Aufgaben:Exercise 3.4Z: GSM Full-Rate Voice Codec: Difference between revisions
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{{quiz-Header|Buchseite= | {{quiz-Header|Buchseite=Mobile_Communications/Similarities_Between_GSM_and_UMTS | ||
}} | }} | ||
[[File:EN_Mob_A_3_4_Z.png|right|frame|LPC | [[File:EN_Mob_A_3_4_Z.png|right|frame|LPC, LTP and RPE parameters in the GSM Full Rate Vocoder]] | ||
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 ( | *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 [[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|Speech Coding]] of the book "Examples of Communication Systems". | |||
=== | ===Questionnaire=== | ||
<quiz display=simple> | <quiz display=simple> | ||
{ | {To which bandwidth $B$ must the speech signal be limited? | ||
|type="{}"} | |type="{}"} | ||
$B \ = \ $ { 4 3% } $\ \rm kHz$ | $B \ = \ $ { 4 3% } $\ \rm kHz$ | ||
{ | {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 | $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 $R_{\rm Out}$ of the GSM–full rate codec? | ||
|type="{}"} | |type="{}"} | ||
$R_{\rm Out} \ = \ $ { 13 3% } $\ \rm kbit/s$ | $R_{\rm Out} \ = \ $ { 13 3% } $\ \rm kbit/s$ | ||
{ | {Which statements apply to the block "LPC"? | ||
|type="[]"} | |type="[]"} | ||
+ LPC | + LPC makes a short-term prediction over one millisecond. | ||
+ | + The $36$ LPC bits specify coefficients that the receiver uses to undo the LPC filtering. | ||
- | - The filter for short-term prediction is recursive. | ||
- | - The LPC output signal is identical to the input signal $s_{\rm R}(t)$. | ||
{ | {Which statements regarding the block "LTP" are true? | ||
|type="[]"} | |type="[]"} | ||
+ LTP | + LTP removes periodic structures of the speech signal. | ||
- | - The long-term prediction is performed once per frame. | ||
+ | + The memory of the LTP predictor is up to $15 \ \rm ms$. | ||
{ | {Which statements apply to the block "RPE"? | ||
|type="[]"} | |type="[]"} | ||
- RPE | - RPE delivers fewer bits than LPC and LTP. | ||
+ RPE | + RPE removes unimportant parts for the subjective impression. | ||
+ RPE | + RPE subdivides each sub-block into four sub-sequences. | ||
- RPE | - RPE selects the sub-sequence with the minimum energy. | ||
</quiz> | </quiz> | ||
=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' | '''(1)''' To satisfy the sampling theorem, the bandwidth $B$ must not exceed $ f_{\rm A}/2 \hspace{0.15cm}\underline{= 4 \ \ \rm kHz}$. | ||
'''(2)''' | '''(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 | ||
:$$R_{\rm In} = \frac{160 \cdot 13}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 104\,{\rm kbit/s}}\hspace{0.05cm}.$$ | :$$R_{\rm In} = \frac{160 \cdot 13}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 104\,{\rm kbit/s}}\hspace{0.05cm}.$$ | ||
'''(3)''' | '''(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 | ||
:$$R_{\rm Out} = \frac{260}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 13\,{\rm kbit/s}}\hspace{0.05cm}.$$ | :$$R_{\rm Out} = \frac{260}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 13\,{\rm kbit/s}}\hspace{0.05cm}.$$ | ||
* | *The compression factor achieved by the full rate speech codec is thus $104/13 = 8$. | ||
'''(4)''' The <u>first two statements</u> are true: | |||
*The 36 LPC bits describe a total of eight filter coefficients of a non-recursive filter, whereby eight $\rm ACF$ 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 ${\rm ln}\big[(1 - r_{k})/(1 + r_{k})\big]$, 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 <u>statements 1 and 3</u>, 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 $\rm (CCF)$ 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)''' | '''(6)''' The statements <u>2 and 3</u> 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. Voice 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 $12$ bits 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 [[Examples_of_Communication_Systems/Speech_Coding#Regular_Pulse_Excitation_–_RPE_Coding|RPE coding]] of the book "Examples of Communication Systems". | |||
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[[Category: | [[Category:Mobile Communications: Exercises|^3.2 Similarities between GSM and UMTS | ||
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[[de:Exercises:Exercise_3.4Z:_GSM_Full-Rate_Voice_Codec]] | |||
Latest revision as of 14:27, 16 March 2026

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 Speech 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) The first two statements are true:
- The 36 LPC bits describe a total of eight filter coefficients of a non-recursive filter, whereby eight $\rm ACF$ 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 ${\rm ln}\big[(1 - r_{k})/(1 + r_{k})\big]$, 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 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 $\rm (CCF)$ 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. Voice 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 $12$ bits 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 coding of the book "Examples of Communication Systems".