View source for Aufgaben:Exercise 3.5Z: Integration of Dirac Functions


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[[File:P_ID515__Sig_Z_3_5_neu.png|right|frame|Integration von Diracfunktionen ]]
Like in&nbsp; [[Aufgaben:3.5_Differentiation_eines_Dreicksignals|Task 3.5]]&nbsp; the spectrum&nbsp; ${Y(f)}$&nbsp; of the signal
:$$y( t ) = \left\{ \begin{array}{c} A \\   - A \\  0 \\  \end{array} \right.\quad \begin{array}{*{20}c}   {{\rm{f \ddot{u}r}}}  \\   {{\rm{f\ddot{u} r}}}  \\   {\rm{else.}}  \\ \end{array}\;\begin{array}{*{20}c}   { - T \le t < 0,}  \\   {0 < t \le T,}  \\   {}  \\\end{array}$$
can be determined. Again,&nbsp; $A = 1 \,\text{V}$&nbsp; and&nbsp; $T = 0.5 \,\text{ms}$ apply.

Assume the time signal&nbsp; ${x(t)}$&nbsp; according to the middle sketch, which is composed of three Dirac pulses at&nbsp; $–T$,&nbsp; $0$&nbsp; and&nbsp; $+T$&nbsp; with the pulse weights&nbsp; ${AT}$,&nbsp; $-2{AT}$&nbsp; und&nbsp; ${AT}$&nbsp;.

The spectral function&nbsp; ${X(f)}$&nbsp; can be given directly by applying the&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Vertauschungssatz|Vertauschungssatz]]&nbsp; if one takes into account that the time function belonging to&nbsp; ${U(f)}$&nbsp; is as follows (see lower sketch):
:$$u( t ) =  - 2A + 2A \cdot \cos ( {2{\rm{\pi }}f_0 t} ).$$








''Hints:'' 
*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_Laws|Fourier Transform Laws]].
*All of these laws are illustrated with examples in the learning video&nbsp; [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Gesetzmäßigkeiten der Fouriertransformation]]&nbsp;.
The following relationship exists between&nbsp; ${x(t)}$&nbsp; and&nbsp; ${y(t)}$&nbsp;:
:$$y( t ) = \frac{1}{T} \cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau  )}\, {\rm d}\tau .$$
*The&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Integration_Theorem|Integration_Theorem]]&nbsp; reads in a correspondingly adapted form:
:$$\frac{1}{T}\cdot \hspace{-0.1cm} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau  )}\,\, {\rm d}\tau\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ X( f ) \cdot \left( {\frac{1}{{{\rm{j}}\cdot 2{\rm{\pi }\cdot }fT}} + \frac{1}{2T}\cdot {\rm \delta} ( f )} \right).$$
 


===Questions===

<quiz display=simple>
{Calculate the spectral function&nbsp; ${X(f)}$. What is its magnitude at the frequencies&nbsp; $f = 0$&nbsp; and&nbsp; $f = 1\, \text{kHz}$?
|type="{}"}
$|{X(f = 0)}| \ = \ $ { 0. }  &nbsp;$\text{mV/Hz}$
$|{X(f = 1\, \text{kHz})}|\ = \ $ { 2 3% } &nbsp;$\text{mV/Hz}$


{Calculate the spectral function&nbsp; ${Y(f)}$. What values result at the frequencies&nbsp; $f = 0$&nbsp; and&nbsp; $f = 1\, \text{kHz}$?
|type="{}"}
$|{Y(f = 0)}|\ = \ $ { 0. }  &nbsp;$\text{mV/Hz}$
$|{Y(f = 1\, \text{kHz})}| \ = \ $  { 0.636 3%  } &nbsp;$\text{mV/Hz}$



</quiz>

===Solution===
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'''(1)'''&nbsp;  In the description of the task you will find the Fourier correspondence between&nbsp; ${u(t)}$&nbsp; and&nbsp; ${U(f)}$. 
*Since both the time functions&nbsp; ${u(t)}$&nbsp; and&nbsp; ${x(t)}$&nbsp; and the corresponding spectra&nbsp; ${U(f)}$&nbsp; and&nbsp; ${X(f)}$&nbsp; are even and real,&nbsp; ${X(f)}$&nbsp; can be easily calculated by applying the "Vertauschungssatz":
:$$X( f ) =  - 2 \cdot A \cdot T  + 2 \cdot A \cdot T \cdot \cos \left( {{\rm{2\pi }}fT} \right).$$
*Because of the relation&nbsp; $\sin^{2}(\alpha) = (1 – \cos(\alpha))/2$&nbsp; it can also be written for this:
:$$X( f ) =  - 4 \cdot A \cdot T \cdot \sin ^2 ( {{\rm{\pi }}fT} ).$$
:*At frequency&nbsp; $f = 0$&nbsp;&nbsp; ${x(t)}$&nbsp; has no spectral components&nbsp; &rArr; &nbsp;  ${X(f = 0)} \;\underline{= 0}$. 
:*For&nbsp; $f = 1 \,\text{kHz}$&nbsp; &ndash; also&nbsp; $f \cdot T = 0.5$&nbsp; &ndash; &nbsp; on the other hand:
:$$X( f   =  1\;{\rm{kHz}} )  =    - 4 \cdot A \cdot T = -2 \cdot 10^{ - 3} \;{\rm{V/Hz}}\;  \Rightarrow \;
|X( {f = 1\;{\rm{kHz}}} )|   \hspace{0.15 cm}\underline{=  2  \;{\rm{mV/Hz}}}{\rm{.}}$$



'''(2)'''&nbsp; The spectrum&nbsp; ${Y(f)}$&nbsp; can be determined from&nbsp; ${X(f)}$&nbsp; by applying the integration theorem.
*Because of&nbsp; ${X(f = 0)} = 0$&nbsp; the Dirac function does not have to be taken into account at the frequency&nbsp; $f = 0$&nbsp; and one obtains:
:$$Y( f ) = \frac{X( f )}{{{\rm{j}} \cdot 2{\rm{\pi }}fT}} = \frac{{ - 4 \cdot A \cdot T \cdot \sin ^2 ( {{\rm{\pi }}fT} )}}{{{\rm{j}}\cdot 2{\rm{\pi }}fT}} = 2{\rm{j}} \cdot A \cdot T \cdot \frac{{\sin ^2 ( {{\rm{\pi }}fT} )}}{{{\rm{\pi }}fT}}.$$
*Of course, the result is the same as in&nbsp; [[Aufgaben:3.5_Differentiation_eines_Dreicksignals|Task 3.5]]:
:*At frequency&nbsp; $f = 0$&nbsp;&nbsp;  ${y(t)}$&nbsp; also has no spectral components&nbsp; &rArr; &nbsp;  ${Y(f = 0)} \;\underline{= 0}$.  
:*For&nbsp; $f = 1\,\text{kHz} \ \ (f \cdot T = 0.5)$&nbsp; one obtains a value smaller by a factor&nbsp; $\pi$&nbsp; compared to&nbsp; $X(f)$&nbsp;:
:$$|Y( {f = 1\;{\rm{kHz}}} )| =  \frac{4 \cdot A \cdot T}{\rm{\pi }} \hspace{0.15 cm}\underline{=  {\rm{0}}{\rm{.636}}  \;{\rm{mV/Hz}}}{\rm{.}}$$
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[[Category:Signal Representation: Exercises|^3.3 Fourier Transform Theorems^]]