Difference between revisions of "Aufgaben:Exercise 3.1Z: Spectrum of the Triangular Pulse"

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{{quiz-Header|Buchseite=Signaldarstellung/Fouriertransformation und -rücktransformation
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{{quiz-Header|Buchseite=Signal_Representation/Fourier_Transform_and_Its_Inverse
 
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[[File:P_ID493__Sig_Z_3_1_neu.png|right|Dreieckimpuls ]]
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[[File:P_ID493__Sig_Z_3_1_neu.png|right|frame|Triangular pulse]]
Betrachtet wird ein Dreieckimpuls ${x(t)}$, der im Bereich $–T ≤ t ≤ T$ durch folgende Gleichung beschrieben wird:
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A triangular pulse  ${x(t)}$  is considered, which is described in the range  $–T ≤ t ≤ T$  by the following equation:
:$$x(t) = A \cdot \left( {1 - \frac{\left| \hspace{0.05cm}t \hspace{0.05cm}\right|}{T}} \right).$$
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:$$x(t) = A \cdot \left( {1 - {\left| \hspace{0.05cm}t \hspace{0.05cm}\right|}/{T}} \right).$$
Die Impulsamplitude sei $A = 1\, \text{V}$, der Zeitparameter $T = 1 \text{ms}$. Für alle Zeiten $| t | > T$ ist ${x(t)} = 0$.
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Let the pulse amplitude be  $A = 1\, \text{V}$  and the time parameter  $T = 1 \text{ ms}$.  For all times  $|\hspace{0.05cm} t \hspace{0.05cm} | > T$  ⇒   ${x(t)} = 0$.
  
Zur Berechnung der Spektralfunktionen ${X(f)}$ können Sie folgende Eigenschaften ausnutzen:
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To calculate the spectral function  ${X(f)}$  you can exploit the following properties:
  
* Die Zeitfunktion ist gerade und damit die Spektralfunktion reell:
+
* The time function is even and thus the spectral function is real:
:$$X\left( f \right) = \int_{ - \infty }^{ + \infty } {x(t)}  \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d}t = 2 \cdot \int_0^{  \infty } {x(t)}  \cdot \cos \left( {2\pi ft} \right){\rm d}t.$$
+
:$$X\left( f \right) = \int_{ - \infty }^{ + \infty } {x(t)}  \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d}t = 2 \cdot \int_0^{  \infty } {x(t)}  \cdot \cos \left( {2\pi ft} \right)\ {\rm d}t.$$
* Für $| t | > T$ besitzt ${x(t)}$ keine Anteile:
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* For  $|\hspace{0.05cm} t \hspace{0.05cm} | > T$   ⇒   ${x(t)}$  has no components:
:$$X\left( f \right) = 2 \cdot \int_0^T {x(t)}  \cdot \cos \left( {2\pi ft} \right){\rm d}t.$$
 
  
''Hinweise:''
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:$$X\left( f \right) = 2 \cdot \int_0^T {x(t)}  \cdot \cos \left( {2\pi ft} \right)\ {\rm d}t.$$
*Die Aufgabe gehört zum Kapitel [[Signaldarstellung/Fouriertransformation_und_-rücktransformation|Fouriertransformation und -rücktransformation]].
 
*Weitere Informationen zu dieser Thematik liefert das Lernvideo [[Unterschiede und Gemeinsamkeiten von kontinuierlichen und diskreten Spektren]].
 
*Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
 
*Zur Lösung dieser Aufgabe können Sie auf die folgenden Formeln zurückgreifen:
 
:$$\int {t \cdot \cos \left( {\omega _0 t} \right){\rm d}t = \frac{{\cos \left( {\omega _0 t} \right)}}{\omega _0 ^2 }}  + \frac{{t \cdot \sin \left( {\omega _0 t} \right)}}{\omega _0 }, \hspace{0.5cm} \sin ^2 \left( \alpha \right) = {1}/{2} \cdot \left( {1 - \cos \left( {2\alpha } \right)} \right).$$
 
  
  
===Fragebogen===
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 +
 
 +
 
 +
 
 +
 
 +
 
 +
''Hints:''
 +
*This exercise belongs to the chapter  [[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  [[Kontinuierliche_und_diskrete_Spektren_(Lernvideo)|Kontinuierliche und diskrete Spektren]]   ⇒   "Continuous and discrete spectra".
 +
 +
*You can use the following formulas to solve this task:
 +
:$$\sin ^2 \left( \alpha  \right) = {1}/{2} \cdot \left( {1 - \cos \left( {2\alpha } \right)} \right),$$
 +
:$$\int {t \cdot \cos \left( {\omega _0 t} \right)\ {\rm d}t = \frac{{\cos \left( {\omega _0 t} \right)}}{\omega _0 ^2 }}  + \frac{{t \cdot \sin \left( {\omega _0 t} \right)}}{\omega _0 }.$$
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Berechnen Sie die Spektralfunktion ${X(f)}$. Welcher Spektralwert ergibt sich bei der Frequenz $f = 500 \,\text{Hz}$?
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{Calculate the spectral function&nbsp; ${X(f)}$.&nbsp; What spectral value results at the frequency $f = 500 \,\text{Hz}$?
 
|type="{}"}
 
|type="{}"}
$X(f = 500 \,\text{Hz})$ &nbsp;= { 0.405 3% } &nbsp;$\text{mV/Hz}$
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$X(f = 500 \,\text{Hz}) \ = \ $ { 0.405 3% } &nbsp;$\text{mV/Hz}$
  
  
{Geben Sie die Spektralfunktion ${X(f)}$ unter Verwendung der Spaltfunktion $\text{si}(x) = \sin(x)/x$ an. Welcher Wert ergibt sich bei der Frequenz $f = 0$?
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{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="{}"}
 
|type="{}"}
$X(f = 0)$ &nbsp;= { 1 3% } &nbsp;$\text{mV/Hz}$
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$X(f = 0) \ = \ $ { 1 3% } &nbsp;$\text{mV/Hz}$
  
  
{Bei welcher Frequenz $f = f_0$ hat das Spektrum ${X(f)}$ die erste Nullstelle?
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{At what frequency $f = f_0$&nbsp; does the spectrum&nbsp; ${X(f)}$&nbsp; have the first zero?
 
|type="{}"}
 
|type="{}"}
$f_0$ &nbsp;= { 1 3% } &nbsp;$\text{kHz}$
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$f_0 \ = \ $ { 1 3% } &nbsp;$\text{kHz}$
  
  
{Welche der beiden Aussagen sind zutreffend?
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{Which of the two statements is true?
 
|type="[]"}
 
|type="[]"}
+ Bei allen Vielfachen von $f_0$ hat das Spektrum Nullstellen.
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+ At all multiples of&nbsp; $f_0$&nbsp; the spectrum has zeros.
- Bei der Frequenz $f = 1.5 \cdot f_0$ ist die Spektralfunktion negativ.
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- At the frequency&nbsp; $f = 1.5 \cdot f_0$&nbsp; the spectral function is negative.
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
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{{ML-Kopf}}
'''1.''' Unter Ausnutzung der genannten Symmetrieeigenschaften gilt mit der Abkürzung $\omega = 2\pi f$:
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'''(1)'''&nbsp; We use the abbreviation&nbsp; $\omega = 2\pi f$.&nbsp; Then taking advantage of the above symmetry properties:
:$$X(f) = 2A \cdot \int_0^T {\left( {1 - \frac{t}{T}} \right)}  \cdot \cos \left( {\omega t} \right){\rm d}t.$$
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:$$X(f) = 2A \cdot \int_0^T {\left( {1 -{t}/{T}} \right)}  \cdot \cos \left( {\omega t} \right)\hspace{0.1cm}{\rm d}t.$$
Dieses Integral setzt sich aus zwei Anteilen zusammen:
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*This integral is composed of two parts:
:$$X_1 (f) = 2A \cdot \int_0^T {\cos } \left( {\omega t} \right){\rm d}t =  \frac{2A}{\omega } \cdot \sin \left( {\omega T} \right),$$
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:$$X_1 (f) = 2A \cdot \int_0^T {\cos } \left( {\omega t} \right)\hspace{0.1cm}{\rm d}t =  \frac{2A}{\omega } \cdot \sin \left( {\omega T} \right),$$
:$$X_2 (f) =  - \frac{2A}{T} \cdot \int_0^T {t \cdot \cos } \left( {\omega t} \right){\rm d}t =  - \frac{2A}{T} \cdot \left. {\left[ {\frac{{\cos \left( {\omega t} \right)}}{\omega ^2 } + \frac{{t \cdot \sin \left( {\omega t} \right)}}{\omega }} \right]} \right|_0^T  .$$
+
:$$X_2 (f) =  - \frac{2A}{T} \cdot \int_0^T {t \cdot \cos } \left( {\omega t} \right)\hspace{0.1cm}{\rm d}t =  - \frac{2A}{T} \cdot \left. {\left[ {\frac{{\cos \left( {\omega t} \right)}}{\omega ^2 } + \frac{{t \cdot \sin \left( {\omega t} \right)}}{\omega }} \right]} \right|_0^T  .$$
Unter Berücksichtigung von oberer und unterer Grenze erhält man:
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*Taking into account the upper and lower limits, one obtains:
 
:$$X_2 \left( f \right) =  - \frac{2A}{T} \cdot \left[ {\frac{{\cos \left( {\omega T} \right)}}{\omega ^2 } - \frac{1}{\omega ^2 } + \frac{{T \cdot \sin \left( {\omega T} \right)}}{\omega }} \right].$$
 
:$$X_2 \left( f \right) =  - \frac{2A}{T} \cdot \left[ {\frac{{\cos \left( {\omega T} \right)}}{\omega ^2 } - \frac{1}{\omega ^2 } + \frac{{T \cdot \sin \left( {\omega T} \right)}}{\omega }} \right].$$
Addiert man die beiden Anteile, so ergibt sich:
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*Adding the two parts, we get:
:$$X(f) = \frac{2A}{\omega ^2  \cdot T}\left[ {1 - \cos \left( {\omega T} \right)} \right] = \frac{A}{2\pi ^2 f^2 T} \cdot \left[ {1 - \cos \left( {2\pi fT} \right)} \right].$$
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:$$X(f) = \frac{2A}{\omega ^2  \cdot T}\cdot \big[ {1 - \cos \left( {\omega T} \right)} \big] = \frac{A}{2\pi ^2 f^2 T} \cdot \big[ {1 - \cos \left( {2\pi fT} \right)} \big].$$
Bei der Frequenz $f = 1/(2T) = 500 \text{Hz}$ ist das Argument der Cosinusfunktion gleich $\pi$ und damit die Cosinusfunktion selbst gleich $–1$. Daraus folgt:
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*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:
:$$X( {f = \frac{1}{2T} = 500\;{\rm Hz}} ) = \frac{4}{\pi^2} \cdot A \cdot T  = \frac{4}{\pi^2} \cdot 1\;{\rm V} \cdot 10^{ - 3}\;{\rm s}\hspace{0.15 cm}\underline{= 0.405 \cdot 10^{ - 3} \;{\rm V/Hz}}.$$
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:$$X( {f ={1}/{2T} = 500\;{\rm Hz}} ) = \frac{4}{\pi^2} \cdot A \cdot T  = \frac{4}{\pi^2} \cdot 1\;{\rm V} \cdot 10^{ - 3}\;{\rm s}\hspace{0.15 cm}\underline{= 0.405 \,{\rm mV/Hz}}.$$
  
'''2.''' Mit der trigonometrischen Umformung
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:$${1}/{2} \cdot (1 - \cos (2 \alpha)) = \sin^2(\alpha)$$
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erhält man für die Spektralfunktion:
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'''(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|Normalized spectrum&nbsp; $X(f)$]]
 
:$$X(f) = A \cdot T \cdot \frac{\sin^2(\pi f T)}{\pi^2 \cdot {f^2  \cdot T^2}} = A \cdot T \cdot {{{\rm si}^2(\pi f T)}}.$$
 
:$$X(f) = A \cdot T \cdot \frac{\sin^2(\pi f T)}{\pi^2 \cdot {f^2  \cdot T^2}} = A \cdot T \cdot {{{\rm si}^2(\pi f T)}}.$$
Bei der Frequenz $f = 0$ ist die si-Funktion gleich $1$. Daraus folgt:
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*At the frequency&nbsp; $f = 0$&nbsp; the&nbsp; $\rm si$-function is equal to&nbsp; $1$. From this follows:
:$$X( {f = 0} ) = A \cdot T \hspace{0.15 cm}\underline{= 10^{ - 3} \;{\rm V/Hz}}.$$
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:$$X( {f = 0} ) = A \cdot T \hspace{0.15 cm}\underline{= 1\,{\rm mV/Hz}}.$$
 +
 
 +
 
 +
'''(3)'''&nbsp; The first zero occurs when the argument of the &nbsp; $\rm si$-function is equal to&nbsp; $\pi$.
 +
*From this follows:&nbsp; $f_0 \cdot T = 1$&nbsp; bzw.&nbsp; $f_0 = 1/T \hspace{0.15 cm}\underline{= 1 \ \text{kHz}}$.
 +
 
 +
 
  
'''3.''' Die erste Nullstelle tritt auf, wenn das Argument der si-Funktion gleich $\pi$ ist. Daraus folgt $f_0 \cdot T = 1$ bzw. $f_0 = 1/T \underline{= 1 \text{kHz}}$.
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'''(4)'''&nbsp; The <u>first statement</u> is correct:
 +
*The spectrum&nbsp; ${X(f)}$&nbsp; is equal to&nbsp; ${\rm si}^2(n \cdot \pi) = 0$&nbsp; at multiples of&nbsp; $f_0$&nbsp; $(f = n \cdot f_0)$.  
 +
*The second statement is false:&nbsp; At no frequency $f$&nbsp; holds:&nbsp; ${X(f)} < 0$&nbsp; (see sketch).
  
'''4.'''  Die Spektralfunktion $\text{X(f)}$ ist bei Vielfachen von $f_0$ ($f = n \cdot f_0$) gleich $si^2(n \cdot \pi) = 0$. Die erste Aussage trifft also zu im Gegensatz zur zweiten: Bei keiner Frequenz $f$ ist $\text{X(f)} < 0$ (siehe Skizze).
 
[[File:P_ID497__Sig_Z_3_1_d_neu.png|center|]]
 
 
{{ML-Fuß}}
 
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[[Category:Aufgaben zu Signaldarstellung|^3. Aperiodische Signale - Impulse^]]
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[[Category:Signal Representation: Exercises|^3.1 Fourier Transform and Its Inverse^]]

Latest revision as of 14:39, 21 April 2021

Triangular pulse

A triangular pulse  ${x(t)}$  is considered, which is described in the range  $–T ≤ t ≤ T$  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  $A = 1\, \text{V}$  and the time parameter  $T = 1 \text{ ms}$.  For all times  $|\hspace{0.05cm} t \hspace{0.05cm} | > T$  ⇒   ${x(t)} = 0$.

To calculate the spectral function  ${X(f)}$  you can exploit the following properties:

  • The time function is even and thus the spectral function is real:
$$X\left( f \right) = \int_{ - \infty }^{ + \infty } {x(t)} \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d}t = 2 \cdot \int_0^{ \infty } {x(t)} \cdot \cos \left( {2\pi ft} \right)\ {\rm d}t.$$
  • For  $|\hspace{0.05cm} t \hspace{0.05cm} | > T$   ⇒   ${x(t)}$  has no components:
$$X\left( f \right) = 2 \cdot \int_0^T {x(t)} \cdot \cos \left( {2\pi ft} \right)\ {\rm d}t.$$





Hints:

  • You can use the following formulas to solve this task:
$$\sin ^2 \left( \alpha \right) = {1}/{2} \cdot \left( {1 - \cos \left( {2\alpha } \right)} \right),$$
$$\int {t \cdot \cos \left( {\omega _0 t} \right)\ {\rm d}t = \frac{{\cos \left( {\omega _0 t} \right)}}{\omega _0 ^2 }} + \frac{{t \cdot \sin \left( {\omega _0 t} \right)}}{\omega _0 }.$$

Questions

1

Calculate the spectral function  ${X(f)}$.  What spectral value results at the frequency $f = 500 \,\text{Hz}$?

$X(f = 500 \,\text{Hz}) \ = \ $

 $\text{mV/Hz}$

2

Give the spectral function  ${X(f)}$  using the  "slitting function"  $\text{si}(x) = \sin(x)/x$.  What value results for  $f = 0$?

$X(f = 0) \ = \ $

 $\text{mV/Hz}$

3

At what frequency $f = f_0$  does the spectrum  ${X(f)}$  have the first zero?

$f_0 \ = \ $

 $\text{kHz}$

4

Which of the two statements is true?

At all multiples of  $f_0$  the spectrum has zeros.
At the frequency  $f = 1.5 \cdot f_0$  the spectral function is negative.


Solution

(1)  We use the abbreviation  $\omega = 2\pi f$.  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:
$$X_1 (f) = 2A \cdot \int_0^T {\cos } \left( {\omega t} \right)\hspace{0.1cm}{\rm d}t = \frac{2A}{\omega } \cdot \sin \left( {\omega T} \right),$$
$$X_2 (f) = - \frac{2A}{T} \cdot \int_0^T {t \cdot \cos } \left( {\omega t} \right)\hspace{0.1cm}{\rm d}t = - \frac{2A}{T} \cdot \left. {\left[ {\frac{{\cos \left( {\omega t} \right)}}{\omega ^2 } + \frac{{t \cdot \sin \left( {\omega t} \right)}}{\omega }} \right]} \right|_0^T .$$
  • Taking into account the upper and lower limits, one obtains:
$$X_2 \left( f \right) = - \frac{2A}{T} \cdot \left[ {\frac{{\cos \left( {\omega T} \right)}}{\omega ^2 } - \frac{1}{\omega ^2 } + \frac{{T \cdot \sin \left( {\omega T} \right)}}{\omega }} \right].$$
  • 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  $f = 1/(2T) = 500 \,\text{Hz}$  the argument of the cosine function is equal to  $\pi$  and the cosine function itself is equal to  $-1$.  It follows:
$$X( {f ={1}/{2T} = 500\;{\rm Hz}} ) = \frac{4}{\pi^2} \cdot A \cdot T = \frac{4}{\pi^2} \cdot 1\;{\rm V} \cdot 10^{ - 3}\;{\rm s}\hspace{0.15 cm}\underline{= 0.405 \,{\rm mV/Hz}}.$$


(2)  With the trigonometric transformation  ${1}/{2} \cdot (1 - \cos (2 \alpha)) = \sin^2(\alpha)$  one obtains for the spectral function:

Normalized spectrum  $X(f)$
$$X(f) = A \cdot T \cdot \frac{\sin^2(\pi f T)}{\pi^2 \cdot {f^2 \cdot T^2}} = A \cdot T \cdot {{{\rm si}^2(\pi f T)}}.$$
  • At the frequency  $f = 0$  the  $\rm si$-function is equal to  $1$. From this follows:
$$X( {f = 0} ) = A \cdot T \hspace{0.15 cm}\underline{= 1\,{\rm mV/Hz}}.$$


(3)  The first zero occurs when the argument of the   $\rm si$-function is equal to  $\pi$.

  • From this follows:  $f_0 \cdot T = 1$  bzw.  $f_0 = 1/T \hspace{0.15 cm}\underline{= 1 \ \text{kHz}}$.


(4)  The first statement is correct:

  • The spectrum  ${X(f)}$  is equal to  ${\rm si}^2(n \cdot \pi) = 0$  at multiples of  $f_0$  $(f = n \cdot f_0)$.
  • The second statement is false:  At no frequency $f$  holds:  ${X(f)} < 0$  (see sketch).