Aufgaben:Exercise 4.2: Rectangular Spectra - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:55:01Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:55, 16 March 2026
Line 70:		Line 70:
	:$$ W(f) = U(f- f_{\rm T}) + U(f+ f_{\rm T}) = U(f)\star \left[\delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$		:$$ W(f) = U(f- f_{\rm T}) + U(f+ f_{\rm T}) = U(f)\star \left[\delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$

	According to the  [[Signal_Representation/~~Fourier_Transform_Theorems~~#Shifting_Theorem\|Shifting Theorem]],  the following then applies to the associated time signal:		According to the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem\|Shifting Theorem]],  the following then applies to the associated time signal:

	:$$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) = 2 u_0\cdot {\rm si} ( \pi {t}/{T_{\rm u}})\cdot {\cos} ( 2 \pi f_{\rm T} t). $$		:$$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) = 2 u_0\cdot {\rm si} ( \pi {t}/{T_{\rm u}})\cdot {\cos} ( 2 \pi f_{\rm T} t). $$
Line 112:		Line 112:
	__NOEDITSECTION__		__NOEDITSECTION__
	[[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.~~2_Rechteckförmige_Spektren~~]]		[[de:Aufgaben:Aufgabe 4.2: Rechteckförmige Spektren]]

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:48:32Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:48, 16 March 2026
Line 15:		Line 15:
	This task also refers to the band-pass signal		This task also refers to the band-pass signal

	:$$d(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 \hspace{0.05cm}t)		:$$d(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 \hspace{0.05cm}t)- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2\hspace{0.05cm} t)$$
	- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2\hspace{0.05cm} t)$$

	whose spectrum was determined in  [[Aufgaben:Exercise_4.1Z:_High-Pass_System\|Exercise 4.1Z]] . Let  $f_2 = 2 \ \rm kHz.$		whose spectrum was determined in  [[Aufgaben:Exercise_4.1Z:_High-Pass_System\|Exercise 4.1Z]] . Let  $f_2 = 2 \ \rm kHz.$
Line 32:		Line 31:
	*Consider the following trigonometric relationship in the solution:		*Consider the following trigonometric relationship in the solution:

	:$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin		:$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin(\alpha + \beta)+ \sin (\alpha - \beta)\big].$$
	(\alpha + \beta)+ \sin (\alpha - \beta)\big].$$


Line 70:		Line 68:
	'''(2)'''  The band-pass spectrum can be represented with  $f_{\rm T} = 4\, \text{kHz}$  as follows:		'''(2)'''  The band-pass spectrum can be represented with  $f_{\rm T} = 4\, \text{kHz}$  as follows:

	:$$ W(f) = U(f- f_{\rm T}) + U(f+ f_{\rm T}) = U(f)\star \left[		:$$ W(f) = U(f- f_{\rm T}) + U(f+ f_{\rm T}) = U(f)\star \left[\delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$
	\delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$

	According to the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|Shifting Theorem]],  the following then applies to the associated time signal:		According to the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|Shifting Theorem]],  the following then applies to the associated time signal:

	:$$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) = 2 u_0		:$$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) = 2 u_0\cdot {\rm si} ( \pi {t}/{T_{\rm u}})\cdot {\cos} ( 2 \pi f_{\rm T} t). $$
	\cdot {\rm si} ( \pi {t}/{T_{\rm u}})\cdot {\cos} ( 2 \pi f_{\rm T} t). $$

	The graph shows		The graph shows
Line 90:		Line 86:
	The time  $t=62.5 \,{\rm µ} \text{s}$  corresponds exactly to a quarter of the period of the signal  $c(t)$:		The time  $t=62.5 \,{\rm µ} \text{s}$  corresponds exactly to a quarter of the period of the signal  $c(t)$:

	:$$ w(t = 62.5 \hspace{0.05cm}{\rm µ s}) = 2 u_0 \cdot {\rm si} ( \pi \cdot \frac{62.5 \hspace{0.05cm}{\rm µ s}}		:$$ w(t = 62.5 \hspace{0.05cm}{\rm µ s}) = 2 u_0 \cdot {\rm si} ( \pi \cdot \frac{62.5 \hspace{0.05cm}{\rm µ s}}{500 \hspace{0.05cm}{\rm µ s}})\cdot {\cos} ( 2 \pi \cdot 4\hspace{0.05cm}{\rm kHz}\cdot62.5 \hspace{0.05cm}{\rm µ s}) $$
	{500 \hspace{0.05cm}{\rm µ s}})		:$$ \Rightarrow \hspace{0.3cm}w(t =4\hspace{0.05cm}{\rm V}\cdot{\rm si} ( {\pi}/{8}) \cdot \cos ( {\pi}/{4})\hspace{0.15 cm}\underline{ = 0}.$$
	\cdot {\cos} ( 2 \pi \cdot 4\hspace{0.05cm}{\rm kHz}\~~cdot~~
	62.5 \hspace{0.05cm}{\rm µ s}) $$
	:$$ \Rightarrow \hspace{0.3cm}w(t =
	4\hspace{0.05cm}{\rm V}\cdot{\rm si} ( {\pi}/{8}) \cdot \cos ( {\pi}/{4})\hspace{0.15 cm}\underline{ = 0}.$$


Line 103:		Line 95:
	*This proof is somewhat more complex in the time domain.  With  $f_2 = 2 \,\text{kHz}$  can be written for the signal considered here:		*This proof is somewhat more complex in the time domain.  With  $f_2 = 2 \,\text{kHz}$  can be written for the signal considered here:

	:$$w(t ) = 4\hspace{0.05cm}{\rm V}		:$$w(t ) = 4\hspace{0.05cm}{\rm V}\cdot {\rm si} ( \pi f_2 t)\cdot {\cos} ( 4 \pi f_2 t) =({4\hspace{0.05cm}{\rm V}})/({\pi f_2 t})\cdot \sin (\pi f_2 t) \cdot \cos ( 4 \pi f_2 t) .$$
	\cdot {\rm si} ( \pi f_2 t)
	\cdot {\cos} ( 4 \pi f_2 t) =
	({4\hspace{0.05cm}{\rm V}})/({\pi f_2 t})\cdot \sin (\pi f_2 t) \cdot \cos ( 4 \pi f_2 t) .$$

	*Because of the trigonometric relationship		*Because of the trigonometric relationship

	:$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin		:$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin(\alpha + \beta)+ \sin (\alpha - \beta)\big]$$
	(\alpha + \beta)+ \sin (\alpha - \beta)\big]$$

	:the above equation can be transformed:		:the above equation can be transformed:

	:$$w(t ) =		:$$w(t ) =\frac{2\hspace{0.05cm}{\rm V}}{\pi f_2 t}\cdot \big [\sin (5\pi f_2 t) + \sin (-3\pi f_2 t)\big ]= 10\hspace{0.05cm}{\rm V} \cdot \frac{\sin (5\pi f_2 t)}{5\pi f_2 t}-6\hspace{0.05cm}{\rm V} \cdot \frac{\sin (3\pi f_2 t)}{3\pi f_2 t}.$$
	\frac{2\hspace{0.05cm}{\rm V}}{\pi f_2 t}\cdot \big [\sin (5\pi f_2 t) + \sin (-3\pi f_2 t)\big ]
	= 10\hspace{0.05cm}{\rm V} \cdot \frac{\sin (5\pi f_2 t)}{5\pi f_2 t}-
	6\hspace{0.05cm}{\rm V} \cdot \frac{\sin (3\pi f_2 t)}{3\pi f_2 t}.$$

	*This shows that both signals are actually identical   ⇒   Proposed solution 1:		*This shows that both signals are actually identical   ⇒   Proposed solution 1:

	:$$w(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 t)		:$$w(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 t)- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2 t) = d(t).$$
	- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2 t) = d(t).$$
	{{ML-Fuß}}		{{ML-Fuß}}

Maintenance script: Add German interlanguage link

2026-03-16T12:33:27Z

Add German interlanguage link

← Older revision		Revision as of 14:33, 16 March 2026
Line 128:		Line 128:
	__NOEDITSECTION__		__NOEDITSECTION__
	[[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:4.2_Rechteckförmige_Spektren]]

Guenter at 13:49, 5 May 2021

2021-05-05T13:49:48Z

← Older revision		Revision as of 15:49, 5 May 2021
Line 28:		Line 28:
	''Hints:''		''Hints:''
	*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]].		*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]].
	*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.

	*Consider the following trigonometric relationship in the solution:		*Consider the following trigonometric relationship in the solution:
Line 62:		Line 62:
	{{ML-Kopf}}		{{ML-Kopf}}

	'''(1)'''  The time  $T_u$   ⇒   first zero of the LP signal  $u(t)$  – is equal to the reciprocal of the width of the rectangular spectrum, i.e.   $1/(2\, \text{kHz} ) \hspace{0.15 cm}\underline{= 0.5 \, \text{ms}}$.		'''(1)'''  The time  $T_u$   ⇒   first zero of the low-pass signal  $u(t)$  – is equal to the reciprocal of the width of the rectangular spectrum, i.e.   $1/(2\, \text{kHz} ) \hspace{0.15 cm}\underline{= 0.5 \, \text{ms}}$.
	*The pulse amplitude is equal to the rectangular area as shown in the sample solution for  [[Aufgaben:Aufgabe_4.1:_Tiefpass-_und_Bandpass-Signale\|~~task~~ 4.1]] . From this follows  $u_0\hspace{0.15 cm}\underline{= 2 \, \text{V}}$.		*The pulse amplitude is equal to the rectangular area as shown in the sample solution for  [[Aufgaben:Aufgabe_4.1:_Tiefpass-_und_Bandpass-Signale\|Exercise 4.1]] .  From this follows  $u_0\hspace{0.15 cm}\underline{= 2 \, \text{V}}$.



	[[File:P_ID704__Sig_A_4_2_b_neu.png\|250px\|right\|frame\|Multiplication with cosine]]		[[File:P_ID704__Sig_A_4_2_b_neu.png\|250px\|right\|frame\|Multiplication with a cosine function]]
	'''(2)'''  The band-pass spectrum can be represented with  $f_{\rm T} = 4\, \text{kHz}$  as follows:		'''(2)'''  The band-pass spectrum can be represented with  $f_{\rm T} = 4\, \text{kHz}$  as follows:

Line 73:		Line 73:
	\delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$		\delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$

	According to the  [[Signal_Representation/~~Fourier_Transform_Laws~~#~~Verschiebungssatz~~\|~~Verschiebungssatz~~]] , the following then applies to the associated time signal:		According to the  [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem\|Shifting Theorem]],  the following then applies to the associated time signal:

	:$$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) = 2 u_0		:$$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) = 2 u_0
Line 80:		Line 80:
	The graph shows		The graph shows
	*above the low–pass signal $u(t)$,		*above the low–pass signal $u(t)$,
	*then the oscillation $c(t) = 2 · \cos(2 \pi ~~fTt~~$ ),		*then the oscillation $c(t) = 2 · \cos(2 \pi f_{\rm T}t$ ),
	*below the band-pass signal  $w(t) = u(t) \cdot c(t)$.		*below the band-pass signal  $w(t) = u(t) \cdot c(t)$.

Line 100:		Line 100:

	'''(3)'''  Proposed <u>solution 1 is correct</u>:		'''(3)'''  Proposed <u>solution 1 is correct</u>:
	*If we compare the spectral function  $W(f)$  of this task with the spectrum  $D(f)$  in the sample solution to  [[Aufgaben:4.~~1_TP~~-~~_und_BP~~-~~Signale~~\|~~task~~ 4.1]], we see that  $w(t)$  and  $d(t)$  are identical ~~signals~~.		*If we compare the spectral function  $W(f)$  of this task with the spectrum  $D(f)$  in the sample solution to  [[Aufgaben:Exercise_4.1:_Low-Pass_and_Band-Pass_Signals\|Exercise 4.1]], we see that  $w(t)$  and  $d(t)$  are identical.
	*~~Etwas aufwändiger ist dieser Beweis im Zeitbereich~~. ~~Mit~~  $f_2 = 2 \,\text{kHz}$  ~~kann für das hier betrachtete Signal geschrieben werden~~:		*This proof is somewhat more complex in the time domain.  With  $f_2 = 2 \,\text{kHz}$  can be written for the signal considered here:

	:$$w(t ) = 4\hspace{0.05cm}{\rm V}		:$$w(t ) = 4\hspace{0.05cm}{\rm V}
Line 108:		Line 108:
	({4\hspace{0.05cm}{\rm V}})/({\pi f_2 t})\cdot \sin (\pi f_2 t) \cdot \cos ( 4 \pi f_2 t) .$$		({4\hspace{0.05cm}{\rm V}})/({\pi f_2 t})\cdot \sin (\pi f_2 t) \cdot \cos ( 4 \pi f_2 t) .$$

	*~~Wegen der trigonometrischen Beziehung~~		*Because of the trigonometric relationship

	:$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin		:$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin

Guenter at 13:34, 5 May 2021

2021-05-05T13:34:51Z

← Older revision		Revision as of 15:34, 5 May 2021
Line 18:		Line 18:
	- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2\hspace{0.05cm} t)$$		- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2\hspace{0.05cm} t)$$

	whose spectrum was determined in  [[Aufgaben:~~Aufgabe_4~~.1:~~_Tiefpass~~-~~_und_Bandpass-Signale~~\|~~task~~ 4.1]] . Let  $f_2 = 2 \ \rm kHz.$		whose spectrum was determined in  [[Aufgaben:Exercise_4.1Z:_High-Pass_System\|Exercise 4.1Z]] . Let  $f_2 = 2 \ \rm kHz.$


Line 28:		Line 28:
	''Hints:''		''Hints:''
	*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]].		*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]].
			*In this task, the function $\rm si(x) = \rm sin(x)/x = \rm sinc(x/π)$ is used.

	*Consider the following trigonometric relationship in the solution:		*Consider the following trigonometric relationship in the solution:

Line 38:		Line 39:

	<quiz display=simple>		<quiz display=simple>
	{What are the values ~~of the parameters~~  $u_0$  and  $T_u$  of the low-pass signal?		{What are the parameter values  $u_0$  and  $T_u$  of the low-pass signal?
	\|type="{}"}		\|type="{}"}
	$u_0\ = \ $ { 2 3% }  $\text{V}$		$u_0\ = \ $ { 2 3% }  $\text{V}$
	$T_u\ = \ $ { 0.5 3% }  $\text{ms}$		$T_u\ = \ $ { 0.5 3% }  $\text{ms}$

	{Calculate the band-pass signal  $w(t)$. What are the signal values at  $t = 0$  and  $t = 62.5 \, {\rm µ}\text{s}$?		{Calculate the band-pass signal  $w(t)$.  What are the signal values at  $t = 0$  and  $t = 62.5 \, {\rm µ}\text{s}$?
	\|type="{}"}		\|type="{}"}
	$w(t=0)\ = \ $ { 4 3% }  $\text{V}$		$w(t=0)\ = \ $ { 4 3% }  $\text{V}$
	$w(t=62.5 \,{\rm µ} \text{s})\ = \ $ { 0. }  $\text{V}$		$w(t=62.5 \,{\rm µ} \text{s})\ = \ $ { 0. }  $\text{V}$

	{Which statements are true regarding the band-pass signals  $d(t)$  and  $w(t)$ ? Justify your result in the time domain.		{Which statements are true regarding the band-pass signals  $d(t)$  and  $w(t)$ ?  Justify your result in the time domain.
	\|type="[]"}		\|type="()"}
	+ The signals  $d(t)$  and  $w(t)$  are identical.		+ The signals  $d(t)$  and  $w(t)$  are identical.
	- $d(t)$  and  $w(t)$  differ by a constant factor.		- $d(t)$  and  $w(t)$  differ by a constant factor.

Guenter at 13:27, 5 May 2021

2021-05-05T13:27:29Z

← Older revision		Revision as of 15:27, 5 May 2021
Line 1:		Line 1:

	{{quiz-Header\|Buchseite=~~Signal Representation~~/~~Differences and Similarities of LP and BP Signals~~		{{quiz-Header\|Buchseite=Signal_Representation/Differences_and_Similarities_of_Low-Pass_and_Band-Pass_Signals
	}}		}}

	[[File:P_ID695__Sig_A_4_2_neu.png\|250px\|right\|frame\|~~Rectangular~~ low–pass and band-pass spectra]]		[[File:P_ID695__Sig_A_4_2_neu.png\|250px\|right\|frame\|Given low–pass and band-pass spectra]]
	We consider two signals  $u(t)$  and  $w(t)$  with rectangular spectra  $U(f)$  and  $W(f)$ respectively.		We consider two signals  $u(t)$  and  $w(t)$  with rectangular spectra  $U(f)$  and  $W(f)$ respectively.
	*It is obvious that		*It is obvious that
Line 27:		Line 27:

	''Hints:''		''Hints:''
	*~~The task~~ 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]].

	*Consider the following trigonometric relationship in the solution:		*Consider the following trigonometric relationship in the solution:

Guenter at 12:20, 5 May 2021

2021-05-05T12:20:19Z

← Older revision		Revision as of 14:20, 5 May 2021
Line 126:		Line 126:

	__NOEDITSECTION__		__NOEDITSECTION__
	[[Category:Signal Representation: Exercises\|^4.1 Differences~~/Similarities of LP~~ and ~~BP Signals~~^]]		[[Category:Signal Representation: Exercises\|^4.1 Differences between Low-Pass and Band-Pass^]]

Guenter: Guenter moved page Aufgaben:Exercise 4.2: Rectangular-Shaped Spectra to Aufgaben:Exercise 4.2: Rectangular Spectra

2021-05-05T09:31:16Z

Guenter moved page Aufgaben:Exercise 4.2: Rectangular-Shaped Spectra to Aufgaben:Exercise 4.2: Rectangular Spectra

← Older revision	Revision as of 11:31, 5 May 2021
(No difference)

Javier: Text replacement - "bandpass" to "band-pass"

2021-05-03T15:13:21Z

Text replacement - "bandpass" to "band-pass"

← Older revision		Revision as of 17:13, 3 May 2021
Line 3:		Line 3:
	}}		}}

	[[File:P_ID695__Sig_A_4_2_neu.png\|250px\|right\|frame\|Rectangular low–pass and ~~bandpass~~ spectra]]		[[File:P_ID695__Sig_A_4_2_neu.png\|250px\|right\|frame\|Rectangular low–pass and band-pass spectra]]
	We consider two signals  $u(t)$  and  $w(t)$  with rectangular spectra  $U(f)$  and  $W(f)$ respectively.		We consider two signals  $u(t)$  and  $w(t)$  with rectangular spectra  $U(f)$  and  $W(f)$ respectively.
	*It is obvious that		*It is obvious that
Line 10:		Line 10:

	:is a low-pass signal whose two parameters  $u_0$  and  $T_u$  are to be determined in subtask  '''(1)''' .		:is a low-pass signal whose two parameters  $u_0$  and  $T_u$  are to be determined in subtask  '''(1)''' .
	*In contrast, the spectrum  $W(f)$ shows that  $w(t)$  describes a ~~bandpass~~ signal.		*In contrast, the spectrum  $W(f)$ shows that  $w(t)$  describes a band-pass signal.


	This task also refers to the ~~bandpass~~ signal		This task also refers to the band-pass signal

	:$$d(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 \hspace{0.05cm}t)		:$$d(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 \hspace{0.05cm}t)
Line 43:		Line 43:
	$T_u\ = \ $ { 0.5 3% }  $\text{ms}$		$T_u\ = \ $ { 0.5 3% }  $\text{ms}$

	{Calculate the ~~bandpass~~ signal  $w(t)$. What are the signal values at  $t = 0$  and  $t = 62.5 \, {\rm µ}\text{s}$?		{Calculate the band-pass signal  $w(t)$. What are the signal values at  $t = 0$  and  $t = 62.5 \, {\rm µ}\text{s}$?
	\|type="{}"}		\|type="{}"}
	$w(t=0)\ = \ $ { 4 3% }  $\text{V}$		$w(t=0)\ = \ $ { 4 3% }  $\text{V}$
	$w(t=62.5 \,{\rm µ} \text{s})\ = \ $ { 0. }  $\text{V}$		$w(t=62.5 \,{\rm µ} \text{s})\ = \ $ { 0. }  $\text{V}$

	{Which statements are true regarding the ~~bandpass~~ signals  $d(t)$  and  $w(t)$ ? Justify your result in the time domain.		{Which statements are true regarding the band-pass signals  $d(t)$  and  $w(t)$ ? Justify your result in the time domain.
	\|type="[]"}		\|type="[]"}
	+ The signals  $d(t)$  and  $w(t)$  are identical.		+ The signals  $d(t)$  and  $w(t)$  are identical.
Line 67:		Line 67:

	[[File:P_ID704__Sig_A_4_2_b_neu.png\|250px\|right\|frame\|Multiplication with cosine]]		[[File:P_ID704__Sig_A_4_2_b_neu.png\|250px\|right\|frame\|Multiplication with cosine]]
	'''(2)'''  The ~~bandpass~~ spectrum can be represented with  $f_{\rm T} = 4\, \text{kHz}$  as follows:		'''(2)'''  The band-pass spectrum can be represented with  $f_{\rm T} = 4\, \text{kHz}$  as follows:

	:$$ W(f) = U(f- f_{\rm T}) + U(f+ f_{\rm T}) = U(f)\star \left[		:$$ W(f) = U(f- f_{\rm T}) + U(f+ f_{\rm T}) = U(f)\star \left[
Line 80:		Line 80:
	*above the low–pass signal $u(t)$,		*above the low–pass signal $u(t)$,
	*then the oscillation $c(t) = 2 · \cos(2 \pi fTt$ ),		*then the oscillation $c(t) = 2 · \cos(2 \pi fTt$ ),
	*below the ~~bandpass~~ signal  $w(t) = u(t) \cdot c(t)$.		*below the band-pass signal  $w(t) = u(t) \cdot c(t)$.

Guenter at 08:50, 23 April 2021

2021-04-23T08:50:58Z

← Older revision		Revision as of 10:50, 23 April 2021
Line 126:		Line 126:

	__NOEDITSECTION__		__NOEDITSECTION__
	[[Category:Signal Representation: Exercises\|^4.1 Differences ~~and~~ Similarities of LP and BP Signals^]]		[[Category:Signal Representation: Exercises\|^4.1 Differences/Similarities of LP and BP Signals^]]