Aufgaben:Exercise 4.1: Low-Pass and Band-Pass Signals - Revision history

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	:$$F_x = \int_{- \infty}^{+\infty}\hspace{-0.4cm}x(t)\hspace{0.1cm}{\rm d}t , \hspace{0.5cm}F_d = \int_{- \infty}^{+\infty}\hspace{-0.4cm}d(t)\hspace{0.1cm}{\rm d}t .$$		:$$F_x = \int_{- \infty}^{+\infty}\hspace{-0.4cm}x(t)\hspace{0.1cm}{\rm d}t , \hspace{0.5cm}F_d = \int_{- \infty}^{+\infty}\hspace{-0.4cm}d(t)\hspace{0.1cm}{\rm d}t .$$

	On the other hand, for the corresponding signal energies with  [[Signal_Representation/~~Equivalent_Low_Pass_Signal_and_Its_Spectral_Function~~#~~Power_and_Energy_of_a_Bandpass_Signal~~\|Parseval's theorem]]:		On the other hand, for the corresponding signal energies with  [[Signal_Representation/Equivalent_Low-Pass_Signal_and_its_Spectral_Function#Power_and_energy_of_a_band-pass_signal\|Parseval's theorem]]:

	:$$E_x = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|x(t)\|^2\hspace{0.1cm}{\~~rmd~~}t = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|X(f)\|^2\hspace{0.1cm}{\~~rmd~~}f ,$$		:$$E_x = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|x(t)\|^2\hspace{0.1cm}{\rm d}t = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|X(f)\|^2\hspace{0.1cm}{\rm d}f ,$$

	:$$E_d = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|d(t)\|^2\hspace{0.1cm}{\~~rmd~~}t = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|D(f)\|^2\hspace{0.1cm}{\~~rmd~~}f .$$		:$$E_d = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|d(t)\|^2\hspace{0.1cm}{\rm d}t = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|D(f)\|^2\hspace{0.1cm}{\rm d}f .$$


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	:$$E_x = (10^{-3} \hspace{0.1cm}{\rm V/Hz})^2 \cdot 2 \cdot 5\hspace{0.1cm}{\rm kHz} \hspace{0.15 cm}\underline{= 0.01 \hspace{0.1cm}{\rm V^2s}},$$		:$$E_x = (10^{-3} \hspace{0.1cm}{\rm V/Hz})^2 \cdot 2 \cdot 5\hspace{0.1cm}{\rm kHz} \hspace{0.15 cm}\underline{= 0.01 \hspace{0.1cm}{\rm V^2s}},$$

	:$$E_d = (10^{-3} \hspace{0.1cm}{\rm V/Hz})^2 \cdot 2 \cdot 2\hspace{0.1cm}{\rm kHz} \hspace{0.15 cm}\underline{= 0.004 \hspace{0.1cm}{\~~rmV~~^2s}}.$$		:$$E_d = (10^{-3} \hspace{0.1cm}{\rm V/Hz})^2 \cdot 2 \cdot 2\hspace{0.1cm}{\rm kHz} \hspace{0.15 cm}\underline{= 0.004 \hspace{0.1cm}{\rm V^2s}}.$$

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	[[de:Aufgaben:4.~~1_TP~~-~~_und_BP~~-Signale]]		[[de:Aufgaben:Aufgabe 4.1: Tiefpass- und Bandpass-Signale]]

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	Three signal curves are sketched on the right, the first two having the following curve:		Three signal curves are sketched on the right, the first two having the following curve:

	:$$x(t) = 10\hspace{0.05cm}{\rm V} \cdot {\rm si} ( \pi \cdot		:$$x(t) = 10\hspace{0.05cm}{\rm V} \cdot {\rm si} ( \pi \cdot{t}/{T_x}) ,$$
	{t}/{T_x}) ,$$

	:$$y(t) = 6\hspace{0.05cm}{\rm V} \cdot {\rm si}( \pi \cdot		:$$y(t) = 6\hspace{0.05cm}{\rm V} \cdot {\rm si}( \pi \cdot{t}/{T_y}) .$$
	{t}/{T_y}) .$$

	$T_x = 100 \,{\rm µ}\text{s}$  and  $T_y = 166.67 \,{\rm µ}\text{s}$  indicate the first zero of  $x(t)$  and  $y(t)$ respectively.		$T_x = 100 \,{\rm µ}\text{s}$  and  $T_y = 166.67 \,{\rm µ}\text{s}$  indicate the first zero of  $x(t)$  and  $y(t)$ respectively.
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	On the other hand, for the corresponding signal energies with  [[Signal_Representation/Equivalent_Low_Pass_Signal_and_Its_Spectral_Function#Power_and_Energy_of_a_Bandpass_Signal\|Parseval's theorem]]:		On the other hand, for the corresponding signal energies with  [[Signal_Representation/Equivalent_Low_Pass_Signal_and_Its_Spectral_Function#Power_and_Energy_of_a_Bandpass_Signal\|Parseval's theorem]]:

	:$$E_x = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|x(t)\|^2\hspace{0.1cm}{\rm		:$$E_x = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|x(t)\|^2\hspace{0.1cm}{\rmd}t = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|X(f)\|^2\hspace{0.1cm}{\rmd}f ,$$
	d}t = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|X(f)\|^2\hspace{0.1cm}{\rm
	d}f ,$$

	:$$E_d = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|d(t)\|^2\hspace{0.1cm}{\rm		:$$E_d = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|d(t)\|^2\hspace{0.1cm}{\rmd}t = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|D(f)\|^2\hspace{0.1cm}{\rmd}f .$$
	d}t = \int_{- \infty}^{+\infty}\hspace{-0.4cm}\|D(f)\|^2\hspace{0.1cm}{\rm
	d}f .$$


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	*The inverse Fourier transform of a rectangular spectrum  $X(f)$  leads to an  $\rm si$–shaped time function $x(t)$:		*The inverse Fourier transform of a rectangular spectrum  $X(f)$  leads to an  $\rm si$–shaped time function $x(t)$:

	:$$X(f)=\left\{ {X_0 \; \rm f\ddot{u}r\; \|\it f\| < \rm B, \atop {\rm 0 \;\;\; \rm sonst}}\right. \;\;		:$$X(f)=\left\{ {X_0 \; \rm f\ddot{u}r\; \|\it f\| < \rm B, \atop {\rm 0 \;\;\; \rm sonst}}\right. \;\;\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \;\;x(t) = 2 \cdot X_0 \cdot B \cdot {\rm si} ( 2\pi B t) .$$
	\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \;\;x(t) = 2 \cdot X_0 \cdot B \cdot {\rm si} ( 2\pi B t) .$$

	*In this task, the  function $\rm si(x) = \rm sin(x)/x = \rm sinc(x/π)$  is used.		*In this task, the  function $\rm si(x) = \rm sin(x)/x = \rm sinc(x/π)$  is used.
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	*The absolute, two-sided bandwidth  $2 \cdot B_x$  is equal to the reciprocal of the first zero.  It follows that:		*The absolute, two-sided bandwidth  $2 \cdot B_x$  is equal to the reciprocal of the first zero.  It follows that:

	:$$B_x = \frac{1}{2 \cdot T_x} = \frac{1}{2 \cdot 0.1		:$$B_x = \frac{1}{2 \cdot T_x} = \frac{1}{2 \cdot 0.1\hspace{0.1cm}{\rm ms}}\hspace{0.15 cm}\underline{ = 5 \hspace{0.1cm}{\rm kHz}}.$$
	\hspace{0.1cm}{\rm ms}}\hspace{0.15 cm}\underline{ = 5 \hspace{0.1cm}{\rm kHz}}.$$

	*Since the signal value at  $t = 0$  is equal to the rectangular area, the constant height is given by:		*Since the signal value at  $t = 0$  is equal to the rectangular area, the constant height is given by:

	:$$X(f=0) = \frac{x(t=0)}{2 B_x} = \frac{10		:$$X(f=0) = \frac{x(t=0)}{2 B_x} = \frac{10\hspace{0.1cm}{\rm V}}{10 \hspace{0.1cm}{\rm kHz}} \hspace{0.15 cm}\underline{= 1\hspace{0.1cm}{\rm mV/Hz}}.$$
	\hspace{0.1cm}{\rm V}}{10 \hspace{0.1cm}{\rm kHz}} \hspace{0.15 cm}\underline{= 1
	\hspace{0.1cm}{\rm mV/Hz}}.$$


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	'''(4)'''  According to the fundamental laws of the Fourier transform, the integral over the time function is equal to the spectral value at  $f = 0$.  It follows: :		'''(4)'''  According to the fundamental laws of the Fourier transform, the integral over the time function is equal to the spectral value at  $f = 0$.  It follows: :

	:$$F_x = X(f=0) = \frac{x(t=0)}{2 \cdot B_x} = 10^{-3}		:$$F_x = X(f=0) = \frac{x(t=0)}{2 \cdot B_x} = 10^{-3}\hspace{0.1cm}{\rm V/Hz}\hspace{0.15 cm}\underline{= 0.001 \hspace{0.1cm}{\rm Vs}},$$
	\hspace{0.1cm}{\rm V/Hz}\hspace{0.15 cm}\underline{= 0.001 \hspace{0.1cm}{\rm Vs}},$$

	:$$F_d = D(f=0) \hspace{0.15 cm}\underline{= 0}.$$		:$$F_d = D(f=0) \hspace{0.15 cm}\underline{= 0}.$$
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	'''(5)'''  In both cases, the calculation of the signal energy is easier in the frequency domain than in the time domain, because here the integration can be reduced to an area calculation of rectangles:		'''(5)'''  In both cases, the calculation of the signal energy is easier in the frequency domain than in the time domain, because here the integration can be reduced to an area calculation of rectangles:

	:$$E_x = (10^{-3} \hspace{0.1cm}{\rm V/Hz})^2 \cdot 2 \cdot 5		:$$E_x = (10^{-3} \hspace{0.1cm}{\rm V/Hz})^2 \cdot 2 \cdot 5\hspace{0.1cm}{\rm kHz} \hspace{0.15 cm}\underline{= 0.01 \hspace{0.1cm}{\rm V^2s}},$$
	\hspace{0.1cm}{\rm kHz} \hspace{0.15 cm}\underline{= 0.01 \hspace{0.1cm}{\rm V^2s}},$$

	:$$E_d = (10^{-3} \hspace{0.1cm}{\rm V/Hz})^2 \cdot 2 \cdot 2		:$$E_d = (10^{-3} \hspace{0.1cm}{\rm V/Hz})^2 \cdot 2 \cdot 2\hspace{0.1cm}{\rm kHz} \hspace{0.15 cm}\underline{= 0.004 \hspace{0.1cm}{\rmV^2s}}.$$
	\hspace{0.1cm}{\rm kHz} \hspace{0.15 cm}\underline{= 0.004 \hspace{0.1cm}{\rm
	V^2s}}.$$

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	[[Category:Signal Representation: Exercises\|^4.1 Differences between Low-Pass and Band-Pass^]]		[[Category:Signal Representation: Exercises\|^4.1 Differences between Low-Pass and Band-Pass^]]
			[[de:Aufgaben:4.1_TP-_und_BP-Signale]]

Guenter at 13:38, 5 May 2021

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	\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \;\;x(t) = 2 \cdot X_0 \cdot B \cdot {\rm si} ( 2\pi B t) .$$		\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \;\;x(t) = 2 \cdot X_0 \cdot B \cdot {\rm si} ( 2\pi B t) .$$

	*In this task, the function $\rm si(x) = \rm sin(x)/x = \rm sinc(x/π)$ is used.		*In this task, the  function $\rm si(x) = \rm sin(x)/x = \rm sinc(x/π)$  is used.

Guenter at 12:19, 5 May 2021

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	[[Category:Signal Representation: Exercises\|^4.1 Differences between Low-Pass and Band-Pass^]]		[[Category:Signal Representation: Exercises\|^4.1 Differences between Low-Pass and Band-Pass^]]

Guenter at 12:17, 5 May 2021

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	[[Category:Signal Representation: Exercises\|^4.1 ~~Basic information on low~~-~~pass~~ and Band-Pass ~~Signals~~^]]		[[Category:Signal Representation: Exercises\|^4.1 Differences between Low-Pass and Band-Pass^]]

Guenter at 12:14, 5 May 2021

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	[[Category:Signal Representation: Exercises\|^4.1 ~~Differences and Similarities of Low~~-~~Pass~~ and Band-Pass Signals^]]		[[Category:Signal Representation: Exercises\|^4.1 Basic information on low-pass and Band-Pass Signals^]]

Guenter at 12:12, 5 May 2021

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	===Solution===		===Solution===
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	'''(1)'''  ~~Die si–shaped~~ time function  $x(t)$  suggests a rectangular spectrum  $X(f)$ .		'''(1)'''  The  $\rm si$–shaped time function  $x(t)$  suggests a rectangular spectrum  $X(f)$ .
	*The absolute, two-sided bandwidth  $2 \cdot B_x$  is equal to the reciprocal of the first zero. It follows that:		*The absolute, two-sided bandwidth  $2 \cdot B_x$  is equal to the reciprocal of the first zero.  It follows that:

	:$$B_x = \frac{1}{2 \cdot T_x} = \frac{1}{2 \cdot 0.1		:$$B_x = \frac{1}{2 \cdot T_x} = \frac{1}{2 \cdot 0.1
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	[[File:P_ID701__Sig_A_4_1_c_neu.png\|right\|frame\|Rectangular BP spectrum]]		[[File:P_ID701__Sig_A_4_1_c_neu.png\|right\|frame\|Rectangular band-pass spectrum]]
	'''(3)'''  From  $d(t) = x(t) - y(t)$  follows because of the linearity of the Fourier ~~transformation~~:   $D(f) = X(f) - Y(f).$		'''(3)'''  From  $d(t) = x(t) - y(t)$  follows because of the linearity of the Fourier transform:   $D(f) = X(f) - Y(f).$

	*The difference of the two equal ~~square wave~~ functions leads to a rectangular band-pass spectrum between  $3 \,\text{kHz}$  and  $5 \,\text{kHz}$.		*The difference of the two equal high rectangular functions leads to a rectangular band-pass spectrum between  $3 \,\text{kHz}$  and  $5 \,\text{kHz}$.
	*The (one-sided) bandwidth is thus  $B_d \;\underline{= 2 \,\text{kHz}}$. In this frequency interval  $D(f) = 1 \,\text{mV/Hz}$. Outside, i.e. also at  $f = 0$, ~~gilt~~  $D(f)\;\underline{ = 0}$ applies.		*The (one-sided) bandwidth is thus  $B_d \;\underline{= 2 \,\text{kHz}}$.  In this frequency interval  $D(f) = 1 \,\text{mV/Hz}$.  Outside, i.e. also at  $f = 0$,   $D(f)\;\underline{ = 0}$ applies.




	'''(4)'''  According to the fundamental laws of the Fourier ~~transformation~~, the integral over the time function is equal to the spectral value at  $f = 0$. ~~Daraus folgt~~:		'''(4)'''  According to the fundamental laws of the Fourier transform, the integral over the time function is equal to the spectral value at  $f = 0$.  It follows: :

	:$$F_x = X(f=0) = \frac{x(t=0)}{2 \cdot B_x} = 10^{-3}		:$$F_x = X(f=0) = \frac{x(t=0)}{2 \cdot B_x} = 10^{-3}
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	[[Category:Signal Representation: Exercises\|^4.1 Differences/Similarities of LP and BP Signals^]]		[[Category:Signal Representation: Exercises\|^4.1 Differences and Similarities of Low-Pass and Band-Pass Signals^]]

Guenter at 11:58, 5 May 2021

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	{t}/{T_y}) .$$		{t}/{T_y}) .$$

	$T_x = 100 \,{\rm µ}\text{s}$  and  $T_y = 166.67 \,{\rm µ}\text{s}$  indicate the first zero of  $x(t)$  ~~bzw.~~  $y(t)$ respectively.		$T_x = 100 \,{\rm µ}\text{s}$  and  $T_y = 166.67 \,{\rm µ}\text{s}$  indicate the first zero of  $x(t)$  and  $y(t)$ respectively.
	The signal  $d(t)$  results from the difference of the two upper signals (lower graph):		The signal  $d(t)$  results from the difference of the two upper signals (lower graph):


	:$$d(t) = x(t)-y(t) .$$		:$$d(t) = x(t)-y(t) .$$

	In ~~the~~ subtask  '''(4)'''  the integral areas of the ~~impulsive signals~~  $x(t)$  and  $d(t)$  are asked for. For these holds:		In subtask  '''(4)'''  the integral areas of the pulses  $x(t)$  and  $d(t)$  are asked for.  For these holds:

	:$$F_x = \int_{- \infty}^{+\infty}\hspace{-0.4cm}x(t)\hspace{0.1cm}{\rm d}t , \hspace{0.5cm}F_d = \int_{- \infty}^{+\infty}\hspace{-0.4cm}d(t)\hspace{0.1cm}{\rm d}t .$$		:$$F_x = \int_{- \infty}^{+\infty}\hspace{-0.4cm}x(t)\hspace{0.1cm}{\rm d}t , \hspace{0.5cm}F_d = \int_{- \infty}^{+\infty}\hspace{-0.4cm}d(t)\hspace{0.1cm}{\rm d}t .$$
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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Signal_Representation/~~Differences_and_Similarities_of_LP_and_BP_Signals~~\|Differences and Similarities of LP and BP Signals]].		*This exercise belongs to the chapter  [[Signal_Representation/Differences_and_Similarities_of_Low-Pass_and_Band-Pass_Signals\|Differences and Similarities of Low-Pass and Band-Pass Signals]].

	*The Fourier ~~retransform~~ of a rectangular spectrum  $X(f)$  leads to an  $\rm si$–shaped time function $x(t)$:		*The inverse Fourier transform of a rectangular spectrum  $X(f)$  leads to an  $\rm si$–shaped time function $x(t)$:

	:$$X(f)=\left\{ {X_0 \; \rm f\ddot{u}r\; \|\it f\| < \rm B, \atop {\rm 0 \;\;\; \rm sonst}}\right. \;\;		:$$X(f)=\left\{ {X_0 \; \rm f\ddot{u}r\; \|\it f\| < \rm B, \atop {\rm 0 \;\;\; \rm sonst}}\right. \;\;
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	<quiz display=simple>		<quiz display=simple>
	{What is the spectrum  $X(f)$  of the signal  $x(t)$? What are the magnitudes of  $X(f = 0)$  and the physical, one-sided bandwidth  $B_x$  of  $x(t)$?		{What is the spectrum  $X(f)$  of the signal  $x(t)$?  What are the magnitudes of  $X(f = 0)$  and the physical, one-sided bandwidth  $B_x$  of  $x(t)$?
	\|type="{}"}		\|type="{}"}
	$X(f=0)\ = \ $ { 1 3% }  $\text{mV/Hz}$		$X(f=0)\ = \ $ { 1 3% }  $\text{mV/Hz}$
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	{Calculate the spectrum  $D(f)$  of the difference signal  $d(t) = x(t) - y(t)$. How large are  $D(f = 0)$  and the (one-sided) bandwidth  $B_d$?		{Calculate the spectrum  $D(f)$  of the difference signal  $d(t) = x(t) - y(t)$.  How large are  $D(f = 0)$  and the (one-sided) bandwidth  $B_d$?
	\|type="{}"}		\|type="{}"}
	$D(f=0)\ = \ $ { 0. }  $\text{mV/Hz}$		$D(f=0)\ = \ $ { 0. }  $\text{mV/Hz}$

Javier: Javier moved page Aufgaben:Exercise 4.1: Lowpass and Bandpass Signals to Aufgaben:Exercise 4.1: Low-Pass and Band-Pass Signals

2021-05-04T07:30:00Z

Javier moved page Aufgaben:Exercise 4.1: Lowpass and Bandpass Signals to Aufgaben:Exercise 4.1: Low-Pass and Band-Pass Signals

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