Aufgaben:Exercise 2.2: Distortion Power - Revision history

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	*The following is obtained for the signal–to–distortion–power ratio with the result from  '''(1)''' :		*The following is obtained for the signal–to–distortion–power ratio with the result from  '''(1)''' :
	$$\rho_{\rm V1} = \frac{ P_{x}}{P_{\rm V1}}= \frac{ {1 \, \~~rmV~~^2}}{0.005 \, \rm V^2}\hspace{0.05cm}\rm = 200\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}\hspace{0.15cm}\underline{ = {23.01 \, \rm dB}}.$$		$$\rho_{\rm V1} = \frac{ P_{x}}{P_{\rm V1}}= \frac{ {1 \, \rm V^2}}{0.005 \, \rm V^2}\hspace{0.05cm}\rm = 200\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}\hspace{0.15cm}\underline{ = {23.01 \, \rm dB}}.$$


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	'''(5)'''  According to the information sheet the following holds:		'''(5)'''  According to the information sheet the following holds:
	:$$\rho_{\rm V2} = \frac{ \alpha^2 \cdot P_{x}}{P_{\rm V2}}= \frac{ 0.5^2 \cdot {1 \, \~~rmV~~^2}}{0.005 \, \rm V^2}\hspace{0.05cm}\rm = 50\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2} \hspace{0.15cm}\underline{= {16.99 \, \rm dB}}.$$		:$$\rho_{\rm V2} = \frac{ \alpha^2 \cdot P_{x}}{P_{\rm V2}}= \frac{ 0.5^2 \cdot {1 \, \rm V^2}}{0.005 \, \rm V^2}\hspace{0.05cm}\rm = 50\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2} \hspace{0.15cm}\underline{= {16.99 \, \rm dB}}.$$

	*Despite the same distortion power $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2}$  is less than  $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}$  by about $6 \ \rm dB$ .		*Despite the same distortion power $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2}$  is less than  $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}$  by about $6 \ \rm dB$ .
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.1 Classification of the Distortions^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^2.1 Classification of the Distortions^]]
	[[de:Aufgaben:2.~~2_Verzerrungsleistung~~]]		[[de:Aufgaben:Aufgabe 2.2: Verzerrungsleistung]]

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	:$$\varepsilon(t) = y(t) - \alpha \cdot x(t - \tau) .$$		:$$\varepsilon(t) = y(t) - \alpha \cdot x(t - \tau) .$$
	The parameters $\alpha$  and  $\tau$  are to be determined such that the distortion power (the mean squared error) is minimal. For this, the following holds:		The parameters $\alpha$  and  $\tau$  are to be determined such that the distortion power (the mean squared error) is minimal. For this, the following holds:
	:$$P_{\rm V} = \overline{\varepsilon^2(t)} = \frac{1}{T_{\rm M}} \cdot \int\limits_{ ( T_{\rm M})}		:$$P_{\rm V} = \overline{\varepsilon^2(t)} = \frac{1}{T_{\rm M}} \cdot \int\limits_{ ( T_{\rm M})}{\varepsilon^2(t) }\hspace{0.1cm}{\rm d}t$$
	{\varepsilon^2(t) }\hspace{0.1cm}{\rm d}t$$

	These definitions already take into account that a frequency-independent damping just as a runtime which is constant for all frequencies does not contribute to the distortion.		These definitions already take into account that a frequency-independent damping just as a runtime which is constant for all frequencies does not contribute to the distortion.
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	[[File:P_ID915__LZI_A_2_2_a.png\|right\|frame\|Resulting error signals]]		[[File:P_ID915__LZI_A_2_2_a.png\|right\|frame\|Resulting error signals]]
	'''(1)'''  The error signal  $\varepsilon_1(t)$ shown in the graph is obtained with the given parameters $\alpha = 1$  and  $\tau= 0$ . The distortion power is thus equal to:		'''(1)'''  The error signal  $\varepsilon_1(t)$ shown in the graph is obtained with the given parameters $\alpha = 1$  and  $\tau= 0$ . The distortion power is thus equal to:
	:$$P_{\rm V1} = \frac{ {1 \, \rm ms}}{4 \, \rm ms} \cdot \big[ ({0.1 \, \rm V})^2 +		:$$P_{\rm V1} = \frac{ {1 \, \rm ms}}{4 \, \rm ms} \cdot \big[ ({0.1 \, \rm V})^2 +({-0.1 \, \rm V})^2\big]\hspace{0.3cm}\Rightarrow \hspace{0.3cm}P_{\rm V1} \hspace{0.15cm}\underline{ = 5 \cdot 10^{-3} \, \rm V^2}. $$
	({-0.1 \, \rm V})^2\big]\hspace{0.3cm}\Rightarrow \hspace{0.3cm}P_{\rm V1} \hspace{0.15cm}\underline{ = 5 \cdot 10^{-3} \, \rm V^2}. $$


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	*The following is obtained for the signal–to–distortion–power ratio with the result from  '''(1)''' :		*The following is obtained for the signal–to–distortion–power ratio with the result from  '''(1)''' :
	$$\rho_{\rm V1} = \frac{ P_{x}}{P_{\rm V1}}= \frac{ {1 \, \rm		$$\rho_{\rm V1} = \frac{ P_{x}}{P_{\rm V1}}= \frac{ {1 \, \rmV^2}}{0.005 \, \rm V^2}\hspace{0.05cm}\rm = 200\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}\hspace{0.15cm}\underline{ = {23.01 \, \rm dB}}.$$
	V^2}}{0.005 \, \rm V^2}\hspace{0.05cm}\rm = 200\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}\hspace{0.15cm}\underline{ = {23.01 \, \rm dB}}.$$


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	'''(5)'''  According to the information sheet the following holds:		'''(5)'''  According to the information sheet the following holds:
	:$$\rho_{\rm V2} = \frac{ \alpha^2 \cdot P_{x}}{P_{\rm V2}}= \frac{ 0.5^2 \cdot {1 \, \rm		:$$\rho_{\rm V2} = \frac{ \alpha^2 \cdot P_{x}}{P_{\rm V2}}= \frac{ 0.5^2 \cdot {1 \, \rmV^2}}{0.005 \, \rm V^2}\hspace{0.05cm}\rm = 50\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2} \hspace{0.15cm}\underline{= {16.99 \, \rm dB}}.$$
	V^2}}{0.005 \, \rm V^2}\hspace{0.05cm}\rm = 50\hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2} \hspace{0.15cm}\underline{= {16.99 \, \rm dB}}.$$

	*Despite the same distortion power $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2}$  is less than  $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}$  by about $6 \ \rm dB$ .		*Despite the same distortion power $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2}$  is less than  $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}$  by about $6 \ \rm dB$ .

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.1 Classification of the Distortions^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^2.1 Classification of the Distortions^]]
			[[de:Aufgaben:2.2_Verzerrungsleistung]]

Oezer at 20:51, 12 September 2021

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	*Despite the same distortion power $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2}$  is less than  $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}$  by about $6 \ \rm dB$ .		*Despite the same distortion power $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V2}$  is less than  $10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm V1}$  by about $6 \ \rm dB$ .
	*The signal $y_2(t)$  is thus significantly less ~~favorable~~ in terms of SNR than $y_1(t)$.		*The signal $y_2(t)$  is thus significantly less favourable in terms of SNR than $y_1(t)$.
	*It is considered that now the power of the output signal is only a quarter of the input power due to  $\alpha = 0.5$ .		*It is considered that now the power of the output signal is only a quarter of the input power due to  $\alpha = 0.5$ .
	*If this attenuation at the output was to be compensated by amplifying it by $1/\alpha$, the distortion power would indeed increase by $\alpha^2$.		*If this attenuation at the output was to be compensated by amplifying it by $1/\alpha$, the distortion power would indeed increase by $\alpha^2$.

Oezer at 20:49, 12 September 2021

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	:and afterwards the mean squared error for very (infinitely) many $\alpha$–  and  $\tau$–values would have to be determined, in doing so the integration interval is to be adjusted to $\tau$  in each case.		:and afterwards the mean squared error for very (infinitely) many $\alpha$–  and  $\tau$–values would have to be determined, in doing so the integration interval is to be adjusted to $\tau$  in each case.
	*~~Also then~~ the smallest possible result would be obtained for $\alpha \; \underline{= 0.5}$  and  $\tau \; \underline{= 1 \ \rm ms}$ . However, for this optimization of $\alpha$  and  $\tau$  the useage of a computer program should be granted.		*Then, the smallest possible result would also be obtained for $\alpha \; \underline{= 0.5}$  and  $\tau \; \underline{= 1 \ \rm ms}$ . However, for this optimization of $\alpha$  and  $\tau$  the useage of a computer program should be granted.

Oezer at 20:48, 12 September 2021

2021-09-12T20:48:08Z

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	:$$\varepsilon_2(t) = y_2(t) - \alpha \cdot x(t - \tau)$$		:$$\varepsilon_2(t) = y_2(t) - \alpha \cdot x(t - \tau)$$

	:for very (infinitely) many $\alpha$–  and  $\tau$–values ~~and afterwards the mean squared error~~ would have to be determined, in doing so the integration interval is to be adjusted to $\tau$  in each case.		:and afterwards the mean squared error for very (infinitely) many $\alpha$–  and  $\tau$–values would have to be determined, in doing so the integration interval is to be adjusted to $\tau$  in each case.
	*Also then the smallest possible result would be obtained for $\alpha \; \underline{= 0.5}$  and  $\tau \; \underline{= 1 \ \rm ms}$ . However, for this optimization of $\alpha$  and  $\tau$  the useage of a computer program should be granted.		*Also then the smallest possible result would be obtained for $\alpha \; \underline{= 0.5}$  and  $\tau \; \underline{= 1 \ \rm ms}$ . However, for this optimization of $\alpha$  and  $\tau$  the useage of a computer program should be granted.

Oezer at 20:45, 12 September 2021

2021-09-12T20:45:01Z

← Older revision		Revision as of 22:45, 12 September 2021
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	:$$\varepsilon_2(t) = y_2(t) - \alpha \cdot x(t - \tau)$$		:$$\varepsilon_2(t) = y_2(t) - \alpha \cdot x(t - \tau)$$

	:for very (infinitely) ~~manynbsp~~;$\alpha$–  and  $\tau$–values and afterwards the mean squared error would have to be determined, in doing so the integration interval is to be adjusted to $\tau$  in each case.		:for very (infinitely) many $\alpha$–  and  $\tau$–values and afterwards the mean squared error would have to be determined, in doing so the integration interval is to be adjusted to $\tau$  in each case.
	*Also then the smallest possible result would be obtained for $\alpha \; \underline{= 0.5}$  and  $\tau \; \underline{= 1 \ \rm ms}$ . However, for this optimization of $\alpha$  and  $\tau$  the useage of a computer program should be granted.		*Also then the smallest possible result would be obtained for $\alpha \; \underline{= 0.5}$  and  $\tau \; \underline{= 1 \ \rm ms}$ . However, for this optimization of $\alpha$  and  $\tau$  the useage of a computer program should be granted.

Oezer at 20:44, 12 September 2021

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	'''(3)'''  The sketch on the information sheet makes it clear that even without the distortions occuring – but due to attenuation and runtime alone – the signal $y(t)$  would differ significantly from $x(t)$ .		'''(3)'''  The sketch on the information sheet makes it clear that even without the distortions occuring – but due to attenuation and runtime alone – the signal $y(t)$  would differ significantly from $x(t)$ .
	*The following would arise as a result: $y(t) = 0.5 \cdot x(t-1\ {\rm ms}) $ .		*The following would arise as a result:  $y(t) = 0.5 \cdot x(t-1\ {\rm ms}) $ .

	*If someone does not immediately perceive these values from the graph, then first the error signal		*If someone does not immediately perceive these values from the graph, then first the error signal

Oezer at 20:43, 12 September 2021

2021-09-12T20:43:41Z

← Older revision		Revision as of 22:43, 12 September 2021
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	'''(3)'''  The sketch on the information sheet makes it clear that even without the distortions occuring – but due to attenuation and runtime alone – the signal $y(t)$  would differ significantly from $x(t)$ .		'''(3)'''  The sketch on the information sheet makes it clear that even without the distortions occuring – but due to attenuation and runtime alone – the signal $y(t)$  would differ significantly from $x(t)$ .
	*The following would arise as a result $y(t) = 0.5 \cdot x(t-1\ {\rm ms}) $ .		*The following would arise as a result: $y(t) = 0.5 \cdot x(t-1\ {\rm ms}) $ .

	*If someone does not immediately perceive these values from the graph, then first the error signal		*If someone does not immediately perceive these values from the graph, then first the error signal

Oezer at 20:40, 12 September 2021

2021-09-12T20:40:39Z

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	*Use the interval  $0$ ... $4 \hspace{0.08cm} \rm ms$  for $y_1(t)$  and the interval  $1 \hspace{0.08cm} {\rm ms}$ ... $5 \hspace{0.08cm} \rm ms$ for   $y_2(t)$ .		*Use the interval  $0$ ... $4 \hspace{0.08cm} \rm ms$  for $y_1(t)$  and the interval  $1 \hspace{0.08cm} {\rm ms}$ ... $5 \hspace{0.08cm} \rm ms$ for   $y_2(t)$ .
	*Thus, the measurement time is $T_{\rm M} = 4 \hspace{0.08cm} \rm ms$ in both cases.		*Thus, the measurement time is $T_{\rm M} = 4 \hspace{0.08cm} \rm ms$ in both cases.
	*It is obvious that with respect to $y_1(t)$  ~~die Parameter~~  $\alpha = 1$  and  $\tau = 0$  respectively result in the minimum distortion power.		*It is obvious that with respect to $y_1(t)$  the parameters  $\alpha = 1$  and  $\tau = 0$  respectively result in the minimum distortion power.