Difference between revisions of "Aufgaben:Exercise 3.4Z: GSM Full-Rate Voice Codec"
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− | The numbers shown in the | + | The numbers shown in the graphic indicate the number of bits generated by the three units of this Full-Rate speech codec per frame of $20$ millisecond duration each. |
It should be noted that LTP and RPE, unlike LPC, do not work frame by frame, but with sub-blocks of $5$ milliseconds. However, this has no influence on solving the task. | It should be noted that LTP and RPE, unlike LPC, do not work frame by frame, but with sub-blocks of $5$ milliseconds. However, this has no influence on solving the task. | ||
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This results from the analog speech signal $s(t)$ by | This results from the analog speech signal $s(t)$ by | ||
− | *a suitable limitation to the bandwidth | + | *a suitable limitation to the bandwidth $B$, |
− | *sampling at the sampling rate | + | *sampling at the sampling rate $f_{\rm A} = 8 \ \rm kHz$, |
− | *quantization with | + | *quantization with $13 \ \rm bit$, |
*following segmentation into blocks of each $20 \ \rm ms$. | *following segmentation into blocks of each $20 \ \rm ms$. | ||
+ | |||
The further tasks of preprocessing will not be discussed in detail here. | The further tasks of preprocessing will not be discussed in detail here. | ||
− | |||
− | |||
− | |||
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''Notes:'' | ''Notes:'' | ||
− | * | + | *The task belongs to the chapter [[Mobile_Communications/Similarities_Between_GSM_and_UMTS|Similarities between GSM and UMTS]]. |
− | + | *Reference is also made to the Chapter [[Examples_of_Communication_Systems/Voice_Coding|Voice Coding]] of the book "Examples of Communication Systems". | |
− | *Reference is also made to the Chapter [[Examples_of_Communication_Systems/ | ||
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<quiz display=simple> | <quiz display=simple> | ||
− | {To which bandwidth | + | {To which bandwidth $B$ must the speech signal be limited? |
|type="{}"} | |type="{}"} | ||
$B \ = \ $ { 4 3% } $\ \rm kHz$ | $B \ = \ $ { 4 3% } $\ \rm kHz$ | ||
− | {Of | + | {Of how many samples $(N_{\rm R})$ is there a speech frame? How large is the input data rate $R_{\rm In}$? |
|type="{}"} | |type="{}"} | ||
$N_{\rm R} \hspace{0.18cm} = \ $ { 160 3% } $\ \rm samples$ | $N_{\rm R} \hspace{0.18cm} = \ $ { 160 3% } $\ \rm samples$ | ||
$R_{\rm In} \hspace{0.15cm} = \ $ { 104 3% } $\ \rm kbit/s$ | $R_{\rm In} \hspace{0.15cm} = \ $ { 104 3% } $\ \rm kbit/s$ | ||
− | {What is the output data rate | + | {What is the output data rate $R_{\rm Out}$ of the GSM–full rate codec? |
|type="{}"} | |type="{}"} | ||
$R_{\rm Out} \ = \ $ { 13 3% } $\ \rm kbit/s$ | $R_{\rm Out} \ = \ $ { 13 3% } $\ \rm kbit/s$ | ||
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+ The $36$ LPC bits specify coefficients that the receiver uses to undo the LPC filtering. | + The $36$ LPC bits specify coefficients that the receiver uses to undo the LPC filtering. | ||
- The filter for short-term prediction is recursive. | - The filter for short-term prediction is recursive. | ||
− | - The LPC output signal is identical to the input $s_{\rm R}(t)$. | + | - The LPC output signal is identical to the input signal $s_{\rm R}(t)$. |
− | {Which statements regarding the block | + | {Which statements regarding the block "LTP" are true? |
|type="[]"} | |type="[]"} | ||
+ LTP removes periodic structures of the speech signal. | + LTP removes periodic structures of the speech signal. | ||
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+ RPE removes unimportant parts for the subjective impression. | + RPE removes unimportant parts for the subjective impression. | ||
+ RPE subdivides each sub-block into four sub-sequences. | + RPE subdivides each sub-block into four sub-sequences. | ||
− | - RPE selects the | + | - RPE selects the sub-sequence with the minimum energy. |
</quiz> | </quiz> |
Revision as of 13:58, 21 January 2021
This codec called "GSM Full-Rate Vocoder" (which was standardized for the GSM system in 1991) stands for a joint realization of coder and decoder and combines three methods for the compression of speech signals:
- Linear Predictive Coding $\rm (LPC)$,
- Long Term Prediction $\rm (LTP)$, and
- Regular Pulse Excitation $\rm (RPE)$.
The numbers shown in the graphic indicate the number of bits generated by the three units of this Full-Rate speech codec per frame of $20$ millisecond duration each.
It should be noted that LTP and RPE, unlike LPC, do not work frame by frame, but with sub-blocks of $5$ milliseconds. However, this has no influence on solving the task.
The input signal in the above graphic is the digitalized speech signal $s_{\rm R}(n)$.
This results from the analog speech signal $s(t)$ by
- a suitable limitation to the bandwidth $B$,
- sampling at the sampling rate $f_{\rm A} = 8 \ \rm kHz$,
- quantization with $13 \ \rm bit$,
- following segmentation into blocks of each $20 \ \rm ms$.
The further tasks of preprocessing will not be discussed in detail here.
Notes:
- The task belongs to the chapter Similarities between GSM and UMTS.
- Reference is also made to the Chapter Voice Coding of the book "Examples of Communication Systems".
Questionnaire
Solution
(1) To satisfy the sampling theorem, the bandwidth $B$ must not exceed $ f_{\rm A}/2 \hspace{0.15cm}\underline{= 4 \ \ \rm kHz}$.
(2) The given sampling rate $f_{\rm A} = 8 \ \rm kHz$ results in a distance between individual samples of $T_{\rm A} = 0.125 \ \rm ms$.
- Thus a speech frame of $(20 {\rm ms})$ consists of $N_{\rm R} = 20/0.125 = \underline{160 \ \rm samples}$, each quantized with $13 \ \rm Bit$.
- The data rate is thus
- $$R_{\rm In} = \frac{160 \cdot 13}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 104\,{\rm kbit/s}}\hspace{0.05cm}.$$
(3) The graph shows that per speech frame $36 \ {\rm (LPC)} + 36 \ {\rm (LTP)} + 188 \ {\rm (RPE)} = 260 \ \ \rm Bit$ are output.
- From this the output data rate is calculated as
- $$R_{\rm Out} = \frac{260}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 13\,{\rm kbit/s}}\hspace{0.05cm}.$$
- The compression factor achieved by the full rate speech codec is thus $104/13 = $8.
(4) Only the first two statements are true:
- The 36 LPC–bits describe a total of eight filter coefficients of a non-recursive filter, whereby eight acf–values are determined from the short-term analysis and where these are converted into reflection factors $r_{k}$ after the so-called Schur recursion.
- From these the eight LAR–coefficients are calculated according to the function ${\rm ln}[(1 - r_{k})/(1 + r_{k})]$, quantized with a different number of bits and sent to the receiver.
- The LPC output signal has a significantly lower amplitude than its input $s_{\rm R}(n)$, and it has a significantly reduced dynamic range and a flatter spectrum.
(5) Correct are the the statements 1 and 3, but not the second:
- The LTP–analysis and –filtering is done blockwise every $5 \ \rm ms$ (40 samples), i.e. four times per speech frame.
- The cross correlation function (CCF) between the current sub-block and the three previous sub-blocks is formed.
- For each sub-block, an LTP–delay and an LTP–gain are determined which best match the sub-block.
- A correction signal of the following component „RPE” is also taken into account.
- For the long-term prediction, as with the LPC, the output is reduced in redundancy compared to the input.
(6) The statements 2 and 3 are correct:
- The fact that statement 1 is wrong can be seen from the graphic on the data page, because $188$ of the $260$ output bits come from the RPE. Language would be understandable with RPE alone (without LPC and LTP).
- Regarding the last statement: The RPE is of course looking for the subsequence with the 'maximum energy. The RPE pulses are a subsequence (13 of 40 samples) of three bits per subframe of $5 \ \rm ms$ and accordingly $12 \ \rm Bit$ per $20 \ \rm ms$ frame.
- The "RPE pulse" thus occupies $13 \cdot 12 = 156$ of the $260$ output bits.
More details about the RPE block can be found on the page RPE–Codierung des Buches „Beispiele von Nachrichtensystemen”.