Decoding method and device
A decoding method and encoding technology, applied in the field of decoding methods and devices, can solve problems such as waste
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example 1
[0098] The obtained 12-bit code word group (step 20) is 0100 1101 1101, and there is one leading bit "0" (that is, n=1) (step 21). Query the index word group table table X according to the value n (step 23), and obtain the index word group as 0000 1011 0110 1000. After dismantling this index word group, we get
[0099] A=base_index=5;
[0100] B=class_sign="10"(2);
[0101] C=suffix_len1=3;
[0102] D = suffix_len2 = 2;
[0103] E=suffix_len3=0
[0104] Shift the coded word to the left by a total of two bits (n+1=2) (step 22), so the shifted coded word is 0011 0111 0100 (the right side is supplemented with "0"). Since B=class_sign is "10", it is determined (step 24) that it belongs to the first class, so step 25A is performed to check the most significant bit of the shifted coding block. After checking as bit "0", it is determined that it belongs to the first paragraph, so enter step 26A1 to perform the calculation of formula (9), (10) to obtain
[0105] index=A+code(10...
example 2
[0108] The obtained 12-bit code word group (step 20) is 0000 1100 1001, and there are four leading bits "0" (that is, n=4) (step 21). Query the index word group table table X according to the value n (step 23), and obtain the index word group as 0110 0011 1110 1011. After dismantling this index word group, we get
[0109] A=base_index=49;
[0110] B=class_sign="11" (3);
[0111] C=suffix_len1=3;
[0112] D = suffix_len2 = 2;
[0113] E=suffix_len3=3
[0114] Shift the coded word to the left by a total of five bits (n+1=5) (step 22), so the shifted coded word is 1001 0010 0000. Since B=class_sign is "11", it is determined (step 24) that it belongs to the second class, so step 25B is performed to check the most significant two bits of the shifted code word. After checking the position "10", it is determined that it belongs to the third paragraph, so enter step 26B3 to perform the calculation of formula (1), (2) to obtain
[0115] index=A+(1<
example 3
[0118] The obtained 12-bit code word group (step 20) is 0000 0011 0001, and there are six leading bits "0" (that is, n=6) (step 21). Query the index word group table table X according to the value n (step 23), and obtain the index word group as 1011 1100 1110 0000. After dismantling this index word group, we get
[0119] A=base_index=94;
[0120] B=class_sign="01"(1);
[0121] C=suffix_len1=3;
[0122] D = suffix_len2 = 0;
[0123] E=suffix_len3=0
[0124] Shift the coded word to the left by a total of seven bits (n+1=7) (step 22), so the shifted coded word is 1000 1000 0000. Because B=class_sign is " 01 ", so judge (step 24) belongs to the 3rd category, therefore enter step 26C and carry out the operation of formula (11), (12) to obtain
[0125] index=A+code(11:12-C)=94+4=98;
[0126] 1en=n+1+C=6+1+3=10.
[0127] The code table table I-IX used in the embodiment of the present invention is the code table B-16 / B-17 for the video standard MPEG-4, which contains 110 (that...
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