Communication method and device

Abstract

The present invention provides a method and apparatus for reducing the cost and complexity of baseband chips capable of handling low-frequency and high-frequency signal processing. [Solution] The first device generates a first orthogonal frequency division multiplexing (OFDM) signal supporting first and second frequency bands S301, transmits the generated signal on a first channel in the first frequency band S302, and the second device receives and processes the signal on the first channel S303. The amount of subcarriers in the first channel is greater than or equal to the amount of subcarriers in the second channel in the second frequency band, the lowest frequency in the first frequency band is higher than the highest frequency in the second, the first and second channels contain 2 to the power of M fundamental channels, the bandwidth of the first and second fundamental channels is the minimum supported channel bandwidth, the bandwidth of the first fundamental channel is greater than the second bandwidth, and M is an integer greater than or equal to 0.

Description

[Technical Field]

[0001] Cross-reference of related applications This application claims priority to Chinese Patent Application No. 202210369283.1, entitled “COMMUNICATION METHOD AND APPARATUS,” filed with the China National Intellectual Property Administration on 8 April 2022, which is incorporated herein by reference in its entirety.

[0002] This application relates to the field of communication technology, and more particularly to communication methods and apparatus. [Background technology]

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

[0004] The frame structures of Physical Layer Protocol Data Units (PHYs) differ for the different protocols mentioned above. In particular, the frame structure of PPDUs for low-frequency band protocols differs significantly from that of PPDUs for high-frequency band protocols. Therefore, the methods for processing low-frequency signals also differ from those for processing high-frequency signals. In this case, the hardware for processing the signals (e.g., dedicated baseband chips) also differs significantly.

[0005] In light of this, baseband chips are sometimes designed separately for high-frequency and low-frequency signals, but this is costly. Designing converged baseband chips that can be used to process both low-frequency and high-frequency signals is complex. [Overview of the project] [Means for solving the problem]

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

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

[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 Therefore, 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 this application, the correspondence between bandwidth and the amount of subcarriers in the first frequency band is similar to the correspondence between bandwidth and the amount of subcarriers in the second frequency band, and as a result, the first and second frequency bands can share signal processing devices, for example, an IDFT device or a DFT device can be 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 carried out by a receiving device, which may be an access point, a station, a chip, or a circuit. The method includes the steps of receiving a first OFDM signal on a first channel in a first frequency band, and processing a first OFDM symbol. The amount of subcarriers corresponding to the first channel is greater than or equal to the amount of subcarriers corresponding to a second channel in a second frequency band, the lowest frequency of the first frequency band is higher than the highest frequency of the second frequency band, and the first channel is 2 M It includes a first basic channel and a second channel. M It includes n second fundamental channels, the bandwidth of the first fundamental channel is the minimum channel bandwidth supported in the first frequency band, the bandwidth of the second fundamental channel is the 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.

[0010] In this embodiment of the present application, the correspondence between bandwidth and subcarrier quantity (or IDFT size / DFT size) in the second frequency band is reused in the first frequency band, and as a result, the first and second frequency bands can share signal processing devices, for example, an IDFT device or a DFT device can be reused. Device costs are 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 within a transmitting device, and the transmitting device may be an access point or a station. The communication device has the function of implementing any method provided in the first aspect. The communication device may be implemented by hardware, or by hardware running corresponding software. The hardware or software includes one or more units or modules corresponding to the above functions.

[0012] In possible designs, the communication device includes a processor, which is configured to support the communication device in performing the corresponding functions of the distance measuring initiator in the manner described above. The communication device may further include memory, which may be coupled to the processor and store the program instructions and data necessary for the communication device. Optionally, the communication device may further include interface circuitry, which is configured to support communication between the communication device and a device such as a receiving device.

[0013] For example, a communication device may have the function of implementing the method provided in the first embodiment, and a processor may be configured to generate a first OFDM signal. An interface circuit may be configured to send the first OFDM signal on a first channel in a first frequency band, wherein the amount of subcarriers corresponding to the first channel is greater than or equal to the amount of subcarriers corresponding to a second channel in a second frequency band, the lowest frequency of the first frequency band is higher than the highest frequency of the second frequency band, and the first channel is 2 M It includes a first basic channel and a second channel. M It includes n second fundamental channels, the bandwidth of the first fundamental channel is the minimum channel bandwidth supported in the first frequency band, the bandwidth of the second fundamental channel is the 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.

[0014] In possible designs, the communication device includes corresponding functional modules, each functional module configured to perform the steps in the method described above. The functions may be implemented by hardware, or by hardware running corresponding software. The hardware or software includes one or more modules corresponding to the functions described above.

[0015] In a possible design, the structure of the communication device includes a processing unit (or processing module) and a transceiver unit (or transceiver module). These units may implement the 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, the communication device has the function of implementing 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, the amount of subcarriers corresponding to the first channel is greater than or equal to the amount of subcarriers corresponding to a second channel in a second frequency band, the lowest frequency of the first frequency band is higher than the highest frequency of the second frequency band, the first channel includes 2 M first basic channels, the second channel includes 2 M second basic channels, the bandwidth of the first basic channel is the minimum channel bandwidth supported in the first frequency band, the bandwidth of the second basic channel is the minimum channel bandwidth supported in the second frequency band, the bandwidth of the first basic channel is greater than the bandwidth of the second basic channel, and M is 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 the receiving device, and the receiving device may be an access point or a station. The communication device has the function of implementing any method provided in the second aspect. The communication device may be implemented by hardware or by hardware that executes corresponding software. The hardware or software includes one or more units or modules corresponding to the above functions.

[0018] In possible designs, the communication device includes a processor, which is configured to support the communication device in performing the corresponding functions of the terminal device in the manner described above. The communication device may further include memory, which may be coupled to the processor and store the program instructions and data necessary for the communication device. Optionally, the communication device may further include interface circuitry, which is configured to support communication between the communication device and a device such as a transmitting device.

[0019] For example, a communication device may have the function of implementing the method provided in the second embodiment, and the interface circuit may be configured to receive a first OFDM signal on a first channel in a first frequency band, wherein the amount of subcarriers corresponding to the first channel is greater than or equal to the amount of subcarriers corresponding to a second channel in a second frequency band, the lowest frequency of the first frequency band is higher than the highest frequency of the second frequency band, and the first channel is 2 M It includes a first basic channel and a second channel. M The first basic channel includes n second basic channels, the bandwidth of the first basic channel being the minimum channel bandwidth supported in the first frequency band, the bandwidth of the second basic channel being the minimum channel bandwidth supported in the second frequency band, the bandwidth of the first basic channel being greater than the bandwidth of the second basic 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 possible designs, the communication device includes corresponding functional modules, each functional module configured to perform the steps in the method described above. The functions may be implemented by hardware, or by hardware running corresponding software. The hardware or software includes one or more modules corresponding to the functions described above.

[0021] In possible designs, the structure of the communication device includes a processing unit (or processing module) and a transceiver unit (or transceiver module). These units may perform the corresponding functions in the example methods described above. For further details, please refer to the description of the methods provided in the second embodiment. Further details are again not described herein.

[0022] For example, a communication device may have the function of implementing the method provided in the second embodiment, and the transceiver module may be configured to receive a first OFDM signal on a first channel in a first frequency band, wherein the amount of subcarriers corresponding to the first channel is greater than or equal to the amount of subcarriers corresponding to a second channel in a second frequency band, the lowest frequency of the first frequency band is higher than the highest frequency of the second frequency band, and the first channel is 2 M It includes a first basic channel and a second channel. M The first basic channel includes n second basic channels, the bandwidth of the first basic channel being the minimum channel bandwidth supported in the first frequency band, the bandwidth of the second basic channel being the minimum channel bandwidth supported in the second frequency band, the bandwidth of the first basic channel being greater than the bandwidth of the second basic 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 possible designs, the subcarrier spacing of the first channel is greater than or equal to a threshold. In this configuration, the effect of frequency offset on signal transmission on the first channel can be reduced, and signal transmission performance can 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 configuration, signals in the first frequency band and signals in the second frequency band can be transmitted by reusing the same crystal oscillator. This can reduce hardware complexity and hardware overhead.

[0026] In a possible design, the bandwidth of the first fundamental channel is 320 MHz. In this configuration, finer-grained channel division allows for the utilization of spectral resources in the first frequency band as much as possible. This reduces resource waste and improves resource utilization.

[0027] In a possible design, the amount of subcarriers corresponding to the first channel is as follows: If the bandwidth of the first channel is 320 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 640 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 1280 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 2560 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 5120 MHz, then the first channel corresponds to 1024 subcarriers. It satisfies one of the following conditions.

[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 is as follows: If the bandwidth of the first channel is 640 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 1280 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 2560 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 5120 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 10240MHz, then the first channel corresponds to 1024 subcarriers. It satisfies one of the following conditions.

[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 is as follows: If the bandwidth of the first channel is 80 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 160 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 320 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 640 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 1280 MHz, then the first channel corresponds to 1024 subcarriers. It satisfies one of the following conditions.

[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 is as follows: If the bandwidth of the first channel is 160 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 320 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 640 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 1280 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 2560 MHz, then the first channel corresponds to 1024 subcarriers. It satisfies one of the following conditions.

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

[0035] In a possible design, the bandwidth of the first fundamental channel is 270 MHz. By using finer-grained channel division, spectral resources in the first frequency band can be utilized 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 is as follows: If the bandwidth of the first channel is 270 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 540 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 1080 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 2160 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 4320 MHz, then the first channel corresponds to 1024 subcarriers. It satisfies one of the following conditions.

[0037] In a possible design, the bandwidth of the first fundamental channel is 540 MHz. By using finer-grained channel division, spectral resources in the first frequency band can be utilized 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 is as follows: If the bandwidth of the first channel is 540 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 1080 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 2160 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 4320 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 8640MHz, then the first channel corresponds to 1024 subcarriers. It satisfies one of the following conditions.

[0039] In a possible design, the bandwidth of the first fundamental channel is obtained by performing K bipositions on a 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. A portion of the bandwidth is selected from the minimum channel bandwidth supported in the millimeter-wave band, thereby further reducing inter-channel interference.

[0040] In a possible design, the bandwidth of the first fundamental channel is 250 MHz. By using finer-grained channel division, spectral resources in the first frequency band can be utilized 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 is as follows: If the bandwidth of the first channel is 250 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 500 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 1000 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 2000 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 4000 MHz, then the first channel corresponds to 1024 subcarriers. It satisfies one of the following conditions.

[0042] In a possible design, the bandwidth of the first fundamental channel is 500 MHz. By using finer-grained channel division, spectral resources in the first frequency band can be utilized 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 is as follows: If the bandwidth of the first channel is 500 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 1000 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 2000 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 4000 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 8000 MHz, then the first channel corresponds to 1024 subcarriers. It satisfies one of the following conditions.

[0044] In possible designs, the first frequency band is either 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 signals from a communication device other than this communication device and transmit those signals to the processor, or to transmit signals from the processor to a communication device other than this communication device. The processor is configured to implement the method in any one of the first aspect and possible designs of the first aspect by means of 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 signals from a communication device other than this communication device and transmit those signals to the processor, or to send signals from the processor to a communication device other than this communication device. The processor is configured to implement the method in any one of the second aspect and possible designs of the second aspect by means of logic circuits or by executing code instructions.

[0047] According to the 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, a method in any one of the first or second aspects and possible designs of the first or second aspect is carried out.

[0048] According to the eighth aspect, a computer program product for storing instructions is provided. When an instruction is executed by a processor, a method in any one of the first or second aspects and possible designs of the first or second aspect is carried out.

[0049] According to the ninth aspect, a chip system is provided. The chip system includes a processor and may further include memory configured to implement the methods in the first aspect and the possible designs of the first aspect. The chip system may include a chip, or a chip and other separate devices.

[0050] According to a tenth aspect, a chip system is provided. The chip system includes a processor and may further include memory configured to implement the methods in the second aspect and in possible designs of the second aspect. The chip system may include a chip, or a chip and other separate devices.

[0051] According to the eleventh aspect, a communication system is provided. The communication system includes the device in the first aspect (e.g., a transmitting device) and / or the device in the second aspect (e.g., a receiving device). [Brief explanation of the drawing]

[0052] [Figure 1] This is a diagram showing the architecture of a communication system according to one embodiment of this application. [Figure 2] This is a diagram showing the structure of a DFT-s-OFDM transmitter according to one embodiment of this application. [Figure 3]This is a schematic flowchart of a communication method according to one embodiment of this application. [Figure 4] This is a diagram showing channel splitting according to one embodiment of the present application. [Figure 5] This is a diagram showing channel splitting according to one embodiment of the present application. [Figure 6] This is a diagram showing channel splitting according to one embodiment of the present application. [Figure 7] This is a diagram showing channel splitting according to one embodiment of the present application. [Figure 8] This is a diagram showing channel splitting according to one embodiment of the present application. [Figure 9] This is a diagram showing channel splitting according to one embodiment of the present application. [Figure 10] This is a diagram showing channel splitting according to one embodiment of the present application. [Figure 11] This is a diagram showing channel splitting according to one embodiment of the present application. [Figure 12] This is a diagram showing the structure of a communication device according to one embodiment of this application. [Figure 13] This is a diagram showing the structure of a communication device according to one embodiment of this application. [Modes for carrying out the invention]

[0053] The embodiments of this application will be described in detail below with reference to the attached drawings.

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

[0055] Based on the principle of forward compatibility, next-generation WLAN technology supports both high-frequency and low-frequency bands and adapts to WLAN technologies (such as 802.11n / 802.11ac / 802.11ax / 802.11be) on the low-frequency band. However, the frame structure of the Physical Layer Protocol Data Unit (PHY protocol data unit, PPDU) differs for the different protocols mentioned above. In particular, the frame structure of the PPDU for protocols used in the low-frequency band differs significantly from that of the PPDU for protocols used in the high-frequency band. Therefore, the methods for processing low-frequency signals also differ from those for processing high-frequency signals. In this case, the hardware for processing the signals (e.g., a dedicated circuit baseband chip) also differs significantly. Regarding this problem, baseband chips can be designed separately for high-frequency and low-frequency signals, but this is costly. Designing a converged baseband chip that can be used to process both low-frequency and high-frequency signals is complex.

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

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

[0058] While embodiments of this application are primarily described using an example of a deployed WLAN network, particularly one to which the IEEE 802.11 system standard applies, those skilled in the art will readily understand that the various embodiments of this application can be extended to other networks using various standards or protocols, such as Bluetooth (Bluetooth®), high-performance radio LAN (HIPERLAN) (a wireless standard similar to the IEEE 802.11 standard, mainly used in Europe), wide area networks (WANs), personal area networks (PANs), or other networks known or to be developed in the future. Therefore, regardless of the coverage area and wireless access protocol used, the various embodiments provided in this application are applicable to any suitable wireless network.

[0059] Alternatively, embodiments of this application are applicable to wireless local area network systems, such as the Internet of Things (IoT) or Vehicle to Everything (V2X) networks. Certainly, embodiments of this 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, and new communication systems emerging in future communication developments. The above-mentioned communication systems applicable to this application are merely illustrative examples and are not limited to those applicable to this application. This is explained uniformly in this specification and further details are not provided 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, access point stations are referred to as access points (APs) and non-access point stations are referred to as stations (STAs). In Figure 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 carried out between the APs and STAs, and the STAs may be in a fixed location or may be mobile. The number 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 between STAs (e.g., data communication between STA102 and STA103).

[0061] An AP, also known as an access point or hotspot, provides wireless access services, enabling other wireless devices to access and provide data access. In other words, an AP can be an access point used by terminal devices (such as mobile phones) to access a wired (or wireless) network, and is primarily deployed in homes, buildings, and parks. Typical coverage radius is tens to hundreds of meters. Of course, access points can also be deployed outdoors. An AP is equivalent to a bridge that connects wired and wireless networks. APs are primarily 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) with a Wi-Fi chip. An access point may be a device that supports the 802.11be standard. Alternatively, the access point may be a device that supports multiple wireless local area network standards of the 802.11 family, such as next-generation standards of 802.11ax, 802.11ac, 802.11n, 802.11g, 802.11b, 802.11a, and 802.11be. The access point in this application may be a high-efficiency (HE) AP or EHT AP, or an access point applicable to future generations of 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 it may be a module or unit that completes some of the functions of a base station, for example, a central unit (CU), a distributed unit (DU), a router, a switch, or a network bridge. The specific technologies and device forms used by the AP are not limited in this application.

[0062] An 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. A station may also be called a terminal, user equipment (UE), mobile station, or mobile terminal. For example, a station may be a mobile phone, tablet computer, set-top box, smart television, smart wearable device, in-vehicle communication device, or computer that supports Wi-Fi communication capabilities. Optionally, a station may support the 802.11be standard. Alternatively, a station may support multiple WLAN standards in 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 the next generation of Wi-Fi 8.

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

[0064] For example, access points and stations may be devices used in the Internet of Things for vehicles, Internet of Things nodes, sensors in the Internet of Things, smart cameras, smart remote controls, smart water or electricity meters in smart homes, sensors in smart cities, etc.

[0065] APs and STAs may communicate with each other on the 2.4 gigahertz (GHz) frequency band, the 5 GHz or 6 GHz frequency band, or 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 the 57–66 GHz frequency band, or the 60 GHz frequency band may include 56.16–73.44 GHz. It should be understood that as communication advances, the 45 GHz frequency band and the 60 GHz frequency band may each further include other frequency bands. This is not particularly limited in this specification.

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

[0067] In possible implementations, the communication device may be a Discrete Fourier Transform Orthogonal Frequency Division Multiplexing (DFT-s-OFDM) transmitter. Figure 2 shows the structure of a DFT-s-OFDM transmitter according to this 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 transmitted, and the mapping module may be configured to map the signal output by the IDFT module to subcarriers. It can be understood that the communication device may alternatively be a DFT-s-OFDM receiver. For the structure of a DFT-s-OFDM receiver, refer to the above diagram illustrating the structure of a DFT-s-OFDM transmitter. The signal processing performed by a DFT-s-OFDM receiver is the reverse process of the signal processing performed by a DFT-s-OFDM transmitter.

[0068] In embodiments of this application, “at least one” means one or more, and “multiple” means two or more. The term “and / or” indicates a relationship between related objects, and indicates that three relationships may exist. For example, A and / or B may indicate the following cases: only A exists, both A and B exist, only B exists, however A and B may be singular or plural. The letter “ / ” usually indicates an “or” relationship between related objects. “At least one of the following members (parts)” or a similar expression means any combination of these members, including a single member (part) or any combination of multiple members (parts). For example, at least one member (part) of a, b, or c may be a, b, c, ab, ac, bc, or abc, however a, b, and c may be singular or plural.

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

[0070] The network architectures and service scenarios described in the embodiments of this application are intended to more clearly illustrate the technical solutions in the embodiments of this application and do not constitute a limitation on the technical solutions provided in the embodiments of this application. Those skilled in the art will see that, with the evolution of network architectures and the emergence of new service scenarios, the technical solutions provided in the embodiments of this application are also applicable to similar technical problems.

[0071] Note that in the following description, an example is used in which the first device is a transmitting device and the second device is a receiving device. The first device may be an AP and the second device may be an STA, in other words, the AP sends signals to the STA, or the first device may be an STA and the second device may be an AP, in other words, the STA sends signals to the AP. In the following description, only the first and second devices are used as executables for illustrative purposes. Optionally, the operation of the first device may be alternatively performed by a processor, chip, or functional module within the first device, and the operation of the second device may be alternatively performed by a processor, chip, or functional module within the second device. This is not limited to this application.

[0072] In the embodiments of this application, the channel bandwidth may be in Hertz. Further details are again not described below.

[0073] The method provided in this application may be applied to a first frequency band. The following further relates to a second frequency band, where 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 may be a 60 GHz frequency band, or the first frequency band may be a 45 GHz frequency band, and the second frequency band may be a 2.45 GHz frequency band, or the second frequency band may be a 5 GHz / 6 GHz frequency band, etc. To facilitate understanding of this solution, this application will be illustrated by using an example where 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 other frequency bands is similar to the method for the 45 GHz and 60 GHz frequency bands, and details are again not described herein.

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

[0075] In embodiments of this 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 between 1 and 10 millimeters. For example, the millimeter-wave band may be the 30–300 GHz frequency band. It should be understood that the 30–300 GHz frequency band is merely an example of the millimeter-wave band for illustrative purposes, and may be a different frequency band. This is not particularly limited herein.

[0076] Embodiment 1

[0077] Figure 3 is a schematic flowchart of the communication method according to this application. The method includes the following steps.

[0078] S301: The first device generates the first OFDM signal.

[0079] In possible implementations, the first device may perform an IDFT on the signal to be transmitted based on a first size in order to acquire a first OFDM signal. Correspondingly, the second device may perform a DFT on the received first OFDM signal based on a first size in order to convert the received first OFDM signal into a signal in the frequency domain. The first size may be the 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. Correspondingly, the second device performs a DFT on the received SC signal based on the first size in order 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 higher than the highest frequency of the second frequency band, and the first channel is 2 M It includes a first basic channel and a second channel. M It includes n second fundamental channels, the bandwidth of the first fundamental channel is the minimum channel bandwidth supported in the first frequency band, the bandwidth of the second fundamental channel is the 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.

[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 configuration, the effect of frequency offset on signal transmission on the first channel may be reduced, and signal transmission performance may be improved.

[0083] It should be noted that "first basic channel" is merely an illustrative name and represents the channel corresponding to the minimum channel bandwidth supported in the first frequency band. For example, if the supported bandwidth 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 with a bandwidth of 320 MHz in the first frequency band may be understood as the first basic channel in this application. The understanding of the second basic channel is similar to that of the first basic channel. For example, if the supported bandwidth in the second basic 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 with a bandwidth of 20 MHz in the second frequency band may be understood as the second basic channel in this application.

[0084] The relevant explanation for the first channel is described in detail below.

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

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

[0087] Please note that S303 may be an optional step.

[0088] The following describes the amount of subcarriers in the first channel by using an example of 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, or 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, or 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 one of the following conditions:

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

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

[0095] Optionally, the number of subcarriers corresponding to the first channel may be greater than 64. For example, the first channel may correspond to 128 subcarriers, 256 subcarriers, or other subcarriers.

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

[0097] Optionally, the number of subcarriers corresponding to the first channel may be greater than 128. For example, the first channel may correspond to 256 subcarriers, 512 subcarriers, or other subcarriers.

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

[0099] Optionally, the number of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, or other numbers.

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

[0101] Optionally, the first channel corresponds to 512 subcarriers. For example, the first channel could alternatively correspond to 1024 subcarriers, or something similar.

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

[0103] For example, the correspondence between each supported bandwidth in the first frequency band and the amount of subcarriers (or IDFT size / DFT size) 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, or 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 one of the following conditions:

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

[0108] Optionally, the number 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, and so on.

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

[0110] Optionally, the number 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, and so on.

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

[0112] Optionally, the number of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, or other numbers.

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

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

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

[0116] For example, the correspondence between each bandwidth supported in the first frequency band and the amount of subcarriers (or IDFT size / DFT size) 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, or 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 one of the following conditions:

[0120] If the bandwidth of the first channel is 40 MHz, the first channel corresponds to 64 subcarriers. This implementation can be achieved by directly extending the subcarrier spacing of the second signal in a 20 MHz bandwidth. Optionally, the number 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.

[0121] When the bandwidth of the first channel is 80 MHz, the first channel corresponds to 128 sub - carriers. This implementation can be achieved by directly expanding the sub - carrier spacing of the second signal in a 40 MHz bandwidth. Optionally, the amount of sub - carriers corresponding to the first channel may be greater than 128. For example, the first channel may correspond to 256 sub - carriers, 512 sub - carriers, etc.

[0122] When the bandwidth of the first channel is 160 MHz, the first channel corresponds to 256 sub - carriers. This implementation can be achieved by directly expanding the sub - carrier spacing of the second signal in an 80 MHz bandwidth. Optionally, the amount of sub - carriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 sub - carriers, 1024 sub - carriers, etc.

[0123] When the bandwidth of the first channel is 320 MHz, the first channel corresponds to 512 sub - carriers. This implementation can be achieved by directly expanding the sub - carrier spacing of the second signal in a 160 MHz bandwidth. Optionally, the amount of sub - carriers corresponding to the first channel may be greater than 512. For example, the first channel may correspond to 1024 sub - carriers, etc.

[0124] When the bandwidth of the first channel is 640 MHz, the first channel corresponds to 1024 sub - carriers. This implementation can be achieved by directly expanding the sub - carrier spacing of the second signal in a 320 MHz bandwidth.

[0125] For example, the correspondence between each bandwidth supported in the first frequency band and the amount of sub - carriers (or IDFT size / DFT size) can 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 basic 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 of the first channel may satisfy any one of the following.

[0129] When the bandwidth of the first channel is 80 MHz, the first channel corresponds to 64 subcarriers. This implementation may be achieved by a direct extension of the subcarrier spacing of the second signal 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.

[0130] When the bandwidth of the first channel is 160 MHz, the first channel corresponds to 128 subcarriers. This implementation may be achieved by a direct extension of the subcarrier spacing of the second signal in a 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 may be achieved by a direct extension of the subcarrier spacing of the second signal in an 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] If the bandwidth of the first channel is 640 MHz, the first channel corresponds to 512 subcarriers. This implementation can be achieved by directly extending the subcarrier spacing of the second signal in a 160 MHz bandwidth. Optionally, the number of subcarriers corresponding to the first channel may be greater than 512. For example, the first channel may correspond to 1024 subcarriers, for instance.

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

[0134] For example, the correspondence between each bandwidth supported in the first frequency band and the amount of subcarriers (or IDFT size / DFT size) 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, or 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 one of the following conditions:

[0138] If the bandwidth of the first channel is 160 MHz, the first channel corresponds to 64 subcarriers. This implementation can be achieved by directly extending the subcarrier spacing of the second signal in 802.11ax / be with a bandwidth of 20 MHz. Optionally, the number 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.

[0139] If the bandwidth of the first channel is 320 MHz, the first channel corresponds to 128 subcarriers. This implementation can be achieved by directly extending the subcarrier spacing of the second signal in a 40 MHz bandwidth. Optionally, the number 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.

[0140] If the bandwidth of the first channel is 640 MHz, the first channel corresponds to 256 subcarriers. This implementation can be achieved by directly extending the subcarrier spacing of the second signal in an 80 MHz bandwidth. Optionally, the number 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.

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

[0142] When the bandwidth of the first channel is 2560 MHz, the first channel corresponds to 1024 sub - carriers. This implementation can be achieved by directly extending the sub - carrier spacing of the second signal at a 320 MHz bandwidth.

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

[0144]

Table 5

[0145] Example 2: The bandwidth of the first basic channel is obtained by separately performing R bisections on the frequency band obtained by dividing the first frequency band based on the minimum channel bandwidth supported in the millimeter - wave band of the IEEE802.11ad / ay standard, where R is an integer greater than or equal to 0. For example, assume that the minimum channel bandwidth supported in the millimeter - wave band is 2160 MHz. In this case, the bandwidth of the first basic channel can 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, that is, the bandwidth of the first basic channel is 270 MHz. The first frequency band can 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 sub - carriers of the first channel can 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 directly extending the subcarrier spacing of the second signal in a 20 MHz bandwidth.

[0149] Optionally, the number 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, and so on.

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

[0151] Optionally, the number 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, and so on.

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

[0153] Optionally, the number of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, or other numbers.

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

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

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

[0157] For example, the correspondence between each supported bandwidth in the first frequency band and the amount of subcarriers (or IDFT size / DFT size) 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, or 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 one of the following conditions:

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

[0162] Optionally, the number 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, and so on.

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

[0164] Optionally, the number 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, and so on.

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

[0166] Optionally, the number of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, or other numbers.

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

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

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

[0170] For example, the correspondence between each supported bandwidth in the first frequency band and the amount of subcarriers (or IDFT size / DFT size) 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, assume that the minimum channel bandwidth supported in the millimeter-wave band is 2160 MHz. In this case, the first subbands could be 1000 MHz, 1800 MHz, 1940 MHz, 2000 MHz, 2100 MHz, etc. In possible implementations, the bandwidth of the first subbands could be an integer multiple of 20 MHz.

[0173] For the sake of clarity, an example where the first subband is 2000 MHz is used for the 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, or 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 one of the following conditions:

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

[0177] Optionally, the number 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, and so on.

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

[0179] Optionally, the number 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, and so on.

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

[0181] Optionally, the number of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, or other numbers.

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

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

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

[0185] For example, the correspondence between each supported bandwidth in the first frequency band and the amount of subcarriers (or IDFT size / DFT size) 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, or 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 one of the following conditions:

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

[0190] Optionally, the number 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, and so on.

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

[0192] Optionally, the number 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, and so on.

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

[0194] Optionally, the number of subcarriers corresponding to the first channel may be greater than 256. For example, the first channel may correspond to 512 subcarriers, 1024 subcarriers, or other numbers.

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

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

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

[0198] For example, the correspondence between each supported bandwidth in the first frequency band and the amount of subcarriers (or IDFT size / DFT size) 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 Therefore, 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 bandwidth and IDFT size / DFT size (or amount of subcarriers) in the high frequency band (e.g., the first frequency band) is similar to the correspondence between bandwidth and amount of subcarriers in the low frequency band (e.g., the second frequency band), and as a result, the high and low frequency bands can share signal processing devices, for example, IDFT or DFT devices can be reused. Thus, device costs are low.

[0201] Embodiment 1 describes the amount of subcarriers (or IDFT size / DFT size) corresponding to each bandwidth supported in the first frequency band described above. Below, a channel division method for the first frequency band is described as an example. For details, please refer to Embodiments 2, 3, and 4.

[0202] The channel splitting method provided in this application may be implemented in combination with Embodiment 1. Specifically, the channel splitting method provided in Embodiment 2 may be implemented in combination with Example 1 of Embodiment 1, or, if the first frequency band is split into channels in the manner of Embodiment 2, it may be understood that the correspondence between each channel bandwidth and the amount of subcarriers refers to Example 1 of Embodiment 1. The channel splitting method provided in Embodiment 3 may be implemented in combination with Example 2 of Embodiment 1, or, if the first frequency band is split into channels in the manner of Embodiment 3, it may be understood that the correspondence between each channel bandwidth and the amount of subcarriers refers to Example 2 of Embodiment 1. The channel splitting method provided in Embodiment 4 may be implemented in combination with Example 3 of Embodiment 1, or, if the first frequency band is split into channels in the manner of Embodiment 4, it may be understood that 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 (e.g., the low-frequency minimum bandwidth as defined in 802.11be) is 20 MHz. In the channel splitting method provided in Embodiment 2 of this application, the bandwidth of the first frequency band may be split based on integer multiples of 20 MHz, for example, 40 MHz, 80 MHz, 160 MHz, 320 MHz, or 640 MHz. In Embodiment 2 of this 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), and as a result, signals in the high-frequency band and signals in the low-frequency band can be transmitted by reusing the same crystal oscillator. This can reduce hardware complexity and hardware overhead.

[0205] The following describes a channel division method for a first frequency band, using an example relating to a specific frequency band.

[0206] Example 4: The minimum channel bandwidth supported in the first frequency band is 320 MHz, or 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 bandwidths of 320 MHz, 640 MHz, 1280 MHz, 2560 MHz, and 5120 MHz.

[0207] For the amount of subcarriers (or IDFT size / DFT size) corresponding to each channel bandwidth in the channel splitting configuration of Example 4, see the relevant description in Example 1 of Embodiment 1. Further details are again not provided herein.

[0208] For example, the first frequency band is the 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 Figure 4.

[0209] A channel splitting configuration for the 57–66 GHz frequency band is described herein. In this splitting configuration, a 20 MHz bandwidth remains in the forward and backward portions of the first frequency band, as shown in Figure 5.

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

[0211] 640MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 57.340 + (n-1) × 0.640, where 1 ≤ n ≤ 14, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.020 + (n-1) × 0.640, where 1 ≤ n ≤ 14, and the formula for calculating the highest 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, where 1 ≤ n ≤ 7, and the formula for calculating the lowest frequency is f L (n)[GHz]=57.020+(n-1)×1.280, where 1≦n≦7, and the formula for calculating the highest 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, where 1 ≤ n ≤ 3, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.020 + (n-1) × 2.560, where 1 ≤ n ≤ 3, and the formula for calculating the highest frequency is f H (n)[GHz] = 59.580 + (n-1) × 2.560, where 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, where n=1, and the formula for calculating the lowest frequency is f L(n)[GHz]=57.020+(n-1)×5.120, where n=1, and the formula for calculating the highest frequency is f H (n)[GHz] = 62.140 + (n-1) × 5.120, where n = 1.

[0215] In the channel splitting configuration provided in this application, n is the sequence number or index number of the channel. Further details are again not described 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 17 320 MHz channels, 8 640 MHz channels, 3 1280 MHz channels, or 1 2560 MHz channel.

[0217] Channel division methods for the 42.3–47.0 GHz frequency band and the 47.2–48.4 GHz frequency band are described herein. In this division method, the remaining bandwidth in both the forward and backward portions of the 42.3–47.0 GHz frequency band is 110 MHz, and the remaining bandwidth in both the forward and backward 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, where 1 ≤ n ≤ 14, and the formula for calculating the lowest frequency is f L (n)[GHz]=42.410+(n-1)×0.320, where 1≦n≦14, and the formula for calculating the highest frequency is f H (n)[GHz] = 42.730 + (n-1) × 0.320, where 1 ≤ n ≤ 14. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C(n)[GHz] = 47.480 + (n-15) × 0.320, where 15 ≤ n ≤ 17, and the formula for calculating the lowest frequency is f L (n)[GHz] = 47.320 + (n-15) × 0.320, where 15 ≤ n ≤ 17, and the formula for calculating the highest 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 7 640MHz channels, and the 47.2~48.4GHz frequency band contains 1 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, where 1 ≤ n ≤ 7, and the formula for calculating the lowest frequency is f L (n)[GHz]=42.410+(n-1)×0.640, where 1≦n≦7, and the formula for calculating the highest frequency is f H (n)[GHz] = 43.050 + (n-1) × 0.640, where 1 ≤ n ≤ 7. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz] = 47.640 + (n-8) × 0.640, where n = 8, and the formula for calculating the lowest frequency is f L (n)[GHz] = 47.320 + (n-8) × 0.640, where n = 8, and the formula for calculating the highest frequency is f H (n)[GHz] = 47.960 + (n-8) × 0.640, where 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 lowest frequency is f L (n)[GHz] = 42.410 + (n-1) × 1.280, where 1 ≤ n ≤ 3, and the formula for calculating the highest 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, where n=1, and the formula for calculating the lowest frequency is f L (n)[GHz]=42.410+(n-1)×2.560, where n=1, and the formula for calculating the highest frequency is f H (n)[GHz] = 44.970 + (n-1) × 2.560, where 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 this division scheme, there is no remaining bandwidth in the forward and backward 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, where 1 ≤ n ≤ 54, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 0.320, where 1 ≤ n ≤ 54, and the formula for calculating the highest frequency is f H (n)[GHz] = 56.480 + (n-1) × 0.320, where 1 ≤ n ≤ 54.

[0225] 640MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 56.480 + (n-1) × 0.640, where 1 ≤ n ≤ 27, and the formula for calculating the lowest frequency is f L(n)[GHz] = 56.160 + (n-1) × 0.640, where 1 ≤ n ≤ 27, and the formula for calculating the highest 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, where 1 ≤ n ≤ 13, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 1.280, where 1 ≤ n ≤ 13, and the formula for calculating the highest 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, where 1 ≤ n ≤ 6, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 2.560, where 1 ≤ n ≤ 6, and the formula for calculating the highest 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, where 1 ≤ n ≤ 3, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 5.120, where 1 ≤ n ≤ 3, and the formula for calculating the highest frequency is f H (n)[GHz] = 61.280 + (n-1) × 5.120, where 1 ≤ n ≤ 3.

[0229] Example 5: The minimum channel bandwidth supported in the first frequency band is 80 MHz, or 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 bandwidths of 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 splitting configuration of Example 5, refer to the relevant description in Example 4 of Embodiment 1. Further details are again not provided herein.

[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 configuration, a 20 MHz bandwidth remains in both 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 lowest frequency is f L (n)[GHz] = 57.020 + (n-1) × 0.080, 1 ≤ n ≤ 112, and the formula for calculating the highest frequency is f H (n)[GHz] = 57.100 + (n-1) × 0.080, where 1 ≤ n ≤ 112.

[0233] 160MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 57.100 + (n-1) × 0.160, where 1 ≤ n ≤ 56, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.020 + (n-1) × 0.160, where 1 ≤ n ≤ 56, and the formula for calculating the highest 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, where 1 ≤ n ≤ 28, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.020 + (n-1) × 0.320, where 1 ≤ n ≤ 28, and the formula for calculating the highest frequency is f H (n)[GHz] = 57.340 + (n-1) × 0.320, where 1 ≤ n ≤ 28.

[0235] 640MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 57.340 + (n-1) × 0.640, where 1 ≤ n ≤ 14, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.020 + (n-1) × 0.640, where 1 ≤ n ≤ 14, and the formula for calculating the highest 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, where 1 ≤ n ≤ 7, and the formula for calculating the lowest frequency is f L (n)[GHz]=57.020+(n-1)×1.280, where 1≦n≦7, and the formula for calculating the highest 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 73 80 MHz channels, 36 160 MHz channels, 17 320 MHz channels, 8 640 MHz channels, or 3 1280 MHz channels.

[0238] Channel division patterns for the 42.3 - 47.0 GHz frequency band and the 47.2 - 48.4 GHz frequency band are described herein. In this division method, the remaining bandwidths in both the front and rear parts of the 42.3 - 47.0 GHz frequency band are 30 MHz, and there is no remaining bandwidth in the front and rear parts of the 47.2 - 48.4 GHz frequency band.

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

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

[0241] 320 MHz channel division: The frequency band of 42.3 - 47.0 GHz includes 14 320 MHz channels, and the frequency band of 47.2 - 48.4 GHz includes 3 320 MHz channels. For the frequency band of 42.3 - 47.0 GHz, the formula for calculating the center frequency is f C (n) [GHz] = 42.490 + (n - 1) × 0.320, where 1 ≤ n ≤ 14. The formula for calculating the lowest frequency is f L (n) [GHz] = 42.330 + (n - 1) × 0.320, where 1 ≤ n ≤ 14. The formula for calculating the highest frequency is f H (n) [GHz] = 42.650 + (n - 1) × 0.320, where 1 ≤ n ≤ 14. For the frequency band of 47.2 - 48.4 GHz, the formula for calculating the center frequency is f C (n) [GHz] = 47.360 + (n - 15) × 0.320, where 15 ≤ n ≤ 17. The formula for calculating the lowest frequency is f L (n) [GHz] = 47.200 + (n - 15) × 0.320, where 15 ≤ n ≤ 17. The formula for calculating the highest 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 7 640MHz channels, and the 47.2~48.4GHz frequency band contains 1 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, where 1 ≤ n ≤ 7, and the formula for calculating the lowest frequency is f L (n)[GHz] = 42.330 + (n-1) × 0.640, where 1 ≤ n ≤ 7, and the formula for calculating the highest frequency is f H (n)[GHz] = 42.970 + (n-1) × 0.640, where 1 ≤ n ≤ 7. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz] = 47.520 + (n-8) × 0.640, where n = 8, and the formula for calculating the lowest frequency is f L (n)[GHz] = 47.200 + (n-8) × 0.640, where n = 8, and the formula for calculating the highest frequency is f H (n)[GHz] = 47.840 + (n-8) × 0.640, where n = 8.

[0243] 1280MHz channel division: The formula for calculating the center frequency is f C (n)[GHz] = 42.970 + (n-1) × 1.280, where 1 ≤ n ≤ 3, and the formula for calculating the lowest frequency is f L (n)[GHz] = 42.330 + (n-1) × 1.280, where 1 ≤ n ≤ 3, and the formula for calculating the highest 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 this division scheme, there is no remaining bandwidth in the forward and backward 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, where 1 ≤ n ≤ 216, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 0.080, where 1 ≤ n ≤ 216, and the formula for calculating the highest frequency is f H (n)[GHz] = 56.240 + (n-1) × 0.080, where 1 ≤ n ≤ 216.

[0247] 160MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 56.240 + (n-1) × 0.160, where 1 ≤ n ≤ 10⁸, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 0.160, where 1 ≤ n ≤ 10⁸, and the formula for calculating the highest frequency is f H (n)[GHz] = 56.320 + (n-1) × 0.160, where 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, where 1 ≤ n ≤ 54, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 0.320, where 1 ≤ n ≤ 54, and the formula for calculating the highest frequency is f H (n)[GHz] = 56.480 + (n-1) × 0.320, where 1 ≤ n ≤ 54.

[0249] 640MHz channel splitting: The formula for calculating the center frequency is f C(n)[GHz] = 56.480 + (n-1) × 0.640, where 1 ≤ n ≤ 27, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 0.640, where 1 ≤ n ≤ 27, and the formula for calculating the highest 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, where 1 ≤ n ≤ 13, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 1.280, where 1 ≤ n ≤ 13, and the formula for calculating the highest 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, or 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 bandwidths of 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 splitting configuration of Example 6, please refer to the relevant description in Example 5 of Embodiment 1. Further details are again not provided herein.

[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 configuration, a 20 MHz bandwidth remains in both the front and rear portions of the first frequency band.

[0254] 160MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 57.100 + (n-1) × 0.160, where 1 ≤ n ≤ 56, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.020 + (n-1) × 0.160, where 1 ≤ n ≤ 56, and the formula for calculating the highest 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, where 1 ≤ n ≤ 28, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.020 + (n-1) × 0.320, where 1 ≤ n ≤ 28, and the formula for calculating the highest frequency is f H (n)[GHz] = 57.340 + (n-1) × 0.320, where 1 ≤ n ≤ 28.

[0256] 640MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 57.340 + (n-1) × 0.640, where 1 ≤ n ≤ 14, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.020 + (n-1) × 0.640, where 1 ≤ n ≤ 14, and the formula for calculating the highest 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, where 1 ≤ n ≤ 7, and the formula for calculating the lowest frequency is f L (n)[GHz]=57.020+(n-1)×1.280, where 1≦n≦7, and the formula for calculating the highest 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, where 1 ≤ n ≤ 3, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.020 + (n-1) × 2.560, where 1 ≤ n ≤ 3, and the formula for calculating the highest frequency is f H (n)[GHz] = 59.580 + (n-1) × 2.560, where 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 35 160 MHz channels, 17 320 MHz channels, 8 640 MHz channels, 3 1280 MHz channels, or 1 2560 MHz channel.

[0260] Channel division methods for the 42.3–47.0 GHz frequency band and the 47.2–48.4 GHz frequency band are described herein. In this division method, the remaining bandwidth in both the forward and backward portions of the 42.3–47.0 GHz frequency band is 110 MHz, and the remaining bandwidth in both the forward and backward 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, where 1 ≤ n ≤ 28, and the formula for calculating the lowest frequency is f L (n)[GHz] = 42.410 + (n-1) × 0.160, where 1 ≤ n ≤ 28, and the formula for calculating the highest frequency is fH (n)[GHz] = 42.570 + (n-1) × 0.160, where 1 ≤ n ≤ 28. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz] = 47.320 + (n-29) × 0.160, where 29 ≤ n ≤ 35, and the formula for calculating the lowest frequency is f L (n)[GHz] = 47.240 + (n-29) × 0.160, where 29 ≤ n ≤ 35, and the formula for calculating the highest 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, where 1 ≤ n ≤ 14, and the formula for calculating the lowest frequency is f L (n)[GHz]=42.410+(n-1)×0.320, where 1≦n≦14, and the formula for calculating the highest frequency is f H (n)[GHz] = 42.730 + (n-1) × 0.320, where 1 ≤ n ≤ 14. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz] = 47.400 + (n-15) × 0.320, where 15 ≤ n ≤ 17, and the formula for calculating the lowest frequency is f L (n)[GHz] = 47.240 + (n-15) × 0.320, where 15 ≤ n ≤ 17, and the formula for calculating the highest 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 7 640MHz channels, and the 47.2~48.4GHz frequency band contains 1 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, where 1 ≤ n ≤ 7, and the formula for calculating the lowest frequency is f L (n)[GHz]=42.410+(n-1)×0.640, where 1≦n≦7, and the formula for calculating the highest frequency is f H (n)[GHz] = 43.050 + (n-1) × 0.640, where 1 ≤ n ≤ 7. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz] = 47.560 + (n-8) × 0.640, where n = 8, and the formula for calculating the lowest frequency is f L (n)[GHz] = 47.240 + (n-8) × 0.640, where n = 8, and the formula for calculating the highest frequency is f H (n)[GHz] = 47.880 + (n-8) × 0.640, where 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 lowest frequency is f L (n)[GHz] = 42.410 + (n-1) × 1.280, where 1 ≤ n ≤ 3, and the formula for calculating the highest 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, where n=1, and the formula for calculating the lowest frequency is f L (n)[GHz]=42.410+(n-1)×2.560, where n=1, and the formula for calculating the highest frequency is f H(n)[GHz] = 44.970 + (n-1) × 2.560, where 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 this division scheme, there is no remaining bandwidth in the forward and backward portions of the first frequency band.

[0268] 160MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 56.240 + (n-1) × 0.160, where 1 ≤ n ≤ 10⁸, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 0.160, where 1 ≤ n ≤ 10⁸, and the formula for calculating the highest frequency is f H (n)[GHz] = 56.320 + (n-1) × 0.160, where 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, where 1 ≤ n ≤ 54, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 0.320, where 1 ≤ n ≤ 54, and the formula for calculating the highest frequency is f H (n)[GHz] = 56.480 + (n-1) × 0.320, where 1 ≤ n ≤ 54.

[0270] 640MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 56.480 + (n-1) × 0.640, where 1 ≤ n ≤ 27, and the formula for calculating the lowest frequency is f L(n)[GHz] = 56.160 + (n-1) × 0.640, where 1 ≤ n ≤ 27, and the formula for calculating the highest 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, where 1 ≤ n ≤ 13, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 1.280, where 1 ≤ n ≤ 13, and the formula for calculating the highest 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, where 1 ≤ n ≤ 6, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 2.560, where 1 ≤ n ≤ 6, and the formula for calculating the highest 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, or 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 bandwidths of 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 splitting configuration of Example d, please refer to the relevant description in Example 2 of Embodiment 1. Further details are again not provided herein.

[0275] For example, the first frequency band is the 57–66 GHz frequency band. The first frequency band may be divided into 14 640 MHz channels, 7 1280 MHz channels, 3 2560 MHz channels, or 1 5120 MHz channel, as shown in Figure 6. In this division configuration, a 20 MHz bandwidth remains in 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 splitting: The formula for calculating the center frequency is f C (n)[GHz] = 57.340 + (n-1) × 0.640, where 1 ≤ n ≤ 14, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.020 + (n-1) × 0.640, where 1 ≤ n ≤ 14, and the formula for calculating the highest 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, where 1≦n≦7, and the formula for calculating low frequencies is f L (n)[GHz] = 57.020 + (n-1) × 1.280, where 1 ≤ n ≤ 7, and the formula for calculating high frequencies 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 low frequencies is f L (n)[GHz]=57.020+(n-1)×2.560, where 1≦n≦3, and the formula for calculating high frequencies is f H(n)[GHz] = 59.580 + (n-1) × 2.560, where 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, where n=1, and the formula for calculating low frequencies is f L (n)[GHz]=57.020+(n-1)×5.120, where n=1, and the formula for calculating high frequencies is f H (n)[GHz] = 62.140 + (n-1) × 5.120, where 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] Channel division methods for the 42.3–47.0 GHz frequency band and the 47.2–48.4 GHz frequency band are described herein. In this division method, the remaining bandwidth in both the forward and backward portions of the 42.3–47.0 GHz frequency band is 110 MHz, and the remaining bandwidth in both the forward and backward 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 7 640MHz channels, and the 47.2~48.4GHz frequency band contains 1 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, where 1 ≤ n ≤ 7, and the formula for calculating the lowest frequency is f L (n)[GHz]=42.410+(n-1)×0.640, where 1≦n≦7, and the formula for calculating the highest frequency is f H(n)[GHz] = 43.050 + (n-1) × 0.640, where 1 ≤ n ≤ 7. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz] = 47.640 + (n-8) × 0.640, where n = 8, and the formula for calculating the lowest frequency is f L (n)[GHz] = 47.320 + (n-8) × 0.640, where n = 8, and the formula for calculating the highest frequency is f H (n)[GHz] = 47.960 + (n-8) × 0.640, where 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 lowest frequency is f L (n)[GHz] = 42.410 + (n-1) × 1.280, where 1 ≤ n ≤ 3, and the formula for calculating the highest 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, where n=1, and the formula for calculating the lowest frequency is f L (n)[GHz]=42.410+(n-1)×2.560, where n=1, and the formula for calculating the highest frequency is f H (n)[GHz] = 44.970 + (n-1) × 2.560, where 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 27 640 MHz channels, 13 1280 MHz channels, 6 2560 MHz channels, 3 5120 MHz channels, or 1 10240 MHz channel.

[0287] A channel division scheme for the 56.16–73.44 GHz frequency band is described herein. In this division scheme, there is no remaining bandwidth in the forward and backward portions of the first frequency band.

[0288] 640MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 56.480 + (n-1) × 0.640, where 1 ≤ n ≤ 27, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 0.640, where 1 ≤ n ≤ 27, and the formula for calculating the highest 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, where 1 ≤ n ≤ 13, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 1.280, where 1 ≤ n ≤ 13, and the formula for calculating the highest 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, where 1 ≤ n ≤ 6, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 2.560, where 1 ≤ n ≤ 6, and the formula for calculating the highest 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, where 1 ≤ n ≤ 3, and the formula for calculating the lowest frequency is fL (n)[GHz] = 56.160 + (n-1) × 5.120, where 1 ≤ n ≤ 3, and the formula for calculating the highest frequency is f H (n)[GHz] = 61.280 + (n-1) × 5.120, where 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 splitting configuration provided in this application, n is the sequence number or index number of the channel. Further details are again not described below.

[0294] The division patterns in Examples 4 through 7 are merely illustrative examples, and it should be understood that the positions of the channel's center frequency, lowest frequency, and highest frequency are not particularly limited. The channel division patterns can be used under the condition that the bandwidth is divided based on integer multiples of 20 MHz and the subcarrier spacing is appropriate.

[0295] In the channel division method provided in Embodiment 2 of this application, spectral resources in the first frequency band can be used as much as possible by finer-grained channel division. This reduces resource waste and improves resource utilization. In addition, the first frequency band is divided into channels based on integer multiples of the minimum channel bandwidth of the second frequency band, so that signals in the first frequency band and signals in the second frequency band can be transmitted by reusing the same crystal oscillator. Furthermore, the correspondence between bandwidth and IDFT size / DFT size (or amount of subcarriers) in the second frequency band is reused in the first frequency band, so that the first and second frequency bands can share signal processing devices, for example, an IDFT device or a DFT device can be reused. Accordingly, according to Embodiment 2 described above, hardware complexity can be reduced, hardware device overhead can be reduced, and device costs can be reduced.

[0296] Embodiment 3

[0297] In the channel splitting method provided in Embodiment 3 of this application, the bandwidth of a first frequency band can be split based on the minimum channel bandwidth of a millimeter-wave band. In other words, the channel boundaries in Embodiment 3 coincide with the channel boundaries of a millimeter-wave band in the first frequency band. In Embodiment 3, to obtain a channel in the first frequency band, R bifurcations are performed on the minimum channel bandwidth of a millimeter-wave band, or T minimum channel bandwidths and performances of 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 2160 MHz. In the channel splitting method provided in Embodiment 3 of this application, the bandwidth of the first frequency band may be split based on 2160 MHz, in other words, the channel boundaries of Embodiment 3 coincide with the channel boundaries of the millimeter-wave band in the first frequency band. In Embodiment 3, to obtain channels in the first frequency band, R bifurcations are performed on the 2160 MHz bandwidth, or T 2160 MHz channels are aggregated to obtain channels 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 splitting method provided in Embodiment 3 of this application, the bandwidth of the first frequency band may be split based on 540 MHz, in other words, the channel boundaries of Embodiment 3 coincide with the channel boundaries of the millimeter-wave band in the first frequency band. In Embodiment 3, to obtain channels in the first frequency band, R bifurcations are performed on the 540 MHz bandwidth, or T 540 MHz channels are aggregated to obtain channels 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 to accommodate millimeter-wave band devices. In Embodiment 3, the bandwidth supported in the first frequency band may be an integer multiple of the minimum channel bandwidth of the second frequency band, or it may not be an integer multiple of the minimum channel bandwidth of the second frequency band. Optionally, if the channel bandwidth 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, 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, millimeter-wave band devices using a low-frequency band (i.e., a second frequency band) can be adapted, and as a result, signals in the high-frequency band, signals in the low-frequency band, and signals in the millimeter-wave band can be transmitted by reusing the same crystal oscillator. This can reduce hardware complexity and hardware overhead.

[0302] The following describes a channel division method for a first frequency band, using an example relating to a specific frequency band.

[0303] Example 8: The minimum channel bandwidth supported in the first frequency band is 270 MHz, or 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 bandwidths of 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 splitting configuration of Example 8, refer to the relevant description in Example 6 of Embodiment 1, Example 2. Further details are again not provided herein.

[0305] For example, the first frequency band is the 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 Figure 8.

[0306] A channel splitting scheme for the 57–66 GHz frequency band is described herein. In this splitting scheme, as shown in Figure 9, the remaining bandwidths in the forward and backward portions of the first frequency band are 240 MHz and 120 MHz, respectively.

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

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

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

[0310] 2160MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 58.320 + (n-1) × 2.160, where 1 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.240 + (n-1) × 2.160, where 1 ≤ n ≤ 4, and the formula for calculating the highest 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, where 1 ≤ n ≤ 2, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.240 + (n-1) × 4.320, where 1 ≤ n ≤ 2, and the formula for calculating the highest 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 20 270 MHz channels, 10 540 MHz channels, 5 1080 MHz channels, 2 2160 MHz channels, or 1 4320 MHz channel. Optionally, in this channel division configuration, the channel division boundaries coincide with the channel boundaries of the millimeter-wave band in the 42.3–47.0 GHz frequency band and the 47.2–48.4 GHz frequency band.

[0313] Channel division methods for the 42.3–47.0 GHz frequency band and the 47.2–48.4 GHz frequency band are described herein. In this division method, the remaining bandwidths in the forward and backward portions of the 42.3–47.0 GHz frequency band are 90 MHz and 290 MHz, respectively, and the remaining bandwidths in the forward and backward portions 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, where 1 ≤ n ≤ 16, and the formula for calculating the lowest frequency is f L (n)[GHz] = 42.390 + (n-1) × 0.270, where 1 ≤ n ≤ 16, and the formula for calculating the highest frequency is f H (n)[GHz] = 42.660 + (n-1) × 0.270, where 1 ≤ n ≤ 16. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C(n)[GHz] = 47.385 + (n-9) × 0.270, where 17 ≤ n ≤ 20, and the formula for calculating the lowest frequency is f L (n)[GHz] = 47.250 + (n-9) × 0.270, where 17 ≤ n ≤ 20, and the formula for calculating the highest 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 8 540MHz channels, and the 47.2~48.4GHz frequency band contains 2 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, where 1 ≤ n ≤ 8, and the formula for calculating the lowest frequency is f L (n)[GHz] = 42.390 + (n-1) × 0.540, where 1 ≤ n ≤ 8, and the formula for calculating the highest frequency is f H (n)[GHz] = 42.930 + (n-1) × 0.540, where 1 ≤ n ≤ 8. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz] = 47.520 + (n-9) × 0.540, where 9 ≤ n ≤ 10, and the formula for calculating the lowest frequency is f L (n)[GHz] = 47.250 + (n-9) × 0.540, where 9 ≤ n ≤ 10, and the formula for calculating the highest 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, where 1 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L(n)[GHz] = 42.390 + (n-1) × 1.080, where 1 ≤ n ≤ 4, and the formula for calculating the highest frequency is f H (n)[GHz] = 43.470 + (n-1) × 1.080, where 1 ≤ n ≤ 4. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz] = 47.790 + (n-5) × 1.080, where n = 5, and the formula for calculating the lowest frequency is f L (n)[GHz] = 47.250 + (n-5) × 1.080, where n = 5, and the formula for calculating the highest frequency is f H (n)[GHz] = 47.330 + (n-5) × 1.080, where n = 5.

[0317] 2160MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 43.470 + (n-1) × 2.160, where 1 ≤ n ≤ 2, and the formula for calculating the lowest frequency is f L (n)[GHz] = 42.390 + (n-1) × 2.160, where 1 ≤ n ≤ 2, and the formula for calculating the highest 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, where n=1, and the formula for calculating the lowest frequency is f L (n)[GHz] = 42.390 + (n-1) × 4.320, where n=1, and the formula for calculating the highest frequency is f H (n)[GHz] = 46.710 + (n-1) × 4.320, where n = 1.

[0319] For example, the first frequency band is the 56.16–73.44 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 this division scheme, there is no remaining bandwidth in the forward and backward portions of the first frequency band.

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

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

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

[0324] 2160MHz channel splitting: The formula for calculating the center frequency is f C(n)[GHz] = 57.240 + (n-1) × 2.160, where 1 ≤ n ≤ 8, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 2.160, where 1 ≤ n ≤ 8, and the formula for calculating the highest frequency is f H (n)[GHz] = 58.320 + (n-1) × 2.160, where 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, where 1 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 4.320, where 1 ≤ n ≤ 4, and the formula for calculating the highest 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, or 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 bandwidths of 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 splitting configuration of Example 9, refer to the relevant description in Example 7 of Embodiment 1, Example 2. Further details are again not provided herein.

[0328] For example, the first frequency band is the 57-66 GHz frequency band. The first frequency band may be divided into 16 540 MHz channels, 8 1080 MHz channels, 4 2160 MHz channels, 2 4320 MHz channels, or 1 8640 MHz channel.

[0329] A channel division scheme for the 57–66 GHz frequency band is described herein. In this division scheme, the remaining bandwidths in the forward and backward portions of the first frequency band are 240 MHz and 120 MHz, respectively.

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

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

[0332] 2160MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 58.320 + (n-1) × 2.160, where 1 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.240 + (n-1) × 2.160, where 1 ≤ n ≤ 4, and the formula for calculating the highest 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, where 1 ≤ n ≤ 2, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.240 + (n-1) × 4.320, where 1 ≤ n ≤ 2, and the formula for calculating the highest 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, where n=1, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.240 + (n-1) × 8640, where n=1, and the formula for calculating the highest frequency is f H (n)[GHz] = 65.880 + (n-1) × 8640, where 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 10 540 MHz channels, 5 1080 MHz channels, 2 2160 MHz channels, or 1 4320 MHz channel. Optionally, in this channel division configuration, the channel division boundaries coincide with the channel boundaries of the millimeter-wave band in the 42.3–47.0 GHz frequency band and the 47.2–48.4 GHz frequency band.

[0336] Channel division methods for the 42.3–47.0 GHz frequency band and the 47.2–48.4 GHz frequency band are described herein. In this division method, the remaining bandwidths in the forward and backward portions of the 42.3–47.0 GHz frequency band are 90 MHz and 290 MHz, respectively, and the remaining bandwidths in the forward and backward portions 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 8 540MHz channels, and the 47.2~48.4GHz frequency band contains 2 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, where 1 ≤ n ≤ 8, and the formula for calculating the lowest frequency is f L (n)[GHz] = 42.390 + (n-1) × 0.540, where 1 ≤ n ≤ 8, and the formula for calculating the highest frequency is f H (n)[GHz] = 42.930 + (n-1) × 0.540, where 1 ≤ n ≤ 8. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz] = 47.520 + (n-9) × 0.540, where 9 ≤ n ≤ 10, and the formula for calculating the lowest frequency is f L (n)[GHz] = 47.250 + (n-9) × 0.540, where 9 ≤ n ≤ 10, and the formula for calculating the highest 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, where 1 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L (n)[GHz] = 42.390 + (n-1) × 1.080, where 1 ≤ n ≤ 4, and the formula for calculating the highest frequency is f H (n)[GHz] = 43.470 + (n-1) × 1.080, where 1 ≤ n ≤ 4. For the 47.2~48.4GHz frequency band, the formula for calculating the center frequency is f C (n)[GHz] = 47.790 + (n-5) × 1.080, where n = 5, and the formula for calculating the lowest frequency is f L(n)[GHz] = 47.250 + (n-5) × 1.080, where n = 5, and the formula for calculating the highest frequency is f H (n)[GHz] = 47.330 + (n-5) × 1.080, where n = 5.

[0339] 2160MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 43.470 + (n-1) × 2.160, where 1 ≤ n ≤ 2, and the formula for calculating the lowest frequency is f L (n)[GHz] = 42.390 + (n-1) × 2.160, where 1 ≤ n ≤ 2, and the formula for calculating the highest 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, where n=1, and the formula for calculating the lowest frequency is f L (n)[GHz] = 42.390 + (n-1) × 4.320, where n=1, and the formula for calculating the highest frequency is f H (n)[GHz] = 46.710 + (n-1) × 4.320, where 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 this division scheme, there is no remaining bandwidth in the forward and backward portions of the first frequency band.

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

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

[0345] 2160MHz channel splitting: The formula for calculating the center frequency is f C (n)[GHz] = 57.240 + (n-1) × 2.160, where 1 ≤ n ≤ 8, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 2.160, where 1 ≤ n ≤ 8, and the formula for calculating the highest frequency is f H (n)[GHz] = 58.320 + (n-1) × 2.160, where 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, where 1 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.160 + (n-1) × 4.320, where 1 ≤ n ≤ 4, and the formula for calculating the highest 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, where n=1,2, and the formula for calculating the lowest frequency is f L (n)[GHz]=56.160+(n-1)×8640, where n=1,2, and the formula for calculating the highest frequency is f H (n)[GHz] = 64.800 + (n-1) × 8640, where n = 1, 2.

[0348] The division patterns in Examples 8 and 9 are merely illustrative examples, and it should be understood that the locations of the channel's center frequency, lowest frequency, and highest frequency are not particularly limited. The channel division patterns can be used under the condition 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 configuration provided in Embodiment 2 of this application, spectral resources in the first frequency band can be utilized as much as possible through finer-grained channel division. This reduces resource waste and improves resource utilization. In addition, according to Embodiment 3, signals in the first frequency band, signals in the second frequency band, and signals in the millimeter-wave band can be transmitted by reusing the same crystal oscillator. Furthermore, the correspondence between bandwidth and IDFT size / DFT size (or amount of subcarriers) in the second frequency band is reused in the first frequency band, and as a result, the first and second frequency bands can share signal processing devices, for example, an IDFT device or a DFT device can be reused. Thus, according to Embodiment 3 above, hardware complexity can be reduced, hardware device overhead can be reduced, and device costs can be reduced.

[0350] Embodiment 4

[0351] In Embodiment 4 of this application, the bandwidth is also divided based on the minimum channel bandwidth of the millimeter-wave band. In other words, the channel boundaries of Embodiment 4 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 Embodiment 4 of this application, the bandwidth is divided based on 2160 MHz, in other words, the channel boundaries of Embodiment 4 coincide with the channel boundaries 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 Embodiment 4 of this application, the bandwidth is divided based on 540 MHz, in other words, the channel boundaries of Embodiment 4 coincide with the channel boundaries of the millimeter-wave band in the first frequency band.

[0354] The difference between Embodiment 4 and Embodiment 3 is, for example, that in Embodiment 3, bifurcation or aggregation is performed based on the minimum channel bandwidth of the millimeter-wave band, whereas in Embodiment 4, bifurcation or aggregation is performed based on a first subband, the bandwidth of the first subband being smaller than the minimum channel bandwidth of the millimeter-wave band, and the first subband being selected from the minimum channel bandwidths supported in the millimeter-wave band protocols IEEE 802.11ad / ay or 802.11aj, and the center frequencies of the millimeter-wave band and the first subband overlap. The overlap of the center frequencies of the millimeter-wave band and the first subband 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 540 MHz, and the bandwidths of the first subbands are 360 ​​MHz, 480 MHz, 500 MHz, 520 MHz, etc. The minimum channel bandwidth in the millimeter-wave band is 2160 MHz, and the bandwidths of the first subband are 1000 MHz, 1800 MHz, 1940 MHz, 2000 MHz, 2100 MHz, etc. In possible implementations, the bandwidth of the first subband can be an integer multiple of 20 MHz.

[0355] For ease of understanding, the following example uses a millimeter-wave band with a minimum channel bandwidth of 2160 MHz and a first subband of 2000 MHz for illustrative purposes. 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 it may not be an integer multiple of the minimum channel bandwidth of the second frequency band. Optionally, if the channel bandwidth 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, multiple sampling clocks are implemented by using an external crystal oscillator and internal fractional frequency multiplication.

[0356] According to the solution provided in Embodiment 4, a millimeter-wave band device using a low-frequency band (i.e., a second frequency band) can be adapted, and as a result, signals in the high-frequency band, signals in the low-frequency band, and signals in the millimeter-wave band can be transmitted by reusing the same crystal oscillator. This can reduce hardware complexity and hardware overhead.

[0357] The following describes a channel division method for a first frequency band, using an example relating to a specific frequency band.

[0358] Example 10: The minimum channel bandwidth supported in the first frequency band is 250 MHz, or in other words, the bandwidth of the first fundamental channel is 250 MHz. The bandwidths supported in the first frequency band may include one or more bandwidths of 250 MHz, 500 MHz, 1000 MHz, 2000 MHz, and 4000 MHz. In this example, the 2000 MHz center frequency overlaps with the 2160 MHz center frequency obtained by division based on the millimeter wave bandwidth. The 1000 MHz channel is obtained by performing bifurcation once on the 2000 MHz channel, the 500 MHz channel is obtained by performing bifurcation twice on the 2000 MHz channel, and the 250 MHz channel is obtained by performing bifurcation three times on the 2000 MHz channel. 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 splitting configuration of Example 10, please refer to the relevant description in Example 8 of Embodiment 1, Example 3. Further details are again not provided herein.

[0360] For example, the first frequency band is the 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 Figure 10.

[0361] A channel division scheme for the 57–66 GHz frequency band is described herein. In this division scheme, the remaining bandwidths in the front and rear portions of the first frequency band are 320 MHz and 200 MHz, respectively.

[0362] 250MHz channel division: To obtain four 2160MHz frequency bands, which are successively represented as the first, second, third, and fourth 2160MHz frequency bands, the 57-66GHz frequency band is divided based on the millimeter wave band. To obtain four 2000MHz frequency bands, which are successively represented as the first, second, third, and fourth 2000MHz frequency bands, 2000MHz is selected from each of the above 2160MHz frequency bands. 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 should be understood that the selection of the 2000 MHz frequency band in this embodiment is to be made by referring to the above form, and that further details will not be explained again later.

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

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

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

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

[0367] 500MHz channel division: To obtain four 2160MHz frequency bands, which are successively represented as the first, second, third, and fourth 2160MHz frequency bands, the 57-66GHz frequency band is divided based on the millimeter wave band. To obtain four 2000MHz frequency bands, which are successively represented as the first, second, third, and fourth 2000MHz frequency bands, 2000MHz is selected from each of the above 2160MHz frequency bands. 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 500MHz channels, the first 2000MHz frequency band is bifurcated twice, and the formula for calculating the center frequencies of the four 500MHz channels is: f C (n)[GHz] = 57.570 + (n-1) × 0.500, where 1 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.320 + (n-1) × 0.500, where 1 ≤ n ≤ 4, and the formula for calculating the highest frequency is f H (n)[GHz] = 57.820 + (n-1) × 0.500, where 1 ≤ n ≤ 4.

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

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

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

[0372] 1000MHz channel division: To obtain four 2160MHz frequency bands, which are successively represented as the first, second, third, and fourth 2160MHz frequency bands, the 57-66GHz frequency band is divided based on the millimeter wave band. To obtain four 2000MHz frequency bands, which are successively represented as the first, second, third, and fourth 2000MHz frequency bands, 2000MHz is selected from each of the above 2160MHz frequency bands. 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 1000MHz channels, the first 2000MHz frequency band is bifurcated once, and the center frequencies of the two 1000MHz channels are f C (n)[GHz] = 57.820 + (n-1) × 1.000, where 1 ≤ n ≤ 2, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.320 + (n-1) × 1.000, where 1 ≤ n ≤ 2, and the formula for calculating the highest frequency is f H (n)[GHz] = 58.320 + (n-1) × 1.000, where 1 ≤ n ≤ 2.

[0374] To obtain two 1000MHz channels, a second 2000MHz frequency band is bifurcated once, and the center frequencies of the two 1000MHz channels are f C (n)[GHz] = 59.980 + (n-3) × 1.000, where 3 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L (n)[GHz] = 59.480 + (n-3) × 1.000, where 3 ≤ n ≤ 4, and the formula for calculating the highest frequency is f H (n)[GHz] = 60.480 + (n-3) × 1.000, where 3 ≤ n ≤ 4.

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

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

[0377] 2000MHz channel division: To obtain four 2160MHz frequency bands, which are successively represented as the first, second, third, and fourth 2160MHz frequency bands, the 57-66GHz frequency band is divided based on the millimeter wave band. To obtain four 2000MHz frequency bands, which are successively represented as the first, second, third, and fourth 2000MHz frequency bands, 2000MHz is selected from each of the above 2160MHz frequency bands. 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, where 1 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.320 + (n-1) × 2.160, where 1 ≤ n ≤ 4, and the formula for calculating the highest frequency is f H (n)[GHz] = 59.320 + (n-1) × 2.160, where 1 ≤ n ≤ 4.

[0378] 4000MHz channel division: To obtain four 2160MHz frequency bands, which are successively represented as the first, second, third, and fourth 2160MHz frequency bands, the 57-66GHz frequency band is divided based on the millimeter wave band. To obtain four 2000MHz frequency bands, which are successively represented as the first, second, third, and fourth 2000MHz frequency bands, 2000MHz is selected from each of the above 2160MHz frequency bands. 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, two at a time. It can also be understood that to obtain the first 4000MHz channel, the first and second 2000MHz bands are aggregated, and to obtain the second 4000MHz channel, the third and fourth 2000MHz bands are aggregated. The first 4000MHz channel has a center frequency of 59.400, a minimum frequency of 57.320, and a maximum frequency of 61.480. The second 4000MHz channel has a center frequency of 63.720, a minimum frequency of 61.640, and a maximum frequency of 65.800.

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

[0380] 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 20 250 MHz channels, 10 500 MHz channels, 5 1000 MHz channels, 2 2000 MHz channels, or 1 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 Embodiment 3, respectively. In this division method, the remaining bandwidths in the forward and backward portions of the 42.3–47.0 GHz frequency band are 110 MHz and 310 MHz, respectively, and the remaining bandwidths in the forward and backward portions of the 47.2–48.4 GHz frequency band are 70 MHz and 90 MHz, respectively.

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

[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 540MHz frequency bands, 47.2–48.4GHz is divided based on the millimeter wave band, and to obtain two 500MHz frequency bands, 500MHz is selected from each 540MHz frequency band. 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 250MHz channels, each 500MHz frequency band is bifurcated 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: To obtain eight 540MHz frequency bands, 42.3~47.0GHz is divided based on the millimeter wave band, and to obtain eight 500MHz frequency bands, 500MHz is selected from each 540MHz frequency band. The center frequency of one of the 500MHz frequency bands selected from the 500MHz frequency bands overlaps with the center frequency of the 540MHz frequency band.

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

[0389] To obtain two 540MHz frequency bands, 47.2–48.4GHz is divided based on the millimeter wave band, and to obtain two 500MHz frequency bands, 500MHz is selected from each 540MHz frequency band. 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 250MHz channels, each 500MHz frequency band is bifurcated once.

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

[0391] 1000MHz channel division: To obtain eight 540MHz frequency bands, 42.3~47.0GHz is divided based on the millimeter wave band, and to obtain eight 500MHz frequency bands, 500MHz is selected from each 540MHz frequency band. 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 in pairs.

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

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

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

[0395] 2000MHz channel division: To obtain eight 540MHz frequency bands, 42.3~47.0GHz is divided based on the millimeter wave band, and to obtain eight 500MHz frequency bands, 500MHz is selected from each 540MHz frequency band. 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 in groups of four.

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

[0397] 4000MHz channel division: To obtain eight 540MHz frequency bands, 42.3~47.0GHz is divided based on the millimeter wave band, and to obtain eight 500MHz frequency bands, 500MHz is selected from each 540MHz frequency band. The center frequency of one 500MHz frequency band selected from each 500MHz frequency band overlaps with the center frequency of the 540MHz frequency band. To obtain one 4000MHz channel, eight 500MHz frequency bands are aggregated. The 4000MHz channel has a center frequency of 44.55, a minimum frequency of 42.43, and a maximum 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 64 250 MHz channels, 32 500 MHz channels, 16 1000 MHz channels, 8 2000 MHz channels, 4 4000 MHz channels, or 2 8000 MHz channels.

[0399] A channel division scheme for the 56.16–73.44 GHz frequency band is described herein. In this division scheme, the remaining bandwidths in the front and rear portions of the first frequency band are 80 MHz and 80 MHz, respectively.

[0400] 250MHz 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 eight 250MHz channels, each 2000MHz is bifurcated three times.

[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 250MHz channel in each 2000MHz frequency band is:

number

number

number

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

number

number

number

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

number

number

number

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

number

number

number

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

number

number

number

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

number

number

number

[0409] 500MHz 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 four 500MHz channels, each 2000MHz is bifurcated twice.

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

number

number

number

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

number

number

number

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

number

number

number

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

number

number

number

[0414] 1000MHz channel splitting: To obtain eight 2160MHz frequency bands, the 56.16~73.44GHz frequency band is split 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 one 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. To obtain two 1000MHz channels, each 2000MHz is bifurcated once.

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

number

number

number

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

number

number

number

[0417] 2000MHz 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.

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

[0419] 4000MHz 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.

[0420] To obtain four 4000MHz channels, eight 2000MHz frequency bands are aggregated in ascending order of frequency in pairs, and each 4000MHz channel is obtained by aggregating two non-consecutive 2000MHz channels. The formula for calculating the center frequencies of the four 4000MHz channels is: f C (n)[GHz] = 58.320 + (n-1) × 4.320, where 1 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L (n)[GHz] = 56.240 + (n-1) × 4.320, where 1 ≤ n ≤ 4, and the formula for calculating the highest 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, or in other words, the bandwidth of the first fundamental channel is 500 MHz. The bandwidths supported in the first frequency band may include one or more bandwidths of 500 MHz, 1000 MHz, 2000 MHz, 4000 MHz, and 8000 MHz. In this example, the 2000 MHz center frequency overlaps with the 2160 MHz center frequency obtained by division based on the millimeter wave bandwidth. 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. Please understand that the two 2000MHz channels included in the 4000MHz bandwidth are not consecutive, and the four 2000MHz channels included in the 8000MHz bandwidth are not consecutive.

[0422] Optionally, for the amount of subcarriers (or IDFT size / DFT size) corresponding to each channel bandwidth in the channel splitting configuration of Example 11, please refer to the relevant description in Example 9 of Embodiment 1, Example 3. Further details are again not provided herein.

[0423] For example, the first frequency band is the 57–66 GHz frequency band. The first frequency band may be divided into 16 500 MHz channels, 8 1000 MHz channels, 4 2000 MHz channels, 2 4000 MHz channels, or 1 8000 MHz channel, as shown in Figure 11.

[0424] A channel division scheme for the 57–66 GHz frequency band is described herein. In this division scheme, the remaining bandwidths in the front and rear portions of the first frequency band are 320 MHz and 200 MHz, respectively.

[0425] 500MHz channel splitting: To obtain four 2160MHz frequency bands, the 57-66GHz frequency band is split based on the millimeter wave band, and to obtain four 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.

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

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

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

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

[0430] 1000MHz channel splitting: To obtain four 2160MHz frequency bands, the 57-66GHz frequency band is split based on the millimeter wave band, and to obtain four 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.

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

[0432] To obtain two 1000MHz channels, a second 2000MHz frequency band is bifurcated once, and the center frequencies of the two 1000MHz channels are f C (n)[GHz] = 59.980 + (n-3) × 1.000, where 3 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L (n)[GHz] = 59.480 + (n-3) × 1.000, where 3 ≤ n ≤ 4, and the formula for calculating the highest frequency is f H (n)[GHz] = 60.480 + (n-3) × 1.000, where 3 ≤ n ≤ 4.

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

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

[0435] 2000MHz channel division: To obtain four 2160MHz frequency bands, the 57-66GHz frequency band is divided based on the millimeter wave band, and to obtain four 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. The formula for calculating the center frequencies of the four 2000MHz channels is: f C (n)[GHz] = 58.320 + (n-1) × 2.160, where 1 ≤ n ≤ 4, and the formula for calculating the lowest frequency is f L (n)[GHz] = 57.320 + (n-1) × 2.160, where 1 ≤ n ≤ 4, and the formula for calculating the highest frequency is f H (n)[GHz] = 59.320 + (n-1) × 2.160, where 1 ≤ n ≤ 4.

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

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

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

[0439] 8000MHz Channel Division: To obtain four 2160MHz frequency bands, the 57-66GHz frequency band is divided based on the millimeter wave band, and to obtain four 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 an 8000MHz channel, four 2000MHz frequency bands are aggregated. The 8000MHz channel has a center frequency of 61.560, a minimum frequency of 57.320, and a maximum frequency of 65.800. This channel is formed by aggregating four non-contiguous 2000MHz channels.

[0440] 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 10 500 MHz channels, 5 1000 MHz channels, 2 2000 MHz channels, or 1 4000 MHz channel. In addition, the center frequencies of the 500 MHz channels and 1000 MHz channels must overlap with the center frequencies of the 540 MHz channels and 1080 MHz channels in the 42.3–47.0 GHz frequency band and 47.2–48.4 GHz frequency band of Embodiment 3, respectively. In this division method, the remaining bandwidths in the forward and backward portions of the 42.3–47.0 GHz frequency band are 110 MHz and 310 MHz, respectively, and the remaining bandwidths in the forward and backward portions of the 47.2–48.4 GHz frequency band are 70 MHz and 90 MHz, respectively.

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

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

[0443] To obtain two 540MHz frequency bands, 47.2–48.4GHz is divided based on the millimeter wave band, and to obtain two 500MHz frequency bands, 500MHz is selected from each 540MHz frequency band. 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 250MHz channels, each 500MHz frequency band is bifurcated once.

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

[0445] 1000MHz channel division: To obtain eight 540MHz frequency bands, 42.3~47.0GHz is divided based on the millimeter wave band, and to obtain eight 500MHz frequency bands, 500MHz is selected from each 540MHz frequency band. 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 in pairs.

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

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

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

[0449] 2000MHz channel splitting: To obtain eight 540MHz frequency bands, 42.3~47.0GHz is split based on the millimeter wave band, and to obtain eight 500MHz frequency bands, 500MHz is selected from each 540MHz frequency band. 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 in groups of four.

[0450] One 2000MHz channel has a center frequency of 43.47 Hz, a minimum frequency of 42.43 Hz, and a maximum frequency of 44.51 Hz. The other 2000MHz channel has a center frequency of 45.63 Hz, a minimum frequency of 44.59 Hz, and a maximum frequency of 46.67 Hz. Each 2000MHz channel is formed by aggregating two non-contiguous 1000MHz 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 this division scheme, the remaining bandwidths in the front and rear portions of the first frequency band are 80 MHz and 80 MHz, respectively.

[0454] 500MHz 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 four 500MHz channels, each 2000MHz is bifurcated twice.

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

number

number

number

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

number

number

number

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

number

number

number

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

number

number

number

[0459] 1000MHz channel splitting: To obtain eight 2160MHz frequency bands, the 56.16~73.44GHz frequency band is split 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 one 2000MHz frequency band selected from one 2160MHz frequency band overlaps with the center frequency of the 2160MHz frequency band. To obtain two 1000MHz channels, each 2000MHz is bifurcated once.

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

number

number

number

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

number

number

number

[0462] 2000MHz 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.

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

[0464] 4000MHz 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.

[0465] To obtain four 4000MHz channels, eight 2000MHz frequency bands are aggregated in ascending order of frequency in pairs, and each 4000MHz channel is obtained by aggregating two non-consecutive 2000MHz channels. The formula for calculating the center frequencies of the four 4000MHz channels is: f C (n)[GHz] = 58.320 + (n-1) × 4.320, where 1 ≤ n ≤ 4, and the formula for calculating the lowest frequency is fL (n)[GHz] = 56.240 + (n-1) × 4.320, where 1 ≤ n ≤ 4, and the formula for calculating the highest 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 one 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 in groups of four, and each 8000MHz channel is obtained by aggregating four non-contiguous 2000MHz channels.

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

[0468] The division patterns in Examples 10 and 11 are merely illustrative examples, and it should be understood that the locations of the channel's center frequency, lowest frequency, and highest frequency are not particularly limited. The channel division pattern can be used on the condition that the bandwidth is divided based on the minimum channel bandwidth of the millimeter-wave band, and the bandwidth obtained by bipartite is smaller than the minimum channel bandwidth of the millimeter-wave band.

[0469] In the channel division configuration provided in Embodiment 4 of this application, finer-grained channel division allows for the utilization of spectral resources in the first frequency band as much as possible. This reduces resource waste and improves resource utilization. In Embodiment 4, a portion of the bandwidth is selected from the smallest channel bandwidth supported in the millimeter-wave band, and as a result, inter-channel interference 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. Furthermore, the correspondence between the bandwidth and IDFT size / DFT size (or amount of subcarriers) in the second frequency band is reused in the first frequency band, and as a result, the first and second frequency bands can share signal processing devices, for example, the IDFT device or DFT device can be reused. Therefore, according to Embodiment 4 described above, hardware complexity can be reduced, hardware device overhead can be reduced, and device costs can be reduced.

[0471] Based on the same inventive concept as the method embodiment, one embodiment of this application provides a communication device. The structure of the communication device may be shown in Figure 12. The communication device includes a transceiver module 1201 and a processing module 1202. Optionally, the transceiver module 1201 may include a transport module (Tx module) and a receive module (Rx module). The transport module is configured to send signals, and the receive module is configured to receive signals. The communication device may further include a radio frequency unit (RF unit). The radio frequency unit may be configured to transmit signals to a receiver device. The transport module may transmit to the radio frequency unit to send signals to be sent. The receive module may receive signals from the radio frequency unit and send the signals 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 Figure 3. The device may be the first device, or a chip or chipset within the first device, or a part of a chip configured to perform the function of the associated method. The processing module 1202 is configured to generate a first OFDM signal. The transceiver module 1201 is configured to transmit the first OFDM signal on a first channel in a first frequency band, wherein the amount of subcarriers corresponding to the first channel is greater than or equal to the amount of subcarriers corresponding to a second channel in a second frequency band, the lowest frequency of the first frequency band is higher than the highest frequency of the second frequency band, and the first channel is 2 M It includes a first basic channel and a second channel. MIt includes n second fundamental channels, the bandwidth of the first fundamental channel is the minimum channel bandwidth supported in the first frequency band, the bandwidth of the second fundamental channel is the 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.

[0473] In one implementation, the communication device may be specifically configured to implement the method implemented by the second device in the embodiment shown in Figure 3. The device may be the second device, or a chip or chipset within the second device, or a part of a chip configured to implement the function 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, wherein 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 a second frequency band, the lowest frequency of the first frequency band is higher than the highest frequency of the second frequency band, and the first channel is 2 M It includes a first basic channel and a second channel. M The first basic channel includes n second basic channels, the bandwidth of the first basic channel is the minimum channel bandwidth supported in the first frequency band, the bandwidth of the second basic channel is the minimum channel bandwidth supported in the second frequency band, the bandwidth of the first basic channel is greater than the bandwidth of the second basic 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 further details, please refer to the relevant descriptions of Embodiments 1 to 4. Further details are again not provided herein.

[0475] The modularization in the embodiments of this application is merely an example and represents only a logical functional division; other divisions may be used during actual implementation. In addition, the functional modules in the embodiments of this application may be incorporated into a single processor, or each module may exist physically independently, or two or more modules may be incorporated into a single module. The incorporated modules may be implemented in hardware form or in the form of software functional modules. For further information regarding the functionality or implementation of the modules in the embodiments of this application, please refer to the relevant descriptions in the method embodiments.

[0476] In possible implementations, the communication device may be shown in Figure 13. The device may be a communication device or a chip within 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. Figure 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 input / output devices (not shown in the figure).

[0477] The processor 1301 is configured to execute program code stored in memory 1303 and is particularly configured to perform actions of processing module 1202. Further details are again not described herein. The communication interface 1302 is particularly configured to perform actions of transceiver module 1201. Further details are again not described herein.

[0478] The processor 1301 may be a CPU, a digital processing unit, or the like. The processor 1301 may be configured to process communication protocols and data, control the entire communication device, execute software programs, and process data from software programs; for example, it may be configured to perform baseband-related processing, but is not limited to this. The communication interface 1302 may be configured to receive and transmit signals; for example, it may be configured to perform radio frequency reception and transmission, but is not limited to this. The above devices may each be disposed on separate chips, 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 transceiver may be integrated on the same chip, or the digital baseband processor may be disposed on a separate chip. With the continued development of integrated circuit technology, more and more devices can be integrated on the same chip. For example, a digital baseband processor may be integrated on the same chip along with multiple application processors (for example, a geometry processor and a multimedia processor, but is not limited to this). A chip is sometimes referred to as a system on a chip. Whether devices are arranged independently on different chips or integrated on one or more chips typically depends on the specific requirements of the product design. The specific implementation of devices is not limited to the embodiments of this invention.

[0479] The communication interface 1302 may be an interface circuit such as a transceiver or transceiver circuit, or a transceiver chip. Optionally, the communication interface 1302 may also include a radio frequency circuit and an antenna. The radio frequency circuit is mainly configured to perform conversions between baseband signals and radio frequency signals and to process radio frequency signals. The antenna is mainly configured to receive or transmit radio frequency signals in the form of electromagnetic waves. An input / output device such as a touchscreen, display, or keyboard is mainly configured to receive data entered by the user and output data to the user.

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

[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 the data of the software program. When data needs to be transmitted wirelessly, the processor 1301 performs baseband processing on the data to be transmitted 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 transmits the radio frequency signal in the form of electromagnetic waves through the 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 alternative implementations, the radio frequency circuitry and antennas may be located independently of the processor performing baseband processing. For example, in a distributed scenario, the radio frequency circuitry and antennas may be located remotely, independently of the communication equipment.

[0483] The specific connection medium between the communication interface 1302, the processor 1301, and the memory 1303 is not limited to the embodiments of this application. In this embodiment of this application, the memory 1303, the processor 1301, and the communication interface 1302 are connected through a bus 1304 in Figure 13. The bus is represented by using a thick line in Figure 13. The modes of connection between other components are merely illustrative examples and are not limited thereto. The bus may be classified as an address bus, a data bus, a control bus, etc. For ease of representation, only one thick line is used in the representation in Figure 13, but this does not mean that there is only one bus or only one type of bus.

[0484] Optionally, the communication device may be a standalone device or part of a larger device. For example, the communication device may be (1) an independent integrated circuit (IC), chip, or chip system or subsystem, (2) A set of one or more ICs, which optionally includes a set of ICs configured to store data and instructions. (3) Application-specific integrated circuits (ASICs) such as modems, (4) Modules that can be embedded in another device, (5) Receivers, intelligent terminals, wireless devices, handsets, mobile units, in-vehicle devices, cloud devices, artificial intelligence devices, etc. (6) Another device, etc. That's fine.

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

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

[0487] Those skilled in the art should understand that embodiments of this application may be provided as methods, systems, or computer program products. Accordingly, this application may take the form of hardware-only embodiments, software-only embodiments, or embodiments having a combination of software and hardware. In addition, this application may take the form of a computer program product implemented 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] This application will be described with reference to the methods, devices (systems), and computer program product flowcharts and / or block diagrams described herein. It should be understood that computer program instructions may be used to perform each step and / or block in the flowcharts and / or block diagrams, as well as 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 dedicated computer, an embedded processor, or another programmable data processing device for generating machines, and as a result, instructions executed by the computer or the processor of the other programmable data processing device will generate a device for performing a specific function in one or more steps in the flowchart and / or one or more blocks in the block diagram.

[0489] These computer program instructions may be stored in computer-readable memory that can instruct a computer or another programmable data processing device to operate in a specific manner, and as a result, the instructions stored in computer-readable memory generate an artifact that includes an instruction unit. The instruction unit performs a specific function in one or more steps in a flowchart and / or in one or more blocks in a block diagram.

[0490] Computer program instructions may alternatively be loaded onto a computer or another programmable data processing device, resulting in a series of operations and steps being performed on the computer or the other programmable device, thereby generating computer execution processing. Therefore, instructions executed on the computer or the other programmable device provide steps for performing specific functions in one or more steps in a flowchart and / or in one or more blocks in a block diagram.

[0491] It will be apparent that a person skilled in the art can make various changes and modifications to this application without departing from its scope. This application is intended to encompass such changes and modifications, provided that they fall within the scope of protection provided by the following claims and their equivalents in the art. [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. A method of communication, The steps include generating a first orthogonal frequency division multiplexing (OFDM) signal, A step of transmitting the first OFDM signal on a first channel in a first frequency band, wherein the amount of subcarriers corresponding to the first channel is greater than or equal to the amount of subcarriers corresponding to a second channel in a second frequency band, the lowest frequency of the first frequency band is higher than the highest frequency of the second frequency band, and the first channel is 2 M It includes a first basic channel, and the second channel is 2 M The system includes a second basic channel, wherein the bandwidth of the first basic channel is the minimum channel bandwidth supported in the first frequency band, the bandwidth of the second basic channel is the minimum channel bandwidth supported in the second frequency band, the bandwidth of the first basic channel is greater than the bandwidth of the second basic channel, and M is an integer greater than or equal to 0, and Methods that include...

2. A method of communication, A step of receiving a first orthogonal frequency division multiplexing (OFDM) signal on a first channel in a first frequency band, wherein the amount of subcarriers corresponding to the first channel is greater than or equal to the amount of subcarriers corresponding to a second channel in a second frequency band, the lowest frequency of the first frequency band is higher than the highest frequency of the second frequency band, and the first channel is 2 M It includes a first basic channel, and the second channel is 2 M A step comprising a second basic channel, wherein the bandwidth of the first basic channel is the minimum channel bandwidth supported in the first frequency band, the bandwidth of the second basic channel is the minimum channel bandwidth supported in the second frequency band, the bandwidth of the first basic channel is greater than the bandwidth of the second basic channel, and M is an integer greater than or equal to 0. Steps to process the first OFDM symbol and Methods that include...

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

4. The method according to any one of claims 1 to 3, wherein the bandwidth of the first basic channel is N times the bandwidth of the second basic channel, and N is an integer greater than 1.

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

6. The amount of the subcarrier corresponding to the first channel is If the bandwidth of the first channel is 320 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 640 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 1280 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 2560 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 5120 MHz, then the first channel corresponds to 1024 subcarriers. The method according to claim 5, which satisfies any one of the following conditions.

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

8. The amount of the subcarrier corresponding to the first channel is If the bandwidth of the first channel is 640 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 1280 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 2560 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 5120 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 10240 MHz, then the first channel corresponds to 1024 subcarriers. The method according to claim 7, which satisfies any one of the following conditions.

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

10. The amount of the subcarrier corresponding to the first channel is If the bandwidth of the first channel is 80 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 160 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 320 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 640 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 1280 MHz, then the first channel corresponds to 1024 subcarriers. The method according to claim 9, which satisfies any one of the following conditions.

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

12. The amount of the subcarrier corresponding to the first channel is If the bandwidth of the first channel is 160 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 320 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 640 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 1280 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 2560 MHz, then the first channel corresponds to 1024 subcarriers. The method according to claim 11, which satisfies any one of the following.

13. The method according to any one of claims 1 to 3, wherein the bandwidth of the first basic channel is obtained by separately performing R bisections on a frequency band obtained by dividing the first frequency band based on the minimum channel bandwidth supported in the millimeter wave band, where R is an integer greater than or equal to 0.

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

15. The amount of the subcarrier corresponding to the first channel is If the bandwidth of the first channel is 270 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 540 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 1080 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 2160 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 4320 MHz, then the first channel corresponds to 1024 subcarriers. The method according to claim 14, which satisfies any one of the following conditions.

16. The method according to claim 13, wherein the bandwidth of the first basic channel is 540 MHz.

17. The amount of the subcarrier corresponding to the first channel is If the bandwidth of the first channel is 540 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 1080 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 2160 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 4320 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 8640 MHz, then the first channel corresponds to 1024 subcarriers. The method according to claim 16, which satisfies any one of the following.

18. The method according to any one of claims 1 to 3, wherein the bandwidth of the first basic channel is obtained by performing K bisections on a first subband, the bandwidth of the first subband being less than the minimum channel bandwidth supported in the millimeter-wave band, and K is an integer greater than or equal to 0.

19. The method according to claim 18, wherein the bandwidth of the first basic channel is 250 MHz.

20. The amount of the subcarrier corresponding to the first channel is If the bandwidth of the first channel is 250 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 500 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 1000 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 2000 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 4000 MHz, then the first channel corresponds to 1024 subcarriers. The method according to claim 19, which satisfies any one of the following conditions.

21. The method according to claim 18, wherein the bandwidth of the first basic channel is 500 MHz.

22. The amount of the subcarrier corresponding to the first channel is If the bandwidth of the first channel is 500 MHz, then the first channel corresponds to 64 subcarriers. If the bandwidth of the first channel is 1000 MHz, then the first channel corresponds to 128 subcarriers. If the bandwidth of the first channel is 2000 MHz, then the first channel corresponds to 256 subcarriers. If the bandwidth of the first channel is 4000 MHz, then the first channel corresponds to 512 subcarriers, or If the bandwidth of the first channel is 8000 MHz, then the first channel corresponds to 1024 subcarriers. The method according to claim 21, which satisfies any one of the following.

23. The first frequency band is either a 60 GHz frequency band or a 45 GHz frequency band. The second frequency band is either a 2.45 GHz frequency band or a 6 GHz frequency band. The method according to any one of claims 1 to 22.

24. A communication device, Equipped with memory and a processor, The memory is configured to store instructions, The processor is configured to execute the instruction and carry out the method according to claim 1 or any one of claims 3 to 23. Communication device.

25. A communication device, Equipped with memory and a processor, The memory is configured to store instructions, The processor is configured to execute the instruction and carry out the method described in any one of claims 2 to 23. Communication device.

26. A computer-readable storage medium configured to store computer instructions, wherein when the computer instructions are executed on a computer, the computer can carry out the method according to any one of claims 1 to 23.

27. It's a tip, Equipped with memory and a processor, The memory is configured to store instructions, The processor is configured to execute the instructions such that a communication device comprising the chip carries out the method according to claim 1 or any one of claims 3 to 23. Tip.

28. It's a tip, Equipped with memory and a processor, The memory is configured to store instructions, The processor is configured to execute the instructions such that a communication device comprising the chip carries out the method described in any one of claims 2 to 23. Tip.

Images