Aufgaben:Exercise 5.2Z: Two-Way Channel - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastic Systems Theory^]]		[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastic Systems Theory^]]
	[[de:~~Zusatzaufgaben~~:5.~~2_Zweiwegekanal~~]]		[[de:Aufgaben:Aufgabe 5.2Z: Zweiwegekanal]]

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	'''(4)'''  By comparing the drawn function  $h(t) \star h(-t)$  with the result of subtask  '''(2)''':		'''(4)'''  By comparing the drawn function  $h(t) \star h(-t)$  with the result of subtask  '''(2)''':
	:$$C_0 = 1 + \alpha ^2 \hspace{0.15cm} \underline{= 1.25},		:$$C_0 = 1 + \alpha ^2 \hspace{0.15cm} \underline{= 1.25},\hspace{0.5cm}C_3 = \alpha \hspace{0.15cm} \underline{= 0.5},\hspace{0.5cm}\tau _3 = \tau _2 - \tau _1 \hspace{0.15cm} \underline{= 4\;{\rm{ms}}}.$$
	\hspace{0.5cm}C_3 = \alpha \hspace{0.15cm} \underline{= 0.5},
	\hspace{0.5cm}\tau _3 = \tau _2 - \tau _1 \hspace{0.15cm} \underline{= 4\;{\rm{ms}}}.$$

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	[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastic Systems Theory^]]		[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastic Systems Theory^]]
			[[de:Zusatzaufgaben:5.2_Zweiwegekanal]]

Guenter at 15:51, 10 February 2022

2022-02-10T15:51:27Z

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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  $H(f)$  is the Fourier transform to  $h(t)$.		'''(1)'''  $H(f)$  is the Fourier transform to  $h(t)$.
	*Using the shifting theorem, this is  $(\tau_1 = 0)$:		*Using the shifting theorem,  this is with   $\tau_1 = 0$:
	:$$H(f) = 1 + \alpha \cdot {\rm{e}}^{ - {\rm{j2\pi }}f\tau _2 } = 1 + \alpha \cdot \cos ( {{\rm{2\pi }}f\tau _2 } ) - {\rm{j}} \cdot \alpha \cdot \sin ( {{\rm{2\pi }}f\tau _2 } ).$$		:$$H(f) = 1 + \alpha \cdot {\rm{e}}^{ - {\rm{j2\pi }}f\tau _2 } = 1 + \alpha \cdot \cos ( {{\rm{2\pi }}f\tau _2 } ) - {\rm{j}} \cdot \alpha \cdot \sin ( {{\rm{2\pi }}f\tau _2 } ).$$

	*If  $H(f)$  is periodic with~~ ~~ $f_0$,  then for all integer values of  $i$  must hold:		*If  $H(f)$  is periodic with $f_0$,  then for all integer values of  $i$  must hold:
	:$$H( {f + i \cdot f_0 } ) = H( f ).$$		:$$H( {f + i \cdot f_0 } ) = H( f ).$$
	*With  $f_0 = 1/\tau_2\hspace{0.15cm} \underline{= 0.25 \hspace{0.05cm}\rm kHz}$  this condition is satisfied.		*With  $f_0 = 1/\tau_2\hspace{0.15cm} \underline{= 0.25 \hspace{0.05cm}\rm kHz}$  this condition is satisfied.
	:$$H( {f + i \cdot f_0 } ) = 1 + \alpha \cdot \cos ( {{\rm{2\pi }}f\tau _2 + i{\rm{2\pi }}f_0 \tau _2 } ) - {\rm{j}} \cdot \alpha \cdot \sin ( {{\rm{2\pi }}f\tau _2 + i{\rm{2\pi }}f_0 \tau _2 } ) = 1 + \alpha \cdot \cos ( {{\rm{2\pi }}f\tau _2 } ) - {\rm{j}} \cdot \alpha \cdot \sin ( {{\rm{2\pi }}f\tau _2 } ).$$		:$$H( {f + i \cdot f_0 } ) = 1 + \alpha \cdot \cos ( {{\rm{2\pi }}f\tau _2 + i{\rm{2\pi }}f_0 \tau _2 } ) - {\rm{j}} \cdot \alpha \cdot \sin ( {{\rm{2\pi }}f\tau _2 + i{\rm{2\pi }}f_0 \tau _2 } ) = 1 + \alpha \cdot \cos ( {{\rm{2\pi }}f\tau _2 } ) - {\rm{j}} \cdot \alpha \cdot \sin ( {{\rm{2\pi }}f\tau _2 } ).$$



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	:$$\left\| {H( f )} \right\|^2 = \left( {1 + \alpha \cdot \cos ( A )} \right)^2 + \left( {\alpha \cdot \sin ( A )} \right)^2 .$$		:$$\left\| {H( f )} \right\|^2 = \left( {1 + \alpha \cdot \cos ( A )} \right)^2 + \left( {\alpha \cdot \sin ( A )} \right)^2 .$$

	*Here the angle argument is abbreviated as  $A = 2\pi f \tau$.    After multiplying ~~out~~, we get because of  $\cos^2(A) + \sin^2(A) = 1$:		*Here the angle argument is abbreviated as  $A = 2\pi f \tau$.    After multiplying,  we get because of  $\cos^2(A) + \sin^2(A) = 1$:
	:$$\left\| {H(f)} \right\|^2 = 1 + \alpha ^2 + 2\alpha \cdot \cos ( A ).$$		:$$\left\| {H(f)} \right\|^2 = 1 + \alpha ^2 + 2\alpha \cdot \cos ( A ).$$

	*At the frequency~~ ~~ $f = 0$  $($and thus   $A = 0)$,  the ~~general~~ result or with  $\alpha = 0.5$ is:		*At the frequency $f = 0$  $($and thus   $A = 0)$,  the result with  $\alpha = 0.5$  is:
	:$$\left\| {H( {f = 0} )} \right\|^2 = \left( {1 + \alpha } \right)^2 = 1.5^2\hspace{0.15cm} \underline{ = 2.25}.$$		:$$\left\| {H( {f = 0} )} \right\|^2 = \left( {1 + \alpha } \right)^2 = 1.5^2\hspace{0.15cm} \underline{ = 2.25}.$$


			'''(3)'''  Now the transmission system can be composed of two subsystems  (see diagram):
	'''(3)'''  Now the transmission system can be composed of two subsystems (see diagram):		[[File:P_ID551__Sto_Z_5_2_c.png\|frame\|Splitting the impulse response into two subsystems]]
	[[File:P_ID551__Sto_Z_5_2_c.png\|frame\|~~Division of~~ the impulse response into two subsystems]]

	*The transfer function  $H_1(f)$  is calculated as in subtask  '''(2)'''.		*The transfer function  $H_1(f)$  is calculated as in subtask  '''(2)'''.
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	:$$H_2 (f) = {\rm{e}}^{ - {\rm{j2\pi }}f\tau _1 } \quad \Rightarrow \quad \left\| {H_2 (f)} \right\| = 1\quad \Rightarrow \quad \left\| {H_2 (f)} \right\|^2 = 1.$$		:$$H_2 (f) = {\rm{e}}^{ - {\rm{j2\pi }}f\tau _1 } \quad \Rightarrow \quad \left\| {H_2 (f)} \right\| = 1\quad \Rightarrow \quad \left\| {H_2 (f)} \right\|^2 = 1.$$

	*This means:   Due to the additional ~~running~~ time,  $\left\| {H(f)} \right\|^2$  is not changed compared to subtask  '''(2)'''.		*This means:   Due to the additional delay time,  $\left\| {H(f)} \right\|^2$  is not changed compared to subtask  '''(2)'''.
	* At the frequency~~ ~~ $f = 0$,    $\left\| {H(f = 0)} \right\|^2\hspace{0.15cm} \underline{ = 2.25}$ is still valid.		* At the frequency $f = 0$:    $\left\| {H(f = 0)} \right\|^2\hspace{0.15cm} \underline{ = 2.25}$  is still valid.




	'''(4)'''  By comparing the drawn function  $h(t) \star h(-t)$  with the result of subtask  '''(2)'''  ~~we obtain:~~		'''(4)'''  By comparing the drawn function  $h(t) \star h(-t)$  with the result of subtask  '''(2)''':
	:$$C_0 = 1 + \alpha ^2 \hspace{0.15cm} \underline{= 1.25},		:$$C_0 = 1 + \alpha ^2 \hspace{0.15cm} \underline{= 1.25},
	\hspace{0.5cm}C_3 = \alpha \hspace{0.15cm} \underline{= 0.5},		\hspace{0.5cm}C_3 = \alpha \hspace{0.15cm} \underline{= 0.5},
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			'''(5)'''  The power-spectral density of the output signal  $y(t)$  is limited to the range of  $\pm B$  and results in
	'''(5)'''  The ~~PDS~~ of the output signal  $y(t)$  is limited to the range of  $\pm B$  and results in
	:$${\it \Phi}_y(f) = {N_0}/{2} \cdot \|H(f)\|^2 = N_0/{2} \cdot {\left( {1 + \alpha ^2 + 2\alpha \cdot \cos ( {2{\rm{\pi }}f\tau _3 } )} \right)}.$$		:$${\it \Phi}_y(f) = {N_0}/{2} \cdot \|H(f)\|^2 = N_0/{2} \cdot {\left( {1 + \alpha ^2 + 2\alpha \cdot \cos ( {2{\rm{\pi }}f\tau _3 } )} \right)}.$$

	*Taking advantage of symmetry properties, we thus obtain for the power:		*Taking advantage of symmetry properties,  we thus obtain for the power:
	:$$P_y = N_0 \cdot \int_0^B {\left( {1 + \alpha ^2 + 2\alpha \cdot \cos ( {2{\rm{\pi }}f\tau _3 } )} \right)}\hspace{0.1cm} {\rm{d}}f.$$		:$$P_y = N_0 \cdot \int_0^B {\left( {1 + \alpha ^2 + 2\alpha \cdot \cos ( {2{\rm{\pi }}f\tau _3 } )} \right)}\hspace{0.1cm} {\rm{d}}f.$$

	*$B = 10 \hspace{0.08cm} \rm kHz$  is an integer multiple of the frequency period  $f_0 = 1/\tau_2= 250 \hspace{0.08cm}\rm Hz$  $($cf. solution to subtask  '''1'''$)$.		*$B = 10 \hspace{0.08cm} \rm kHz$  is an integer multiple of the frequency period  $f_0 = 1/\tau_2= 250 \hspace{0.08cm}\rm Hz$  $($cf. solution to subtask  '''1'''$)$.  Therefore,  the cosine function does not contribute to the integral,  and we obtain:
	*Therefore, the cosine function does not contribute to the integral, and we obtain:
	:$$P_y = N_0 \cdot B \cdot \left( {1 + \alpha ^2 } \right) = 1.25 \cdot P_x \hspace{0.15cm} \underline{ = 12.5\;{\rm{mW}}}.$$		:$$P_y = N_0 \cdot B \cdot \left( {1 + \alpha ^2 } \right) = 1.25 \cdot P_x \hspace{0.15cm} \underline{ = 12.5\;{\rm{mW}}}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 15:38, 10 February 2022

2022-02-10T15:38:05Z

Hwang at 16:47, 13 January 2022

2022-01-13T16:47:18Z

← Older revision		Revision as of 18:47, 13 January 2022
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	''Notes:''		''Notes:''
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Stochastic_System_Theory Stochastic System Theory]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Stochastic_System_Theory\| Stochastic System Theory]].
	*Always use the value  $\alpha = 0.5$ for the numerical calculations.		*Always use the value  $\alpha = 0.5$ for the numerical calculations.
	*For the subtasks  '''(1)'''  and  '''(2)'''  let  $\tau_1 = 0$  and  $\tau_2 = 4\hspace{0.08cm}\rm ms$ be valid.		*For the subtasks  '''(1)'''  and  '''(2)'''  let  $\tau_1 = 0$  and  $\tau_2 = 4\hspace{0.08cm}\rm ms$ be valid.

Hwang at 16:46, 13 January 2022

2022-01-13T16:46:38Z

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Guenter at 10:30, 12 January 2022

2022-01-12T10:30:12Z

← Older revision		Revision as of 12:30, 12 January 2022
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	[[Category:Theory of Stochastic Signals: Exercises\|^5.1 ~~Stochastische Systemtheorie~~^]]		[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastic Systems Theory^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 5.2Z: Zweiwegekanal to Aufgaben:Exercise 5.2Z: Two-Way Channel

2022-01-12T08:56:25Z

Guenter moved page Aufgaben:Aufgabe 5.2Z: Zweiwegekanal to Aufgaben:Exercise 5.2Z: Two-Way Channel

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Javier: Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

2021-03-23T12:43:52Z

Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

← Older revision		Revision as of 14:43, 23 March 2021
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	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^5.1 Stochastische Systemtheorie^]]		[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastische Systemtheorie^]]

← Older revision		Revision as of 17:38, 10 February 2022
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	}}		}}

	[[File:P_ID517__Sto_Z_5_2.png\|right\|frame\|Two–way channel impulse ~~response~~  $h(t)$~~  and~~  $h(t) * h( { - t} )$]]		[[File:P_ID517__Sto_Z_5_2.png\|right\|frame\|Two–way channel impulse responses  $h(t)$,  $h(t) * h( { - t} )$]]
	~~It is known from~~ a transmission system that ~~the~~ following relationship exists between the input signal  $x(t)$  and the output signal  $y(t)$ :		From a transmission system it is known that following relationship exists between the input signal  $x(t)$  and the output signal  $y(t)$ :
	:$$y(t) = x( {t - \tau _1 } ) + \alpha \cdot x( {t - \tau _2 } ).$$		:$$y(t) = x( {t - \tau _1 } ) + \alpha \cdot x( {t - \tau _2 } ).$$

	The corresponding impulse response  $h(t)$  is sketched above.		The corresponding impulse response  $h(t)$  is sketched above.

	In the sketch below, the function		In the sketch below,  the function
	:$$h(t) * h( { - t} )\hspace{0.25cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\hspace{0.25cm}\left\| {H(f)} \right\|^2$$		:$$h(t) * h( { - t} )\hspace{0.25cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\hspace{0.25cm}\left\| {H(f)} \right\|^2$$

	is shown, where the parameters  $C_0$,  $C_3$  and  $\tau_3$  depend on  $\alpha$,  $\tau_1$  and  $\tau_2$   ⇒   see subtask  '''(4)'''.		is shown,  where the parameters  $C_0$,  $C_3$  and  $\tau_3$  depend on  $\alpha$,  $\tau_1$  and  $\tau_2$   ⇒   see subtask  '''(4)'''.

	Let the input signal  $x(t)$  be band-limited white noise		Let the input signal  $x(t)$  be band-limited white noise
	*with power density  $N_0 = 10^{-6} \hspace{0.08cm} \rm W/Hz$		*with power density  $N_0 = 10^{-6} \hspace{0.08cm} \rm W/Hz$
	*and ~~the~~ bandwidth  $B = 10 \hspace{0.08cm} \rm kHz$,		*and bandwidth  $B = 10 \hspace{0.08cm} \rm kHz$,


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			Notes:


	''Notes:''
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Stochastic_System_Theory\| Stochastic System Theory]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Stochastic_System_Theory\| Stochastic System Theory]].
	*~~Always use~~ the value  $\alpha = 0.5$ for ~~the~~ numerical calculations.		*Use always the value  $\alpha = 0.5$  for numerical calculations.
	*For the subtasks  '''(1)'''  and  '''(2)'''  let  $\tau_1 = 0$  and  $\tau_2 = 4\hspace{0.08cm}\rm ms$ be valid.		*For the subtasks  '''(1)'''  and  '''(2)''',  let  $\tau_1 = 0$  and  $\tau_2 = 4\hspace{0.08cm}\rm ms$  be valid.
	*For later tasks, assume  $\tau_1 = 1\hspace{0.08cm}\rm ms$  and  $\tau_2 = 5\hspace{0.08cm}\rm ms$.		*For later tasks,  assume  $\tau_1 = 1\hspace{0.08cm}\rm ms$  and  $\tau_2 = 5\hspace{0.08cm}\rm ms$.


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	<quiz display=simple>		<quiz display=simple>
	{Calculate the frequency response  $H(f)$  for  $\tau_1 = 0$  and $\tau_2 = 4\hspace{0.08cm}\rm ms$.  Show that  $H(f)$  is a function ~~periodic~~ with  $f_0$.    What is the ~~magnitude~~ of~~ ~~ $f_0$?		{Calculate the frequency response  $H(f)$  for  $\tau_1 = 0$  and $\tau_2 = 4\hspace{0.08cm}\rm ms$.  Show that  $H(f)$  is a periodic function with  $f_0$.    What is the value of $f_0$?
	\|type="{}"}		\|type="{}"}
	$f_0 \ = \ $ { 0.25 3% } $\ \rm kHz$		$f_0 \ = \ $ { 0.25 3% } $\ \rm kHz$


	{What is the size of  $\|H(f)\|^2$  with  $\tau_1 = 0$,  $\tau_2 = 4\hspace{0.08cm}\rm ms$  and  $\alpha = 0.5$?  Enter the value at~~ ~~ $f = 0$.		{What is the size of  $\|H(f)\|^2$  with  $\tau_1 = 0$,  $\tau_2 = 4\hspace{0.08cm}\rm ms$  and  $\alpha = 0.5$?  Enter the value at $f = 0$.
	\|type="{}"}		\|type="{}"}
	$\|H(f = 0)\|^2 \ = \ $ { 2.25 3% }		$\|H(f = 0)\|^2 \ = \ $ { 2.25 3% }


	{How does  $\|H(f)\|^2$  change with  $\tau_1 = 1\hspace{0.08cm}\rm ms$  and  $\tau_2 = 5\hspace{0.08cm}\rm ms$?  Let the attenuation constant still be  $\alpha = 0.5$.  Enter the value at~~ ~~ $f = 0$.		{How does  $\|H(f)\|^2$  change with  $\tau_1 = 1\hspace{0.08cm}\rm ms$  and  $\tau_2 = 5\hspace{0.08cm}\rm ms$?  Let the attenuation constant still be  $\alpha = 0.5$.  Enter the value at $f = 0$.
	\|type="{}"}		\|type="{}"}
	$\|H(f = 0)\|^2 \ = \ $ { 2.25 }		$\|H(f = 0)\|^2 \ = \ $ { 2.25 }