Aufgaben:Exercise 1.2: ISDN and PCM - Revision history

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	[[Category:Examples of Communication Systems: Exercises\|^1.1 General Description of ISDN^]]		[[Category:Examples of Communication Systems: Exercises\|^1.1 General Description of ISDN^]]
	[[de:Aufgaben:1.~~2_ISDN_und_PCM~~]]		[[de:Aufgaben:Aufgabe 1.2: ISDN und PCM]]

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	[[Category:Examples of Communication Systems: Exercises\|^1.1 General Description of ISDN^]]		[[Category:Examples of Communication Systems: Exercises\|^1.1 General Description of ISDN^]]
			[[de:Aufgaben:1.2_ISDN_und_PCM]]

Hwang at 15:36, 23 January 2023

2023-01-23T15:36:14Z

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	[[File:EN_Bei_A_1_2.png\|right\|frame\|Components of PCM transmitter]]		[[File:EN_Bei_A_1_2.png\|right\|frame\|Components of PCM transmitter]]
	The conversion of the analog ~~voice~~ signal  $q(t)$  into the binary signal  $q_{\rm C}(t)$  is done at  $\rm ISDN$  ("Integrated Services Digital Network")  according to the guidelines of  "pulse code modulation"  $\rm (PCM)$  by		The conversion of the analog speech signal  $q(t)$  into the binary signal  $q_{\rm C}(t)$  is done at  $\rm ISDN$  ("Integrated Services Digital Network")  according to the guidelines of  "pulse code modulation"  $\rm (PCM)$  by
	*sampling in the interval  $T_{\rm A} = 1/f_{\rm A}$,		*sampling in the interval  $T_{\rm A} = 1/f_{\rm A}$,

Hwang at 14:35, 9 November 2022

2022-11-09T14:35:35Z

← Older revision		Revision as of 16:35, 9 November 2022
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	:$$\rho_{v} \approx \rho_{\rm Q} = 2^{2M} = 2^{16} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v} \approx 48\, {\rm dB}\hspace{0.05cm}.$$		:$$\rho_{v} \approx \rho_{\rm Q} = 2^{2M} = 2^{16} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v} \approx 48\, {\rm dB}\hspace{0.05cm}.$$

	*With larger noise interference,  $\rho_{v}$  can further  (significantly)  reduced by the transmission errors.		*With larger noise interference,  $\rho_{v}$  can further  (significantly)  be reduced by the transmission errors.

	*In contrast,  sampling results in no loss of quality if the sampling theorem is obeyed.		*In contrast,  sampling results in no loss of quality if the sampling theorem is obeyed.

Guenter at 11:23, 18 October 2022

2022-10-18T11:23:22Z

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	}}		}}

	[[File:EN_Bei_A_1_2.png\|right\|frame\|Components of PCM transmitter ~~'''Korrektur'''~~]]		[[File:EN_Bei_A_1_2.png\|right\|frame\|Components of PCM transmitter]]
	The conversion of the analog voice signal  $q(t)$  into the binary signal  $q_{\rm C}(t)$  is done at  $\rm ISDN$  ("Integrated Services Digital Network")  according to the guidelines of  "pulse code modulation"  $\rm (PCM)$  by		The conversion of the analog voice signal  $q(t)$  into the binary signal  $q_{\rm C}(t)$  is done at  $\rm ISDN$  ("Integrated Services Digital Network")  according to the guidelines of  "pulse code modulation"  $\rm (PCM)$  by
	*sampling in the interval  $T_{\rm A} = 1/f_{\rm A}$,		*sampling in the interval  $T_{\rm A} = 1/f_{\rm A}$,

Guenter at 16:17, 17 October 2022

2022-10-17T16:17:53Z

← Older revision		Revision as of 18:17, 17 October 2022
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	}}		}}

	[[File:~~P_ID1578__Bei_A_1_2~~.png\|right\|frame\|Components of PCM transmitter '''Korrektur''']]		[[File:EN_Bei_A_1_2.png\|right\|frame\|Components of PCM transmitter '''Korrektur''']]
	The conversion of the analog voice signal  $q(t)$  into the binary signal  $q_{\rm C}(t)$  is done at  $\rm ISDN$  ("Integrated Services Digital Network")  according to the guidelines of  "pulse code modulation"  $\rm (PCM)$  by		The conversion of the analog voice signal  $q(t)$  into the binary signal  $q_{\rm C}(t)$  is done at  $\rm ISDN$  ("Integrated Services Digital Network")  according to the guidelines of  "pulse code modulation"  $\rm (PCM)$  by
	*sampling in the interval  $T_{\rm A} = 1/f_{\rm A}$,		*sampling in the interval  $T_{\rm A} = 1/f_{\rm A}$,

Guenter at 13:38, 15 October 2022

2022-10-15T13:38:00Z

← Older revision		Revision as of 15:38, 15 October 2022
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	'''(2)'''  For the bit rate,  $R_{\rm B} = N \cdot f_{\rm A}$.		'''(2)'''  For the bit rate,  $R_{\rm B} = N \cdot f_{\rm A}$.
	*Thus, from  $R_{\rm B} = 64 \ \rm kbit/s$  and  $N = 8$,  we get  $f_{\rm A} \hspace{0.15cm}\underline{= 8 \ \rm kHz}$.		*Thus,  from  $R_{\rm B} = 64 \ \rm kbit/s$  and  $N = 8$,  we get  $f_{\rm A} \hspace{0.15cm}\underline{= 8 \ \rm kHz}$.



	'''(3)'''  Due to the bandwidth limitation, the highest frequency contained in the signal  $q(t)$  is equal to  $3.4 \ \rm kHz$.		'''(3)'''  Due to the bandwidth limitation,  the highest frequency contained in the signal  $q(t)$  is equal to  $3.4 \ \rm kHz$.
	*Therefore, according to the sampling theorem,  $f_{\rm A} ≥ 6.8 \ \rm kHz$  should hold.
			*Therefore,  according to the sampling theorem,  $f_{\rm A} ≥ 6.8 \ \rm kHz$  should hold.

	*With  $f_{\rm A} = 8 \ \rm kHz$  the condition is fulfilled   ⇒   $\underline {\rm YES}$.		*With  $f_{\rm A} = 8 \ \rm kHz$  the condition is fulfilled   ⇒   $\underline {\rm YES}$.


	'''(4)'''  The <u>last statement</u> is correct:		'''(4)'''  The  <u>last statement</u>  is correct:
	*Even if the influence of the AWGN noise is small $($small noise power density  $N_{0})$, the sink SNR  $\rho_{v}$  cannot fall below a limit given by the quantization noise:		*Even if the influence of the AWGN noise is small  $($small noise power density  $N_{0})$,  the sink SNR  $\rho_{v}$  cannot fall below a limit given by the quantization noise:
	:$$\rho_{v} \approx \rho_{\rm Q} = 2^{2M} = 2^{16} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v} \approx 48\, {\rm dB}\hspace{0.05cm}.$$		:$$\rho_{v} \approx \rho_{\rm Q} = 2^{2M} = 2^{16} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v} \approx 48\, {\rm dB}\hspace{0.05cm}.$$

	*With larger noise interference,  $\rho_{v}$  is further (significantly) reduced by the transmission errors ~~then present~~.		*With larger noise interference,  $\rho_{v}$  can further  (significantly)  reduced by the transmission errors.
	*In contrast, sampling results in no loss of quality if the sampling theorem is obeyed.
	*Sampling can then be completely undone if the source signal  $q(t)$  is bandlimited and the signal reconstruction is correctly dimensioned:  ideal low-pass.		*In contrast,  sampling results in no loss of quality if the sampling theorem is obeyed.

			*Sampling can then be completely undone if the source signal  $q(t)$  is bandlimited and the signal reconstruction is correctly dimensioned   ⇒   ideal low-pass.

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 13:22, 15 October 2022

2022-10-15T13:22:30Z

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	[[File:P_ID1578__Bei_A_1_2.png\|right\|frame\|Components of PCM transmitter '''Korrektur''']]		[[File:P_ID1578__Bei_A_1_2.png\|right\|frame\|Components of PCM transmitter '''Korrektur''']]
	The conversion of the analog voice signal  $q(t)$  into the binary signal  $q_{\rm C}(t)$  is done at ISDN  ("Integrated Services Digital Network")  according to the guidelines of pulse code modulation  $\rm (PCM)$  by		The conversion of the analog voice signal  $q(t)$  into the binary signal  $q_{\rm C}(t)$  is done at  $\rm ISDN$  ("Integrated Services Digital Network")  according to the guidelines of  "pulse code modulation"  $\rm (PCM)$  by
	*sampling in the interval  $T_{\rm A} = 1/f_{\rm A}$,		*sampling in the interval  $T_{\rm A} = 1/f_{\rm A}$,

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	The net data rate of a ~~so-called~~  $\rm B$ channel  ("Bearer Channel") is  $64 \ \rm kbit/s$  and corresponds to the bit rate of the redundancy-free binary signal  $q_{\rm C}(t)$.  However,  because of the subsequent redundant channel coding and the inserted signaling bits,  the gross data rate – i.e., the transmission rate of the transmitted signal  $s(t)$  – is greater.		The net data rate of a  $\rm B$ channel  ("Bearer Channel") is  $64 \ \rm kbit/s$  and corresponds to the bit rate of the redundancy-free binary signal  $q_{\rm C}(t)$.

			However,  because of the subsequent redundant channel coding and the inserted signaling bits,  the gross data rate – i.e.,  the transmission rate of the transmitted signal  $s(t)$  – is greater.

	A measure for the quality of the entire ISDN transmission system is the sink SNR		A measure for the quality of the entire ISDN transmission system is the sink SNR
	:$$\rho_{v} = \frac{P_q}{P_{\varepsilon}} = \frac{\overline{q(t)^2}}{\overline{[\upsilon(t) - q(t)]^2}}$$		:$$\rho_{v} = \frac{P_q}{P_{\varepsilon}} = \frac{\overline{q(t)^2}}{\overline{[\upsilon(t) - q(t)]^2}}$$

	as the ratio of the powers of the analog signal  $q(t)$  bandlimited to the range  $300 \ {\rm Hz}\ \text{...}\ 3400 \ {\rm Hz}$  ~~and the error signal  $\varepsilon (t) = v (t) - q(t)$.~~		as the ratio of the powers
			*of the analog signal  $q(t)$  bandlimited to the range  $300 \ {\rm Hz}\ \text{...}\ 3400 \ {\rm Hz}$

	~~An ideal signal reconstruction with an ideal rectangular low-pass filter is assumed here for~~ the ~~sink~~ signal  $v (t)$.~~ ~~		*and the error signal  $\varepsilon (t) = v (t) - q(t)$.


			An ideal signal reconstruction with an ideal rectangular low-pass filter is assumed here for the sink signal  $v (t)$.




			<u>Notes:</u>
			*The exercise refers to the chapter  [[Examples_of_Communication_Systems/General_Description_of_ISDN\|"General Description of ISDN"]] of this book.

	~~''Notes:''~~		*Reference is also made to the chapter  [[Modulation_Methods/Pulse_Code_Modulation\|"Pulse Code Modulation"]]  of the book  "Modulation Methods".
	*The exercise refers to the chapter  [[Examples_of_Communication_Systems/General_Description_of_ISDN\|"General Description of ISDN"]] of this book.
	*Reference is also made to the chapter  [[Modulation_Methods/Pulse_Code_Modulation\|"Pulse Code Modulation"]]  of the book "Modulation Methods".


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	<quiz display=simple>		<quiz display=simple>

	{With how many bits  $(N)$  is each (quantized) sample represented?		{With how many bits  $(N)$  is each quantized sample represented?
	\|type="{}"}		\|type="{}"}
	$N \ = \ $ { 8 3% }		$N \ = \ $ { 8 3% }
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	{Is the sink SNR  $\rho_{v}$  at ISDN limited by the following effects?		{Is the sink SNR  $\rho_{v}$  at ISDN limited by the following effects?
	\|type="[]"}		\|type="[]"}
	- Sampling (if sampling theorem is satisfied),		- Sampling  (if sampling theorem is satisfied),
	+ AWGN noise (transmission error).		+ AWGN noise  (transmission error).

Guenter at 16:32, 14 October 2022

2022-10-14T16:32:44Z

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	}}		}}

	[[File:P_ID1578__Bei_A_1_2.png\|right\|frame\|Components of ~~the~~ PCM transmitter]]		[[File:P_ID1578__Bei_A_1_2.png\|right\|frame\|Components of PCM transmitter '''Korrektur''']]
	The conversion of the analog voice signal  $q(t)$  into the binary signal  $q_{\rm C}(t)$  is done at ISDN  (''Integrated Services Digital Network'')  according to the guidelines of pulse code modulation (PCM) by		The conversion of the analog voice signal  $q(t)$  into the binary signal  $q_{\rm C}(t)$  is done at ISDN  ("Integrated Services Digital Network")  according to the guidelines of pulse code modulation  $\rm (PCM)$  by
	*sampling in the interval  $T_{\rm A} = 1/f_{\rm A}$,		*sampling in the interval  $T_{\rm A} = 1/f_{\rm A}$,

	*quantization to  $M = 256$  discrete values,		*quantization to  $M = 256$  discrete values,
	*binary PCM ~~coding~~ with  $N$  bits per quantization value.
			*binary PCM encoding with  $N$  bits per quantization value.


	The net data rate of a so-called  $\rm B$ channel  (''Bearer Channel'') is  $64 \ \rm kbit/s$  and corresponds to the bit rate of the redundancy-free binary signal  $q_{\rm C}(t)$. However, because of the subsequent redundant channel coding and the inserted signaling bits, the gross data rate – i.e., the transmission rate of the transmitted signal  $s(t)$  – is greater.		The net data rate of a so-called  $\rm B$ channel  ("Bearer Channel") is  $64 \ \rm kbit/s$  and corresponds to the bit rate of the redundancy-free binary signal  $q_{\rm C}(t)$.  However,  because of the subsequent redundant channel coding and the inserted signaling bits,  the gross data rate – i.e., the transmission rate of the transmitted signal  $s(t)$  – is greater.


	A measure for the quality of the entire (ISDN) transmission system is the sink SNR		A measure for the quality of the entire ISDN transmission system is the sink SNR
	:$$\rho_{v} = \frac{P_q}{P_{\varepsilon}} = \frac{\overline{q(t)^2}}{\overline{[\upsilon(t) - q(t)]^2}}$$		:$$\rho_{v} = \frac{P_q}{P_{\varepsilon}} = \frac{\overline{q(t)^2}}{\overline{[\upsilon(t) - q(t)]^2}}$$

Hwang at 08:36, 2 October 2022

2022-10-02T08:36:51Z

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