Aufgaben:Exercise 4.1: PCM System 30/32 - Revision history

Guenter at 15:23, 7 April 2022

2022-04-07T15:23:45Z

Guenter at 14:57, 7 April 2022

2022-04-07T14:57:39Z

Guenter at 14:47, 7 April 2022

2022-04-07T14:47:27Z

Noah at 20:42, 4 April 2022

2022-04-04T20:42:27Z

Noah at 15:50, 19 March 2022

2022-03-19T15:50:38Z

Guenter at 13:49, 16 March 2022

2022-03-16T13:49:21Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.1: PCM–System 30/32 to Aufgaben:Exercise 4.1: PCM System 30/32

2022-03-16T13:40:26Z

Guenter moved page Aufgabe 4.1: PCM–System 30/32 to Exercise 4.1: PCM System 30/32

Guenter at 16:16, 9 November 2021

2021-11-09T16:16:20Z

Javier: Text replacement - "”" to """

2021-05-28T14:37:39Z

Text replacement - "”" to """

Javier: Text replacement - "„" to """

2021-05-28T14:31:59Z

Text replacement - "„" to """

@@ Line 30: / Line 30: @@
-{How is the sample value&nbsp; $-0.182$&nbsp; represented? With
+{How is the sample value&nbsp; "$-0.182$"&nbsp; represented? With
 |type="()"}
 - the bit sequence 1,
@@ Line 69: / Line 69: @@
 *With the value range&nbsp; $±1$&nbsp; each quantization interval has width&nbsp; ${\it Δ} = 1/128$.
 *The index&nbsp; $μ = 183$&nbsp; thus represents the interval from&nbsp; $183/128 - 1 = 0.4297$&nbsp; to&nbsp; $184/128 - 1 = 0.4375$.
-* $μ = 104$&nbsp; denotes the interval from&nbsp; $-0.1875$&nbsp; to&nbsp; $-0.1797$.
+* $μ = 104$&nbsp; denotes the interval from&nbsp; "$-0.1875$"&nbsp; to&nbsp; $-0.1797$.
 *The sample $-0.182$ is thus represented by&nbsp; <u>bit sequence 2</u>.

@@ Line 63: / Line 63: @@
-'''(2)'''&nbsp; Numbering the quantization intervals from&nbsp; $0$&nbsp; to&nbsp; $255$, the "bit sequence 1" represents.
+'''(2)'''&nbsp; Numbering the quantization intervals from&nbsp; $0$&nbsp; to&nbsp; $255$,&nbsp; the&nbsp; "bit sequence 1"&nbsp; represents
 :$$ \mu_1 = 2^7 + 2^5 +2^4 +2^2 +2^1 +2^0 = 255 -2^6 -2^3 = 183\hspace{0.05cm},$$
-and the "bit sequence 2" for
+and the&nbsp; "bit sequence 2"&nbsp; represents
 :$$\mu_2 = 2^6 + 2^5 +2^3 = 104\hspace{0.05cm}.$$
 *With the value range&nbsp; $±1$&nbsp; each quantization interval has width&nbsp; ${\it Δ} = 1/128$.
 *The index&nbsp; $μ = 183$&nbsp; thus represents the interval from&nbsp; $183/128 - 1 = 0.4297$&nbsp; to&nbsp; $184/128 - 1 = 0.4375$.
 * $μ = 104$&nbsp; denotes the interval from&nbsp; $-0.1875$&nbsp; to&nbsp; $-0.1797$.
-*The sample $-0.182$ is thus represented by the <u>bit sequence 2</u>.
+*The sample $-0.182$ is thus represented by&nbsp; <u>bit sequence 2</u>.
@@ Line 77: / Line 77: @@
-'''(4)'''&nbsp; During duration&nbsp; $T_{\rm A}$&nbsp; binary symbols are transmitted&nbsp; $Z \cdot N$&nbsp;:
+'''(4)'''&nbsp; During duration&nbsp; $T_{\rm A}$ &nbsp; &rArr;  &nbsp; $Z \cdot N$&nbsp; binary symbols are transmitted:
 :$$T_{\rm A} = Z \cdot N \cdot {T_{\rm B} } = 32 \cdot 8 \cdot 0.488\,{\rm &micro; s} \hspace{0.15cm}\underline {= 125\,{\rm &micro; s}} \hspace{0.05cm}.$$
@@ Line 85: / Line 85: @@
-'''(6)'''&nbsp; The sampling theorem would already be given by&nbsp; $f_{\rm A} ≥ 2 \cdot f_\text{N, max} = 6.8 \ \rm kHz$&nbsp; Thus the <u>last proposed solution</u> is correct.
+'''(6)'''&nbsp; The sampling theorem would already be given by&nbsp; $f_{\rm A} ≥ 2 \cdot f_\text{N, max} = 6.8 \ \rm kHz$.&nbsp; Thus the <u>last proposed solution</u> is correct.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Mod_A_4_1.png|right|frame|binary display with dual code]]
+[[File:EN_Mod_A_4_1.png|right|frame|Binary display with dual code]]
-For many years, the PCM system 30/32 was used in Germany, which has the following specifications:
+For many years,&nbsp; the&nbsp; '''PCM system 30/32'''&nbsp; was used in Germany,&nbsp; which has the following specifications:
-* It allows digital transmission of 30 voice channels in time division multiplex together with one each of synchronization and dial character channels &nbsp; ⇒ &nbsp; the total number of channels is&nbsp; $Z = 32$.
+* It allows digital transmission of 30 voice channels in time division multiplex together with one channel each of synchronization and dial character &nbsp; ⇒ &nbsp; the total number of channels is&nbsp; $Z = 32$.
-* Each individual voice channel is bandlimited to the frequency range of&nbsp; $300 \ \rm Hz$&nbsp; to&nbsp; $3400 \ \rm Hz$&nbsp; .
+* Each individual voice channel is bandlimited to the frequency range of&nbsp; $300 \ \rm Hz$&nbsp; to&nbsp; $3400 \ \rm Hz$.
-* Each individual sample is represented by&nbsp; $N = 8$&nbsp; bits, assuming the so-called dual code.
+* Each individual sample is represented by&nbsp; $N = 8$&nbsp; bits,&nbsp; assuming the so-called&nbsp; "dual code".
-* The total bit rate is&nbsp; $R_{\rm B} = 2.048 \rm Mbit/s$.
+* The total bit rate is&nbsp; $R_{\rm B} = 2.048\ \rm Mbit/s$.
 The graph shows the binary representation of two arbitrarily selected samples.
@@ Line 24: / Line 18: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Pulse_Code_Modulation|Pulse Code Modulation]].
 *Reference is made in particular to the page&nbsp; [[Modulation_Methods/Pulse_Code_Modulation#PCM_encoding_and_decoding|PCM encoding and decoding]].
+*For the solution of subtask&nbsp; '''(2)'''&nbsp; it is to be assumed:&nbsp; All speech signals are normalized and limited to the amplitude range&nbsp; $±1$.
@@ Line 51: / Line 44: @@
 $T_{\rm A} \ = \ $ { 125 3% } $\ \rm &micro; s$
-{What is the sampling rate&nbsp; $f_{\rm A}$?
+{What is the sampling rate $f_{\rm A}$?
 |type="{}"}
 $f_{\rm A} \ = \ $ { 8 3% } $\ \rm kHz$

@@ Line 84: / Line 84: @@
-'''(4)'''&nbsp; During duration&nbsp; $T_{\rm A}$&nbsp; binary symbols are transmitted&nbsp; $Z - N$&nbsp;:
+'''(4)'''&nbsp; During duration&nbsp; $T_{\rm A}$&nbsp; binary symbols are transmitted&nbsp; $Z \cdot N$&nbsp;:
 :$$T_{\rm A} = Z \cdot N \cdot {T_{\rm B} } = 32 \cdot 8 \cdot 0.488\,{\rm &micro; s} \hspace{0.15cm}\underline {= 125\,{\rm &micro; s}} \hspace{0.05cm}.$$
@@ Line 92: / Line 92: @@
-'''(6)'''&nbsp; The sampling theorem would already be given by&nbsp; $f_{\rm A} ≥ 2 - f_\text{N, max} = 6.8 \ \rm kHz$&nbsp; Thus the <u>last proposed solution</u> is correct.
+'''(6)'''&nbsp; The sampling theorem would already be given by&nbsp; $f_{\rm A} ≥ 2 \cdot f_\text{N, max} = 6.8 \ \rm kHz$&nbsp; Thus the <u>last proposed solution</u> is correct.
 {{ML-Fuß}}

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Modulationsverfahren/Pulscodemodulation
+{{quiz-Header|Buchseite=Modulation_Methods/Pulse_Code_Modulation
 }}
-[[File:EN_Mod_A_4_1.png|right|frame|Binärdarstellung mit Dualcode]]
+[[File:EN_Mod_A_4_1.png|right|frame|binary display with dual code]]
-Über viele Jahre wurde in Deutschland das PCM–System 30/32 eingesetzt, das folgende Spezifikationen aufweist:
+For many years, the PCM system 30/32 was used in Germany, which has the following specifications:
-* Es erlaubt die digitale Übertragung von 30 Sprachkanälen im Zeitmultiplex zusammen mit je einem Sychronisations– und Wählzeichenkanal &nbsp; ⇒ &nbsp; die Gesamtkanalzahl ist&nbsp; $Z = 32$.
+* It allows digital transmission of 30 voice channels in time division multiplex together with one each of synchronization and dial character channels &nbsp; ⇒ &nbsp; the total number of channels is&nbsp; $Z = 32$.
-* Jeder einzelne Sprachkanal ist auf den Frequenzbereich von&nbsp; $300 \ \rm Hz$&nbsp; bis&nbsp; $3400 \ \rm  Hz$&nbsp; bandbegrenzt.
+* Each individual voice channel is bandlimited to the frequency range of&nbsp; $300 \ \rm Hz$&nbsp; to&nbsp; $3400 \ \rm Hz$&nbsp; .
-* Jeder einzelne Abtastwert wird durch&nbsp; $N = 8$&nbsp; Bit dargestellt, wobei vom so genannten Dualcode ausgegangen wird.
+* Each individual sample is represented by&nbsp; $N = 8$&nbsp; bits, assuming the so-called dual code.
-* Die Gesamtbitrate beträgt&nbsp; $R_{\rm B} = 2.048 \ \rm  Mbit/s$.
+* The total bit rate is&nbsp; $R_{\rm B} = 2.048 \rm Mbit/s$.
-Die Grafik zeigt die Binärdarstellung zweier willkürlich ausgewählter Abtastwerte.
+The graph shows the binary representation of two arbitrarily selected samples.
@@ Line 22: / Line 22: @@
-''Hinweise:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Modulation_Methods/Pulscodemodulation|Pulscodemodulation]].
+*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Pulse_Code_Modulation|Pulse Code Modulation]].
-*Bezug genommen wird insbesondere auf die Seite&nbsp;  [[Modulation_Methods/Pulscodemodulation#PCM.E2.80.93Codierung_und_.E2.80.93Decodierung|PCM-Codierung und -Decodierung]].
+*Reference is made in particular to the page&nbsp; [[Modulation_Methods/Pulse_Code_Modulation#PCM_encoding_and_decoding|PCM encoding and decoding]].
-*Für die Lösung der Teilaufgabe&nbsp; '''(2)'''&nbsp; ist vorauszusetzen:&nbsp; Alle Sprachsignale sind normiert und auf den Bereich&nbsp; $±1$&nbsp; amplitudenbegrenzt.
+*For the solution of the subtask&nbsp; '''(2)'''&nbsp; it is to be assumed:&nbsp; All speech signals are normalized and limited to the range&nbsp; $±1$&nbsp; amplitude.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wie groß ist die Quantisierungsstufenzahl&nbsp; $M$?
+{What is the quantization step number&nbsp; $M$?
 |type="{}"}
 $M \ = \ $ { 256 }
-{Wie wird der Abtastwert&nbsp; $-0.182$&nbsp; dargestellt? Mit
+{How is the sample value&nbsp; $-0.182$&nbsp; represented? With
 |type="()"}
-- der Bitfolge 1,
+- the bit sequence 1,
-+ der Bitfolge 2,
++ the bit sequence 2,
-- keiner von beiden.
+- neither of them.
-{Wie groß ist die Bitdauer&nbsp; $T_{\rm B}$?
+{What is the bit duration&nbsp; $T_{\rm B}$?
 |type="{}"}
 $T_{\rm B} \ = \ $ { 0.488 3% } $\ \rm &micro; s$
-{In welchem Abstand&nbsp; $T_{\rm A}$&nbsp; werden die Sprachsignale abgetastet?
+{At what distance&nbsp; $T_{\rm A}$&nbsp; are the speech signals sampled?
 |type="{}"}
 $T_{\rm A} \ = \ $ { 125 3% } $\ \rm &micro; s$
-{Wie groß ist die Abtastrate&nbsp; $f_{\rm A}$?
+{What is the sampling rate&nbsp; $f_{\rm A}$?
 |type="{}"}
 $f_{\rm A} \ = \ $ { 8 3% } $\ \rm kHz$
-{Welche der folgenden Aussagen ist richtig?
+{Which of the following statements is correct?
 |type="()"}
-- Das Abtasttheorem wird nicht erfüllt.
+- The sampling theorem is not satisfied.
-- Das Abtasttheorem wird gerade noch erfüllt.
+- The sampling theorem is just fulfilled.
-+ Die Abtastfrequenz ist größer als der kleinstmögliche Wert.
++ The sampling frequency is greater than the smallest possible value.
@@ Line 65: / Line 65: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Mit&nbsp; $N = 8$&nbsp; Bit können insgesamt&nbsp; $2^8$&nbsp; Quantisierungsintervalle dargestellt werden &nbsp; ⇒ &nbsp; $\underline{M = 256}$.
+'''(1)'''&nbsp; With&nbsp; $N = 8$&nbsp; bits a total of&nbsp; $2^8$&nbsp; quantization intervals can be represented &nbsp; ⇒ &nbsp; $\underline{M = 256}$.
-'''(2)'''&nbsp;  Nummeriert man die Quantisierungsintervalle von&nbsp; $0$&nbsp; bis&nbsp; $255$, so steht die "Bitfolge 1" für
+'''(2)'''&nbsp; Numbering the quantization intervals from&nbsp; $0$&nbsp; to&nbsp; $255$, the "bit sequence 1" represents.
 :$$ \mu_1 = 2^7 + 2^5 +2^4 +2^2 +2^1 +2^0 = 255 -2^6 -2^3 = 183\hspace{0.05cm},$$
-und die "Bitfolge 2" für
+and the "bit sequence 2" for
 :$$\mu_2 = 2^6 + 2^5 +2^3 = 104\hspace{0.05cm}.$$
-*Mit dem Wertebereich&nbsp; $±1$&nbsp; hat jedes Quantisierungsintervall die Breite&nbsp; ${\it Δ} = 1/128$.
+*With the value range&nbsp; $±1$&nbsp; each quantization interval has width&nbsp; ${\it Δ} = 1/128$.
-*Der Index&nbsp; $μ = 183$&nbsp; steht somit für das Intervall von&nbsp; $183/128 - 1 = 0.4297$&nbsp; bis&nbsp; $184/128 - 1 = 0.4375$.
+*The index&nbsp; $μ = 183$&nbsp; thus represents the interval from&nbsp; $183/128 - 1 = 0.4297$&nbsp; to&nbsp; $184/128 - 1 = 0.4375$.
-* $μ = 104$&nbsp;  kennzeichnet das Intervall von&nbsp; $-0.1875$&nbsp; bis&nbsp; $-0.1797$.
+* $μ = 104$&nbsp; denotes the interval from&nbsp; $-0.1875$&nbsp; to&nbsp; $-0.1797$.
-*Der Abtastwert $–0.182$ wird somit durch die <u>Bitfolge 2</u> dargestellt.
+*The sample $-0.182$ is thus represented by the <u>bit sequence 2</u>.
-'''(3)'''&nbsp;  Die Bitdauer&nbsp; $T_{\rm B}$&nbsp; ist der Kehrwert der Bitrate&nbsp; $R_{\rm B}$:
+'''(3)'''&nbsp; The bit duration&nbsp; $T_{\rm B}$&nbsp; is the reciprocal of the bit rate&nbsp; $R_{\rm B}$:
 :$$T_{\rm B} = \frac{1}{R_{\rm B} }= \frac{1}{2.048 \cdot 10^6\,{\rm 1/s} } \hspace{0.15cm}\underline {= 0.488\,{\rm &micro; s}} \hspace{0.05cm}.$$
-'''(4)'''&nbsp;  Während der Zeitdauer&nbsp; $T_{\rm A}$&nbsp; werden&nbsp; $Z · N$&nbsp; Binärsymbole übertragen:
+'''(4)'''&nbsp; During duration&nbsp; $T_{\rm A}$&nbsp; binary symbols are transmitted&nbsp; $Z - N$&nbsp;:
 :$$T_{\rm A} = Z \cdot N \cdot {T_{\rm B} } = 32 \cdot 8 \cdot 0.488\,{\rm &micro; s} \hspace{0.15cm}\underline {= 125\,{\rm &micro; s}} \hspace{0.05cm}.$$
-'''(5)'''&nbsp;  Den Kehrwert von&nbsp; $T_{\rm A}$&nbsp; bezeichnet man als die Abtastrate:
+'''(5)'''&nbsp; The reciprocal of&nbsp; $T_{\rm A}$&nbsp; is called the sampling rate:
 :$$f_{\rm A} = \frac{1}{T_{\rm A} } \hspace{0.15cm}\underline {= 8\,{\rm kHz}} \hspace{0.05cm}.$$
-'''(6)'''&nbsp;  Das Abtasttheorem wäre bereits mit&nbsp; $f_{\rm A} ≥ 2 · f_\text{N, max} = 6.8 \ \rm kHz$&nbsp; erfüllt.&nbsp; Richtig ist somit der <u>letzte Lösungsvorschlag</u>.
+'''(6)'''&nbsp; The sampling theorem would already be given by&nbsp; $f_{\rm A} ≥ 2 - f_\text{N, max} = 6.8 \ \rm kHz$&nbsp; Thus the <u>last proposed solution</u> is correct.
 {{ML-Fuß}}

← Older revision		Revision as of 13:49, 16 March 2022
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−	[[Category:Modulation Methods: Exercises\|^4.1 ~~Pulscodemodulation~~^]]	+	[[Category:Modulation Methods: Exercises\|^4.1 Pulse Code Modulation^]]

@@ Line 70: / Line 70: @@
-'''(2)'''&nbsp;  Nummeriert man die Quantisierungsintervalle von&nbsp; $0$&nbsp; bis&nbsp; $255$, so steht die &bdquo;Bitfolge 1&rdquo; für
+'''(2)'''&nbsp;  Nummeriert man die Quantisierungsintervalle von&nbsp; $0$&nbsp; bis&nbsp; $255$, so steht die "Bitfolge 1&rdquo; für
 :$$ \mu_1 = 2^7 + 2^5 +2^4 +2^2 +2^1 +2^0 = 255 -2^6 -2^3 = 183\hspace{0.05cm},$$
-und die &bdquo;Bitfolge 2&rdquo; für
+und die "Bitfolge 2&rdquo; für
 :$$\mu_2 = 2^6 + 2^5 +2^3 = 104\hspace{0.05cm}.$$
 *Mit dem Wertebereich&nbsp; $±1$&nbsp; hat jedes Quantisierungsintervall die Breite&nbsp; ${\it Δ} = 1/128$.