Aufgaben:Exercise 2.5: Distortion and Equalization - Revision history

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	*The strongest deviations between  $x_2(t)$  and  $y_2(t)$  will occur at the triangle peaks since the missing high frequencies have the strongest effect here.		*The strongest deviations between  $x_2(t)$  and  $y_2(t)$  will occur at the triangle peaks since the missing high frequencies have the strongest effect here.
	*For example, for the time  $\underline{t= 0}$ one obtains:		*For example, for the time  $\underline{t= 0}$ one obtains:
	:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +{3}/{72}\right)= 0.844\,{\~~rmV~~} \hspace{0.3cm}\Rightarrow\hspace{0.3cm}\varepsilon_{\rm max} = \|y_2(t=0)- x_2(t=0)\| \hspace{0.15cm}\underline{= 0.156\,{\~~rmV~~}}.$$		:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +{3}/{72}\right)= 0.844\,{\rm V} \hspace{0.3cm}\Rightarrow\hspace{0.3cm}\varepsilon_{\rm max} = \|y_2(t=0)- x_2(t=0)\| \hspace{0.15cm}\underline{= 0.156\,{\rm V}}.$$



	'''(5)'''  With the basic frequency  $f_0 = 2 \ \rm kHz$  and the values  $H(3f_0) = 0.75$,  $H(5f_0) = 0.25$,  $H(7f_0) = 0$  the following is obtained:		'''(5)'''  With the basic frequency  $f_0 = 2 \ \rm kHz$  and the values  $H(3f_0) = 0.75$,  $H(5f_0) = 0.25$,  $H(7f_0) = 0$  the following is obtained:
	:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +\frac{3}{4}\cdot \frac{1}{9} + \frac{1}{4} \cdot\frac{1}{25}\right)= 0.886\,{\~~rmV~~}\hspace{0.5cm}\Rightarrow \hspace{0.5cm}\varepsilon_{\rm max} \hspace{0.15cm}\underline{= 0.114\,{\~~rmV~~}}.$$		:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +\frac{3}{4}\cdot \frac{1}{9} + \frac{1}{4} \cdot\frac{1}{25}\right)= 0.886\,{\rm V}\hspace{0.5cm}\Rightarrow \hspace{0.5cm}\varepsilon_{\rm max} \hspace{0.15cm}\underline{= 0.114\,{\rm V}}.$$


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	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]
	[[de:Aufgaben:2.~~5_Verzerrung_und_Entzerrung~~]]		[[de:Aufgaben:Aufgabe 2.5: Verzerrung und Entzerrung]]

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	[[File:P_ID907__LZI_A_2_5.png\|right\|frame\|Trapezoidal spectrum (top), <br>associated impulse response)]]		[[File:P_ID907__LZI_A_2_5.png\|right\|frame\|Trapezoidal spectrum (top), <br>associated impulse response)]]
	A communication system with input  $x(t)$  and output  $y(t)$,  which is fully described by the trapezoidal frequency response  $H(f)$  according to the top graph, is considered.  Using the roll-off factor  $r = 0.5$  and the equivalent bandwidth  $\Delta f = 16 \ \rm kHz$  the corresponding impulse response,  which is computable by applying the inverse Fourier transform, is:		A communication system with input  $x(t)$  and output  $y(t)$,  which is fully described by the trapezoidal frequency response  $H(f)$  according to the top graph, is considered.  Using the roll-off factor  $r = 0.5$  and the equivalent bandwidth  $\Delta f = 16 \ \rm kHz$  the corresponding impulse response,  which is computable by applying the inverse Fourier transform, is:
	:$$h(t) = \Delta f \cdot {\rm si}(\pi \cdot \Delta f \cdot t )\cdot		:$$h(t) = \Delta f \cdot {\rm si}(\pi \cdot \Delta f \cdot t )\cdot{\rm si}(\pi \cdot r \cdot \Delta f \cdot t) = \Delta f \cdot {\rm sinc}(\Delta f \cdot t )\cdot{\rm sinc}(r \cdot \Delta f \cdot t).$$
	{\rm si}(\pi \cdot r \cdot \Delta f \cdot t
	) = \Delta f \cdot {\rm sinc}(\Delta f \cdot t )\cdot
	{\rm sinc}(r \cdot \Delta f \cdot t
	).$$
	Here the following functions which can be converted into each other are used:		Here the following functions which can be converted into each other are used:
	:$${\rm si}(x) = \sin(x)/x,\hspace{0.5cm}{\rm sinc}(x) = \sin(\pi x)/(\pi x).$$		:$${\rm si}(x) = \sin(x)/x,\hspace{0.5cm}{\rm sinc}(x) = \sin(\pi x)/(\pi x).$$
	The available input signals are:		The available input signals are:
	*The sum of two harmonic oscillations:		*The sum of two harmonic oscillations:
	:$$x_1(t) = {1\, \rm V} \cdot \cos(\omega_1 \cdot t) + {1\, \rm V} \cdot \sin(\omega_2 \~~cdot~~		:$$x_1(t) = {1\, \rm V} \cdot \cos(\omega_1 \cdot t) + {1\, \rm V} \cdot \sin(\omega_2 \cdott).$$
	t).$$
	:Here, the following holds:   $\omega_1 = 2\pi \cdot 2000 \ {\rm 1/s}$  and  $\omega_2 \gt \omega_1$.		:Here, the following holds:   $\omega_1 = 2\pi \cdot 2000 \ {\rm 1/s}$  and  $\omega_2 \gt \omega_1$.
	*A periodic triangular signal:		*A periodic triangular signal:
	:$$x_2(t) = \frac{8\, \rm V}{\pi^2} \cdot \big[\cos(\omega_0 t) + {1}/{9} \cdot \cos(3\omega_0 t)		:$$x_2(t) = \frac{8\, \rm V}{\pi^2} \cdot \big[\cos(\omega_0 t) + {1}/{9} \cdot \cos(3\omega_0 t)+ {1}/{25} \cdot \cos(5\omega_0 t) + \hspace{0.05cm}\text{...}\big].$$
	+ {1}/{25} \cdot \cos(5\omega_0 t) + \hspace{0.05cm}\text{...}\big].$$
	:It should be noted that the basic frequency is  $f_0 = 2 \ \rm kHz$  or  $3\ \rm kHz$.  At time  $t = 0$  the signal value in both cases is  $1 \ \rm V$.		:It should be noted that the basic frequency is  $f_0 = 2 \ \rm kHz$  or  $3\ \rm kHz$.  At time  $t = 0$  the signal value in both cases is  $1 \ \rm V$.
	*A rectangular pulse  $x_3(t)$  with amplitude  $A = 1 \ \rm V$  and duration  $T = 1 \ \rm ms$. <br>Since its spectrum  $X_3(f)$  extends to infinity,   $H(f)$  always results in linear distortions here.		*A rectangular pulse  $x_3(t)$  with amplitude  $A = 1 \ \rm V$  and duration  $T = 1 \ \rm ms$. <br>Since its spectrum  $X_3(f)$  extends to infinity,   $H(f)$  always results in linear distortions here.
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	'''(4)'''  The output signal  $y_2(t)$  has the following form taking into account the basic frequency  $f_0 = 3 \ \rm kHz$:		'''(4)'''  The output signal  $y_2(t)$  has the following form taking into account the basic frequency  $f_0 = 3 \ \rm kHz$:
	:$$y_2(t)= \frac{8\,{\rm V}}{\pi^2} \left( \cos(\omega_0 t) +		:$$y_2(t)= \frac{8\,{\rm V}}{\pi^2} \left( \cos(\omega_0 t) +\frac{3}{8}\cdot \frac{1}{9} \cdot \cos(3\omega_0 t)\right).$$
	\frac{3}{8}\cdot \frac{1}{9} \cdot \cos(3\omega_0 t)\right)
	.$$

	*The factor  $3/8$  describes  $H(f = 9 \ \rm kHz)$.  All other spectral components at  $15 \ \rm kHz$,  $21 \ \rm kHz$,  etc. are suppressed by the system.		*The factor  $3/8$  describes  $H(f = 9 \ \rm kHz)$.  All other spectral components at  $15 \ \rm kHz$,  $21 \ \rm kHz$,  etc. are suppressed by the system.
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	*The strongest deviations between  $x_2(t)$  and  $y_2(t)$  will occur at the triangle peaks since the missing high frequencies have the strongest effect here.		*The strongest deviations between  $x_2(t)$  and  $y_2(t)$  will occur at the triangle peaks since the missing high frequencies have the strongest effect here.
	*For example, for the time  $\underline{t= 0}$ one obtains:		*For example, for the time  $\underline{t= 0}$ one obtains:
	:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +		:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +{3}/{72}\right)= 0.844\,{\rmV} \hspace{0.3cm}\Rightarrow\hspace{0.3cm}\varepsilon_{\rm max} = \|y_2(t=0)- x_2(t=0)\| \hspace{0.15cm}\underline{= 0.156\,{\rmV}}.$$
	{3}/{72}\right)= 0.844\,{\rm
	V} \hspace{0.3cm}\Rightarrow\hspace{0.3cm}
	\varepsilon_{\rm max} = \|y_2(t=0)- x_2(t=0)\| \hspace{0.15cm}\underline{= 0.156\,{\rm
	V}}.$$



	'''(5)'''  With the basic frequency  $f_0 = 2 \ \rm kHz$  and the values  $H(3f_0) = 0.75$,  $H(5f_0) = 0.25$,  $H(7f_0) = 0$  the following is obtained:		'''(5)'''  With the basic frequency  $f_0 = 2 \ \rm kHz$  and the values  $H(3f_0) = 0.75$,  $H(5f_0) = 0.25$,  $H(7f_0) = 0$  the following is obtained:
	:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +		:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +\frac{3}{4}\cdot \frac{1}{9} + \frac{1}{4} \cdot\frac{1}{25}\right)= 0.886\,{\rmV}\hspace{0.5cm}\Rightarrow \hspace{0.5cm}\varepsilon_{\rm max} \hspace{0.15cm}\underline{= 0.114\,{\rmV}}.$$
	\frac{3}{4}\cdot \frac{1}{9} + \frac{1}{4} \cdot\frac{1}{25}\right)= 0.886\,{\rm
	V}\hspace{0.5cm}
	\Rightarrow \hspace{0.5cm}\varepsilon_{\rm max} \hspace{0.15cm}\underline{= 0.114\,{\rm
	V}}.$$



	'''(6)'''  In the range up to  $4 \ \rm kHz$,  $H_{\rm E}(f) = H(f) = 1$  is to be set.  In contrast, in the range from  $4 \ \rm kHz$  to  $12 \ \rm kHz$ the following holds:		'''(6)'''  In the range up to  $4 \ \rm kHz$,  $H_{\rm E}(f) = H(f) = 1$  is to be set.  In contrast, in the range from  $4 \ \rm kHz$  to  $12 \ \rm kHz$ the following holds:
	:$$H_{\rm E}(f)= \frac{1}{H(f)} =		:$$H_{\rm E}(f)= \frac{1}{H(f)} =\frac{1}{1.5 \cdot \big[1 - f/(12\,{\rm kHz})\big]}\hspace{0.5cm} \Rightarrow \hspace{0.5cm}H_{\rm E}(f = 10\,{\rm kHz})\hspace{0.15cm}\underline{= 4}.$$
	\frac{1}{1.5 \cdot \big[1 - f/(12\,{\rm kHz})\big]}
	\hspace{0.5cm} \Rightarrow \hspace{0.5cm}
	H_{\rm E}(f = 10\,{\rm kHz})\hspace{0.15cm}\underline{= 4}
	.$$

	Here, the denominator expression describes the equation of the straight line of the frequency roll-off.		Here, the denominator expression describes the equation of the straight line of the frequency roll-off.

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]
			[[de:Aufgaben:2.5_Verzerrung_und_Entzerrung]]

Guenter at 14:56, 5 October 2021

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Guenter at 14:46, 5 October 2021

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	{For the test signal  $x_2(t)$  with  $\underline{f_0 = 3 \ \rm kHz}$,  what is the maximum deviation $\varepsilon_{\rm max} = \|y_2(t_0) - x_2(t_0)\|$. <br>At what time $t_0$  does  $\varepsilon_{\rm max}$  occur for the first time?		{For the test signal  $x_2(t)$  with  $\underline{f_0 = 3 \ \rm kHz}$,  what is the maximum deviation  $\varepsilon_{\rm max} = \|y_2(t_0) - x_2(t_0)\|$. <br>At what time  $t_0$  does  $\varepsilon_{\rm max}$  occur for the first time?
	\|type="{}"}		\|type="{}"}
	$\varepsilon_\text{max} \ = \ $ { 0.156 3% } $\ \rm V$		$\varepsilon_\text{max} \ = \ $ { 0.156 3% } $\ \rm V$
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	{What is the maximum deviation $\varepsilon_{\rm max}$  with $\underline{f_0 = 2 \ \rm kHz}$?		{What is the maximum deviation  $\varepsilon_{\rm max}$  with  $\underline{f_0 = 2 \ \rm kHz}$?
	\|type="{}"}		\|type="{}"}
	$\varepsilon_\text{max} \ = \ $ { 0.114 3% } $\ \rm V$		$\varepsilon_\text{max} \ = \ $ { 0.114 3% } $\ \rm V$

Guenter at 14:42, 5 October 2021

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	''Please note:''		''Please note:''
	*The task belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Linear_Distortions\|Linear Distortions]].		*The task belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Linear_Distortions\|Linear Distortions]].
	*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Linear_Distortions#~~Equalization_method~~\|Equalization methods]].		*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Linear_Distortions#Equalization_methods\|Equalization methods]].
	*The term  "overall distortion"  used in the formulation of the questions refers to the input signal  $x(t)$  and the output signal  $z(t)$.		*The term  "overall distortion"  used in the formulation of the questions refers to the input signal  $x(t)$  and the output signal  $z(t)$.

Guenter at 14:42, 5 October 2021

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	''Please note:''		''Please note:''
	*The task belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Linear_Distortions\|Linear Distortions]].		*The task belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Linear_Distortions\|Linear Distortions]].
	*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Linear_Distortions#~~Entzerrungsverfahren~~\|~~Entzerrungsverfahren~~]].		*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Linear_Distortions#Equalization_method\|Equalization methods]].
	*The term "overall distortion" used in the formulation of the questions refers to the input signal  $x(t)$  and the output signal  $z(t)$.		*The term  "overall distortion"  used in the formulation of the questions refers to the input signal  $x(t)$  and the output signal  $z(t)$.

Guenter at 14:39, 5 October 2021

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	+ {1}/{25} \cdot \cos(5\omega_0 t) + \hspace{0.05cm}\text{...}\big].$$		+ {1}/{25} \cdot \cos(5\omega_0 t) + \hspace{0.05cm}\text{...}\big].$$
	:It should be noted that the basic frequency is  $f_0 = 2 \ \rm kHz$  or  $3\ \rm kHz$.  At time  $t = 0$  the signal value in both cases is  $1 \ \rm V$.		:It should be noted that the basic frequency is  $f_0 = 2 \ \rm kHz$  or  $3\ \rm kHz$.  At time  $t = 0$  the signal value in both cases is  $1 \ \rm V$.
	*A rectangular pulse  $x_3(t)$  with amplitude  $A = 1 \ \rm V$  and duration  $T = 1 \ \rm ms$. Since its spectrum  $X_3(f)$  extends to infinity,   $H(f)$  always results in linear distortions here.		*A rectangular pulse  $x_3(t)$  with amplitude  $A = 1 \ \rm V$  and duration  $T = 1 \ \rm ms$. <br>Since its spectrum  $X_3(f)$  extends to infinity,   $H(f)$  always results in linear distortions here.

Guenter at 14:36, 5 October 2021

2021-10-05T14:36:44Z

Guenter: Guenter moved page Aufgaben:Exercise 2.5: Distortion and Equalisation to Aufgaben:Exercise 2.5: Distortion and Equalization

2021-10-05T12:21:24Z

Guenter moved page Aufgaben:Exercise 2.5: Distortion and Equalisation to Aufgaben:Exercise 2.5: Distortion and Equalization

← Older revision	Revision as of 14:21, 5 October 2021
(No difference)

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Proposed solutions 1 and 3</u> are correct:		'''(1)'''  <u>Proposed solutions 1 and 3</u>  are correct:
	*A linear system is already implicitly assumed by specifying a frequency response so that nonlinear distortions cannot occur.		*A linear system is already implicitly assumed by specifying a frequency response so that nonlinear distortions cannot occur.
	*Since  $H(f)$  is purely real, phase distortions can also be ruled out.		*Since  $H(f)$  is purely real, phase distortions can also be ruled out.
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	'''(3)'''  <u>Proposed solution 3</u> is correct:		'''(3)'''  <u>Proposed solution 3</u>  is correct:
	*In this case, the following is obtained for the output signal:		*In this case, the following is obtained for the output signal:
	:$$y_1(t)= 1\,{\rm V}\cdot \cos(2 \pi \cdot f_1 \cdot t) + {1}/{4}\cdot 1\,{\rm V}\cdot \sin(2 \pi \cdot f_2 \cdot t).$$		:$$y_1(t)= 1\,{\rm V}\cdot \cos(2 \pi \cdot f_1 \cdot t) + {1}/{4}\cdot 1\,{\rm V}\cdot \sin(2 \pi \cdot f_2 \cdot t).$$
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	'''(4)'''  The output signal  $y_2(t)$  has the following form taking into account the ~~fundamental~~ frequency  $f_0 = 3 \ \rm kHz$~~ ~~:		'''(4)'''  The output signal  $y_2(t)$  has the following form taking into account the basic frequency  $f_0 = 3 \ \rm kHz$:
	:$$y_2(t)= \frac{8\,{\rm V}}{\pi^2} \left( \cos(\omega_0 t) +		:$$y_2(t)= \frac{8\,{\rm V}}{\pi^2} \left( \cos(\omega_0 t) +
	\frac{3}{8}\cdot \frac{1}{9} \cdot \cos(3\omega_0 t)\right)		\frac{3}{8}\cdot \frac{1}{9} \cdot \cos(3\omega_0 t)\right)
	.$$		.$$

	*The factor  $3/8$  describes  $H(f = 9 \ \rm kHz)$. All other spectral components at  $15 \ \rm kHz$,  $21 \ \rm kHz$,  etc. are suppressed by the system.		*The factor  $3/8$  describes  $H(f = 9 \ \rm kHz)$.  All other spectral components at  $15 \ \rm kHz$,  $21 \ \rm kHz$,  etc. are suppressed by the system.

	*The strongest deviations between  $x_2(t)$  and  $y_2(t)$  will occur at the triangle peaks since the missing high frequencies have the strongest effect here. For example, for the time  $\underline{t= 0}$ one obtains:		*The strongest deviations between  $x_2(t)$  and  $y_2(t)$  will occur at the triangle peaks since the missing high frequencies have the strongest effect here.
			*For example, for the time  $\underline{t= 0}$ one obtains:
	:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +		:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +
	{3}/{72}\right)= 0.844\,{\rm		{3}/{72}\right)= 0.844\,{\rm
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	'''(5)'''  With the ~~fundamental~~ frequency  $f_0 = 2 \ \rm kHz$  and the ~~transmission~~ values  $H(3f_0) = 0.75$,  $H(5f_0) = 0.25$,  $H(7f_0) = 0$  the following is obtained:		'''(5)'''  With the basic frequency  $f_0 = 2 \ \rm kHz$  and the values  $H(3f_0) = 0.75$,  $H(5f_0) = 0.25$,  $H(7f_0) = 0$  the following is obtained:
	:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +		:$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 +
	\frac{3}{4}\cdot \frac{1}{9} + \frac{1}{4} \cdot\frac{1}{25}\right)= 0.886\,{\rm		\frac{3}{4}\cdot \frac{1}{9} + \frac{1}{4} \cdot\frac{1}{25}\right)= 0.886\,{\rm
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	'''(6)'''  In the range up to  $4 \ \rm kHz$,  $H_{\rm E}(f) = H(f) = 1$  is to be set. In contrast, in the range from  $4 \ \rm kHz$  to  $12 \ \rm kHz$ the following holds:		'''(6)'''  In the range up to  $4 \ \rm kHz$,  $H_{\rm E}(f) = H(f) = 1$  is to be set.  In contrast, in the range from  $4 \ \rm kHz$  to  $12 \ \rm kHz$ the following holds:
	:$$H_{\rm E}(f)= \frac{1}{H(f)} =		:$$H_{\rm E}(f)= \frac{1}{H(f)} =
	\frac{1}{1.5 \cdot \big[1 - f/(12\,{\rm kHz})\big]}		\frac{1}{1.5 \cdot \big[1 - f/(12\,{\rm kHz})\big]}

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	}}		}}

	[[File:P_ID907__LZI_A_2_5.png\|right\|frame\|Trapezoidal spectrum (top) ~~and~~ <br>associated impulse response ~~(bottom~~)]]		[[File:P_ID907__LZI_A_2_5.png\|right\|frame\|Trapezoidal spectrum (top), <br>associated impulse response)]]
	A communication system with input  $x(t)$  and output  $y(t)$, which is fully described by the trapezoidal frequency response  $H(f)$  according to the top graph, is considered. Using the roll-off factor  $r = 0.5$  and the equivalent bandwidth  $\Delta f = 16 \ \rm kHz$  the corresponding impulse response, which is computable by applying the inverse Fourier transform, is:		A communication system with input  $x(t)$  and output  $y(t)$,  which is fully described by the trapezoidal frequency response  $H(f)$  according to the top graph, is considered.  Using the roll-off factor  $r = 0.5$  and the equivalent bandwidth  $\Delta f = 16 \ \rm kHz$  the corresponding impulse response,  which is computable by applying the inverse Fourier transform, is:
	:$$h(t) = \Delta f \cdot {\rm si}(\pi \cdot \Delta f \cdot t )\cdot		:$$h(t) = \Delta f \cdot {\rm si}(\pi \cdot \Delta f \cdot t )\cdot
	{\rm si}(\pi \cdot r \cdot \Delta f \cdot t		{\rm si}(\pi \cdot r \cdot \Delta f \cdot t
	) .$$		) = \Delta f \cdot {\rm sinc}(\Delta f \cdot t )\cdot
			{\rm sinc}(r \cdot \Delta f \cdot t
			).$$
			Here the following functions which can be converted into each other are used:
			:$${\rm si}(x) = \sin(x)/x,\hspace{0.5cm}{\rm sinc}(x) = \sin(\pi x)/(\pi x).$$
	The available input signals are:		The available input signals are:
	*The sum of two harmonic oscillations:		*The sum of two harmonic oscillations:
	:$$x_1(t) = {1\, \rm V} \cdot \cos(\omega_1 \cdot t) + {1\, \rm V} \cdot \sin(\omega_2 \cdot		:$$x_1(t) = {1\, \rm V} \cdot \cos(\omega_1 \cdot t) + {1\, \rm V} \cdot \sin(\omega_2 \cdot
	t).$$		t).$$
	:Here, the following holds:  $\omega_1 = 2\pi \cdot 2000 \ {\rm 1/s}$  and  $\omega_2 \gt \omega_1$.		:Here, the following holds:   $\omega_1 = 2\pi \cdot 2000 \ {\rm 1/s}$  and  $\omega_2 \gt \omega_1$.
	*A periodic triangular signal:		*A periodic triangular signal:
	:$$x_2(t) = \frac{8\, \rm V}{\pi^2} \cdot \big[\cos(\omega_0 t) + {1}/{9} \cdot \cos(3\omega_0 t)		:$$x_2(t) = \frac{8\, \rm V}{\pi^2} \cdot \big[\cos(\omega_0 t) + {1}/{9} \cdot \cos(3\omega_0 t)
	+ {1}/{25} \cdot \cos(5\omega_0 t) + \hspace{0.05cm}\text{...}\big].$$		+ {1}/{25} \cdot \cos(5\omega_0 t) + \hspace{0.05cm}\text{...}\big].$$
	:It should be noted that the ~~fundamental~~ frequency is  $f_0 = 2 \ \rm kHz$  or  $3\ \rm kHz$ . At time  $t = 0$  the signal value in both cases is  $1 \ \rm V$.		:It should be noted that the basic frequency is  $f_0 = 2 \ \rm kHz$  or  $3\ \rm kHz$.  At time  $t = 0$  the signal value in both cases is  $1 \ \rm V$.
	*A rectangular pulse  $x_3(t)$  with amplitude  $A = 1 \ \rm V$  and duration  $T = 1 \ \rm ms$. Since its spectrum  $X_3(f)$  extends to infinity,   $H(f)$  always results in linear distortions here.		*A rectangular pulse  $x_3(t)$  with amplitude  $A = 1 \ \rm V$  and duration  $T = 1 \ \rm ms$. Since its spectrum  $X_3(f)$  extends to infinity,   $H(f)$  always results in linear distortions here.