Aufgaben:Exercise 3.1Z: Spectrum of the Triangular Pulse - Revision history

Guenter at 13:39, 21 April 2021

2021-04-21T13:39:25Z

Guenter at 13:28, 21 April 2021

2021-04-21T13:28:10Z

Guenter at 13:27, 21 April 2021

2021-04-21T13:27:27Z

Noah at 13:43, 26 March 2021

2021-03-26T13:43:18Z

Noah at 13:17, 26 March 2021

2021-03-26T13:17:43Z

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

2021-03-23T12:39:51Z

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

Noah at 21:18, 17 January 2021

2021-01-17T21:18:50Z

Noah at 20:55, 17 January 2021

2021-01-17T20:55:02Z

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 3.1Z: Spektrum des Dreieckimpulses to Aufgaben:Exercise 3.1Z: Spectrum of the Triangular Pulse

2020-11-02T14:27:35Z

Oezdemir moved page Aufgabe 3.1Z: Spektrum des Dreieckimpulses to Exercise 3.1Z: Spectrum of the Triangular Pulse

Oezdemir at 14:23, 2 November 2020

2020-11-02T14:23:46Z

@@ Line 58: / Line 58: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Taking advantage of the above symmetry properties, the abbreviation&nbsp; $\omega = 2\pi f$ holds:
+'''(1)'''&nbsp; We use the abbreviation&nbsp; $\omega = 2\pi f$.&nbsp; Then taking advantage of the above symmetry properties:
 :$$X(f) = 2A \cdot \int_0^T {\left( {1 -{t}/{T}} \right)}  \cdot \cos \left( {\omega t} \right)\hspace{0.1cm}{\rm d}t.$$
 *This integral is composed of two parts:
@@ Line 67: / Line 67: @@
 *Adding the two parts, we get:
 :$$X(f) = \frac{2A}{\omega ^2  \cdot T}\cdot \big[ {1 - \cos \left( {\omega T} \right)} \big] = \frac{A}{2\pi ^2 f^2 T} \cdot \big[ {1 - \cos \left( {2\pi fT} \right)} \big].$$
-*At frequency&nbsp; $f = 1/(2T) = 500 \,\text{Hz}$&nbsp; the argument of the cosine function is equal to&nbsp; $\pi$&nbsp; and the cosine function itself is equal to&nbsp; $-1$. It follows:
+*At frequency&nbsp; $f = 1/(2T) = 500 \,\text{Hz}$&nbsp; the argument of the cosine function is equal to&nbsp; $\pi$&nbsp; and the cosine function itself is equal to&nbsp; $-1$.&nbsp; It follows:
 :$$X( {f ={1}/{2T} = 500\;{\rm Hz}} ) = \frac{4}{\pi^2} \cdot A \cdot T  = \frac{4}{\pi^2} \cdot 1\;{\rm V} \cdot 10^{ - 3}\;{\rm s}\hspace{0.15 cm}\underline{= 0.405 \,{\rm mV/Hz}}.$$
@@ Line 73: / Line 73: @@
 '''(2)'''&nbsp; With the trigonometric transformation&nbsp; ${1}/{2} \cdot (1 - \cos (2 \alpha)) = \sin^2(\alpha)$&nbsp; one obtains for the spectral function:
-[[File:P_ID497__Sig_Z_3_1_d_neu.png|right|frame|$\rm si$-squared spectrum]]
+[[File:P_ID497__Sig_Z_3_1_d_neu.png|right|frame|Normalized spectrum&nbsp; $X(f)$]]
 :$$X(f) = A \cdot T \cdot \frac{\sin^2(\pi f T)}{\pi^2 \cdot {f^2  \cdot T^2}} = A \cdot T \cdot {{{\rm si}^2(\pi f T)}}.$$
 *At the frequency&nbsp; $f = 0$&nbsp; the&nbsp; $\rm si$-function is equal to&nbsp; $1$. From this follows:
@@ Line 79: / Line 79: @@
+'''(3)'''&nbsp; The first zero occurs when the argument of the &nbsp; $\rm si$-function is equal to&nbsp; $\pi$.
-'''(3)'''&nbsp; The first zero occurs when the argument of the si function is equal to&nbsp; $\pi$&nbsp;.
+*From this follows:&nbsp; $f_0 \cdot T = 1$&nbsp; bzw.&nbsp; $f_0 = 1/T \hspace{0.15 cm}\underline{= 1 \ \text{kHz}}$.
 '''(4)'''&nbsp; The <u>first statement</u> is correct:
-*The spectrum&nbsp; ${X(f)}$&nbsp; is equal to&nbsp; ${\rm si}^2(n \cdot \pi) = 0$ at multiples of&nbsp; $f_0$&nbsp; $(f = n \cdot f_0)$&nbsp;.
+*The spectrum&nbsp; ${X(f)}$&nbsp; is equal to&nbsp; ${\rm si}^2(n \cdot \pi) = 0$&nbsp; at multiples of&nbsp; $f_0$&nbsp; $(f = n \cdot f_0)$.
-*The second statement is false:&nbsp; At no frequency&nbsp; $f$&nbsp; is&nbsp; ${X(f)} < 0$&nbsp; (see sketch).
+*The second statement is false:&nbsp; At no frequency $f$&nbsp; holds:&nbsp; ${X(f)} < 0$&nbsp; (see sketch).
 {{ML-Fuß}}

@@ Line 24: / Line 24: @@
 ''Hints:''
-*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse|Fourier Transform and Its Inverse]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse|Fourier Transform and its Inverse]].
-*Further information on this topic can be found in the (German language) learning video&nbsp; [[Kontinuierliche_und_diskrete_Spektren_(Lernvideo)|Kontinuierliche_und_diskrete_Spektren]] &nbsp; &rArr; &nbsp; "Continuous and discrete spectra".
+*Further information on this topic can be found in the (German language) learning video&nbsp; [[Kontinuierliche_und_diskrete_Spektren_(Lernvideo)|Kontinuierliche und diskrete Spektren]] &nbsp; &rArr; &nbsp; "Continuous and discrete spectra".
 *You can use the following formulas to solve this task:

@@ Line 4: / Line 4: @@
 [[File:P_ID493__Sig_Z_3_1_neu.png|right|frame|Triangular pulse]]
-A triangular pulses&nbsp; ${x(t)}$, is considered, which is described in the range&nbsp; $–T ≤ t ≤ T$&nbsp; by the following equation:
+A triangular pulse&nbsp; ${x(t)}$&nbsp; is considered, which is described in the range&nbsp; $–T ≤ t ≤ T$&nbsp; by the following equation:
 :$$x(t) = A \cdot \left( {1 - {\left| \hspace{0.05cm}t \hspace{0.05cm}\right|}/{T}} \right).$$
-Let the pulse amplitude be&nbsp; $A = 1\, \text{V}$, the time parameter&nbsp; $T = 1 \text{ ms}$. For all times&nbsp; $|\hspace{0.05cm} t \hspace{0.05cm} | > T$&nbsp; ist&nbsp; ${x(t)} = 0$.
+Let the pulse amplitude be&nbsp; $A = 1\, \text{V}$&nbsp; and the time parameter&nbsp; $T = 1 \text{ ms}$.&nbsp; For all times&nbsp; $|\hspace{0.05cm} t \hspace{0.05cm} | > T$&nbsp; &rArr; &nbsp; ${x(t)} = 0$.
 To calculate the spectral function&nbsp; ${X(f)}$&nbsp; you can exploit the following properties:
 * The time function is even and thus the spectral function is real:
-:$$X\left( f \right) = \int_{ - \infty }^{ + \infty } {x(t)}  \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d}t = 2 \cdot \int_0^{  \infty } {x(t)}  \cdot \cos \left( {2\pi ft} \right){\rm d}t.$$
+:$$X\left( f \right) = \int_{ - \infty }^{ + \infty } {x(t)}  \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d}t = 2 \cdot \int_0^{  \infty } {x(t)}  \cdot \cos \left( {2\pi ft} \right)\ {\rm d}t.$$
-* For&nbsp; $|\hspace{0.05cm} t \hspace{0.05cm} | > T$, ${x(t)}$&nbsp; has no components:
+* For&nbsp; $|\hspace{0.05cm} t \hspace{0.05cm} | > T$ &nbsp; &rArr; &nbsp; ${x(t)}$&nbsp; has no components:
-:$$X\left( f \right) = 2 \cdot \int_0^T {x(t)}  \cdot \cos \left( {2\pi ft} \right){\rm d}t.$$
+:$$X\left( f \right) = 2 \cdot \int_0^T {x(t)}  \cdot \cos \left( {2\pi ft} \right)\ {\rm d}t.$$
@@ Line 25: / Line 25: @@
 ''Hints:''
 *This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse|Fourier Transform and Its Inverse]].
-*Further information on this topic can be found in the learning video&nbsp; [[Kontinuierliche_und_diskrete_Spektren_(Lernvideo)|Continuous and discrete spectra]].
+*Further information on this topic can be found in the (German language) learning video&nbsp; [[Kontinuierliche_und_diskrete_Spektren_(Lernvideo)|Kontinuierliche_und_diskrete_Spektren]] &nbsp; &rArr; &nbsp; "Continuous and discrete spectra".
 *You can use the following formulas to solve this task:
-:$$\int {t \cdot \cos \left( {\omega _0 t} \right)\ {\rm d}t = \frac{{\cos \left( {\omega _0 t} \right)}}{\omega _0 ^2 }}  + \frac{{t \cdot \sin \left( {\omega _0 t} \right)}}{\omega _0 }, \hspace{0.5cm} \sin ^2 \left( \alpha  \right) = {1}/{2} \cdot \left( {1 - \cos \left( {2\alpha } \right)} \right).$$
+:$$\sin ^2 \left( \alpha  \right) = {1}/{2} \cdot \left( {1 - \cos \left( {2\alpha } \right)} \right),$$
 :$$\int {t \cdot \cos \left( {\omega _0 t} \right)\ {\rm d}t = \frac{{\cos \left( {\omega _0 t} \right)}}{\omega _0 ^2 }}  + \frac{{t \cdot \sin \left( {\omega _0 t} \right)}}{\omega _0 }.$$
 ===Questions===
 <quiz display=simple>
-{Calculate the spectral function&nbsp; ${X(f)}$. What spectral value results at the frequency &nbsp; $f = 500 \,\text{Hz}$?
+{Calculate the spectral function&nbsp; ${X(f)}$.&nbsp; What spectral value results at the frequency $f = 500 \,\text{Hz}$?
 |type="{}"}
 $X(f = 500 \,\text{Hz}) \ = \ $ { 0.405 3% } &nbsp;$\text{mV/Hz}$
-{Give the spectral function&nbsp; ${X(f)}$&nbsp; using the slit function&nbsp; $\text{si}(x) = \sin(x)/x$&nbsp;. What value results for&nbsp; $f = 0$?
+{Give the spectral function&nbsp; ${X(f)}$&nbsp; using the&nbsp; "slitting function"&nbsp; $\text{si}(x) = \sin(x)/x$.&nbsp; What value results for&nbsp; $f = 0$?
 |type="{}"}
 $X(f = 0) \ = \ $ { 1 3% } &nbsp;$\text{mV/Hz}$
-{At what frequency&nbsp; $f = f_0$&nbsp; does the spectrum&nbsp; ${X(f)}$&nbsp; have the first zero?
+{At what frequency $f = f_0$&nbsp; does the spectrum&nbsp; ${X(f)}$&nbsp; have the first zero?
 |type="{}"}
 $f_0 \ = \ $ { 1 3% } &nbsp;$\text{kHz}$

@@ Line 73: / Line 73: @@
 '''(2)'''&nbsp; With the trigonometric transformation&nbsp; ${1}/{2} \cdot (1 - \cos (2 \alpha)) = \sin^2(\alpha)$&nbsp; one obtains for the spectral function:
-[[File:P_ID497__Sig_Z_3_1_d_neu.png|right|frame|Spectrum of $\rm si$-squared]]
+[[File:P_ID497__Sig_Z_3_1_d_neu.png|right|frame|$\rm si$-squared spectrum]]
 :$$X(f) = A \cdot T \cdot \frac{\sin^2(\pi f T)}{\pi^2 \cdot {f^2  \cdot T^2}} = A \cdot T \cdot {{{\rm si}^2(\pi f T)}}.$$
 *At the frequency&nbsp; $f = 0$&nbsp; the&nbsp; $\rm si$-function is equal to&nbsp; $1$. From this follows:

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID493__Sig_Z_3_1_neu.png|right|frame|Dreieckimpuls]]
+[[File:P_ID493__Sig_Z_3_1_neu.png|right|frame|Triangular pulse]]
 A triangular pulses&nbsp; ${x(t)}$, is considered, which is described in the range&nbsp; $–T ≤ t ≤ T$&nbsp; by the following equation:
 :$$x(t) = A \cdot \left( {1 - {\left| \hspace{0.05cm}t \hspace{0.05cm}\right|}/{T}} \right).$$
@@ Line 25: / Line 25: @@
 ''Hints:''
 *This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse|Fourier Transform and Its Inverse]].
-*Further information on this topic can be found in the learning video&nbsp; [[Kontinuierliche_und_diskrete_Spektren_(Lernvideo)|Kontinuierliche und diskrete Spektren]].
+*Further information on this topic can be found in the learning video&nbsp; [[Kontinuierliche_und_diskrete_Spektren_(Lernvideo)|Continuous and discrete spectra]].
-*Zur Lösung dieser Aufgabe können Sie auf die folgenden Formeln zurückgreifen:
+*You can use the following formulas to solve this task:
 :$$\int {t \cdot \cos \left( {\omega _0 t} \right)\ {\rm d}t = \frac{{\cos \left( {\omega _0 t} \right)}}{\omega _0 ^2 }}  + \frac{{t \cdot \sin \left( {\omega _0 t} \right)}}{\omega _0 }, \hspace{0.5cm} \sin ^2 \left( \alpha  \right) = {1}/{2} \cdot \left( {1 - \cos \left( {2\alpha } \right)} \right).$$
@@ Line 73: / Line 73: @@
 '''(2)'''&nbsp; With the trigonometric transformation&nbsp; ${1}/{2} \cdot (1 - \cos (2 \alpha)) = \sin^2(\alpha)$&nbsp; one obtains for the spectral function:
-[[File:P_ID497__Sig_Z_3_1_d_neu.png|right|frame|$\rm si$-Quadrat-Spektrum]]
+[[File:P_ID497__Sig_Z_3_1_d_neu.png|right|frame|Spectrum of $\rm si$-squared]]
 :$$X(f) = A \cdot T \cdot \frac{\sin^2(\pi f T)}{\pi^2 \cdot {f^2  \cdot T^2}} = A \cdot T \cdot {{{\rm si}^2(\pi f T)}}.$$
 *At the frequency&nbsp; $f = 0$&nbsp; the&nbsp; $\rm si$-function is equal to&nbsp; $1$. From this follows:

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← Older revision		Revision as of 14:23, 2 November 2020
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