of the first 2000MHz frequency band. It can be understood that the selection of the 2000 MHz frequency band in this embodiment refers to the above format, and the details will not be described again later.

[0363] To obtain eight 250 MHz channels, the first 2000 MHz frequency band is bisected three times, and the formula for calculating the center frequencies of the eight 250 MHz channels is f C (n) [GHz] = 57.445 + (n-1) × 0.250, 1 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L(n) [GHz] = 57.320 + (n-1) × 0.250, 1 ≤ n ≤ 8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.570 + (n-1) × 0.250, 1≦n≦8.

[0364] To obtain eight 250 MHz channels, the second 2000 MHz frequency band is bisected three times, and the formula for calculating the center frequencies of the eight 250 MHz channels is f C (n) [GHz] = 59.605 + (n-9) × 0.250, 9 ≤ n ≤ 16, and the formula for calculating the minimum frequency is f L (n) [GHz] = 59.480 + (n-9) × 0.250, 9 ≤ n ≤ 16, and the formula for calculating the maximum frequency is f H (n) [GHz] = 59.730 + (n-9) × 0.250, where 9≦n≦16.

[0365] The third 2000 MHz frequency band is bisected three times to obtain eight 250 MHz channels, and the formula for calculating the center frequencies of the eight 250 MHz channels is f C (n) [GHz] = 61.765 + (n-17) × 0.250, 17 ≤ n ≤ 24, and the formula for calculating the minimum frequency is f L (n) [GHz] = 61.640 + (n-17) × 0.250, 17 ≤ n ≤ 24, and the formula for calculating the maximum frequency is f H (n) [GHz] = 61.890 + (n-17) × 0.250, where 17≦n≦24.

[0366] The fourth 2000 MHz frequency band is bisected three times to obtain eight 250 MHz channels, and the formula for calculating the center frequencies of the eight 250 MHz channels is f C (n) [GHz] = 63.925 + (n-25) × 0.250, 25 ≤ n ≤ 32, and the formula for calculating the minimum frequency is f L (n)[GHz]=63.800+(n-25)×0.250, 25≦n≦32, and the formula for calculating the maximum frequency is f H(n) [GHz] = 64.050 + (n - 25) × 0.250, where 25≦n≦32.

[0367] 500MHz Channel Division: The 57-66GHz frequency band is divided based on the millimeter wave band to obtain four 2160MHz frequency bands, which are consecutively indicated as the first 2160MHz frequency band, the second 2160MHz frequency band, the third 2160MHz frequency band, and the fourth 2160MHz frequency band. 2000MHz is selected from each of the above 2160MHz frequency bands to obtain four 2000MHz frequency bands, which are consecutively indicated as the first 2000MHz frequency band, the second 2000MHz frequency band, the third 2000MHz frequency band, and the fourth 2000MHz frequency band. The center frequency of the 2000MHz frequency band selected from each 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. For example, the center frequency of the first 2160MHz frequency band overlaps with or is the same as the center frequency of the first 2000MHz frequency band.

[0368] To obtain four 500 MHz channels, the first 2000 MHz frequency band is bisected twice, and the formula for calculating the center frequencies of the four 500 MHz channels is f C (n) [GHz] = 57.570 + (n-1) × 0.500, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.320 + (n-1) × 0.500, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.820 + (n-1) × 0.500, 1≦n≦4.

[0369] To obtain four 500 MHz channels, the second 2000 MHz frequency band is bisected twice, and the formula for calculating the center frequencies of the four 500 MHz channels is f C (n) [GHz] = 59.730 + (n-5) × 0.500, 5 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L(n)[GHz]=59.480+(n-5)×0.500, 5≦n≦8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 59.980 + (n-5) × 0.500, 5≦n≦8.

[0370] The third 2000 MHz frequency band is bisected twice to obtain four 500 MHz channels, and the formula for calculating the center frequencies of the four 500 MHz channels is f C (n) [GHz] = 61.890 + (n-9) × 0.500, 9 ≤ n ≤ 12, and the formula for calculating the minimum frequency is f L (n) [GHz] = 61.640 + (n-9) × 0.500, 9 ≤ n ≤ 12, and the formula for calculating the maximum frequency is f H (n) [GHz] = 62.140 + (n-9) × 0.500, where 9≦n≦12.

[0371] The fourth 2000 MHz frequency band is bisected twice to obtain four 500 MHz channels, and the formula for calculating the center frequencies of the four 500 MHz channels is f C (n) [GHz] = 64.050 + (n-13) × 0.500, 13 ≤ n ≤ 16, and the formula for calculating the minimum frequency is f L (n)[GHz]=63.800+(n-13)×0.500, 13≦n≦16, and the formula for calculating the maximum frequency is f H (n) [GHz] = 64.300 + (n-13) × 0.500, where 13≦n≦16.

[0372] 1000MHz Channel Division: The 57-66GHz frequency band is divided based on the millimeter wave band to obtain four 2160MHz frequency bands, which are consecutively indicated as the first 2160MHz frequency band, the second 2160MHz frequency band, the third 2160MHz frequency band, and the fourth 2160MHz frequency band. 2000MHz is selected from each of the above 2160MHz frequency bands to obtain four 2000MHz frequency bands, which are consecutively indicated as the first 2000MHz frequency band, the second 2000MHz frequency band, the third 2000MHz frequency band, and the fourth 2000MHz frequency band. The center frequency of the 2000MHz frequency band selected from each 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. For example, the center frequency of the first 2160MHz frequency band overlaps with or is the same as the center frequency of the first 2000MHz frequency band.

[0373] To obtain two 1000 MHz channels, the first 2000 MHz frequency band is bisected once, and the center frequencies of the two 1000 MHz channels are f C (n) [GHz] = 57.820 + (n-1) × 1.000, where 1 ≤ n ≤ 2, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.320 + (n-1) × 1.000, 1 ≤ n ≤ 2, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.320 + (n-1) × 1.000, where 1≦n≦2.

[0374] The second 2000 MHz frequency band is bisected once to obtain two 1000 MHz channels, and the center frequencies of the two 1000 MHz channels are f C (n) [GHz] = 59.980 + (n-3) × 1.000, 3 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n)[GHz]=59.480+(n-3)×1.000, 3≦n≦4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 60.480 + (n-3) × 1.000, 3≦n≦4.

[0375] The third 2000 MHz frequency band is bisected once to obtain two 1000 MHz channels, and the center frequencies of the two 1000 MHz channels are f C (n) [GHz] = 62.140 + (n-5) × 1.000, 5 ≤ n ≤ 6, and the formula for calculating the minimum frequency is f L (n)[GHz]=61.640+(n-5)×1.000, 5≦n≦6, and the formula for calculating the maximum frequency is f H (n) [GHz] = 62.640 + (n-5) × 1.000, where 5≦n≦6.

[0376] The fourth 2000 MHz frequency band is bisected once to obtain two 1000 MHz channels, and the center frequencies of the two 1000 MHz channels are f C (n) [GHz] = 64.300 + (n-7) × 1.000, 7 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n)[GHz]=63.800+(n-7)×1.000, 7≦n≦8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 64.800 + (n-7) × 1.000, where 7≦n≦8.

[0377] 2000MHz Channel Division: The 57-66GHz frequency band is divided based on the millimeter wave band to obtain four 2160MHz frequency bands, which are consecutively indicated as the first 2160MHz frequency band, the second 2160MHz frequency band, the third 2160MHz frequency band, and the fourth 2160MHz frequency band. A 2000MHz is selected from each of the above 2160MHz frequency bands to obtain four 2000MHz frequency bands, which are consecutively indicated as the first 2000MHz frequency band, the second 2000MHz frequency band, the third 2000MHz frequency band, and the fourth 2000MHz frequency band. The center frequency of the 2000MHz frequency band selected from each 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. For example, the center frequency of the first 2160MHz frequency band overlaps with or is the same as the center frequency of the first 2000MHz frequency band. The formula for calculating the center frequency of the four 2000MHz channels is f C (n) [GHz] = 58.320 + (n-1) × 2.160, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.320 + (n-1) × 2.160, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 59.320 + (n-1) × 2.160, 1≦n≦4.

[0378] 4000MHz Channel Division: The 57-66GHz frequency band is divided based on the millimeter wave band to obtain four 2160MHz frequency bands, which are consecutively indicated as the first 2160MHz frequency band, the second 2160MHz frequency band, the third 2160MHz frequency band, and the fourth 2160MHz frequency band. 2000MHz is selected from each of the above 2160MHz frequency bands to obtain four 2000MHz frequency bands, which are consecutively indicated as the first 2000MHz frequency band, the second 2000MHz frequency band, the third 2000MHz frequency band, and the fourth 2000MHz frequency band. The center frequency of the 2000MHz frequency band selected from each 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. For example, the center frequency of the first 2160MHz frequency band overlaps with or is the same as the center frequency of the first 2000MHz frequency band. To obtain two 4000MHz channels, four 2000MHz frequency bands are aggregated in ascending order of frequency by two. It can also be understood that a first 2000MHz band and a second 2000MHz band are aggregated to obtain a first 4000MHz channel, and a third 2000MHz band and a fourth 2000MHz band are aggregated to obtain a second 4000MHz channel. The first 4000MHz channel has a center frequency of 59.400, a lowest frequency of 57.320, and a highest frequency of 61.480. The second 4000MHz channel has a center frequency of 63.720, a lowest frequency of 61.640, and a highest frequency of 65.800.

[0379] Each 4000 MHz channel is formed by aggregating two non-contiguous 2000 MHz channels.

[0380] For example, the first frequency band includes a 42.3-47.0 GHz frequency band and a 47.2-48.4 GHz frequency band. The first frequency band may be divided into twenty 250 MHz channels, ten 500 MHz channels, five 1000 MHz channels, two 2000 MHz channels, or one 4000 MHz channel. In addition, the center frequencies of the 500 MHz channel and the 1000 MHz channel must overlap with the center frequencies of the 540 MHz channel and the 1080 MHz channel in the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band of the third embodiment, respectively. In this division method, the remaining bandwidths in the front and rear parts of the 42.3-47.0 GHz frequency band are 110 MHz and 310 MHz, respectively, and the remaining bandwidths in the front and rear parts of the 47.2-48.4 GHz frequency band are 70 MHz and 90 MHz, respectively.

[0381] 250MHz channel division: 42.3~47.0GHz is divided based on the millimeter wave band to obtain eight 540MHz frequency bands, and 500MHz is selected from each 540MHz frequency band to obtain eight 500MHz frequency bands. The center frequency of the 500MHz frequency band selected from one 500MHz frequency band overlaps with the center frequency of the 540MHz frequency band. Each 500MHz frequency band is bisected once to obtain two 250MHz channels.

[0382] The formula for calculating the center frequency of the first 250 MHz channel in each 500 MHz frequency band is:

number

number

number

[0383] The formula for calculating the center frequency of the second 250 MHz channel in each 500 MHz frequency band is:

number

number

number

[0384] To obtain two 540 MHz frequency bands, 47.2-48.4 GHz is divided based on the millimeter wave band, and 500 MHz is selected from each 540 MHz frequency band to obtain two 500 MHz frequency bands. The center frequency of the 500 MHz frequency band selected from one 500 MHz frequency band overlaps with the center frequency of the 540 MHz frequency band. To obtain two 250 MHz channels, each 500 MHz frequency band is bisected once.

[0385] The formula for calculating the center frequency of the first 250 MHz channel in each 500 MHz frequency band is:

number

number

number

[0386] The formula for calculating the center frequency of the second 250 MHz channel in each 500 MHz frequency band is:

number

number

number

[0387] 500MHz channel division: 42.3~47.0GHz is divided based on the millimeter wave band to obtain eight 540MHz frequency bands, and 500MHz is selected from each 540MHz frequency band to obtain eight 500MHz frequency bands. The center frequency of the 500MHz frequency band selected from one 500MHz frequency band overlaps with the center frequency of the 540MHz frequency band.

[0388] The formula for calculating the center frequency of each 500 MHz frequency band is f C (n) [GHz] = 42.660 + (n-1) × 0.540, 1 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.410 + (n-1) × 0.540, 1 ≤ n ≤ 8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 42.910 + (n-1) × 0.540, where 1≦n≦8.

[0389] To obtain two 540 MHz frequency bands, 47.2-48.4 GHz is divided based on the millimeter wave band, and 500 MHz is selected from each 540 MHz frequency band to obtain two 500 MHz frequency bands. The center frequency of the 500 MHz frequency band selected from one 500 MHz frequency band overlaps with the center frequency of the 540 MHz frequency band. To obtain two 250 MHz channels, each 500 MHz frequency band is bisected once.

[0390] The formula for calculating the center frequency of each 500 MHz frequency band is f C (n) [GHz] = 47.520 + (n-9) × 0.540, 9 ≤ n ≤ 10, and the formula for calculating the minimum frequency is f L (n) [GHz] = 47.270 + (n-9) × 0.540, 9 ≤ n ≤ 10, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.770 + (n-9) × 0.540, where 9≦n≦10.

[0391] 1000MHz channel division: 42.3~47.0GHz is divided based on the millimeter wave band to obtain eight 540MHz frequency bands, and 500MHz is selected from each 540MHz frequency band to obtain eight 500MHz frequency bands. The center frequency of the 500MHz frequency band selected from one 500MHz frequency band overlaps with the center frequency of the 540MHz frequency band. To obtain four 1000MHz channels, the eight 500MHz frequency bands are aggregated in ascending order of frequency by two.

[0392] The formula for calculating the center frequency of a 1000 MHz channel is f C (n) [GHz] = 42.930 + (n-1) × 1.080, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.430 + (n-1) × 1.080, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 43.430 + (n-1) × 1.080, where 1≦n≦4.

[0393] To obtain two 540 MHz frequency bands, 47.2-48.4 GHz is divided based on the millimeter wave band, and 500 MHz is selected from each 540 MHz frequency band to obtain two 500 MHz frequency bands. The center frequency of the 500 MHz frequency band selected from one 500 MHz frequency band overlaps with the center frequency of the 540 MHz frequency band. To obtain one 1000 MHz channel, the two 500 MHz frequency bands are aggregated.

[0394] The formula for calculating the center frequency of a 1000 MHz channel is f C (n)[GHz]=47.790+(n-5)×1.080, n=5, and the formula for calculating the minimum frequency is f L (n)[GHz]=47.290+(n-5)×1.080, n=5, and the formula for calculating the maximum frequency is f H (n) [GHz] = 48.290 + (n-5) × 1.080, n = 5.

[0395] 2000MHz channel division: 42.3~47.0GHz is divided based on the millimeter wave band to obtain eight 540MHz frequency bands, and 500MHz is selected from each 540MHz frequency band to obtain eight 500MHz frequency bands. The center frequency of the 500MHz frequency band selected from one 500MHz frequency band overlaps with the center frequency of the 540MHz frequency band. To obtain two 2000MHz channels, the eight 500MHz frequency bands are aggregated in ascending order of frequency by four.

[0396] One 2000 MHz channel has a center frequency of 43.47, a minimum frequency of 42.43, and a maximum frequency of 44.51. The other 2000 MHz channel has a center frequency of 45.63, a minimum frequency of 44.59, and a maximum frequency of 46.67. Each 2000 MHz channel is formed by aggregating two non-contiguous 1000 MHz channels.

[0397] 4000MHz channel division: 42.3~47.0GHz is divided based on the millimeter wave band to obtain eight 540MHz frequency bands, and 500MHz is selected from each 540MHz frequency band to obtain eight 500MHz frequency bands. The center frequency of the 500MHz frequency band selected from one 500MHz frequency band overlaps with the center frequency of the 540MHz frequency band. Eight 500MHz frequency bands are aggregated to obtain one 4000MHz channel. The 4000MHz channel has a center frequency of 44.55, a lowest frequency of 42.43, and a highest frequency of 46.67. This channel is formed by aggregating eight non-contiguous 500MHz channels.

[0398] For example, the first frequency band is the 56.16 to 73.44 GHz frequency band. The first frequency band may be divided into sixty-four 250 MHz channels, thirty-two 500 MHz channels, sixteen 1000 MHz channels, eight 2000 MHz channels, four 4000 MHz channels, or two 8000 MHz channels.

[0399] A channel division scheme for the 56.16-73.44 GHz frequency band is described herein, in which the remaining bandwidth in the front and rear parts of the first frequency band is 80 MHz and 80 MHz, respectively.

[0400] 250MHz channel division: The 56.16~73.44GHz frequency band is divided based on the millimeter wave band to obtain eight 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain eight 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. Each 2000MHz is bisected three times to obtain eight 250MHz channels.

[0401] The formula for calculating the center frequency of the first 250 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0402] The formula for calculating the center frequency of the second 250 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0403] The formula for calculating the center frequency of the third 250 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0404] The formula for calculating the center frequency of the fourth 250 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0405] The formula for calculating the center frequency of the fifth 250 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0406] The formula for calculating the center frequency of the sixth 250 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0407] The formula for calculating the center frequency of the seventh 250 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0408] The formula for calculating the center frequency of the eighth 250 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0409] 500MHz channel division: The 56.16~73.44GHz frequency band is divided based on the millimeter wave band to obtain eight 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain eight 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. Each 2000MHz is bisected twice to obtain four 500MHz channels.

[0410] The formula for calculating the center frequency of the first 500 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0411] The formula for calculating the center frequency of the second 500 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0412] The formula for calculating the center frequency of the third 500 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0413] The formula for calculating the center frequency of the fourth 500 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0414] 1000MHz channel division: The 56.16~73.44GHz frequency band is divided based on the millimeter wave band to obtain eight 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain eight 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. Each 2000MHz is bisected once to obtain two 1000MHz channels.

[0415] The formula for calculating the center frequency of the first 1000 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0416] The formula for calculating the center frequency of the second 1000 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0417] 2000MHz channel division: The 56.16-73.44GHz frequency band is divided based on the millimeter wave band to obtain eight 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain eight 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band.

[0418] The formula for calculating the center frequency of each 2000 MHz frequency band is f C (n) [GHz] = 57.240 + (n-1) × 2.160, 1 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.240 + (n-1) × 2.160, 1 ≤ n ≤ 8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.240 + (n-1) × 2.160, 1≦n≦8.

[0419] 4000MHz channel division: The 56.16-73.44GHz frequency band is divided based on the millimeter wave band to obtain eight 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain eight 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band.

[0420] To obtain four 4000 MHz channels, eight 2000 MHz frequency bands are aggregated in ascending frequency order by two, and each 4000 MHz channel is obtained by aggregating two non-consecutive 2000 MHz channels. The formula for calculating the center frequency of the four 4000 MHz channels is f C (n) [GHz] = 58.320 + (n-1) × 4.320, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.240 + (n-1) × 4.320, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 60.400 + (n-1) × 4.320, where 1≦n≦4.

[0421] Example 11: The minimum channel bandwidth supported in the first frequency band is 500 MHz, in other words, the bandwidth of the first basic channel is 500 MHz. The bandwidth supported in the first frequency band may include one or more of the following bandwidths: 500 MHz, 1000 MHz, 2000 MHz, 4000 MHz, and 8000 MHz. In this example, the center frequency of 2000 MHz overlaps with the center frequency of 2160 MHz obtained by division based on the millimeter wave band. The 1000 MHz channel is obtained by performing bisection on the 2000 MHz channel, and the 500 MHz channel is obtained by performing bisection twice on the 2000 MHz channel. The 4000 MHz channel is obtained by aggregating two 2000 MHz channels. The 8000 MHz channel is obtained by aggregating four 2000 MHz channels. It should be understood that two 2000 MHz channels within a 4000 MHz bandwidth are not contiguous, and four 2000 MHz channels within an 8000 MHz bandwidth are not contiguous.

[0422] Optionally, for the amount of subcarriers (or IDFT size / DFT size) corresponding to each channel bandwidth in the channel division manner of Example 11, refer to the relevant description of Example 9 in Example 3 of Embodiment 1. Details are not described again in this specification.

[0423] For example, the first frequency band is a 57-66 GHz frequency band. The first frequency band may be divided into sixteen 500 MHz channels, eight 1000 MHz channels, four 2000 MHz channels, two 4000 MHz channels, or one 8000 MHz channel, as shown in FIG.

[0424] A channel division scheme for the 57-66 GHz frequency band is described herein, in which the remaining bandwidth in the front and rear parts of the first frequency band is 320 MHz and 200 MHz, respectively.

[0425] 500MHz channel division: The 57-66GHz frequency band is divided based on the millimeter wave band to obtain four 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain four 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band.

[0426] To obtain four 500 MHz channels, the first 2000 MHz frequency band is bisected twice, and the formula for calculating the center frequencies of the four 500 MHz channels is f C (n) [GHz] = 57.570 + (n-1) × 0.500, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.320 + (n-1) × 0.500, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 57.820 + (n-1) × 0.500, 1≦n≦4.

[0427] To obtain four 500 MHz channels, the second 2000 MHz frequency band is bisected twice, and the formula for calculating the center frequencies of the four 500 MHz channels is f C (n) [GHz] = 59.730 + (n-5) × 0.500, 5 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n)[GHz]=59.480+(n-5)×0.500, 5≦n≦8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 59.980 + (n-5) × 0.500, 5≦n≦8.

[0428] The third 2000 MHz frequency band is bisected twice to obtain four 500 MHz channels, and the formula for calculating the center frequencies of the four 500 MHz channels is f C (n) [GHz] = 61.890 + (n-9) × 0.500, 9 ≤ n ≤ 12, and the formula for calculating the minimum frequency is f L (n) [GHz] = 61.640 + (n-9) × 0.500, 9 ≤ n ≤ 12, and the formula for calculating the maximum frequency is f H (n) [GHz] = 62.140 + (n-9) × 0.500, where 9≦n≦12.

[0429] The fourth 2000 MHz frequency band is bisected twice to obtain four 500 MHz channels, and the formula for calculating the center frequencies of the four 500 MHz channels is f C (n) [GHz] = 64.050 + (n-13) × 0.500, 13 ≤ n ≤ 16, and the formula for calculating the minimum frequency is f L (n)[GHz]=63.800+(n-13)×0.500, 13≦n≦16, and the formula for calculating the maximum frequency is f H (n) [GHz] = 64.300 + (n-13) × 0.500, where 13≦n≦16.

[0430] 1000MHz channel division: The 57-66GHz frequency band is divided based on the millimeter wave band to obtain four 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain four 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band.

[0431] To obtain two 1000 MHz channels, the first 2000 MHz frequency band is bisected once, and the center frequencies of the two 1000 MHz channels are f C (n) [GHz] = 57.820 + (n-1) × 1.000, where 1 ≤ n ≤ 2, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.320 + (n-1) × 1.000, 1 ≤ n ≤ 2, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.320 + (n-1) × 1.000, where 1≦n≦2.

[0432] The second 2000 MHz frequency band is bisected once to obtain two 1000 MHz channels, and the center frequencies of the two 1000 MHz channels are f C (n) [GHz] = 59.980 + (n-3) × 1.000, 3 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n)[GHz]=59.480+(n-3)×1.000, 3≦n≦4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 60.480 + (n-3) × 1.000, 3≦n≦4.

[0433] The third 2000 MHz frequency band is bisected once to obtain two 1000 MHz channels, and the center frequencies of the two 1000 MHz channels are f C (n) [GHz] = 62.140 + (n-5) × 1.000, 5 ≤ n ≤ 6, and the formula for calculating the minimum frequency is f L(n)[GHz]=61.640+(n-5)×1.000, 5≦n≦6, and the formula for calculating the maximum frequency is f H (n) [GHz] = 62.640 + (n-5) × 1.000, where 5≦n≦6.

[0434] The fourth 2000 MHz frequency band is bisected once to obtain two 1000 MHz channels, and the center frequencies of the two 1000 MHz channels are f C (n) [GHz] = 64.300 + (n-7) × 1.000, 7 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n)[GHz]=63.800+(n-7)×1.000, 7≦n≦8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 64.800 + (n-7) × 1.000, where 7≦n≦8.

[0435] 2000MHz channel division: The 57-66GHz frequency band is divided based on the millimeter wave band to obtain four 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain four 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. The formula for calculating the center frequency of the four 2000MHz channels is f C (n) [GHz] = 58.320 + (n-1) × 2.160, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 57.320 + (n-1) × 2.160, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 59.320 + (n-1) × 2.160, 1≦n≦4.

[0436] 4000MHz channel division: The 57-66GHz frequency band is divided based on the millimeter wave band to obtain four 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain four 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band.

[0437] Four 2000 MHz frequency bands are aggregated by two in ascending frequency order to obtain two 4000 MHz channels. One 4000 MHz channel has a center frequency of 59.400, a minimum frequency of 57.320, and a maximum frequency of 61.480. The other 4000 MHz channel has a center frequency of 63.720, a minimum frequency of 61.640, and a maximum frequency of 65.800.

[0438] Each 4000 MHz channel is formed by aggregating two non-contiguous 2000 MHz channels.

[0439] 8000MHz channel division: The 57-66GHz frequency band is divided based on the millimeter wave band to obtain four 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain four 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. The four 2000MHz frequency bands are aggregated to obtain an 8000MHz channel. The 8000MHz channel has a center frequency of 61.560, a lowest frequency of 57.320, and a highest frequency of 65.800. This channel is formed by aggregating four non-contiguous 2000MHz channels.

[0440] For example, the first frequency band includes a 42.3-47.0 GHz frequency band and a 47.2-48.4 GHz frequency band. The first frequency band may be divided into ten 500 MHz channels, five 1000 MHz channels, two 2000 MHz channels, or one 4000 MHz channel. In addition, the center frequencies of the 500 MHz channel and the 1000 MHz channel must overlap with the center frequencies of the 540 MHz channel and the 1080 MHz channel in the 42.3-47.0 GHz frequency band and the 47.2-48.4 GHz frequency band of the third embodiment, respectively. In this division method, the remaining bandwidths in the front and rear parts of the 42.3-47.0 GHz frequency band are 110 MHz and 310 MHz, respectively, and the remaining bandwidths in the front and rear parts of the 47.2-48.4 GHz frequency band are 70 MHz and 90 MHz, respectively.

[0441] 500MHz channel division: 42.3~47.0GHz is divided based on the millimeter wave band to obtain eight 540MHz frequency bands, and 500MHz is selected from each 540MHz frequency band to obtain eight 500MHz frequency bands. The center frequency of the 500MHz frequency band selected from one 500MHz frequency band overlaps with the center frequency of the 540MHz frequency band.

[0442] The formula for calculating the center frequency of each 500 MHz frequency band is f C (n) [GHz] = 42.660 + (n-1) × 0.540, 1 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.410 + (n-1) × 0.540, 1 ≤ n ≤ 8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 42.910 + (n-1) × 0.540, where 1≦n≦8.

[0443] To obtain two 540 MHz frequency bands, 47.2-48.4 GHz is divided based on the millimeter wave band, and 500 MHz is selected from each 540 MHz frequency band to obtain two 500 MHz frequency bands. The center frequency of the 500 MHz frequency band selected from one 500 MHz frequency band overlaps with the center frequency of the 540 MHz frequency band. To obtain two 250 MHz channels, each 500 MHz frequency band is bisected once.

[0444] The formula for calculating the center frequency of each 500 MHz frequency band is f C (n) [GHz] = 47.520 + (n-9) × 0.540, 9 ≤ n ≤ 10, and the formula for calculating the minimum frequency is f L (n) [GHz] = 47.270 + (n-9) × 0.540, 9 ≤ n ≤ 10, and the formula for calculating the maximum frequency is f H (n) [GHz] = 47.770 + (n-9) × 0.540, where 9≦n≦10.

[0445] 1000MHz channel division: 42.3~47.0GHz is divided based on the millimeter wave band to obtain eight 540MHz frequency bands, and 500MHz is selected from each 540MHz frequency band to obtain eight 500MHz frequency bands. The center frequency of the 500MHz frequency band selected from one 500MHz frequency band overlaps with the center frequency of the 540MHz frequency band. To obtain four 1000MHz channels, the eight 500MHz frequency bands are aggregated in ascending order of frequency by two.

[0446] The formula for calculating the center frequency of a 1000 MHz channel is f C (n) [GHz] = 42.930 + (n-1) × 1.080, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is f L (n) [GHz] = 42.430 + (n-1) × 1.080, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 43.430 + (n-1) × 1.080, where 1≦n≦4.

[0447] To obtain two 540 MHz frequency bands, 47.2-48.4 GHz is divided based on the millimeter wave band, and 500 MHz is selected from each 540 MHz frequency band to obtain two 500 MHz frequency bands. The center frequency of the 500 MHz frequency band selected from one 500 MHz frequency band overlaps with the center frequency of the 540 MHz frequency band. To obtain one 1000 MHz channel, the two 500 MHz frequency bands are aggregated.

[0448] The formula for calculating the center frequency of a 1000 MHz channel is f C (n)[GHz]=47.790+(n-5)×1.080, n=5, and the formula for calculating the minimum frequency is f L (n)[GHz]=47.290+(n-5)×1.080, n=5, and the formula for calculating the maximum frequency is f H (n) [GHz] = 48.290 + (n-5) × 1.080, n = 5.

[0449] 2000MHz channel division: 42.3~47.0GHz is divided based on the millimeter wave band to obtain eight 540MHz frequency bands, and 500MHz is selected from each 540MHz frequency band to obtain eight 500MHz frequency bands. The center frequency of the 500MHz frequency band selected from one 500MHz frequency band overlaps with the center frequency of the 540MHz frequency band. To obtain two 1000MHz channels, the eight 500MHz frequency bands are aggregated in ascending order of frequency by four.

[0450] One 2000 MHz channel has a center frequency of 43.47, a minimum frequency of 42.43, and a maximum frequency of 44.51. The other 2000 MHz channel has a center frequency of 45.63, a minimum frequency of 44.59, and a maximum frequency of 46.67. Each 2000 MHz channel is formed by aggregating two non-contiguous 1000 MHz channels.

[0451] 4000MHz Channel Division: The center frequency is 44.55, the lowest frequency is 42.43 and the highest frequency is 46.67. This channel is formed by aggregating two non-contiguous 2000MHz channels.

[0452] For example, the first frequency band is the 56.16 to 73.44 GHz frequency band. The first frequency band may be divided into 32 500 MHz channels, 16 1000 MHz channels, 8 2000 MHz channels, 4 4000 MHz channels, or 2 8000 MHz channels.

[0453] A channel division scheme for the 56.16-73.44 GHz frequency band is described herein, in which the remaining bandwidth in the front and rear parts of the first frequency band is 80 MHz and 80 MHz, respectively.

[0454] 500MHz channel division: The 56.16~73.44GHz frequency band is divided based on the millimeter wave band to obtain eight 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain eight 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. Each 2000MHz is bisected twice to obtain four 500MHz channels.

[0455] The formula for calculating the center frequency of the first 500 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0456] The formula for calculating the center frequency of the second 500 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0457] The formula for calculating the center frequency of the third 500 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0458] The formula for calculating the center frequency of the fourth 500 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0459] 1000MHz channel division: The 56.16~73.44GHz frequency band is divided based on the millimeter wave band to obtain eight 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain eight 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. Each 2000MHz is bisected once to obtain two 1000MHz channels.

[0460] The formula for calculating the center frequency of the first 1000 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0461] The formula for calculating the center frequency of the second 1000 MHz channel in each 2000 MHz frequency band is:

number

number

number

[0462] 2000MHz channel division: The 56.16-73.44GHz frequency band is divided based on the millimeter wave band to obtain eight 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain eight 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band.

[0463] The formula for calculating the center frequency of each 2000 MHz frequency band is f C (n) [GHz] = 57.240 + (n-1) × 2.160, 1 ≤ n ≤ 8, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.240 + (n-1) × 2.160, 1 ≤ n ≤ 8, and the formula for calculating the maximum frequency is f H (n) [GHz] = 58.240 + (n-1) × 2.160, 1≦n≦8.

[0464] 4000MHz channel division: The 56.16-73.44GHz frequency band is divided based on the millimeter wave band to obtain eight 2160MHz frequency bands, and 2000MHz is selected from each 2160MHz frequency band to obtain eight 2000MHz frequency bands. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band.

[0465] To obtain four 4000 MHz channels, eight 2000 MHz frequency bands are aggregated in ascending frequency order by two, and each 4000 MHz channel is obtained by aggregating two non-consecutive 2000 MHz channels. The formula for calculating the center frequency of the four 4000 MHz channels is f C (n) [GHz] = 58.320 + (n-1) × 4.320, 1 ≤ n ≤ 4, and the formula for calculating the minimum frequency is fL (n) [GHz] = 56.240 + (n-1) × 4.320, 1 ≤ n ≤ 4, and the formula for calculating the maximum frequency is f H (n) [GHz] = 60.400 + (n-1) × 4.320, where 1≦n≦4.

[0466] 8000MHz channel division: To obtain eight 2160MHz frequency bands, the 56.16-73.44GHz frequency band is divided based on the millimeter wave band, and to obtain eight 2000MHz frequency bands, 2000MHz is selected from each 2160MHz frequency band. The center frequency of the 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. To obtain two 8000MHz channels, the eight 2000MHz frequency bands are aggregated in ascending order of frequency by four, and each 8000MHz channel is obtained by aggregating four non-contiguous 2000MHz channels.

[0467] The formula for calculating the center frequency of the two 8000MHz channels is f C (n) [GHz] = 60.480 + (n-1) × 8.640, 1 ≤ n ≤ 2, and the formula for calculating the minimum frequency is f L (n) [GHz] = 56.240 + (n-1) × 8.640, 1 ≤ n ≤ 2, and the formula for calculating the maximum frequency is f H (n) [GHz] = 64.720 + (n-1) × 8.640, 1≦n≦2.

[0468] It should be understood that the division manners of Examples 10 and 11 are merely illustrative examples, and the locations of the center frequency, the lowest frequency, and the highest frequency of the channel are not particularly limited. The channel division manner may be used, provided that the bandwidth is divided based on the minimum channel bandwidth of the millimeter wave band, and the bandwidth obtained by bisection is smaller than the minimum channel bandwidth of the millimeter wave band.

[0469] In the channel division manner provided in the fourth embodiment of the present application, the spectrum resources in the first frequency band can be used as much as possible by the channel division with smaller granularity. This reduces resource waste and improves resource utilization. In the fourth embodiment, a part of the bandwidth is selected from the minimum channel bandwidth supported in the millimeter wave band, and as a result, the interference between channels can be further reduced.

[0470] In addition, according to embodiment 4, the signal in the first frequency band, the signal in the second frequency band, and the signal in the millimeter wave band can be transmitted by reusing the same crystal oscillator. In addition, the correspondence between the bandwidth and the IDFT size / DFT size (or the amount of subcarriers) in the second frequency band is reused in the first frequency band, so that the first frequency band and the second frequency band can share a signal processing device, for example, the IDFT device or the DFT device is reused. Therefore, according to the above embodiment 4, the hardware complexity can be reduced, the hardware device overhead can be reduced, and the device cost can be reduced.

[0471] Based on the same inventive concept as the method embodiment, an embodiment of the present application provides a communication device. The structure of the communication device can be shown in FIG. 12. The communication device includes a transceiver module 1201 and a processing module 1202. Optionally, the transceiver module 1201 can include a transport module (Tx module) and a receive module (Rx module). The transport module is configured to send a signal, and the receive module is configured to receive a signal. The communication device can further include a radio frequency unit (RF unit). The radio frequency unit can be configured to send a signal to a receiver device. The transmission module can transmit to the radio frequency unit to send the signal to be sent. The receive module can receive the signal from the radio frequency unit and send the signal to the processing module 1202 for further processing.

[0472] In one implementation, the communication device may be specifically configured to perform the method performed by the first device in the embodiment shown in FIG. 3. The device may be the first device, or may be a chip or chipset in the first device, or a part of a chip configured to perform the functions of the associated method. The processing module 1202 is configured to generate a first OFDM signal. The transceiver module 1201 is configured to send a first OFDM signal on a first channel in a first frequency band, where an amount of subcarriers corresponding to the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band, where a lowest frequency of the first frequency band is greater than a highest frequency of the second frequency band, where ... M The first basic channel includes 2 Mnumber of second fundamental channels, the bandwidth of the first fundamental channel being a minimum channel bandwidth supported in the first frequency band, the bandwidth of the second fundamental channel being a minimum channel bandwidth supported in the second frequency band, the bandwidth of the first fundamental channel being greater than the bandwidth of the second fundamental channel, and M being an integer greater than or equal to 0.

[0473] In one implementation, the communication device may be specifically configured to perform the method performed by the second device in the embodiment shown in FIG. 3. The device may be the second device, or may be a chip or chipset in the second device, or a part of a chip configured to perform the functions of the associated method. The transceiver module 1201 is configured to receive a first OFDM signal on a first channel in a first frequency band, where an amount of subcarriers corresponding to the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band, where a lowest frequency of the first frequency band is greater than a highest frequency of the second frequency band, and where the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band. M The first basic channel includes 2 M the first OFDM signal includes M number of second fundamental channels, the bandwidth of the first fundamental channel is a minimum channel bandwidth supported in the first frequency band, the bandwidth of the second fundamental channel is a minimum channel bandwidth supported in the second frequency band, the bandwidth of the first fundamental channel is greater than the bandwidth of the second fundamental channel, and M is an integer greater than or equal to 0. The processing module 1202 is configured to process the first OFDM signal.

[0474] For details, please refer to the relevant descriptions of Embodiment 1 to Embodiment 4. The details will not be described again in this specification.

[0475] The division into modules in the embodiments of the present application is an example, and is merely a logical division of functions, and may be other divisions during actual implementation. In addition, the functional modules in the embodiments of the present application may be integrated into one processor, or each of the modules may exist physically alone, or two or more modules may be integrated into one module. The integrated modules may be implemented in the form of hardware, or in the form of software functional modules. It may be understood that the functions or implementation forms of the modules in the embodiments of the present application may further refer to the relevant descriptions in the method embodiments.

[0476] In a possible implementation, a communication device may be shown in FIG. 13. The device may be a communication device or a chip in a communication device. The communication device may be the first device in the above embodiment or the second device in the above embodiment. The device includes a processor 1301 and a communication interface 1302, and may further include a memory 1303. The processing module 1202 may be the processor 1301. The transceiver module 1201 may be the communication interface 1302. FIG. 13 shows only the main components of the communication device. In addition to the processor 1301 and the communication interface 1302, the communication device may further include a memory 1303 and an input / output device (not shown in the figure).

[0477] The processor 1301 is configured to execute program codes stored in the memory 1303, and is particularly configured to perform the actions of the processing module 1202. Details are not described again herein in this application. The communication interface 1302 is particularly configured to perform the actions of the transceiver module 1201. Details are not described again herein in this application.

[0478] The processor 1301 may be a CPU, a digital processing unit, etc. The processor 1301 may be configured to process communication protocols and communication data, control the entire communication device, execute software programs, and process data of the software programs, for example, but not limited to, performing baseband-related processing. The communication interface 1302 may be configured to receive and transmit signals, for example, but not limited to, performing radio frequency reception and transmission. The above devices may be disposed on separate chips, respectively, or at least some or all of the components may be disposed on the same chip. For example, the processor 1301 may be further divided into an analog baseband processor and a digital baseband processor. The analog baseband processor and the transceiver may be integrated on the same chip, and the digital baseband processor may be disposed on a separate chip. With the continuous development of integrated circuit technology, more and more devices may be integrated on the same chip. For example, a digital baseband processor may be integrated on the same chip with multiple application processors (for example, but not limited to, a geometry processor and a multimedia processor). The chip may be referred to as a system on chip. Whether the devices are independently arranged on different chips or integrated on one or more chips usually depends on the specific requirements of the product design. The specific implementation form of the devices is not limited in the embodiments of the present invention.

[0479] The communication interface 1302 may be a transceiver, an interface circuit such as a transceiver circuit, a transceiver chip, etc. Optionally, the communication interface 1302 may include a radio frequency circuit and an antenna. The radio frequency circuit is mainly configured to perform conversion between a baseband signal and a radio frequency signal and process the radio frequency signal. The antenna is mainly configured to receive or transmit the radio frequency signal in the form of an electromagnetic wave. The input / output device, such as a touch screen, a display, or a keyboard, is mainly configured to receive data input by a user and output data to the user.

[0480] The memory 1303 is configured to store programs executed by the processor 1301. The memory 1303 may be a non-volatile memory such as a hard disk drive (HDD) or a solid-state drive (SSD), or a volatile memory such as a random-access memory (RAM). The memory 1303 may be any other medium that may be used to carry or store the expected program code in the form of instructions or data structures that can be accessed by a computer, but is not limited thereto.

[0481] After the communication device is powered on, the processor 1301 may read the software program in the memory 1303, interpret and execute the instructions of the software program, and process data of the software program. When data needs to be sent wirelessly, the processor 1301 performs baseband processing on the data to be sent, and then outputs the baseband signal to the radio frequency circuit. The radio frequency circuit performs radio frequency processing on the baseband signal, and then sends the radio frequency signal in the form of electromagnetic waves through an antenna. When data is sent to the communication device, the radio frequency circuit receives the radio frequency signal through the antenna, converts the radio frequency signal into a baseband signal, and outputs the baseband signal to the processor 1301. The processor 1301 converts the baseband signal into data and processes the data.

[0482] In another implementation, the radio frequency circuitry and antenna may be located independently of the processor performing the baseband processing, e.g., in a distributed scenario, the radio frequency circuitry and antenna may be located independently and remotely from the communication device.

[0483] The specific connection medium between the communication interface 1302, the processor 1301, and the memory 1303 is not limited in the embodiment of the present application. In this embodiment of the present application, the memory 1303, the processor 1301, and the communication interface 1302 are connected through the bus 1304 in FIG. 13. The bus is represented by using a thick line in FIG. 13. The manner of connection between the other components is only an example for explanation and is not limited thereto. The bus may be classified into an address bus, a data bus, a control bus, and the like. For ease of representation, only one thick line is used in the representation of FIG. 13, but this does not mean that there is only one bus or only one type of bus.

[0484] Optionally, the communication unit may be a separate device or may be part of a larger device. For example, the communication unit may include: (1) A standalone integrated circuit (IC), chip, or chip system or subsystem; (2) a set of one or more ICs, although optionally the set of ICs may alternatively include a storage component configured to store data and instructions; (3) Application-specific integrated circuits (ASICs) such as modems, (4) A module that may be embedded in another device; (5) Receivers, intelligent terminals, wireless devices, handsets, mobile units, vehicle-mounted devices, cloud devices, artificial intelligence devices, etc., or (6) Another device, etc. may be also possible.

[0485] An embodiment of the present invention further provides a computer readable storage medium configured to store computer software instructions that need to be executed for execution of the above processor, the computer readable storage medium including a program that needs to be executed for execution of the above processor.

[0486] An embodiment of the present application further provides a communication system including a communication device configured to perform functions of a first device in the embodiment shown in FIG. 3, and a communication device configured to perform functions of a second device in the embodiment shown in FIG. 3.

[0487] Those skilled in the art should understand that the embodiments of the present application may be provided as a method, a system, or a computer program product. Thus, the present application may use the form of a hardware-only embodiment, a software-only embodiment, or an embodiment having a combination of software and hardware. In addition, the present application may use the form of a computer program product embodied on one or more computer-usable storage media (including, but not limited to, disk memory, CD-ROM, optical memory, etc.) containing computer-usable program code.

[0488] The present application is described with reference to flowcharts and / or block diagrams of methods, devices (systems), and computer program products according to the present application. It should be understood that computer program instructions can be used to implement each step and / or each block in the flowcharts and / or block diagrams, and combinations of steps and / or blocks in the flowcharts and / or block diagrams. These computer program instructions may be provided to a processor of a general-purpose computer, a special-purpose computer, an embedded processor, or another programmable data processing device to create a machine, such that the instructions executed by the processor of the computer or another programmable data processing device create an apparatus for implementing a particular function in one or more steps in the flowcharts and / or in one or more blocks in the block diagrams.

[0489] These computer program instructions may be stored in a computer-readable memory that can direct a computer or another programmable data processing device to operate in a particular manner, such that the instructions stored in the computer-readable memory generate an artifact that includes an instruction apparatus that implements a particular function in one or more procedures in the flowcharts and / or in one or more blocks in the block diagrams.

[0490] The computer program instructions may alternatively be loaded onto a computer or another programmable data processing device, resulting in a sequence of operations and steps being performed on the computer or another programmable device, resulting in a computer-implemented process. Thus, the instructions that run on the computer or another programmable device provide steps for implementing a particular function in one or more procedures in the flowcharts and / or in one or more blocks in the block diagrams.

[0491] It is obvious that those skilled in the art can make various modifications and variations to this application without departing from the scope of this application. This application is intended to cover these modifications and variations of this application, provided that they fall within the scope of protection defined by the following claims and their equivalent technologies. [Explanation of symbols]

[0492] 101 Access Point (AP) 102,103 stations (STA) 1201 Transceiver Module 1202 Processing Module 1301 Processor 1302 Communication Interface 1303 Memory 1304 Bus

Claims

1. 1. A communication method comprising: generating a first orthogonal frequency division multiplexing (OFDM) signal; transmitting the first OFDM signal on a first channel in a first frequency band, wherein an amount of subcarriers corresponding to the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band, a lowest frequency of the first frequency band is greater than a highest frequency of the second frequency band, and the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band; M The first fundamental channel includes two M a second fundamental channel, a bandwidth of the first fundamental channel being a minimum channel bandwidth supported in the first frequency band, a bandwidth of the second fundamental channel being a minimum channel bandwidth supported in the second frequency band, the bandwidth of the first fundamental channel being greater than the bandwidth of the second fundamental channel, and M being an integer greater than or equal to 0; A method comprising:

2. 1. A communication method comprising: Receiving a first Orthogonal Frequency Division Multiplexing (OFDM) signal on a first channel in a first frequency band, wherein an amount of subcarriers corresponding to the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band, a lowest frequency of the first frequency band is greater than a highest frequency of the second frequency band, and the first channel is greater than or equal to an amount of subcarriers corresponding to a second channel in a second frequency band. M The first fundamental channel includes two M a second fundamental channel, a bandwidth of the first fundamental channel being a minimum channel bandwidth supported in the first frequency band, a bandwidth of the second fundamental channel being a minimum channel bandwidth supported in the second frequency band, the bandwidth of the first fundamental channel being greater than the bandwidth of the second fundamental channel, and M being an integer greater than or equal to 0; processing a first OFDM symbol; A method comprising:

3. The method of claim 1 or 2, wherein the subcarrier spacing of the first channel is greater than or equal to a threshold value.

4. 4. The method of claim 1, wherein the bandwidth of the first fundamental channel is N times the bandwidth of the second fundamental channel, where N is an integer greater than 1.

5. The method of claim 4 , wherein the bandwidth of the first fundamental channel is 320 MHz.

6. the amount of the subcarriers corresponding to the first channel is When the bandwidth of the first channel is 320 MHz, the first channel corresponds to 64 subcarriers; When the bandwidth of the first channel is 640 MHz, the first channel corresponds to 128 subcarriers; When the bandwidth of the first channel is 1280 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 2560 MHz, the first channel corresponds to 512 subcarriers; or When the bandwidth of the first channel is 5120 MHz, the first channel corresponds to 1024 subcarriers. The method according to claim 5, wherein any one of the following is satisfied:

7. The method of claim 4 , wherein the bandwidth of the first fundamental channel is 640 MHz.

8. the amount of the subcarriers corresponding to the first channel is When the bandwidth of the first channel is 640 MHz, the first channel corresponds to 64 subcarriers; When the bandwidth of the first channel is 1280 MHz, the first channel corresponds to 128 subcarriers; When the bandwidth of the first channel is 2560 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 5120 MHz, the first channel corresponds to 512 subcarriers; or If the bandwidth of the first channel is 10240 MHz, the first channel corresponds to 1024 subcarriers. The method according to claim 7, wherein any one of the following is satisfied:

9. The method of claim 4 , wherein the bandwidth of the first fundamental channel is 80 MHz.

10. the amount of the subcarriers corresponding to the first channel is When the bandwidth of the first channel is 80 MHz, the first channel corresponds to 64 subcarriers; When the bandwidth of the first channel is 160 MHz, the first channel corresponds to 128 subcarriers; When the bandwidth of the first channel is 320 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 640 MHz, the first channel corresponds to 512 subcarriers; or If the bandwidth of the first channel is 1280 MHz, the first channel corresponds to 1024 subcarriers. The method according to claim 9, wherein any one of the following is satisfied:

11. The method of claim 4 , wherein the bandwidth of the first fundamental channel is 160 MHz.

12. the amount of the subcarriers corresponding to the first channel is When the bandwidth of the first channel is 160 MHz, the first channel corresponds to 64 subcarriers; When the bandwidth of the first channel is 320 MHz, the first channel corresponds to 128 subcarriers; When the bandwidth of the first channel is 640 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 1280 MHz, the first channel corresponds to 512 subcarriers; or When the bandwidth of the first channel is 2560 MHz, the first channel corresponds to 1024 subcarriers. The method according to claim 11, wherein any one of the following is satisfied:

13. 4. The method according to claim 1, wherein the bandwidth of the first fundamental channel is obtained by performing R separate bisections on a frequency band obtained by dividing the first frequency band based on a minimum channel bandwidth supported in a millimeter wave band, where R is an integer greater than or equal to 0.

14. 14. The method of claim 13, wherein the bandwidth of the first fundamental channel is 270 MHz.

15. the amount of the subcarriers corresponding to the first channel is When the bandwidth of the first channel is 270 MHz, the first channel corresponds to 64 subcarriers; When the bandwidth of the first channel is 540 MHz, the first channel corresponds to 128 subcarriers; When the bandwidth of the first channel is 1080 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 2160 MHz, the first channel corresponds to 512 subcarriers; or When the bandwidth of the first channel is 4320 MHz, the first channel corresponds to 1024 subcarriers. The method according to claim 14, wherein any one of the following is satisfied:

16. 14. The method of claim 13, wherein the bandwidth of the first fundamental channel is 540 MHz.

17. the amount of the subcarriers corresponding to the first channel is When the bandwidth of the first channel is 540 MHz, the first channel corresponds to 64 subcarriers; When the bandwidth of the first channel is 1080 MHz, the first channel corresponds to 128 subcarriers; When the bandwidth of the first channel is 2160 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 4320 MHz, the first channel corresponds to 512 subcarriers; or When the bandwidth of the first channel is 8640 MHz, the first channel corresponds to 1024 subcarriers. The method according to claim 16, wherein any one of the following is satisfied:

18. 4. The method of claim 1, wherein the bandwidth of the first fundamental channel is obtained by performing K bisections on a first subband, the bandwidth of the first subband being smaller than a minimum channel bandwidth supported in a millimeter wave band, and K is an integer greater than or equal to 0.

19. 20. The method of claim 18, wherein the bandwidth of the first fundamental channel is 250 MHz.

20. the amount of the subcarriers corresponding to the first channel is When the bandwidth of the first channel is 250 MHz, the first channel corresponds to 64 subcarriers; When the bandwidth of the first channel is 500 MHz, the first channel corresponds to 128 subcarriers; When the bandwidth of the first channel is 1000 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 2000 MHz, the first channel corresponds to 512 subcarriers; or When the bandwidth of the first channel is 4000 MHz, the first channel corresponds to 1024 subcarriers. The method according to claim 19, wherein any one of the following is satisfied:

21. 20. The method of claim 18, wherein the bandwidth of the first fundamental channel is 500 MHz.

22. the amount of the subcarriers corresponding to the first channel is When the bandwidth of the first channel is 500 MHz, the first channel corresponds to 64 subcarriers; When the bandwidth of the first channel is 1000 MHz, the first channel corresponds to 128 subcarriers; When the bandwidth of the first channel is 2000 MHz, the first channel corresponds to 256 subcarriers; If the bandwidth of the first channel is 4000 MHz, the first channel corresponds to 512 subcarriers; or When the bandwidth of the first channel is 8000 MHz, the first channel corresponds to 1024 subcarriers. The method according to claim 21, wherein any one of the following is satisfied:

23. the first frequency band is a 60 GHz frequency band, or the first frequency band is a 45 GHz frequency band; The second frequency band is a 2.45 GHz frequency band, or the second frequency band is a 6 GHz frequency band.

23. The method according to any one of claims 1 to 22.

24. 1. A communication device, comprising: A memory and a processor, the memory configured to store instructions; The processor is configured to execute the instructions to perform a method according to any one of claims 1 or 3 to 23. Communications equipment.

25. 1. A communication device, comprising: A memory and a processor, the memory configured to store instructions; The processor is configured to execute the instructions to perform the method of any one of claims 2 to 23. Communications equipment.

26. 24. A computer readable storage medium configured to store computer instructions that, when executed on a computer, enable the computer to perform a method according to any one of claims 1 to 23.

27. A chip, A memory and a processor, the memory configured to store instructions; The processor is configured to execute the instructions such that a communications device comprising the chip performs a method according to any one of claims 1 or 3 to 23. Tips.

28. A chip, A memory and a processor, the memory configured to store instructions; The processor is configured to execute the instructions such that a communication device comprising the chip performs the method of any one of claims 2 to 23. Tips.

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