Difference between revisions of "Aufgaben:Exercise 3.3Z: Rectangular Pulse and Dirac Delta"

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+ The spectral value  $X_1(f = 0)$  is equal to  $10^{–3} \,\text{V/Hz}$.
 
+ The spectral value  $X_1(f = 0)$  is equal to  $10^{–3} \,\text{V/Hz}$.
+ $X_1(f)$  has zeros at a distance of  $2 \,\text{kHz}$.
+
+ $X_1(f)$  has zeros at the interval of  $2 \,\text{kHz}$.
- $X_1(f)$  has zeros at a distance of  $4 \,\text{kHz}$.
+
- $X_1(f)$  has zeros at the interval of  $4 \,\text{kHz}$.
  
  
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+ The spectral value  $X_2(f = 0)$  ist gleich  $10^{–3} \,\text{V/Hz}$.
 
+ The spectral value  $X_2(f = 0)$  ist gleich  $10^{–3} \,\text{V/Hz}$.
- $X_2(f)$  has zeros at a distance of  $2\, \text{kHz}$.
+
- $X_2(f)$  has zeros at the interval of  $2\, \text{kHz}$.
+ $X_2(f)$  has zeros at a distance of  $4 \,\text{kHz}$.
+
+ $X_2(f)$  has zeros at the interval of  $4 \,\text{kHz}$.
  
  
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===Musterlösung===
 
===Musterlösung===
 
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'''(1)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 1 und 2</u>:
+
'''(1)'''&nbsp;  Proposed <u>solutions 1 and 2</u> are correct:
*Der Spektralwert bei der Frequenz&nbsp; $f = 0$&nbsp; ist nach dem&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Das_erste_Fourierintegral|ersten Fourierintegral]]&nbsp; stets gleich der Fläche unter der Zeitfunktion:
+
*The spectral value at frequency&nbsp; $f = 0$&nbsp; is always equal to the area under the time function according to&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_First_Fourier_Integral|the first Fourier integral]]&nbsp;:
 
:$$X( f ) = \int_{ - \infty }^{ + \infty } {x( t )}  \cdot {\rm{e}}^{ - {\rm{j2\pi }}ft} \hspace{0.1cm} {\rm d}t \hspace{0.5cm} \Rightarrow \hspace{0.5cm} \;X( {f = 0} ) = \int_{ - \infty }^{ + \infty } {x( t )}\hspace{0.1cm}  {\rm d}t.$$
 
:$$X( f ) = \int_{ - \infty }^{ + \infty } {x( t )}  \cdot {\rm{e}}^{ - {\rm{j2\pi }}ft} \hspace{0.1cm} {\rm d}t \hspace{0.5cm} \Rightarrow \hspace{0.5cm} \;X( {f = 0} ) = \int_{ - \infty }^{ + \infty } {x( t )}\hspace{0.1cm}  {\rm d}t.$$
*Im vorliegenden Fall ist die Impulsfläche stets&nbsp; $A \cdot T = 10^{–3} \,\text{Vs} = 1\, \text{mV/Hz}$.  
+
*In the present case, the pulse area is always&nbsp; $A \cdot T = 10^{–3} \,\text{Vs} = 1\, \text{mV/Hz}$.  
*Wegen&nbsp; $T_1 = 500 \,&micro;\text{s}$&nbsp; weist das Spektrum&nbsp; $X_1(f)$&nbsp; Nulldurchgänge im Abstand&nbsp; $f_1 = 1/T_1 = 2 \,\text{kHz}$&nbsp; auf.
+
*Because of&nbsp; $T_1 = 500 \,&micro;\text{s}$&nbsp; the spectrum&nbsp; $X_1(f)$&nbsp; has zero crossings at the interval&nbsp; $f_1 = 1/T_1 = 2 \,\text{kHz}$&nbsp;.
  
  
 +
'''(2)'''&nbsp;  Proposed <u>solutions 1 and 3</u> are correct:
 +
*Due to equal pulse areas, the spectral value is not changed at the frequency&nbsp; $f = 0$&nbsp;.
 +
*The equidistant zero crossings now occur at the interval&nbsp; $f_2 = 1/T_2 = 4 \,\text{kHz}$&nbsp; auf.
  
'''(2)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 1 und 3</u>:
 
*Aufgrund gleicher Impulsflächen wird der Spektralwert bei der Frequenz&nbsp; $f = 0$&nbsp; nicht verändert.
 
*Die äquidistanten Nulldurchgänge treten nun im Abstand&nbsp; $f_2 = 1/T_2 = 4 \,\text{kHz}$&nbsp; auf.
 
  
  
 
+
'''(3)'''&nbsp;  Zero crossings occur at multiples of&nbsp; $f_{10} = 1/T_{10} = 20 \,\text{kHz}$, and the spectral function is:
'''(3)'''&nbsp;  Nullstellen gibt es bei Vielfachen von&nbsp; $f_{10} = 1/T_{10} = 20 \,\text{kHz}$, und die Spektralfunktion lautet:
 
 
:$$X_{10} ( f ) = X_0  \cdot {\mathop{\rm si}\nolimits} ( {{\rm{\pi }}f/f_{10} } ).$$
 
:$$X_{10} ( f ) = X_0  \cdot {\mathop{\rm si}\nolimits} ( {{\rm{\pi }}f/f_{10} } ).$$
*Bei der Frequenz&nbsp; $f = 2 \,\text{kHz}$ ist&nbsp; das Argument der&nbsp; $\rm si$-Funktion gleich&nbsp; $\pi/10$&nbsp; $($oder&nbsp; $18^{\circ})$:
+
*At frequency&nbsp; $f = 2 \,\text{kHz}$ &nbsp; the argument of the&nbsp; $\rm si$-function is equal to&nbsp; $\pi/10$&nbsp; $($or&nbsp; $18^{\circ})$:
 
:$$X_{10} ( {f = 2\;{\rm{kHz}}}) = 10^{ - 3} \;{\rm{V/Hz}} \cdot \frac{{\sin ( {18^\circ } )}}{{{\rm{\pi /10}}}} \hspace{0.15 cm}\underline{= 0.984 \;{\rm{mV/Hz}}}{\rm{.}}$$
 
:$$X_{10} ( {f = 2\;{\rm{kHz}}}) = 10^{ - 3} \;{\rm{V/Hz}} \cdot \frac{{\sin ( {18^\circ } )}}{{{\rm{\pi /10}}}} \hspace{0.15 cm}\underline{= 0.984 \;{\rm{mV/Hz}}}{\rm{.}}$$
  
 
&nbsp;
 
&nbsp;
'''(4)'''&nbsp;  Im Grenzfall&nbsp; $k \rightarrow \infty$&nbsp; geht der dann unendlich hohe und unendlich schmale&nbsp; [[Signal_Representation/Special_Cases_of_Impulse_Signals#Rechteckimpuls|Rechteckimpuls]]&nbsp; in den&nbsp; [[Signal_Representation/Special_Cases_of_Impulse_Signals#Diracimpuls|Diracimpuls]]&nbsp; über.  
+
'''(4)'''&nbsp;  In the limiting case&nbsp; $k \rightarrow \infty$&nbsp; the then infinitely high and infinitely narrow&nbsp; [[Signal_Representation/Special_Cases_of_Impulse_Signals#Rectangular_Impulse|Rectangular impulse]]&nbsp; changes into the&nbsp; [[Signal_Representation/Special_Cases_of_Impulse_Signals#Dirac_Delta_Impulse|Dirac Delta impulse]]&nbsp;.  
*Dessen Spektrum ist für alle Frequenzen konstant.  
+
*Its spectrum is constant for all frequencies.
*Damit gilt auch bei der Frequenz&nbsp; $f = 2 \,\text{kHz}$&nbsp; der Spektralwert&nbsp; $X_{\infty}(f = 2 \,\text{kHz})\hspace{0.15 cm}\underline{=1 \text{ mV/Hz}}$.
+
*Thus the spectral value&nbsp; $X_{\infty}(f = 2 \,\text{kHz})\hspace{0.15 cm}\underline{=1 \text{ mV/Hz}}$ also applies at the frequency &nbsp; $f = 2 \,\text{kHz}$&nbsp;.
 
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Revision as of 19:10, 23 January 2021

Verschiedene Rechteckimpulse

We consider here a multitude of symmetrical rectangular functions  $x_k(t)$. The individual rectangles differ in amplitudes (heights)

$$A_k = k \cdot A$$

and different pulse durations (widths)

$$T_k = T/k.$$

Let  $k$  be any positive value.

  • The rectangular pulse  $x_1(t)$  shown in red has the amplitude   $A_1 = {A} = 2 \,\text{V}$  and the duration  $T_1 = {T} = 500 \,µ\text{s}$.
  • The pulse  $x_2(t)$ shown in blue is half as wide  ⇒   $T_2 =250 \,µ\text{s}$, but twice as high   ⇒   $A_2 = 4 \text{ V}$.





Hints:



Questions

1

Which of the following statements are true regarding the spectrum  $X_1(f)$ ?

The spectral value  $X_1(f = 0)$  is equal to  $10^{–3} \,\text{V/Hz}$.
$X_1(f)$  has zeros at the interval of  $2 \,\text{kHz}$.
$X_1(f)$  has zeros at the interval of  $4 \,\text{kHz}$.

2

Which of the following statements are true regarding the spectrum  $X_2(f)$  zu?

The spectral value  $X_2(f = 0)$  ist gleich  $10^{–3} \,\text{V/Hz}$.
$X_2(f)$  has zeros at the interval of  $2\, \text{kHz}$.
$X_2(f)$  has zeros at the interval of  $4 \,\text{kHz}$.

3

Let  $k = 10$. Calculate the frequency  $f_{10}$  of the first zero and the spectral value at  $f = 2 \,\text{kHz}$.

$f_{10} \ = \ $

 $\text{kHz}$
$X_{10}(f = 2 \text{kHz})\ = \ $

 $\text{mV/Hz}$

4

What is the spectral value at  bei $f = 2 \,\text{kHz}$  in the limiting case  $k \rightarrow \infty$? Interpret the result.

$X_{\infty}(f = 2 \,\text{kHz})\ = \ $

 $\text{mV/Hz}$


Musterlösung

(1)  Proposed solutions 1 and 2 are correct:

  • The spectral value at frequency  $f = 0$  is always equal to the area under the time function according to  the first Fourier integral :
$$X( f ) = \int_{ - \infty }^{ + \infty } {x( t )} \cdot {\rm{e}}^{ - {\rm{j2\pi }}ft} \hspace{0.1cm} {\rm d}t \hspace{0.5cm} \Rightarrow \hspace{0.5cm} \;X( {f = 0} ) = \int_{ - \infty }^{ + \infty } {x( t )}\hspace{0.1cm} {\rm d}t.$$
  • In the present case, the pulse area is always  $A \cdot T = 10^{–3} \,\text{Vs} = 1\, \text{mV/Hz}$.
  • Because of  $T_1 = 500 \,µ\text{s}$  the spectrum  $X_1(f)$  has zero crossings at the interval  $f_1 = 1/T_1 = 2 \,\text{kHz}$ .


(2)  Proposed solutions 1 and 3 are correct:

  • Due to equal pulse areas, the spectral value is not changed at the frequency  $f = 0$ .
  • The equidistant zero crossings now occur at the interval  $f_2 = 1/T_2 = 4 \,\text{kHz}$  auf.


(3)  Zero crossings occur at multiples of  $f_{10} = 1/T_{10} = 20 \,\text{kHz}$, and the spectral function is:

$$X_{10} ( f ) = X_0 \cdot {\mathop{\rm si}\nolimits} ( {{\rm{\pi }}f/f_{10} } ).$$
  • At frequency  $f = 2 \,\text{kHz}$   the argument of the  $\rm si$-function is equal to  $\pi/10$  $($or  $18^{\circ})$:
$$X_{10} ( {f = 2\;{\rm{kHz}}}) = 10^{ - 3} \;{\rm{V/Hz}} \cdot \frac{{\sin ( {18^\circ } )}}{{{\rm{\pi /10}}}} \hspace{0.15 cm}\underline{= 0.984 \;{\rm{mV/Hz}}}{\rm{.}}$$

  (4)  In the limiting case  $k \rightarrow \infty$  the then infinitely high and infinitely narrow  Rectangular impulse  changes into the  Dirac Delta impulse .

  • Its spectrum is constant for all frequencies.
  • Thus the spectral value  $X_{\infty}(f = 2 \,\text{kHz})\hspace{0.15 cm}\underline{=1 \text{ mV/Hz}}$ also applies at the frequency   $f = 2 \,\text{kHz}$ .