Aufgaben:Exercise 3.5: GSM Full Rate Vocoder: Difference between revisions
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{{quiz-Header|Buchseite= | {{quiz-Header|Buchseite=Examples_of_Communication_Systems/Voice_Coding | ||
}} | }} | ||
[[File: | [[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 encoder and decoder and combines three methods for the compression of speech signals: | |||
*Linear Predictive Coding (LPC), | *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 exercise. | |||
The input signal in the above graphic is the digitalized speech signal $s_{\rm block}(n)$. 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 $13 \ \rm bits$, | |||
=== | |||
*following segmentation into blocks of each $20 \ \rm ms$. | |||
The further tasks of preprocessing will not be discussed in detail here. | |||
<u>Hint:</u> This exercise belongs to the chapter [[Examples_of_Communication_Systems/Voice_Coding|"Speech Coding"]]. | |||
===Questions=== | |||
<quiz display=simple> | <quiz display=simple> | ||
{To what bandwidth must the speech signal be limited? | |||
|type="{}"} | |||
$B \ = \ $ { 4 3% } $ \ \rm kHz$ | |||
{How many samples $(N_{\rm block})$ does a speech lock consist of? What is the input data rate $R_{\rm in}$? | |||
|type="{}"} | |||
$N_{\rm block} \hspace{0.25cm} = \ $ { 160 3% } | |||
$R_{\rm in} \hspace{0.22cm} = \ $ { 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} \hspace{0.09cm} = \ $ { 13 3% } $ \ \rm kbit/s$ | ||
{Which statements are true regarding the block "LPC"? | |||
|type="[]"} | |||
+ LPC makes a short-term prediction over one millisecond. | |||
+ The $36$ LPC bits are filter coefficients used at the receiver to undo the LPC filtering. | |||
- The filter for long-term prediction is recursive. | |||
- The LPC output is identical to its input $s_{\rm block}(t)$. | |||
{Which statements are true regarding the block "LTP"? | |||
|type="[]"} | |||
+ Periodic structures of the speech signal are removed. | |||
- Long-term prediction is performed once per block. | |||
+ The memory of the LTP predictor is up to $15 \ \rm ms$. | |||
{Which statements are true for the block "RPE"? | |||
|type="[]"} | |||
- RPE provides less information than LPC and LTP. | |||
+ RPE removes parts that are unimportant for the subjective impression. | |||
+ RPE divides each subblock again into four sub-sequences. | |||
- RPE selects the subsequence with the minimum energy. | |||
</quiz> | </quiz> | ||
=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' | |||
'''(2)''' | '''(1)''' To satisfy the sampling theorem, the bandwidth must not exceed $f_{\rm A}/2 \hspace{0.15cm} \underline{= 4 \ \rm kHz}$. | ||
'''(3)''' | |||
'''(4)''' | |||
'''(5)''' | '''(2)''' From the given sampling rate $f_{\rm A} = 8 \ \rm kHz$ results a distance between individual samples of $T_{\rm A} = 0.125 \ \rm ms$. | ||
*Thus, a speech block $(20 \ \rm ms)$ consists of $N_{\rm block} = 20/0.125\hspace{0.15cm} \underline{= 160 \ \rm samples}$, each quantized with $13 \ \rm bits$. | |||
'''( | |||
*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)''' From the graph, it can be seen that $36$ (LPC) $+\ 36$ (LTP) $+\ 188$ (RPE) $= 260 \ \rm bits$ are output per speech block. | |||
*From this, the output data rate is calculated to be | |||
:$$R_{\rm out} = \frac{260}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 13\,{\rm kbit/s}}\hspace{0.05cm}.$$ | |||
*The compression factor achieved by the full rate speech codec is thus $104/13 = 8$. | |||
'''(4)''' Correct are the <u>statements 1 and 2</u>: | |||
*The $36$ LPC bits describe a total of eight filter coefficients of a non-recursive filter, where eight ACF values are determined from the short-time analysis. | |||
*These are converted into reflection coefficients $r_{k}$ according to 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 passed on to the receiver. | |||
*The LPC output signal has a significantly smaller amplitude compared to its input $s_{\rm block}(n)$, 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 in blocks every $5 \rm ms \ (40 \rm samples)$, i.e. four times per speech block. | |||
*To do this, the cross-correlation function (CCF) is formed between the current and the three preceding sub-blocks. | |||
*For each sub-block, an LTP delay and an LTP gain are determined that best fit the sub-block. | |||
*A correction signal of the subsequent component "RPE" is also taken into account. | |||
*In the case of long-term prediction, as with LPC, the output is redundancy-reduced compared to the input. | |||
'''(6)''' Correct are the <u>statements 2 and 3</u>: | |||
*That statement 1 is false can already be seen from the graph on the statements page, since $188$ of the $260$ output bits come from the RPE. | |||
*To the last statement: The RPE searches for the sub-sequence with the maximum energy. | |||
*This parameter "RPE pulses" alone occupies $188$ of the $260$ output bits. | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
[[Category: | [[Category:Examples of Communication Systems: Exercises|^3.3 Speech Coding^]] | ||
[[de:Aufgaben:Aufgabe 3.5: GSM–Vollraten–Sprachcodec]] | |||
Latest revision as of 17:55, 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 encoder 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 exercise.
The input signal in the above graphic is the digitalized speech signal $s_{\rm block}(n)$. 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 $13 \ \rm bits$,
- following segmentation into blocks of each $20 \ \rm ms$.
The further tasks of preprocessing will not be discussed in detail here.
Hint: This exercise belongs to the chapter "Speech Coding".
Questions
Solution
(1) To satisfy the sampling theorem, the bandwidth must not exceed $f_{\rm A}/2 \hspace{0.15cm} \underline{= 4 \ \rm kHz}$.
(2) From the given sampling rate $f_{\rm A} = 8 \ \rm kHz$ results a distance between individual samples of $T_{\rm A} = 0.125 \ \rm ms$.
- Thus, a speech block $(20 \ \rm ms)$ consists of $N_{\rm block} = 20/0.125\hspace{0.15cm} \underline{= 160 \ \rm samples}$, each quantized with $13 \ \rm bits$.
- 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) From the graph, it can be seen that $36$ (LPC) $+\ 36$ (LTP) $+\ 188$ (RPE) $= 260 \ \rm bits$ are output per speech block.
- From this, the output data rate is calculated to be
- $$R_{\rm out} = \frac{260}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 13\,{\rm kbit/s}}\hspace{0.05cm}.$$
- The compression factor achieved by the full rate speech codec is thus $104/13 = 8$.
(4) Correct are the statements 1 and 2:
- The $36$ LPC bits describe a total of eight filter coefficients of a non-recursive filter, where eight ACF values are determined from the short-time analysis.
- These are converted into reflection coefficients $r_{k}$ according to 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 passed on to the receiver.
- The LPC output signal has a significantly smaller amplitude compared to its input $s_{\rm block}(n)$, 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 in blocks every $5 \rm ms \ (40 \rm samples)$, i.e. four times per speech block.
- To do this, the cross-correlation function (CCF) is formed between the current and the three preceding sub-blocks.
- For each sub-block, an LTP delay and an LTP gain are determined that best fit the sub-block.
- A correction signal of the subsequent component "RPE" is also taken into account.
- In the case of long-term prediction, as with LPC, the output is redundancy-reduced compared to the input.
(6) Correct are the statements 2 and 3:
- That statement 1 is false can already be seen from the graph on the statements page, since $188$ of the $260$ output bits come from the RPE.
- To the last statement: The RPE searches for the sub-sequence with the maximum energy.
- This parameter "RPE pulses" alone occupies $188$ of the $260$ output bits.