Difference between revisions of "Aufgaben:Exercise 3.5: GSM Full Rate Vocoder"
From LNTwww
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The further tasks of preprocessing will not be discussed in detail here. | The further tasks of preprocessing will not be discussed in detail here. | ||
| − | + | You can also take questions and solutions (almost completely?) from | |
| − | https://en.lntwww.de/Aufgaben:Exercise_3.4Z:_GSM_Full-Rate_Voice_Codec | + | [https://en.lntwww.de/Aufgaben:Exercise_3.4Z:_GSM_Full-Rate_Voice_Codec " Exercise 3.4Z"] |
| − | + | Hint: | |
| − | *Diese Aufgabe gehört zum Kapitel [[Examples_of_Communication_Systems/ | + | *Diese Aufgabe gehört zum Kapitel [[Examples_of_Communication_Systems/Voice_Coding|"Voice Coding"]]. |
| − | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
| − | { | + | {To what bandwidth must the speech signal be limited? |
|type="{}"} | |type="{}"} | ||
$B \ = \ $ { 4 3% } $ \ \rm kHz$ | $B \ = \ $ { 4 3% } $ \ \rm kHz$ | ||
| − | { | + | {How many samples $(N_{\rm R})$ does a speech frame consist of? What is the input data rate $R_{\rm In}$? |
|type="{}"} | |type="{}"} | ||
$N_{\rm R} \hspace{0.25cm} = \ $ { 160 3% } | $N_{\rm R} \hspace{0.25cm} = \ $ { 160 3% } | ||
$R_{\rm In} \hspace{0.22cm} = \ $ { 104 3% } $\ \rm kbit/s$ | $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$ | $R_{\rm Out} \hspace{0.09cm} = \ $ { 13 3% } $ \ \rm kbit/s$ | ||
| − | { | + | {Which statements are true regarding the block "LPC"? |
|type="[]"} | |type="[]"} | ||
| − | + LPC | + | + 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 $s_{\rm R}(t)$. |
| − | { | + | {Which statements are true regarding the block "LTP"? |
|type="[]"} | |type="[]"} | ||
| − | + | + | + Periodic structures of the speech signal are removed. |
| − | - | + | - Long-term prediction is performed once per frame. |
| − | + | + | + The memory of the LTP predictor is up to $15 \ \rm ms$. |
| − | { | + | {Which statements are true for the block "RPE"? |
|type="[]"} | |type="[]"} | ||
| − | - RPE | + | - RPE provides less information than LPC and LTP. |
| − | + RPE | + | + RPE removes parts that are unimportant for the subjective impression. |
| − | + RPE | + | + RPE divides each subblock again into four sub-sequences. |
| − | - RPE | + | - RPE selects the subsequence with the minimum energy. |
</quiz> | </quiz> | ||
| − | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' | + | '''(1)''' To satisfy the sampling theorem, the bandwidth must not exceed $f_{\rm A}/2 \hspace{0.15cm} \underline{= 4 \ \rm kHz}$. |
| − | '''(2)''' | + | '''(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 frame $(20 \ \rm ms)$ consists of $N_{\rm R} = 20/0.125\hspace{0.15cm} \underline{= 160 \ \rm samples}$, each quantised with $13 \ \rm bits$. |
| − | * | + | *The data rate is thus |
:$$R_{\rm In} = \frac{160 \cdot 13}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 104\,{\rm kbit/s}}\hspace{0.05cm}.$$ | :$$R_{\rm In} = \frac{160 \cdot 13}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 104\,{\rm kbit/s}}\hspace{0.05cm}.$$ | ||
| − | '''(3)''' | + | '''(3)''' From the graph, it can be seen that $36$ (LPC) $+ 36$ (LTP) $+ 188$ (RPE) $= 260 \ \rm bits$ are output per speech frame. |
| − | * | + | *From this, the output data rate is calculated to be |
:$$R_{\rm Out} = \frac{260}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 13\,{\rm kbit/s}}\hspace{0.05cm}.$$ | :$$R_{\rm Out} = \frac{260}{20 \,{\rm ms}} \hspace{0.15cm} \underline {= 13\,{\rm kbit/s}}\hspace{0.05cm}.$$ | ||
| − | * | + | *The compression factor achieved by the full rate speech codec is thus $104/13 = 8$. |
| − | '''(4)''' | + | '''(4)''' Correct are <u>statements 1 and 2</u>: |
| − | * | + | *The $36$ LPC bits describe a total of eight filter coefficients of a non-recursive filter, where eight AKF values are determined from the short-time analysis and 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 ${\rm ln}[(1 - r_{k})/(1 + r_{k})]$, quantised 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 $s_{\rm R}(n)$, has a significantly reduced dynamic range and a flatter spectrum. |
| − | '''(5)''' | + | '''(5)''' Correct are <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 frame. |
| − | * | + | *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)''' | + | '''(6)''' Correct are <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 subsequence with the maximum energy. |
| − | * | + | *This parameter "RPE pulses" alone occupies $156$ of the $260$ output bits. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
Revision as of 21:36, 21 January 2023
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 $\rm (LPC)$,
- 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.
You can also take questions and solutions (almost completely?) from " Exercise 3.4Z"
Hint:
- Diese Aufgabe gehört zum Kapitel "Voice 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 frame $(20 \ \rm ms)$ consists of $N_{\rm R} = 20/0.125\hspace{0.15cm} \underline{= 160 \ \rm samples}$, each quantised 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 frame.
- 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 statements 1 and 2:
- The $36$ LPC bits describe a total of eight filter coefficients of a non-recursive filter, where eight AKF values are determined from the short-time analysis and 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 ${\rm ln}[(1 - r_{k})/(1 + r_{k})]$, quantised 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 $s_{\rm R}(n)$, has a significantly reduced dynamic range and a flatter spectrum.
(5) Correct are 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 frame.
- 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 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 subsequence with the maximum energy.
- This parameter "RPE pulses" alone occupies $156$ of the $260$ output bits.
