View source for Aufgaben:Exercise 3.1Z: Spectrum of the Triangular Pulse


{{quiz-Header|Buchseite=Signal_Representation/Fourier_Transform_and_Its_Inverse
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[[File:P_ID493__Sig_Z_3_1_neu.png|right|frame|Triangular pulse]]
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; $A = 1\, \text{V}$&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:
:$$X\left( f \right) = \int_{ - \infty }^{ + \infty } {x(t)}  \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d}t = 2 \cdot \int_0^{  \infty } {x(t)}  \cdot \cos \left( {2\pi ft} \right)\ {\rm d}t.$$
* For&nbsp; $|\hspace{0.05cm} t \hspace{0.05cm} | > T$ &nbsp; &rArr; &nbsp; ${x(t)}$&nbsp; has no components:

:$$X\left( f \right) = 2 \cdot \int_0^T {x(t)}  \cdot \cos \left( {2\pi ft} \right)\ {\rm d}t.$$








''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 (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:
:$$\sin ^2 \left( \alpha  \right) = {1}/{2} \cdot \left( {1 - \cos \left( {2\alpha } \right)} \right),$$
:$$\int {t \cdot \cos \left( {\omega _0 t} \right)\ {\rm d}t = \frac{{\cos \left( {\omega _0 t} \right)}}{\omega _0 ^2 }}  + \frac{{t \cdot \sin \left( {\omega _0 t} \right)}}{\omega _0 }.$$

===Questions===

<quiz display=simple>
{Calculate the spectral function&nbsp; ${X(f)}$.&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&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 $f = f_0$&nbsp; does the spectrum&nbsp; ${X(f)}$&nbsp; have the first zero?
|type="{}"}
$f_0 \ = \ $ { 1 3% } &nbsp;$\text{kHz}$


{Which of the two statements is true?
|type="[]"}
+ At all multiples of&nbsp; $f_0$&nbsp; the spectrum has zeros.
- At the frequency&nbsp; $f = 1.5 \cdot f_0$&nbsp; the spectral function is negative.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; Taking advantage of the above symmetry properties, the abbreviation&nbsp; $\omega = 2\pi f$ holds:
:$$X(f) = 2A \cdot \int_0^T {\left( {1 -{t}/{T}} \right)}  \cdot \cos \left( {\omega t} \right)\hspace{0.1cm}{\rm d}t.$$
*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&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:
:$$X( {f ={1}/{2T} = 500\;{\rm Hz}} ) = \frac{4}{\pi^2} \cdot A \cdot T  = \frac{4}{\pi^2} \cdot 1\;{\rm V} \cdot 10^{ - 3}\;{\rm s}\hspace{0.15 cm}\underline{= 0.405 \,{\rm mV/Hz}}.$$



'''(2)'''&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]]
:$$X(f) = A \cdot T \cdot \frac{\sin^2(\pi f T)}{\pi^2 \cdot {f^2  \cdot T^2}} = A \cdot T \cdot {{{\rm si}^2(\pi f T)}}.$$
*At the frequency&nbsp; $f = 0$&nbsp; the&nbsp; $\rm si$-function is equal to&nbsp; $1$. From this follows:
:$$X( {f = 0} ) = A \cdot T \hspace{0.15 cm}\underline{= 1\,{\rm mV/Hz}}.$$



'''(3)'''&nbsp; The first zero occurs when the argument of the 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 second statement is false:&nbsp; At no frequency&nbsp; $f$&nbsp; is&nbsp; ${X(f)} < 0$&nbsp; (see sketch).

{{ML-Fuß}}


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[[Category:Signal Representation: Exercises|^3.1 Fourier Transform and Its Inverse^]]