EFFICIENT CODING OF SIDE INFORMATION IN A LOSSLESS ENCODER

The invention relates to an apparatus for lossless encoding of a digital information signal, for a lossless encoding method, to an apparatus for decoding and to a record carrier.

For "Super Audio CD" (SACD) the DSD signals are losslessly coded, using framing, prediction and entropy coding. Besides the efficiently encoded signals, a large number of parameters, i.e. the side-information, has to be stored on the SACD too. The smaller the storage capacity that is required for the side-information, the better the overall coding gain is. Therefore coding techniques are applied to the side-information too.

A description of the lossless encoding of DSD signals is given in the publication 'Improved lossless coding of 1-bit audio signals', by F. Bruekers et al, preprint 4563(I-6) presented at the 103rd convention of the AES, September 26-29, 1997 in New York. An example of lossless coding is provided in "Improved Lossless Coding of 1-Bit Audio Signals" by Bruekers et al., Fons; Oomen, Wemer; van der Vleuten, René, van de Kerkhof; Preprints of papers presented at the 103rd AES convention, 1997-09-26

The invention aims at providing methods that can be used e.g. in SACD to save on the number of bits that have to be used for storing the side-information.

In the following description those methods will be presented.

These and other aspects of the invention will be further explained hereafter in the figure description, in which

• Figure la shows a circuit diagram of a lossless encoder and figure 1b shows a circuit diagram of a corresponding decoder, using linear prediction and arithmetic coding,
• Figure 2 shows subsequent frames of a multi channel information signal,
• Figure 3 shows the segmentation of time equivalent frames of the multi channel information signal, and
• Figure 4 shows the contents of a frame of the output signal of the encoding apparatus.

The process of lossless encoding and decoding, for the example of 1-bit oversampled audio signals, will be explained briefly hereafter by means of figure 1, which shows an embodiment of the encoder apparatus in figure 1a and shows an embodiment of the decoder apparatus in figure 1b.

The lossless coding in the apparatus of figure 1a is performed on isolated parts (frames) of the audio signal. A typical length of such a frame is 37632 bits. The two possible bit-values of the input signal F, '1' and '0', represent the sample values +1 and -1 respectively. Per frame, the set of coefficients for the prediction filter z^-1.A(z), denoted by 4, is determined in a filter coefficient generator unit 12, by e.g. the autocorrelation method. The sign of the filter output signal, Z, determines the value of the predicted bit Fp , whereas the magnitude of the filter output signal, Z, is an indication for the probability that the prediction is correct. Upon quantizing the filter output signal Z in a quantizer 10, a predicted input signal F_p is obtained, which is ex-ored in a combining unit 2, resulting in a residual signal E. A correct prediction, or F = Fp , is equivalent to E = 0 in the residual signal E. The content of the probability table, p(|.|), is designed per frame such that per possible value of Z, p₀ is the probability that E = 0. For small values of |Z| the probability for a correct prediction is close to 0.5 and for large values of |Z| the probability for a correct prediction is close to 1.0. Clearly the probability for an incorrect prediction, F ≠ Fp or E=1, is p₁ =1-p₀.

The probability tables for the frames (or segments, to be described later) are determined by the unit 13. Using this probability table, supplied by the unit 13 to the unit 8, the unit 8 generates a probability value P₀ in response to its input signal, which is the signal Z.

The arithmetic encoder (AC Enc.) in the apparatus of figure 1a, denoted by 6, codes the sequence of bits of E such that the code (D) requires less bits. For this, the arithmetic coder uses the probability that bit n of signal E, E[n], has a particular value. The number of bits to code the bit E[n]=0 is: $${\mathrm{d}}_{\mathrm{n}}\mathrm{=}\mathrm{-}{}^{\mathrm{2}}\log \left({\mathrm{p}}_{\mathrm{0}}\right)\mathrm{+}\mathrm{M}\left(\mathrm{bits}\right)$$

which is practically not more than 1 bit, since p₀ ≥ 1/2. The number of bits to code the bit E[n]=1 is: $${\mathrm{d}}_{\mathrm{n}}\mathrm{=}\mathrm{-}{}^{\mathrm{2}}\log \left({\mathrm{p}}_{\mathrm{1}}\right)+\mathrm{M}=\mathrm{-}{}^{\mathrm{2}}\log \left(1-{\mathrm{p}}_{\mathrm{0}}\right)\mathrm{+}\mathrm{M}\left(\mathrm{bits}\right)$$

which is not less than I bit. The M in both equations represents the non-optimal behavior of the arithmetic coder, but can be neglected in practice.

A correct prediction (E[n]=0) results in less than 1 bit and an incorrect prediction (E[n]=1) results in more than 1 bit in the code (D). The probability table is designed such that on the average for the complete frame, the number of bits for code D is minimal.

Besides code D, also the coefficients of the prediction filter 4, generated by the coefficient generator unit 12, and the content of the probability table, generated by the probability table determining unit 13, have to be transmitted from encoder to decoder. To that purpose, the encoder apparatus comprises a multiplexer unit 14, which receives the output signal of the coder 6, as well as side information from the generator units 12 and 13. This side information comprises the prediction filter coefficients and the probability table. The multiplexer unit 14 supplies the serial datastream of information to a transmission medium, such as a record carrier.

In the decoder apparatus of figure 1b, exactly the inverse of the encoder process is implemented thus creating a lossless coding system. The demultiplexer unit 20 receives the serial datastream comprising the data D and the side information. It retrieves the data D therefrom and supplies the data D to an arithmetic decoder 22. The arithmetic decoder (AC Dec.) is provided with the identical probabilities as the arithmetic encoder was, to retrieve the correct values of signal E. Therefore the demultiplexer unit retrieves the same prediction filter coefficients and probability table as used the encoder from the serial datastream received and supplies the prediction filter coefficients to the prediction filter 24 and the probability table to the probability value generator unit 26.

The circuit constructions shown in figure 1 are meant for encoding/decoding a single serial datastream of information. Encoding/decoding a multi channel information signal, such as a multi channel digital audio signal, requires the processing described above with reference to figure 1 to be carried out in time multiplex by the circuits of figure 1, or can be carried out in parallel by a plurality of such circuits. Another solution can be found in international patent application IB 99/00313, which corresponds to US ser. no. 09/268252 (PHN 16.805).

It should be noted here that in accordance with the invention, the encoding apparatus may be devoid of the quantizer Q and the combining unit 2. Reference is made to earlier patent publications discussing this.

In SACD the 1-bit audio channels are chopped into frames of constant length and per frame the optimal strategy for coding will be used. Frames can be decoded independently from neighbouring frames. Therefore we can discuss the data structure within a single frame.

Figure 2 shows time equivalent frames B of two channel signals, such as the left and right hand signal component of a digital stereo audio signal, indicated by ...., B(l,m-1), B(l,m), B(l,m+1), ..... for the left hand signal component and by ....., B(r,m-1), B(r,m), B(r,m+1), ..... for the right hand signal component. The frames can be segmented, as will be explained hereafter. If not segmented, the frames will be encoded in their entirety, with one set of filter coefficients and one probability table for the complete frame. If segmented, each segment in a frame can have its own set of filter coefficients and probability table. Furthermore, the segmentation in a frame for the filtercoefficients need not be the same as for the probability tables. As an example, figure 3 shows the two time equivalent frames B(l,m) and B(r,m) of the two channel signals being segmented. The frame B(l,m) has been segmented into three segments fs(l,1), fs(l,2) and fs(l,3) in order to carry out three different prediction filterings in the frame. It should however be noted that the filterings in two segments, such as the segments fs(1,1) and fs(1,3) can be the same. The frame B(l,m) has further been segmented into two segments ps(1,1) and ps(l,2) in order to have two different probability tables for those segments.

The frame B(r,m) has been segmented into three segments fs(r,1), fs(r,2) and fs(r,3) in order to carry out three different prediction filterings in the frame. It should however again be noted that the filterings in two segments, such as the segments fs(r,1) and fs(r,3) can be the same. The frame B(r,m) has further been segmented into four segments ps(r,1), ps(r,2), ps(r,3) and ps(r,4) in order to have four different probability tables for those segments. Again, it should be noted that some of the segments can have the same probability table.

The decision to have the same probability table for different segments can be taken on beforehand by a user of the apparatus, after having carried out a signal analysis on the signals in the segments. Or the apparatus may be capable of carrying out this signal analysis and decide in response thereto. In some situations, a signal analysis carried out on two segments may result in probability tables that differ only slightly. In such situation, it can be decided to have one and the same probability table for both segments. This one probability table could be equal to one of the two probability tables established for the two segments, or could be an averaged version of both tables. An equivalent reasoning is valid for the sets of filter coefficients in the various segments.

To summarize: in order to encode a small portion of audio in an audio channel signal, the coding algorithm in SACD requires both a prediction filter (the filter) and a probability table (the table).

For improving the coding gain it can be efficient to use different filters in different channels. But also within the same channel is can be beneficial to use different filters. That is why the concept of segmentation is introduced. A channel is partitioned into segments and in a segment a particular filter is used. Several segments, also from other channels, may use the same or a different filter. Besides storage of the filters that are used, also information about the segments (segmentation) and information about what filter is used in what segment (mapping) have to be stored.

For the tables, the same idea is applicable, however the segmentation and mapping may be different from the segmentation and mapping for the filters. In case of equal segmentation for both filter and table this is indicated. The same idea is used for the mapping. If the segmentation for the filters is equal for all channels this is indicate too. The same idea is used for the mapping.

First, a description will be given of the contents of a frame of a transmission signal comprising the encoded channel signals and the corresponding side information. Figure 4 shows a schematic drawing of the frame. Apart from synchronization information (not shown), the frame comprises two words w₁ and w₂, followed by segmentation information on the prediction filters. Next a word w₃ is present followed by segmentation information on the probability tables. Next follow two words w₄ and w₅, followed by mapping information on the prediction filters. Next, follows a word w₆, followed by mapping information on the probability tables. Next follow the filter coefficients and the probability tables, as supplied by the generator units 12 and 13, respectively. The frame ends with the data D, supplied by the arithmetic encoder 6.

The word w₁ is in this example one bit long and can have the value '0' or '1', and indicates whether the segment information for the filter coefficients and the probability tables are the same ('1'), or not ('0'). The word w₄ is in this example one bit long and can have the value '0' or '1', and indicates whether the mapping information for the filter coefficients and the probability tables are the same ('1'), or not ('0'). The word w₂, again one bit long, can have the value '0' or '1', and indicates whether the channel signals have the same segmentation information for the prediction filter coefficients ('1'), or not ('0'). The word w₃ (one bit long) can have the value '0' or '1', and indicates whether the channel signals have the same segmentation information for the probability tables ('1'), or not ('0'). The word w₅ can have the value '0' or '1', and indicates whether the channel signals have the same mapping information for the prediction filter coefficients ('1'), or not ('0'). The word w₆ can have the value '0' or '1', and indicates whether the channel signals have the same mapping information for the probability tables ('1'), or not ('0').

First, the representation of the total number of segments S in a frame will be described.

To code a number, e.g. the total number of segments in a frame in a particular channel signal, a kind of run-length coding is applied. It is important that the code is short for small values of S. Since the number of segments in a channel S ≥ 1, S = 0 needs not to be coded. In SACD the following codes are used.

S

code(S)

1

1

2

01

3

001

4

0001

S

0^(s-1)1

Remark: Here the "1" is used as delimiter. It is clear that in general the role of the "0" and "1" can be interchanged. The basic idea of the delimiter is that a certain sequence is violated; the sequence of "0's" is violated by a "1". An alternative is e.g. to "inverse" the next symbol and "no inversion" is used as a delimiter. In this way long constant sequences are avoided. An example of inverting sequences that start with an "1" is (not used in SACD):

S

code(S)

1

0

2

11

3

100

4

1011

5

10100

6

101011

Second, the representation of the segment sizes will be described. The length of a segment will be expressed in number of bytes of the channel signal. The B bytes in a frame of a channel signal are partitioned into S segments. For the first S - 1 segments the number of bytes of each segment has to be specified. For the S^th segment the number of bytes is specified implicitly, it is the remaining number of bytes in the channel. The number of bytes in segment i, equals

B_i so the number of bytes in the last segment is: $$B_{s-1}=B-{\displaystyle \sum _{i=0}^{s-2}}{B}_i$$

Since the number of bytes in the first S - 1 segments are multiples of R, the resolution R ≥ 1, we define:

• B_¡ = b_i R and consequently: $B_{s-1}=B-{\displaystyle \sum _{i=0}^{s-2}}{b}_{i}R$
• The S - 1 values of b_i are stored and R is stored in a channel only if S > 1 and when it is not stored already for another channel.
• The number of bits required to store b_i depends on its possible values.
• 0 ≤ b_i ≤ b_i.max with e.g. $b_{i,\max }=\lfloor \frac{B}{R}\rfloor -{\displaystyle \sum _{j=0}^{i-1}}{b}_j$
• so the required number of bits to store b_i is:
• $$\mathrm{\#bits}\left(b_i\right)=\lfloor {}^2\log \left(b_{i,\max }\right)\rfloor +1$$

This has as advantage that the required number of bits for the segment length may decrease for segments at the end of the frame. If restrictions are imposed on e.g. minimal length of a segment the calculation of the number of bits may be adapted accordingly. The number of bits to store the resolution R is: # bits(R)

Third, the representation of the segmentation information in the serial datastream will be described. Use will be made of the representations given above under table 1. This will be illustrated by some examples.

In order to distinguish between filters and probability tables, the subscripts f and t are used. To distinguish between segments in different channels the double argument is used: (channel number, segment number).

Next follows a first example. For a 2-channel case, we have different segmentations for filters and probability tables, and the segmentation is different for both channels. The following table shows the parameters in the stream.

Value

#bits

comment

(w₁=) 0

1

segmentation information for the filters is different from the segmentation for the probability tables

filter segmentation

(w₂=) 0

1

channels have own filter segmentation information

filter segmentation in channel 0

(y₁ =) 0

1

first bit of code(S_f(0)) indicating that S_f(0)≥2

R_f

#bits(R_f)

resolution for filters

b_f(0,0)

#bits(b_f(0,0))

first segment in channel 0 has length R_fb_f(0,0) bytes

(y₂ =) 1

1

last bit of code(S_f(0)) indicating that S_f(0)=2

filter segmentation in channel 1

(y₁ =) 0

1

first bit of code(S_f(1)) indicating that S_f(1)≥2

b_f(1,0)

#bits(b_f(1,0))

first segment in channel 1 has length R_f b_f(1,0) bytes

(y₂ =) 0

1

second bit of code(S_f(1)) indicating that S_f(1)≥3

b_f(1,1)

#bits(b_f(1,1))

second segment in channel 1 has length R_f b_f(1,1) bytes

(y₃ =) 1

1

last bit of code(S_f(1)) indicating that S_f(1)=3

probability table segmentation

(w₃=) 0

1

channels have own table segmentation specification

probability table segmentation in channel 0

(y₁ =) 1

1

last bit of code(S_t(0)) indicating that S_t(0)=1

probability table segmentation in channel 1

(y₁=)0

1

first bit of code(S_t(1)) indicating that S_t(1)≥2

R_t

#bits(R_t)

resolution for tables

b_t(1,0)

#bits(b_t(1,0))

first segment in channel 1 has length R_t b_t(1,0) bytes

(y₂ =) 0

1

second bit of code(S_t(1)) indicating that S_t(1)≥3

b_t(1,1)

#bits(b_t(1.1))

second segment in channel 1 has length R_t b_t(1,1) bytes

(y₃ =) 1

1

last bit of code(S_t(1)) indicating that S_t(1)=3

In the above table 2, the first combination (y₁,y₂) equal to (0,1) is the codeword code(S) in table 1 above, and indicates that in the channel signal numbered 0 the frame is divided into two segments for the purpose of prediction filtering. Further, the combination (y₁,y₂,y₃) equal to (0,0,1) is the codeword code(S) in table 1 above, and indicates that in the channel signal numbered 1 the frame is divided into three segments for the purpose of prediction filtering. Next, we find a combination (y₁) equal to (1), which is the first codeword in table 1, indicating that the channel signal numbered 0, the frame is not divided for the probability table. Finally, we find a combination (y₁,y₂,y₃) equal to (0,0,1), which indicates that the frame of the second channel signal is divided into three segments, each with a corresponding probability table.

Next, follows another example for a 5-channel case. It is assumed that for this 5-channel case, we have equal segmentation for filters and tables, and the segmentation is equal for all channels.

Value

#bits

comment

(w₁=) 1

1

the segmentation information for the prediction filters and probability tables is the same

filter segmentation

(w₂=) 1

1

channels have equal filter segmentation information

filter segmentation in channel 0

(y₁ =)0

1

first bit of code(S_f(0)) indicating that S_f(0)≥2

R_f

#bits(R_f)

resolution for filters

b_f(0,0)

#bits(b_f(0,0))

first segment in channel 0 has length R_f b_f(0,0) bytes

(y₂ =) 0

1

second bit of code(S_f(0)) indicating that S_f(0)≥3

b_f(0,1)

#bits(b_f(0,1))

second segment in channel 0 has length R_f b_f(0,1) bytes

(y₃ =) 1

1

last bit of code(S_f(0)) indicating that S_f(0)=3

filter segmentation in channel c

b_f(c,0)=b_f(0,0) and b_f(c,1)=b_f(0,1) for 1≤c<5

probability table segmentation in channel c

b_t(c,0)=b_f(0,0) and b_t(c,1)=b_f(0,1) for 0≤c<5

Remark: The single bits of code(S) interleaved in de segmentation information can be interpreted as "another segment will be specified" in case of "0" or "no more segments will be specified" in case of "1".

Next, mapping will be described.

For each of the segments, all segments of all channels are considered together, it has to be specified which filter or table is used. The segments are ordered; first the segments of channel 0 followed by the segments of channel 1 and so on.

The filter or table number for segment s, N(s) is defined as: $$\{\begin{array}{c}N\left(0\right)=0\\ 0\le N\left(s\right)\le N_{\max }\left(s\right)\end{array}$$

with N_max (s), the maximum allowed number for a given segment, defined as: $$N_{\max }\left(s\right)=1+\max \left(N\left(i\right)\right)\mathrm{with}0\le i<s$$

The required number of bits to store N(s) equals: $$\mathit{\#bits}\left(N\left(s\right)\right)=\lfloor {}^2\log \left(N_{\max }\left(s\right)\right)\rfloor +1$$

The number of bits that is required to store a filter or table number according to this method depends on the set of numbers that already has been assigned.

If the tables use the same mapping as the filters, which is not always possible, this is indicated. Also when all channels use the same mapping this is indicated.

With two examples the idea will be illustrated.

Example 3

Assume that in total we have 7 segments (0 through 6), some segments use the same filter and some use a unique filter. Furthermore it is assumed that the tables use the same mapping specification as the filters.

Channel number

Segment number

Filter number

Possible filter numbers

#bits

0

0

0

-

0

1

1

0

0 or 1

1

1

2

1

0 or 1

1

1

3

2

0, 1 or 2

2

1

4

3

0, 1, 2 or 3

2

2

5

3

0, 1, 2, 3 or 4

3

3

6

1

0, 1, 2, 3 or 4

3

Total #bits

12

Segment number 0 uses filter number 0 per definition, so no bits are needed for this specification. Segment number 1 may use an earlier assigned filter (0) or the next higher not yet assigned filter (1), so 1 bit is needed for this specification. Segment number 1 uses filter number 0 in this example. Segment number 2 may use an earlier assigned filter (0) or the next higher not yet assigned filter (1), so 1 bit is needed for this specification. Segment number 2 uses filter number 1 in this example.

Segment number 3 may use an earlier assigned filter (0 or 1) or the next higher not yet assigned filter (2), so 2 bits are needed for this specification. Segment number 3 uses filter number 2 in this example.

Segment number 4 may use an earlier assigned filter (0, 1 or 2) or the next higher not yet assigned filter (3), so 2 bits are needed for this specification. Segment number 4 uses filter number 3 in this example. Segment number 5 may use an earlier assigned filter (0, 1, 2 or 3) or the next higher not yet assigned filter (4), so 3 bits are needed for this specification. Segment number 5 uses filter number 3 in this example.

Segment number 6 may use an earlier assigned filter (0, 1, 2 or 3) or the next higher not yet assigned filter (4), so 3 bits are needed for this specification. Segment number 6 uses filter number 1 in this example.

In total 12 bits are required to store the mapping. The total number of segments (7 segments in this example) is known at this point in the stream.

Value

#bits

comment

(w₄ =) 1

1

probability tables have same mapping information as prediction filters

prediction filter mapping

(w₅ =) 0

1

channels have own filter segmentation information

0

filter number for segment 0 is 0 per definition

N_f(1)

#bits(N_f(1))

filter number for segment 1

N_f(2)

#bits(N_f(2))

filter number for segment 2

N_f(3)

#bits(N_f(3))

filter number for segment 3

N_f(4)

#bits(N_f(4))

filter number for segment 4

N_f(5)

#bits(N_f(5))

filter number for segment 5

N_f(6)

#bits(N_f(6))

filter number for segment 6

probability table mapping

N_t(i)=N_f(i) for 0≤i<7

Another example. Assume that in total we have 6 channels each with 1 segment and each segment uses the same prediction filter and the same probability table.

Value

#bits

comment

(w₄ =) 1

1

probability tables have same mapping information as prediction filters

prediction filter mapping

(w₅ =) 1

1

channels have own filter mapping specification

0

filter number for segment 0 is 0 per definition

prediction filter mapping for segment i

N_f(i)=0 for 1≤i<6

probability table mapping for segment i

N_t(i)=N_f(i) for 0≤i<6

In total 2 bits are required to store the complete mapping.

Remark: A reason to give the indication that a following specification is also used for other application (e.g. for tables the same segmentation is used as for the filters) is that this simplifies the decoder.

Whilst the invention has been described with reference to preferred embodiments thereof, it is to be understood that these are not limitative examples. Thus, various modifications may become apparent to those skilled in the art without departing from the scope of the invention as defined by the claims. As an example, the invention could also have been incorporated in an embodiment in which time equivalent signal blocks are encoded, without making use of segmentation. In such embodiment, the serial datastream obtained, like the datastream of figure 4, will be devoid of the segment information described there for the filters and the probability tables, as well as some of the indicator words, such as the indicator words w₁, w₂ and w₃. Further, the invention lies in each and every novel feature and combination of features.