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

From LNTwww
 
(26 intermediate revisions by 4 users not shown)
Line 1: Line 1:
  
{{quiz-Header|Buchseite=Signaldarstellung/Einige Sonderfälle impulsartiger Signale
+
{{quiz-Header|Buchseite=Signal_Representation/Special_Cases_of_Impulse_Signals
 
}}
 
}}
  
[[File:P_ID507__Sig_Z_3_3.png|right|frame|Verschiedene Rechteckimpulse]]
+
[[File:P_ID507__Sig_Z_3_3.png|right|frame|Various rectangular pulses]]
Wir betrachten hier eine Vielzahl von symmetrischen Rechteckfunktionen $x_k(t)$. Die einzelnen Rechtecke unterscheiden sich durch unterschiedliche Amplituden (Höhen)
+
We consider here a multitude of symmetrical rectangular functions  $x_k(t)$.  The individual rectangles differ in amplitudes (heights)
 
:$$A_k  = k \cdot A$$
 
:$$A_k  = k \cdot A$$
und unterschiedliche Impulsdauern (Breiten)
+
and different pulse durations (widths)
 
:$$T_k = T/k.$$
 
:$$T_k = T/k.$$
Hierbei sei $k$ ein beliebiger positiver Wert.  
+
Let  $k$  be any positive value.
  
*Der im Bild rot dargestellte Rechteckimpuls $x_1(t)$ hat die Amplitude $A_1 = {A} = 2 \,\text{V}$ und die Dauer $T_1 = {T} = 500 \,µ\text{s}$.  
+
*The rectangular pulse  $x_1(t)$  shown in red has the amplitude   $A_1 = {A} = 2 \,\text{V}$  and the duration  $T_1 = {T} = 500 \,µ\text{s}$.  
*Der blau gezeichnete Impuls $x_2(t)$ ist halb so breit   ⇒   $T_2 =250 \,µ\text{s}$, aber doppelt so hoch   ⇒   $A_2 = 4 \text{ V}$.
+
*The pulse  $x_2(t)$ shown in blue is half as wide  ⇒   $T_2 =250 \,µ\text{s}$, but twice as high   ⇒   $A_2 = 4 \text{ V}$.
  
  
Line 17: Line 17:
  
  
''Hinweise:''  
+
 
*Die Aufgabe gehört zum Kapitel [[Signaldarstellung/Einige_Sonderfälle_impulsartiger_Signale|Einige Sonderfälle impulsartiger Signale]].
+
 
 +
 
 +
''Hints:''  
 +
*This task belongs to the chapter  [[Signal_Representation/Special_Cases_of_Pulses|Special Cases of Pulses]].
 +
*Use one of the functions  $\text{si}(x) = \sin(x)/x$  or  $\text{sinc}(x) = \sin(\pi x)/(\pi x)$.  
 
   
 
   
*Sie können Ihre Ergebnisse anhand der beiden  interaktiven Applets [[Applets:Impulse_und_Spektren|Impulse und Spektren]] sowie  [[Applets:Frequenzgang_und_Impulsantwort|Frequenzgang und Impulsantwort]] überprüfen.
+
*You can check your results using the two interactive applets 
 +
:[[Applets:Pulses_and_Spectra|Pulses and Spectra]],
 +
:[[Applets:Frequency_%26_Impulse_Responses|Frequency & Impulse Responses]].
  
  
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welche der folgenden Aussagen treffen bezüglich des Spektrums $X_1(f)$ zu?
+
{Which of the following statements are true regarding the spectrum&nbsp; $X_1(f)$?
 
|type="[]"}
 
|type="[]"}
+ Der Spektralwert $X_1(f = 0)$ ist gleich $10^{–3} \,\text{V/Hz}$.
+
+ The spectral value&nbsp; $X_1(f = 0)$&nbsp; is equal to&nbsp; $10^{–3} \,\text{V/Hz}$.
+ $X_1(f)$ besitzt Nullstellen im Abstand von $2 \,\text{kHz}$.
+
+ $X_1(f)$&nbsp; has zeros at the interval of&nbsp; $2 \,\text{kHz}$.
- $X_1(f)$ besitzt Nullstellen im Abstand von $4 \,\text{kHz}$.
+
- $X_1(f)$&nbsp; has zeros at the interval of&nbsp; $4 \,\text{kHz}$.
  
  
{Welche der folgenden Aussagen treffen bezüglich des Spektrums $X_2(f)$ zu?
+
{Which of the following statements are true regarding the spectrum&nbsp; $X_2(f)$?
 
|type="[]"}
 
|type="[]"}
+ Der Spektralwert $X_2(f = 0)$ ist gleich $10^{–3} \,\text{V/Hz}$.
+
+ The spectral value is&nbsp; $X_2(f = 0)$&nbsp; is equal to&nbsp; $10^{–3} \,\text{V/Hz}$.
- $X_2(f)$ besitzt Nullstellen im Abstand von $2\, \text{kHz}$.
+
- $X_2(f)$&nbsp; has zeros at the interval of&nbsp; $2\, \text{kHz}$.
+ $X_2(f)$ besitzt Nullstellen im Abstand von $4 \,\text{kHz}$.
+
+ $X_2(f)$&nbsp; has zeros at the interval of&nbsp; $4 \,\text{kHz}$.
  
  
{Es gelte $k = 10$. Berechnen Sie die Frequenz $f_{10}$ der ersten Nullstelle und den Spektralwert bei $f = 2 \,\text{kHz}$.
+
{Let&nbsp; $k = 10$.&nbsp; Calculate the frequency&nbsp; $f_{10}$&nbsp; of the first zero and the spectral value at&nbsp; $f = 2 \,\text{kHz}$.
 
|type="{}"}
 
|type="{}"}
 
$f_{10} \ = \ ${ 20 3% } &nbsp;$\text{kHz}$
 
$f_{10} \ = \ ${ 20 3% } &nbsp;$\text{kHz}$
Line 48: Line 54:
  
  
{Wie groß wird der Spektralwert bei $f = 2 \,\text{kHz}$ im Grenzfall $k \rightarrow \infty$? Interpretieren Sie das Ergebnis.
+
{What is the spectral value at&nbsp; $f = 2 \,\text{kHz}$&nbsp; in the limiting case&nbsp; $k \rightarrow \infty$?&nbsp; Interpret the result.
 
|type="{}"}
 
|type="{}"}
 
$X_{\infty}(f = 2 \,\text{kHz})\ = \ $ { 1 3% } &nbsp;$\text{mV/Hz}$
 
$X_{\infty}(f = 2 \,\text{kHz})\ = \ $ { 1 3% } &nbsp;$\text{mV/Hz}$
Line 56: Line 62:
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solutions===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 1 und 2</u>:
+
'''(1)'''&nbsp;  The <u>proposed solutions 1 and 2</u> are correct:
*Der Spektralwert bei der Frequenz $f = 0$ ist nach dem [[Signaldarstellung/Fouriertransformation_und_-rücktransformation#Das_erste_Fourierintegral|ersten Fourierintegral]] 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 $A \cdot T = 10^{–3} \,\text{Vs} = 1\, \text{mV/Hz}$.  
+
*In the present case, the pulse area is always&nbsp; $A \cdot T = 10^{–3} \,\text{Vs} = 1\, \text{mV/Hz}$.  
*Wegen $T_1 = 500 \,&micro;\text{s}$ weist das Spektrum $X_1(f)$ Nulldurchgänge im Abstand $f_1 = 1/T_1 = 2 \,\text{kHz}$ 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;  The <u>proposed 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; $f_2 = 1/T_2 = 4 \,\text{kHz}$.
  
'''(2)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 1 und 3</u>:
 
*Aufgrund gleicher Impulsflächen wird der Spektralwert bei der Frequenz $f = 0$ nicht verändert.
 
*Die äquidistanten Nulldurchgänge treten nun im Abstand $f_2 = 1/T_2 = 4 \,\text{kHz}$ auf.
 
  
  
'''(3)'''&nbsp;  Nullstellen gibt es bei Vielfachen von $f_{10} = 1/T_{10} = 20 \,\text{kHz}$, und die Spektralfunktion lautet:
+
'''(3)'''&nbsp;  Zero crossings occur at multiples of&nbsp; $f_{10} = 1/T_{10} = 20 \,\text{kHz}$, and the spectral function is:
 
:$$X_{10} ( f ) = X_0  \cdot {\mathop{\rm si}\nolimits} ( {{\rm{\pi }}f/f_{10} } ).$$
 
:$$X_{10} ( f ) = X_0  \cdot {\mathop{\rm si}\nolimits} ( {{\rm{\pi }}f/f_{10} } ).$$
Bei der Frequenz $f = 2 \,\text{kHz}$ ist das Argument der $\rm si$-Funktion gleich $\pi/10$ (oder $18^{\circ}$):
+
*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; $18^{\circ})$:
 
:$$X_{10} ( {f = 2\;{\rm{kHz}}}) = 10^{ - 3} \;{\rm{V/Hz}} \cdot \frac{{\sin ( {18^\circ } )}}{{{\rm{\pi /10}}}} \hspace{0.15 cm}\underline{= 0.984 \;{\rm{mV/Hz}}}{\rm{.}}$$
 
:$$X_{10} ( {f = 2\;{\rm{kHz}}}) = 10^{ - 3} \;{\rm{V/Hz}} \cdot \frac{{\sin ( {18^\circ } )}}{{{\rm{\pi /10}}}} \hspace{0.15 cm}\underline{= 0.984 \;{\rm{mV/Hz}}}{\rm{.}}$$
  
'''(4)'''&nbsp;  Im Grenzfall $k \rightarrow \infty$ geht der dann unendlich hohe und unendlich schmale [[Signaldarstellung/Einige_Sonderfälle_impulsartiger_Signale#Rechteckimpuls|Rechteckimpuls]] in den [[Signaldarstellung/Einige_Sonderfälle_impulsartiger_Signale#Diracimpuls|Diracimpuls]] über. Dessen Spektrum ist für alle Frequenzen konstant. Damit gilt auch bei der Frequenz $f = 2 \,\text{kHz}$ der Spektralwert $X_{\infty}(f = 2 \,\text{kHz})\underline{=1 \text{ mV/Hz}}$.
+
&nbsp;
 +
'''(4)'''&nbsp;  In the limiting case&nbsp; $k \rightarrow \infty$&nbsp; the then infinitely high and infinitely narrow&nbsp; [[Signal_Representation/Special_Cases_of_Pulses#Rectangular_pulse|Rectangular pulse]]&nbsp; changes into the&nbsp; [[Signal_Representation/Special_Cases_of_Pulses#Dirac_.28delta.29_impulse|Dirac delta impulse]].  
 +
*Its spectrum is constant for all frequencies.
 +
*Thus the spectral value&nbsp; $X_{\infty}(f = 2 \,\text{kHz})\hspace{0.15 cm}\underline{=1 \text{ mV/Hz}}$&nbsp; also applies at the frequency &nbsp; $f = 2 \,\text{kHz}$&nbsp;.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
 
__NOEDITSECTION__
 
__NOEDITSECTION__
[[Category:Aufgaben zu Signaldarstellung|^3. Aperiodische Signale - Impulse^]]
+
[[Category:Signal Representation: Exercises|^3.2 Special Cases of Pulses^]]

Latest revision as of 17:42, 25 May 2021

Various rectangular pulses

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:

  • This task belongs to the chapter  Special Cases of Pulses.
  • Use one of the functions  $\text{si}(x) = \sin(x)/x$  or  $\text{sinc}(x) = \sin(\pi x)/(\pi x)$.
  • You can check your results using the two interactive applets 
Pulses and Spectra,
Frequency & Impulse Responses.



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  $X_2(f)$?

The spectral value is  $X_2(f = 0)$  is equal to  $10^{–3} \,\text{V/Hz}$.
$X_2(f)$  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  $f = 2 \,\text{kHz}$  in the limiting case  $k \rightarrow \infty$?  Interpret the result.

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

 $\text{mV/Hz}$


Solutions

(1)  The 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)  The 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  $f_2 = 1/T_2 = 4 \,\text{kHz}$.


(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 pulse  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   $f = 2 \,\text{kHz}$ .