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

Guenter at 13:38, 5 May 2021

2021-05-05T13:38:30Z

Guenter at 12:19, 5 May 2021

2021-05-05T12:19:49Z

Guenter at 12:17, 5 May 2021

2021-05-05T12:17:39Z

Guenter at 12:14, 5 May 2021

2021-05-05T12:14:34Z

Guenter at 12:12, 5 May 2021

2021-05-05T12:12:24Z

Guenter at 11:58, 5 May 2021

2021-05-05T11:58:30Z

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 Exercise 4.1: Lowpass and Bandpass Signals to Exercise 4.1: Low-Pass and Band-Pass Signals

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

2021-05-03T15:13:10Z

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

Guenter at 08:49, 23 April 2021

2021-04-23T08:49:06Z

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

2021-03-23T12:39:49Z

Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

← Older revision		Revision as of 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.

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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^]]

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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^]]

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 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die si–shaped time function&nbsp; $x(t)$&nbsp; suggests a rectangular spectrum&nbsp; $X(f)$&nbsp;.
+'''(1)'''&nbsp; The&nbsp; $\rm si$–shaped time function&nbsp; $x(t)$&nbsp; suggests a rectangular spectrum&nbsp; $X(f)$&nbsp;.
-*The absolute, two-sided bandwidth&nbsp; $2 \cdot B_x$&nbsp; is equal to the reciprocal of the first zero. It follows that:
+*The absolute, two-sided bandwidth&nbsp; $2 \cdot B_x$&nbsp; is equal to the reciprocal of the first zero.&nbsp; It follows that:
 :$$B_x = \frac{1}{2 \cdot T_x}  =  \frac{1}{2 \cdot 0.1
@@ Line 104: / Line 104: @@
-[[File:P_ID701__Sig_A_4_1_c_neu.png|right|frame|Rectangular BP spectrum]]
+[[File:P_ID701__Sig_A_4_1_c_neu.png|right|frame|Rectangular band-pass spectrum]]
-'''(3)'''&nbsp;  From&nbsp; $d(t) = x(t) - y(t)$&nbsp; follows because of the linearity of the Fourier transformation: &nbsp; $D(f)  = X(f) - Y(f).$
+'''(3)'''&nbsp;  From&nbsp; $d(t) = x(t) - y(t)$&nbsp; follows because of the linearity of the Fourier transform: &nbsp; $D(f)  = X(f) - Y(f).$
-*The difference of the two equal square wave functions leads to a rectangular band-pass spectrum between&nbsp; $3 \,\text{kHz}$&nbsp; and&nbsp; $5 \,\text{kHz}$.
+*The difference of the two equal high rectangular functions leads to a rectangular band-pass spectrum between&nbsp; $3 \,\text{kHz}$&nbsp; and&nbsp; $5 \,\text{kHz}$.
-*The (one-sided) bandwidth is thus&nbsp; $B_d \;\underline{= 2 \,\text{kHz}}$. In this frequency interval&nbsp; $D(f) = 1 \,\text{mV/Hz}$. Outside, i.e. also at&nbsp; $f = 0$, gilt&nbsp; $D(f)\;\underline{ = 0}$ applies.
+*The (one-sided) bandwidth is thus&nbsp; $B_d \;\underline{= 2 \,\text{kHz}}$.&nbsp; In this frequency interval&nbsp; $D(f) = 1 \,\text{mV/Hz}$.&nbsp; Outside, i.e. also at&nbsp; $f = 0$, &nbsp; $D(f)\;\underline{ = 0}$ applies.
-'''(4)'''&nbsp; According to the fundamental laws of the Fourier transformation, the integral over the time function is equal to the spectral value at&nbsp; $f = 0$. Daraus folgt:
+'''(4)'''&nbsp; According to the fundamental laws of the Fourier transform, the integral over the time function is equal to the spectral value at&nbsp; $f = 0$.&nbsp; It follows: :
 :$$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^]]

@@ Line 13: / Line 13: @@
 {t}/{T_y}) .$$
-$T_x = 100 \,{\rm &micro;}\text{s}$&nbsp; and&nbsp; $T_y = 166.67 \,{\rm &micro;}\text{s}$&nbsp; indicate the first zero of&nbsp; $x(t)$&nbsp; bzw.&nbsp; $y(t)$&nbsp;respectively.
+$T_x = 100 \,{\rm &micro;}\text{s}$&nbsp; and&nbsp; $T_y = 166.67 \,{\rm &micro;}\text{s}$&nbsp; indicate the first zero of&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp;respectively.
 The signal&nbsp; $d(t)$&nbsp; results from the difference of the two upper signals (lower graph):
 :$$d(t)  =  x(t)-y(t)   .$$
-In the subtask&nbsp; '''(4)'''&nbsp; the integral areas of the impulsive signals&nbsp; $x(t)$&nbsp; and&nbsp; $d(t)$&nbsp; are asked for. For these holds:
+In subtask&nbsp; '''(4)'''&nbsp; the integral areas of the pulses&nbsp; $x(t)$&nbsp; and&nbsp; $d(t)$&nbsp; are asked for.&nbsp; For these holds:
 :$$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 .$$
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 ''Hints:''
-*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals|Differences and Similarities of LP and BP Signals]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Differences_and_Similarities_of_Low-Pass_and_Band-Pass_Signals|Differences and Similarities of Low-Pass and Band-Pass Signals]].
-*The Fourier retransform of a rectangular spectrum&nbsp; $X(f)$&nbsp; leads to an&nbsp; $\rm si$&ndash;shaped time function $x(t)$:
+*The inverse Fourier transform of a rectangular spectrum&nbsp; $X(f)$&nbsp; leads to an&nbsp; $\rm si$&ndash;shaped time function $x(t)$:
 :$$X(f)=\left\{ {X_0 \; \rm f\ddot{u}r\; |\it f| < \rm B, \atop {\rm 0 \;\;\; \rm sonst}}\right. \;\;
@@ Line 54: / Line 53: @@
 <quiz display=simple>
-{What is the spectrum&nbsp; $X(f)$&nbsp; of the signal&nbsp; $x(t)$? What are the magnitudes of&nbsp; $X(f = 0)$&nbsp; and the physical, one-sided bandwidth&nbsp; $B_x$&nbsp; of&nbsp; $x(t)$?
+{What is the spectrum&nbsp; $X(f)$&nbsp; of the signal&nbsp; $x(t)$?&nbsp; What are the magnitudes of&nbsp; $X(f = 0)$&nbsp; and the physical, one-sided bandwidth&nbsp; $B_x$&nbsp; of&nbsp; $x(t)$?
 |type="{}"}
 $X(f=0)\ = \ $   { 1 3% } &nbsp;$\text{mV/Hz}$
@@ Line 65: / Line 64: @@
-{Calculate the spectrum&nbsp; $D(f)$&nbsp; of the difference signal&nbsp; $d(t) = x(t) - y(t)$. How large are&nbsp; $D(f = 0)$&nbsp; and the (one-sided) bandwidth&nbsp; $B_d$?
+{Calculate the spectrum&nbsp; $D(f)$&nbsp; of the difference signal&nbsp; $d(t) = x(t) - y(t)$.&nbsp; How large are&nbsp; $D(f = 0)$&nbsp; and the (one-sided) bandwidth&nbsp; $B_d$?
 |type="{}"}
 $D(f=0)\ = \ $ { 0. } &nbsp;$\text{mV/Hz}$

← Older revision	Revision as of 07:30, 4 May 2021
(No difference)

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 '''(3)'''&nbsp;  From&nbsp; $d(t) = x(t) - y(t)$&nbsp; follows because of the linearity of the Fourier transformation: &nbsp; $D(f)  = X(f) - Y(f).$
-*The difference of the two equal square wave functions leads to a rectangular bandpass spectrum between&nbsp; $3 \,\text{kHz}$&nbsp; and&nbsp; $5 \,\text{kHz}$.
+*The difference of the two equal square wave functions leads to a rectangular band-pass spectrum between&nbsp; $3 \,\text{kHz}$&nbsp; and&nbsp; $5 \,\text{kHz}$.
 *The (one-sided) bandwidth is thus&nbsp; $B_d \;\underline{= 2 \,\text{kHz}}$. In this frequency interval&nbsp; $D(f) = 1 \,\text{mV/Hz}$. Outside, i.e. also at&nbsp; $f = 0$, gilt&nbsp; $D(f)\;\underline{ = 0}$ applies.
@@ Line 121: / Line 121: @@
 :$$F_d = D(f=0) \hspace{0.15 cm}\underline{= 0}.$$
-&rArr;&nbsp; For each bandpass signal, the areas of the positive signal components are equal to the areas of the negative components.
+&rArr;&nbsp; For each band-pass signal, the areas of the positive signal components are equal to the areas of the negative components.

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