Aufgaben:Exercise 3.6: Adaptive Multi Rate Codec - Revision history

Guenter at 12:58, 25 January 2023

2023-01-25T12:58:26Z

Guenter at 12:43, 25 January 2023

2023-01-25T12:43:43Z

Guenter at 12:42, 25 January 2023

2023-01-25T12:42:17Z

Guenter at 11:56, 25 January 2023

2023-01-25T11:56:12Z

Guenter: Guenter moved page Aufgaben:Exercise 3.6: Adaptive Multi–Rate Codec to Aufgaben:Exercise 3.6: Adaptive Multi Rate Codec

2023-01-25T09:02:02Z

Guenter moved page Exercise 3.6: Adaptive Multi–Rate Codec to Exercise 3.6: Adaptive Multi Rate Codec

Hwang at 15:39, 23 January 2023

2023-01-23T15:39:49Z

Hwang at 11:35, 23 January 2023

2023-01-23T11:35:26Z

Noah at 19:52, 21 January 2023

2023-01-21T19:52:49Z

Noah at 19:52, 21 January 2023

2023-01-21T19:52:39Z

Noah at 19:51, 21 January 2023

2023-01-21T19:51:33Z

@@ Line 91: / Line 91: @@
 {{ML-Kopf}}
-'''(1)'''&nbsp; With the data rate $12.2 \ \rm kbit/s$, exactly $\underline{244 \ \rm bit}$ results within $20 \ \rm ms$, while for example in $4.75 \ \rm kbit/s$ mode only $95 \ \rm bit$ is transmitted.
+'''(1)'''&nbsp; With the data rate&nbsp; $R_{\rm C} = 12.2 \ \rm kbit/s$,&nbsp; exactly&nbsp; $\underline{244 \ \rm bits}$&nbsp; results within&nbsp; $20 \ \rm ms$,&nbsp; while e.g. in&nbsp; $4.75 \ \rm kbit/s$&nbsp; mode only&nbsp; $95 \ \rm bits$&nbsp; are transmitted.
-'''(2)'''&nbsp; In each subframe, the FCB pulse requires $35 \ \rm bit$ (five tracks of seven bits each) and the FCB gain requires five bits.
+'''(2)'''&nbsp; In each subframe,&nbsp; the FCB pulse requires&nbsp; $35 \ \rm bits$&nbsp; (five tracks of seven bits each)&nbsp; and the FCB gain requires five bits.
 '''(6)'''&nbsp; The fifth track provides&nbsp; $N_{1} \underline{ =-30}$&nbsp; and&nbsp; $N_{2} \underline{ = +5}$
 {{ML-Fuß}}

@@ Line 6: / Line 6: @@
 In the late 1990s,&nbsp; a very flexible,&nbsp; adaptive speech codec was developed and standardized in the form of&nbsp; $\rm AMR$&nbsp; codec.&nbsp; This provides a total of eight different modes with data rates between&nbsp; $4.75 \ \rm kbit/s$&nbsp; and&nbsp; $12.2 \ \rm kbit/s$.
-The AMR codec,&nbsp; like the full rate codec&nbsp; $\rm (FRC)$&nbsp; discussed in&nbsp; [[Aufgaben:Exercise_3.5:_GSM_Full-Rate_Voice_Codec|$\text{Exercise 3.5}$]],&nbsp; includes both a short-term prediction&nbsp; $\rm (LPC)$&nbsp; and a long-term prediction&nbsp; $\rm (LTP)$.&nbsp; However,&nbsp; these two components are realized differently from FRC.
+The AMR codec,&nbsp; like the full rate codec&nbsp; $\rm (FRC)$&nbsp; discussed in&nbsp; [[Aufgaben:Exercise_3.5:_GSM_Full_Rate_Vocoder|$\text{Exercise 3.5}$]],&nbsp; includes both a short-term prediction&nbsp; $\rm (LPC)$&nbsp; and a long-term prediction&nbsp; $\rm (LTP)$.&nbsp; However,&nbsp; these two components are realized differently from FRC.
 The main difference between AMR and FRC is the encoding of the residual signal&nbsp; $($after LPC and LTP$)$:

@@ Line 32: / Line 32: @@
 &rArr; &nbsp; '''Track 1'''&nbsp; includes
-#a positive pulse&nbsp; $({\rm sign} = 1)$&nbsp; at position&nbsp; $\big [1$&nbsp; (first possible position for track 1)&nbsp; $\hspace{0.02cm}\text{plus}\hspace{0.2cm}0$ (bit specification for&nbsp; "000") $= 1\big]$,
+#a positive pulse&nbsp; $({\rm sign} = 1)$&nbsp; at position&nbsp; $\big [1$&nbsp; (first possible position for track 1)&nbsp; $\hspace{0.02cm}\text{plus}\hspace{0.2cm}0$&nbsp; (bit specification for&nbsp; "000") $= 1\big]$,
-#another positive pulse&nbsp; $($since $110 > 000)$&nbsp; at position&nbsp; $\big [1 \hspace{0.2cm}\text{plus}\hspace{0.2cm}5$ (pulse spacing in each track) $\hspace{0.2cm}\text{mal}\hspace{0.2cm}6$ (bit specification for " 110") = $31\hspace{0.05cm}.$
+#another positive pulse&nbsp; $($since $110 > 000)$&nbsp; at position&nbsp; $\big [1 \hspace{0.2cm}\text{plus}\hspace{0.2cm}5$ (pulse spacing in each track) $\hspace{0.02cm}\text{times}\hspace{0.2cm}6$&nbsp; (bit specification for " 110") = $31\hspace{0.05cm}\big].$
 '''Track 2''' includes.
-*a negative pulse (${\rm VZ} = 0$) at $2$ (first possible position for track 2) $\hspace{0.2cm}\text{plus}\hspace{0.2cm}5\hspace{0.2cm}\text{mal}\hspace{0.2cm}4$ (bit specification for " 100") = $22\hspace{0.05cm},$
+#a negative pulse (${\rm sign} = 0$)&nbsp; at position&nbsp; $\big [2$ (first possible position for track 2)&nbsp; $\hspace{0.02cm}\text{plus}\hspace{0.2cm}5\hspace{0.2cm}\text{times}\hspace{0.2cm}4$&nbsp; (bit specification for&nbsp; " 100")&nbsp;  $=22\hspace{0.05cm}\big],$
-*a positive pulse (sign reversal due to $011 > 100$) at position $2 \hspace{0.2cm}\text{plus}\hspace{0.2cm}5\hspace{0.2cm}\text{mal}\hspace{0.2cm}3$ (bit specification for " 011") = $17\hspace{0.05cm}.$
+#a positive pulse&nbsp; $($sign reversal due to&nbsp; $011 > 100)$&nbsp; at position&nbsp; $\big [2 \hspace{0.2cm}\text{plus}\hspace{0.2cm}5\hspace{0.2cm}\text{times}\hspace{0.2cm}3$&nbsp; (bit specification for&nbsp; " 011")&nbsp;  $=17\hspace{0.05cm}\big].$
@@ Line 46: / Line 46: @@
-Hint:
+<u>Hint:</u>
 *This exercise belongs to the chapter&nbsp; [[Examples_of_Communication_Systems/Voice_Coding|"Speech Coding"]].
 *When entering the pulse positions&nbsp; $N_{1}$&nbsp; denotes the first triple of bits and&nbsp; $N_{2}$&nbsp; the second.
-*For example, for track&nbsp; $2$&nbsp; one would have to enter the values&nbsp; $N_{1}=-22$&nbsp; and&nbsp; $N_{2}=+17$&nbsp;.
@@ Line 61: / Line 62: @@
 {How many bits describe a speech frame $($of duration&nbsp; $20 \ \rm ms)$&nbsp; in&nbsp; $12.2 \ \rm kbit/s$ mode?
 |type="{}"}
-$N_{12.2} \ = \ $ { 244 3% } $ \ \rm Bit$
+$N_{12.2} \ = \ $ { 244 3% } $ \ \rm bits$
 {How many bits are needed for FCB pulse and gain per frame?
 |type="{}"}
-$N_{\rm FCB} \ = \ $ { 160 3% } $ \ \rm Bit$
+$N_{\rm FCB} \ = \ $ { 160 3% } $ \ \rm bits$
 { How many bits are left for LPC and LTP?
 |type="{}"}
-$N_{\rm LPC/LTP} \ = \ $ { 84 3% } $ \ \rm Bit$
+$N_{\rm LPC/LTP} \ = \ $ { 84 3% } $ \ \rm bits$
-{What subframe pulse positions and signs does track&nbsp; $3$ describe? <br>Follow the instructions for input on the input page.
+{What subframe pulse positions and signs does track&nbsp; $3$&nbsp; describe?&nbsp; Follow the instructions for input on the information page.
 |type="{}"}
 $N_{1} \ = \ $ { -8.24--7.76 }

@@ Line 4: / Line 4: @@
 [[File:En_Bei_A_3_6.png|right|frame|Tracks of the AMR codec]]
-In the late 1990s, a very flexible, adaptive speech codec was developed and standardised in the form of the AMR codec. This provides a total of eight different modes with data rates between&nbsp; $4.75 \ \rm kbit/s$&nbsp; and&nbsp; $12.2 \ \rm kbit/s$&nbsp;.
+In the late 1990s,&nbsp; a very flexible,&nbsp; adaptive speech codec was developed and standardized in the form of&nbsp; $\rm AMR$&nbsp; codec.&nbsp; This provides a total of eight different modes with data rates between&nbsp; $4.75 \ \rm kbit/s$&nbsp; and&nbsp; $12.2 \ \rm kbit/s$.
-The AMR codec, like the full rate codec (FRC) discussed in&nbsp; [[Aufgaben:Exercise_3.5:_GSM_Full-Rate_Voice_Codec|"Exercise 3.5"]]&nbsp;, includes both a short-term prediction (LPC) and a long-term prediction (LTP). However, these two components are realised differently from FRC.
+The AMR codec,&nbsp; like the full rate codec&nbsp; $\rm (FRC)$&nbsp; discussed in&nbsp; [[Aufgaben:Exercise_3.5:_GSM_Full-Rate_Voice_Codec|$\text{Exercise 3.5}$]],&nbsp; includes both a short-term prediction&nbsp; $\rm (LPC)$&nbsp; and a long-term prediction&nbsp; $\rm (LTP)$.&nbsp; However,&nbsp; these two components are realized differently from FRC.
-The main difference between AMR and FRC is the coding of the residual signal (after LPC and LTP):
+The main difference between AMR and FRC is the encoding of the residual signal&nbsp; $($after LPC and LTP$)$:
-*Instead of "Regular Pulse Excitation" (RPE), the "Algebraic Code Excitation Linear Prediction" (ACELP) procedure is used for the AMR code.
+#Instead of&nbsp; "Regular Pulse Excitation"&nbsp; $\rm (RPE)$,&nbsp; here the&nbsp; "Algebraic Code Excitation Linear Prediction"&nbsp; $\rm (ACELP)$&nbsp; is used.
-*From the fixed codebook (FCB), for each subframe of&nbsp; $5 \ \rm ms$&nbsp; duration, the FCB pulse and FCB gain that best match the residual signal (for which the mean square error of the difference signal becomes minimum) is selected.
+#From the fixed code book&nbsp; $\rm (FCB)$,&nbsp; for each subframe of&nbsp; $5 \ \rm ms$&nbsp; duration,&nbsp; the&nbsp; "FCB pulse"&nbsp; and the&nbsp; "FCB gain"&nbsp; that best match the residual signal&nbsp; $($for which the mean square error of the difference signal becomes minimum$)$&nbsp; is selected.
-Each entry in the fixed codebook identifies a pulse where exactly&nbsp; $10$&nbsp; of&nbsp; $40$&nbsp; positions are occupied by&nbsp; $\pm1$&nbsp;. In this regard it should be noted:
+Each entry in the fixed code book identifies a pulse where exactly&nbsp; $10$&nbsp; of&nbsp; $40$&nbsp; positions are occupied by&nbsp; $\pm1$.&nbsp;
-In the diagram, the&nbsp; $35$&nbsp; bits describing an FCB pulse are given as an example.
+*In each track there are exactly two values&nbsp; $\pm1$,&nbsp; while all the other six values are&nbsp; zero.&nbsp;
-'''Track 1''' includes.
+&rArr; &nbsp; '''Track 1'''&nbsp; includes
-*a positive pulse&nbsp; $({\rm VZ} = 1)$&nbsp; at&nbsp; $1$&nbsp; (first possible position for track 1)&nbsp; $\hspace{0.2cm}\text{plus}\hspace{0.2cm}0$ (bit specification for " 000") $= 1$,
+#a positive pulse&nbsp; $({\rm sign} = 1)$&nbsp; at position&nbsp; $\big [1$&nbsp; (first possible position for track 1)&nbsp; $\hspace{0.02cm}\text{plus}\hspace{0.2cm}0$ (bit specification for&nbsp; "000") $= 1\big]$,
-*another positive pulse (since $110 > 000$) at position $1 \hspace{0.2cm}\text{plus}\hspace{0.2cm}5$ (pulse spacing in each track) $\hspace{0.2cm}\text{mal}\hspace{0.2cm}6$ (bit specification for " 110") = $31\hspace{0.05cm}.$
+#another positive pulse&nbsp; $($since $110 > 000)$&nbsp; at position&nbsp; $\big [1 \hspace{0.2cm}\text{plus}\hspace{0.2cm}5$ (pulse spacing in each track) $\hspace{0.2cm}\text{mal}\hspace{0.2cm}6$ (bit specification for " 110") = $31\hspace{0.05cm}.$

← Older revision		Revision as of 15:39, 23 January 2023
Line 110:		Line 110:


−	[[Category:Examples of Communication Systems: Exercises\|^3.3 ~~Voice~~ Coding^]]	+	[[Category:Examples of Communication Systems: Exercises\|^3.3 Speech Coding^]]

@@ Line 96: / Line 96: @@
 *Because $001 < 011$, the second pulse has the same sign.
 *The two amounts result in
-:$$|N_1| \ = \ 3 \hspace{0.1cm}{\rm(da \hspace{0.1cm} track \hspace{0.1cm}3)} + 5\cdot 1 \hspace{0.1cm} {\rm(bit indication \hspace{0.1cm} 001)} = 8\hspace{0.05cm}, $$
+:$$|N_1| \ = \ 3 \hspace{0.1cm}{\rm(da \hspace{0.1cm} track \hspace{0.1cm}3)} + 5\cdot 1 \hspace{0.1cm} {\rm(bit\:specification \hspace{0.1cm} 001)} = 8\hspace{0.05cm}, $$
-:$$ |N_2| \ = \ 3 \hspace{0.1cm}{\rm(da \hspace{0.1cm} track \hspace{0.1cm}3)} + 5\cdot 3 \hspace{0.1cm} {\rm(bit specification \hspace{0.1cm} 011)} = 18\hspace{0.05cm}.$$
+:$$ |N_2| \ = \ 3 \hspace{0.1cm}{\rm(da \hspace{0.1cm} track \hspace{0.1cm}3)} + 5\cdot 3 \hspace{0.1cm} {\rm(bit\:specification \hspace{0.1cm} 011)} = 18\hspace{0.05cm}.$$
 *Therefore, to be entered for the third track is $N_{1} \underline{ = -8}$ and $N_{2} \underline{ = -18}.$

@@ Line 51: / Line 51: @@
 <quiz display=simple>
-{Wie viele Bit beschreiben einen Sprachrahmen $($der Dauer&nbsp; $20 \ \rm ms)$&nbsp; im&nbsp; $12.2 \ \rm kbit/s$–Modus?
+{How many bits describe a speech frame $($of duration&nbsp; $20 \ \rm ms)$&nbsp; in&nbsp; $12.2 \ \rm kbit/s$ mode?
 |type="{}"}
 $N_{12.2} \ = \ $ { 244 3% } $ \ \rm Bit$
-{Wie viele Bit werden für FCB–Puls und –Verstärkung pro Rahmen benötigt?
+{How many bits are needed for FCB pulse and gain per frame?
 |type="{}"}
 $N_{\rm FCB} \ = \ $ { 160 3% } $ \ \rm Bit$
-{Wie viele Bit verbleiben somit für LPC und LTP?
+{ How many bits are left for LPC and LTP?
 |type="{}"}
 $N_{\rm LPC/LTP} \ = \ $ { 84 3% } $ \ \rm Bit$
-{Welche Impulspositionen des Unterrahmens und Vorzeichen beschreibt die Spur&nbsp; $3$? <br>Beachten Sie die Hinweise zur Eingabe auf der Angabenseite.
+{What subframe pulse positions and signs does track&nbsp; $3$ describe? <br>Follow the instructions for input on the input page.
 |type="{}"}
 $N_{1} \ = \ $ { -8.24--7.76 }
 $N_{2} \ = \ $ { -18.54--17.46 }
-{Welche Impulspositionen inklusive Vorzeichen beschreiben die Spur&nbsp; $4$?
+{What pulse positions including sign describe the track&nbsp; $4$?
 |type="{}"}
 $N_{1} \ = \ $ { 39 3% }
 $N_{2} \ = \ $ { -14.42--13.58 }
-{Welche Impulspositionen inklusive Vorzeichen beschreiben die Spur&nbsp; $5$?
+{What pulse positions including sign describe the track&nbsp; $5$?
 |type="{}"}
 $N_{1} \ = \ $ { -30.9--29.1 }
@@ Line 79: / Line 79: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Mit der Datenrate $12.2 \ \rm kbit/s$ ergeben sich innerhalb von $20 \ \rm ms$ genau $\underline{244 \ \rm Bit}$, während zum Beispiel im  $4.75 \ \rm kbit/s$–Modus nur $95 \ \rm Bit$ übertragen werden.
+'''(1)'''&nbsp; With the data rate $12.2 \ \rm kbit/s$, exactly $\underline{244 \ \rm bit}$ results within $20 \ \rm ms$, while for example in $4.75 \ \rm kbit/s$ mode only $95 \ \rm bit$ is transmitted.
-'''(2)'''&nbsp;  In jedem Unterrahmen benötigt der FCB–Puls $35 \ \rm Bit$ (fünf Spuren zu je sieben Bit) und die FCB–Verstärkung fünf Bit.
+'''(2)'''&nbsp; In each subframe, the FCB pulse requires $35 \ \rm bit$ (five tracks of seven bits each) and the FCB gain requires five bits.
-*Bei vier Unterrahmen kommt man so auf $N_{\rm FCB} \underline{= 160 \ \rm Bit}$.
+*With four subframes, this gives $N_{\rm FCB} \underline{= 160 \ \rm bits}$.
-'''(3)'''&nbsp; Hierfür verbleiben die Differenz aus (1) und (2), also $N_{\rm LPC/LTP}\underline{ = 84 \ \rm Bit}$.
+'''(3)'''&nbsp; This leaves the difference from (1) and (2), i.e. $N_{\rm LPC/LTP}\underline{ = 84 \rm bits}$.
-'''(4)'''&nbsp;  Das Vorzeichenbit "$0$" deutet auf einen negativen ersten Impuls hin.
+'''(4)'''&nbsp; The sign bit "$0$" indicates a negative first pulse.
-*Wegen $001 < 011$ hat der zweite Impuls das gleiche Vorzeichen.
+*Because $001 < 011$, the second pulse has the same sign.
-*Die beiden Beträge ergeben sich zu
+*The two amounts result in
-:$$|N_1| \ = \ 3 \hspace{0.1cm}{\rm(da \hspace{0.1cm} Spur \hspace{0.1cm}3)} + 5\cdot 1 \hspace{0.1cm} {\rm(Bitangabe \hspace{0.1cm} 001)} = 8\hspace{0.05cm}, $$
+:$$|N_1| \ = \ 3 \hspace{0.1cm}{\rm(da \hspace{0.1cm} track \hspace{0.1cm}3)} + 5\cdot 1 \hspace{0.1cm} {\rm(bit indication \hspace{0.1cm} 001)} = 8\hspace{0.05cm}, $$
-:$$ |N_2| \ = \ 3 \hspace{0.1cm}{\rm(da \hspace{0.1cm} Spur \hspace{0.1cm}3)} + 5\cdot 3 \hspace{0.1cm} {\rm(Bitangabe \hspace{0.1cm} 011)} = 18\hspace{0.05cm}.$$
+:$$ |N_2| \ = \ 3 \hspace{0.1cm}{\rm(da \hspace{0.1cm} track \hspace{0.1cm}3)} + 5\cdot 3 \hspace{0.1cm} {\rm(bit specification \hspace{0.1cm} 011)} = 18\hspace{0.05cm}.$$
-*Einzugeben sind deshalb für die dritte Spur  $N_{1} \underline{ = -8}$ und $N_{2} \underline{ = -18}.$
+*Therefore, to be entered for the third track is $N_{1} \underline{ = -8}$ and $N_{2} \underline{ = -18}.$
-'''(5)'''&nbsp;  In analoger Weise erhält man für die Spur $4$ die Werte&nbsp; $N_{1} \underline{ = +39}$&nbsp; und&nbsp; $N_{2} \underline{ = -14}$.
+'''(5)'''&nbsp; In an analogous way, for track $4$ we obtain the values&nbsp; $N_{1} \underline{ = +39}$&nbsp; and&nbsp; $N_{2} \underline{ = -14}$.
-'''(6)'''&nbsp; Die fünfte Spur liefert&nbsp; $N_{1} \underline{ =-30}$&nbsp; und&nbsp; $N_{2} \underline{ = +5}$
+'''(6)'''&nbsp; The fifth track provides&nbsp; $N_{1} \underline{ =-30}$&nbsp; and&nbsp; $N_{2} \underline{ = +5}$
 {{ML-Fuß}}