Difference between revisions of "Aufgaben:Exercise 4.2: Rectangular Spectra"

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{{quiz-Header|Buchseite=Signaldarstellung/Unterschiede und Gemeinsamkeiten von TP- und BP-Signalen
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{{quiz-Header|Buchseite=Signal_Representation/Differences_and_Similarities_of_Low-Pass_and_Band-Pass_Signals
 
}}
 
}}
  
[[File:P_ID695__Sig_A_4_2_neu.png|250px|right|frame|Rechteckförmige Tiefpass- und Bandpass-Spektren]]
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[[File:P_ID695__Sig_A_4_2_neu.png|250px|right|frame|Given low–pass and band-pass spectra]]
Wir betrachten zwei Signale $u(t)$ und $w(t)$ mit jeweils rechteckförmigen Spektren $U(f)$ bzw. $W(f)$.  
+
We consider two signals  $u(t)$  and  $w(t)$  with rectangular spectra  $U(f)$  and  $W(f)$ respectively.
*Es ist offensichtlich, dass
+
*It is obvious that
 
   
 
   
 
:$$u(t)  =  u_0  \cdot {\rm si} ( \pi \cdot {t}/{T_{ u}})$$
 
:$$u(t)  =  u_0  \cdot {\rm si} ( \pi \cdot {t}/{T_{ u}})$$
  
:ein TP–Signal ist, dessen zwei Parameter $u_0$ und $T_u$ in der Teilaufgabe (1) zu bestimmen sind.  
+
:is a low-pass signal whose two parameters  $u_0$  and  $T_u$  are to be determined in subtask  '''(1)''' .
*Dagegen zeigt das Spektrum $W(f)$, dass $w(t)$ ein BP–Signal beschreibt.
+
*In contrast, the spectrum  $W(f)$ shows that  $w(t)$  describes a band-pass signal.
  
  
In dieser Aufgabe wird außerdem auf das BP–Signal
+
This task also refers to the band-pass signal
 
   
 
   
 
:$$d(t)  =  10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 \hspace{0.05cm}t)
 
:$$d(t)  =  10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 \hspace{0.05cm}t)
 
- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2\hspace{0.05cm} t)$$
 
- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2\hspace{0.05cm} t)$$
  
Bezug genommen, dessen Spektrum in [[Aufgaben:Aufgabe_4.1:_Tiefpass-_und_Bandpass-Signale|Aufgabe 4.1] ermittelt wurde. Es sei $f_2 = 2 \ \rm kHz.$
+
whose spectrum was determined in  [[Aufgaben:Exercise_4.1Z:_High-Pass_System|Exercise 4.1Z]] . Let  $f_2 = 2 \ \rm kHz.$
  
  
  
  
''Hinweise:''  
+
 
*Die Aufgabe gehört zum  Kapitel [[Signaldarstellung/Unterschiede_und_Gemeinsamkeiten_von_TP-_und_BP-Signalen|Unterschiede und Gemeinsamkeiten von Tiefpass- und Bandpass-Signalen]].
+
 
*Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
+
 
*Berücksichtigen Sie bei der Lösung die folgende trigonometrische Beziehung:
+
''Hints:''  
 +
*This exercise belongs to the chapter  [[Signal_Representation/Differences_and_Similarities_of_Low-Pass_and_Band-Pass_Signals|Differences and Similarities of Low-Pass and Band-Pass Signals]].
 +
*In this task, the function  $\rm si(x) = \rm sin(x)/x = \rm sinc(x/π)$  is used.
 +
 
 +
*Consider the following trigonometric relationship in the solution:
  
 
:$$\sin (\alpha) \cdot \cos (\beta)  =  {1}/{2} \cdot \big[ \sin
 
:$$\sin (\alpha) \cdot \cos (\beta)  =  {1}/{2} \cdot \big[ \sin
Line 32: Line 36:
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welche Werte besitzen die Parameter $u_0$ und $T_u$ des TP-Signals?
+
{What are the parameter values&nbsp; $u_0$&nbsp; and&nbsp; $T_u$&nbsp; of the low-pass signal?
 
|type="{}"}
 
|type="{}"}
 
$u_0\ = \ $  { 2 3% }  &nbsp;$\text{V}$
 
$u_0\ = \ $  { 2 3% }  &nbsp;$\text{V}$
 
$T_u\ = \ $  { 0.5 3% } &nbsp;$\text{ms}$
 
$T_u\ = \ $  { 0.5 3% } &nbsp;$\text{ms}$
  
{Berechnen Sie das Bandpass–Signal $w(t)$. Wie groß sind die Signalwerte bei $t = 0$ und $t = 62.5 \, {\rm &micro;}\text{s}$?
+
{Calculate the band-pass signal&nbsp; $w(t)$.&nbsp; What are the signal values at&nbsp; $t = 0$&nbsp; and&nbsp; $t = 62.5 \, {\rm &micro;}\text{s}$?
 
|type="{}"}
 
|type="{}"}
 
$w(t=0)\ = \ $ { 4 3% } &nbsp;$\text{V}$
 
$w(t=0)\ = \ $ { 4 3% } &nbsp;$\text{V}$
$w(t=62.5 \,\mu \text{s})\ = \ $ { 0. } &nbsp;$\text{V}$
+
$w(t=62.5 \,{\rm &micro;}  \text{s})\ = \ $ { 0. } &nbsp;$\text{V}$
  
{Welche Aussagen sind bezüglich der Bandpass–Signale $d(t)$ und $w(t)$ zutreffend? Begründen Sie Ihr Ergebnis im Zeitbereich.
+
{Which statements are true regarding the band-pass signals&nbsp; $d(t)$&nbsp; and&nbsp; $w(t)$&nbsp;?&nbsp; Justify your result in the time domain.
|type="[]"}
+
|type="()"}
+ Die Signale $d(t)$ und $w(t)$ sind identisch.
+
+ The signals&nbsp; $d(t)$&nbsp; and&nbsp; $w(t)$&nbsp; are identical.
- $d(t)$ und $w(t)$ unterscheiden sich durch einen konstanten Faktor.
+
- $d(t)$&nbsp; and&nbsp; $w(t)$&nbsp; differ by a constant factor.
- $d(t)$ und $w(t)$ haben unterschiedliche Form.
+
- $d(t)$&nbsp; und&nbsp; $w(t)$&nbsp; have different shapes.
  
 
</quiz>
 
</quiz>
  
  
===Musterlösung===
+
===Solution===
  
 
{{ML-Kopf}}
 
{{ML-Kopf}}
  
'''1.'''  Die Zeit $T_u$, welche die erste Nullstelle des TP-Signals $u(t)$ angibt, ist gleich dem Kehrwert der Breite des Rechteckspektrums, also $1/(2\, \text{kHz} ) = 0.5 \, \text{ms}$. Die Impulsamplitude ist, wie in der Musterlösung zur [[Aufgaben:4.1_TP-_und_BP-Signale|Aufgabe 4.1]] ausführlich dargelegt wurde, gleich der Rechteckfläche. Daraus folgt $u_0= 2 \, \text{V}$.
+
'''(1)'''&nbsp; The time&nbsp; $T_u$ &nbsp; &rArr; &nbsp; first zero of the low-pass signal&nbsp; $u(t)$&nbsp; &ndash; is equal to the reciprocal of the width of the rectangular spectrum, i.e. &nbsp; $1/(2\, \text{kHz} ) \hspace{0.15 cm}\underline{= 0.5 \, \text{ms}}$.  
 +
*The pulse amplitude is equal to the rectangular area as shown in the sample solution for&nbsp; [[Aufgaben:Aufgabe_4.1:_Tiefpass-_und_Bandpass-Signale|Exercise 4.1]]&nbsp;.&nbsp; From this follows&nbsp; $u_0\hspace{0.15 cm}\underline{= 2 \, \text{V}}$.
 +
 
 +
 
  
[[File:P_ID704__Sig_A_4_2_b_neu.png|250px|right|Multiplikation mit Cosinus]]
+
[[File:P_ID704__Sig_A_4_2_b_neu.png|250px|right|frame|Multiplication with a cosine function]]
'''2.''' Das BP-Spektrum kann mit $f_{\rm T} = 4\, \text{kHz}$ wie folgt dargestellt werden:
+
'''(2)'''&nbsp;  The band-pass spectrum can be represented with&nbsp; $f_{\rm T} = 4\, \text{kHz}$&nbsp; as follows:
 
   
 
   
$$ W(f)  = U(f- f_{\rm T}) + U(f+ f_{\rm T}) =  U(f)\star \left[
+
:$$ W(f)  = U(f- f_{\rm T}) + U(f+ f_{\rm T}) =  U(f)\star \left[
 
\delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$
 
\delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$
  
Entsprechend dem [[Signaldarstellung/Gesetzmäßigkeiten_der_Fouriertransformation#Verschiebungssatz|Verschiebungssatz]] gilt dann für das dazugehörige Zeitsignal:
+
According to the&nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem|Shifting Theorem]],&nbsp; the following then applies to the associated time signal:
 
   
 
   
$$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) =  2 u_0
+
:$$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) =  2 u_0
  \cdot {\rm si} ( \pi \frac{t}{T_{\rm u}})\cdot {\cos} ( 2 \pi f_{\rm T} t). $$
+
  \cdot {\rm si} ( \pi {t}/{T_{\rm u}})\cdot {\cos} ( 2 \pi f_{\rm T} t). $$
  
Die Grafik zeigt
+
The graph shows
*oben das TP-Signal $u(t)$,
+
*above the low&ndash;pass signal $u(t)$,
*dann die Schwingung $c(t) = 2 · \cos(2 \pi fTt$ ),
+
*then the oscillation $c(t) = 2 · \cos(2 \pi f_{\rm T}t$ ),
*unten das BP-Signal $w(t) = u(t) \cdot c(t)$.
+
*below the band-pass signal&nbsp; $w(t) = u(t) \cdot c(t)$.
  
Insbesondere erhält man zum Zeitpunkt $t = 0$:
+
 
 +
In particular, at time&nbsp; $t = 0$ one obtains:
 
   
 
   
$$w(t = 0)  =  2 \cdot u_0 \hspace{0.15 cm}\underline{= 4 \hspace{0.05cm}{\rm V}}.$$
+
:$$w(t = 0)  =  2 \cdot u_0 \hspace{0.15 cm}\underline{= 4 \hspace{0.05cm}{\rm V}}.$$
  
Der Zeitpunkt $t=62.5 \,\mu \text{s}$ entspricht genau einer viertel Periodendauer des Signals $c(t)$:
+
The time&nbsp; $t=62.5 \,{\rm &micro;} \text{s}$&nbsp; corresponds exactly to a quarter of the period of the signal&nbsp; $c(t)$:
 
   
 
   
$$ w(t = 62.5 \hspace{0.05cm}{\rm \mu s})  =  2 u_0 \cdot{\rm si} ( \pi \frac{62.5 \hspace{0.05cm}{\rm \mu s}}
+
:$$ w(t = 62.5 \hspace{0.05cm}{\rm &micro; s})  =  2 u_0 \cdot {\rm si} ( \pi \cdot \frac{62.5 \hspace{0.05cm}{\rm &micro;  s}}
  {500 \hspace{0.05cm}{\rm \mu s}})
+
  {500 \hspace{0.05cm}{\rm &micro;  s}})
 
  \cdot {\cos} ( 2 \pi \cdot 4\hspace{0.05cm}{\rm kHz}\cdot
 
  \cdot {\cos} ( 2 \pi \cdot 4\hspace{0.05cm}{\rm kHz}\cdot
  62.5 \hspace{0.05cm}{\rm \mu s}) =  
+
  62.5 \hspace{0.05cm}{\rm &micro;  s}) $$
 +
:$$ \Rightarrow  \hspace{0.3cm}w(t =  
 
  4\hspace{0.05cm}{\rm V}\cdot{\rm si} ( {\pi}/{8}) \cdot \cos ( {\pi}/{4})\hspace{0.15 cm}\underline{ = 0}.$$
 
  4\hspace{0.05cm}{\rm V}\cdot{\rm si} ( {\pi}/{8}) \cdot \cos ( {\pi}/{4})\hspace{0.15 cm}\underline{ = 0}.$$
  
'''3.''' Vergleicht man die Spektralfunktion $W(f)$ dieser Aufgabe mit dem Spektrum $D(f)$ in der Musterlösung zu [[Aufgaben:4.1_TP-_und_BP-Signale|Aufgabe 4.1]] , so erkennt man, dass $w(t)$ und $d(t)$ identische Signale sind. Etwas aufwändiger ist dieser Beweis im Zeitbereich. Mit $f_2 = 2 \,\text{kHz}$ kann für das hier betrachtete Signal geschrieben werden:
+
 
 +
 
 +
'''(3)'''&nbsp;  Proposed <u>solution 1 is correct</u>:
 +
*If we compare the spectral function&nbsp; $W(f)$&nbsp; of this task with the spectrum&nbsp; $D(f)$&nbsp; in the sample solution to&nbsp; [[Aufgaben:Exercise_4.1:_Low-Pass_and_Band-Pass_Signals|Exercise 4.1]], we see that&nbsp; $w(t)$&nbsp; and&nbsp; $d(t)$&nbsp; are identical.
 +
*This proof is somewhat more complex in the time domain.&nbsp; With&nbsp; $f_2 = 2 \,\text{kHz}$&nbsp; can be written for the signal considered here:
 
   
 
   
$$w(t )  =  4\hspace{0.05cm}{\rm V}
+
:$$w(t )  =  4\hspace{0.05cm}{\rm V}
 
  \cdot {\rm si} ( \pi f_2 t)
 
  \cdot {\rm si} ( \pi f_2 t)
 
  \cdot {\cos} ( 4 \pi f_2 t)  =  
 
  \cdot {\cos} ( 4 \pi f_2 t)  =  
 
({4\hspace{0.05cm}{\rm V}})/({\pi f_2 t})\cdot \sin (\pi f_2 t) \cdot \cos ( 4 \pi f_2 t) .$$
 
({4\hspace{0.05cm}{\rm V}})/({\pi f_2 t})\cdot \sin (\pi f_2 t) \cdot \cos ( 4 \pi f_2 t) .$$
  
Wegen der trigonometrischen Beziehung
+
*Because of the trigonometric relationship
 
   
 
   
$$\sin (\alpha) \cdot \cos (\beta)  =  {1}/{2} \cdot \left[ \sin
+
:$$\sin (\alpha) \cdot \cos (\beta)  =  {1}/{2} \cdot \big[ \sin
(\alpha + \beta)+ \sin (\alpha - \beta)\right]$$
+
(\alpha + \beta)+ \sin (\alpha - \beta)\big]$$
  
kann obige Gleichung umgeformt werden:
+
:the above equation can be transformed:
 
   
 
   
$$w(t )  =
+
:$$w(t )  =
  \frac{2\hspace{0.05cm}{\rm V}}{\pi f_2 t}\cdot \left[\sin (5\pi f_2 t) + \sin (-3\pi f_2 t)\right]  
+
  \frac{2\hspace{0.05cm}{\rm V}}{\pi f_2 t}\cdot \big [\sin (5\pi f_2 t) + \sin (-3\pi f_2 t)\big ]  
 
  = 10\hspace{0.05cm}{\rm V} \cdot \frac{\sin (5\pi f_2 t)}{5\pi f_2 t}-
 
  = 10\hspace{0.05cm}{\rm V} \cdot \frac{\sin (5\pi f_2 t)}{5\pi f_2 t}-
 
  6\hspace{0.05cm}{\rm V} \cdot \frac{\sin (3\pi f_2 t)}{3\pi f_2 t}.$$
 
  6\hspace{0.05cm}{\rm V} \cdot \frac{\sin (3\pi f_2 t)}{3\pi f_2 t}.$$
  
Damit ist gezeigt, dass beide Signale tatsächlich identisch sind ⇒  <u>Lösungsvorschlag 1</u>:
+
*This shows that both signals are actually identical &nbsp; ⇒  &nbsp; Proposed solution 1:
 
   
 
   
$$w(t)  =  10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 t)
+
:$$w(t)  =  10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 t)
 
- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2 t) = d(t).$$
 
- 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2 t) = d(t).$$
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
 
__NOEDITSECTION__
 
__NOEDITSECTION__
[[Category:Aufgaben zu Signaldarstellung|^4. Bandpassartige Signale^]]
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[[Category:Signal Representation: Exercises|^4.1 Differences between Low-Pass and Band-Pass^]]

Latest revision as of 14:49, 5 May 2021

Given low–pass and band-pass spectra

We consider two signals  $u(t)$  and  $w(t)$  with rectangular spectra  $U(f)$  and  $W(f)$ respectively.

  • It is obvious that
$$u(t) = u_0 \cdot {\rm si} ( \pi \cdot {t}/{T_{ u}})$$
is a low-pass signal whose two parameters  $u_0$  and  $T_u$  are to be determined in subtask  (1) .
  • In contrast, the spectrum  $W(f)$ shows that  $w(t)$  describes a band-pass signal.


This task also refers to the band-pass signal

$$d(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 \hspace{0.05cm}t) - 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2\hspace{0.05cm} t)$$

whose spectrum was determined in  Exercise 4.1Z . Let  $f_2 = 2 \ \rm kHz.$




Hints:

  • Consider the following trigonometric relationship in the solution:
$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin (\alpha + \beta)+ \sin (\alpha - \beta)\big].$$


Questions

1

What are the parameter values  $u_0$  and  $T_u$  of the low-pass signal?

$u_0\ = \ $

 $\text{V}$
$T_u\ = \ $

 $\text{ms}$

2

Calculate the band-pass signal  $w(t)$.  What are the signal values at  $t = 0$  and  $t = 62.5 \, {\rm µ}\text{s}$?

$w(t=0)\ = \ $

 $\text{V}$
$w(t=62.5 \,{\rm µ} \text{s})\ = \ $

 $\text{V}$

3

Which statements are true regarding the band-pass signals  $d(t)$  and  $w(t)$ ?  Justify your result in the time domain.

The signals  $d(t)$  and  $w(t)$  are identical.
$d(t)$  and  $w(t)$  differ by a constant factor.
$d(t)$  und  $w(t)$  have different shapes.


Solution

(1)  The time  $T_u$   ⇒   first zero of the low-pass signal  $u(t)$  – is equal to the reciprocal of the width of the rectangular spectrum, i.e.   $1/(2\, \text{kHz} ) \hspace{0.15 cm}\underline{= 0.5 \, \text{ms}}$.

  • The pulse amplitude is equal to the rectangular area as shown in the sample solution for  Exercise 4.1 .  From this follows  $u_0\hspace{0.15 cm}\underline{= 2 \, \text{V}}$.


Multiplication with a cosine function

(2)  The band-pass spectrum can be represented with  $f_{\rm T} = 4\, \text{kHz}$  as follows:

$$ W(f) = U(f- f_{\rm T}) + U(f+ f_{\rm T}) = U(f)\star \left[ \delta(f- f_{\rm T})+ \delta(f+ f_{\rm T})\right].$$

According to the  Shifting Theorem,  the following then applies to the associated time signal:

$$w(t) = 2 \cdot u(t) \cdot {\cos} ( 2 \pi f_{\rm T} t) = 2 u_0 \cdot {\rm si} ( \pi {t}/{T_{\rm u}})\cdot {\cos} ( 2 \pi f_{\rm T} t). $$

The graph shows

  • above the low–pass signal $u(t)$,
  • then the oscillation $c(t) = 2 · \cos(2 \pi f_{\rm T}t$ ),
  • below the band-pass signal  $w(t) = u(t) \cdot c(t)$.


In particular, at time  $t = 0$ one obtains:

$$w(t = 0) = 2 \cdot u_0 \hspace{0.15 cm}\underline{= 4 \hspace{0.05cm}{\rm V}}.$$

The time  $t=62.5 \,{\rm µ} \text{s}$  corresponds exactly to a quarter of the period of the signal  $c(t)$:

$$ w(t = 62.5 \hspace{0.05cm}{\rm µ s}) = 2 u_0 \cdot {\rm si} ( \pi \cdot \frac{62.5 \hspace{0.05cm}{\rm µ s}} {500 \hspace{0.05cm}{\rm µ s}}) \cdot {\cos} ( 2 \pi \cdot 4\hspace{0.05cm}{\rm kHz}\cdot 62.5 \hspace{0.05cm}{\rm µ s}) $$
$$ \Rightarrow \hspace{0.3cm}w(t = 4\hspace{0.05cm}{\rm V}\cdot{\rm si} ( {\pi}/{8}) \cdot \cos ( {\pi}/{4})\hspace{0.15 cm}\underline{ = 0}.$$


(3)  Proposed solution 1 is correct:

  • If we compare the spectral function  $W(f)$  of this task with the spectrum  $D(f)$  in the sample solution to  Exercise 4.1, we see that  $w(t)$  and  $d(t)$  are identical.
  • This proof is somewhat more complex in the time domain.  With  $f_2 = 2 \,\text{kHz}$  can be written for the signal considered here:
$$w(t ) = 4\hspace{0.05cm}{\rm V} \cdot {\rm si} ( \pi f_2 t) \cdot {\cos} ( 4 \pi f_2 t) = ({4\hspace{0.05cm}{\rm V}})/({\pi f_2 t})\cdot \sin (\pi f_2 t) \cdot \cos ( 4 \pi f_2 t) .$$
  • Because of the trigonometric relationship
$$\sin (\alpha) \cdot \cos (\beta) = {1}/{2} \cdot \big[ \sin (\alpha + \beta)+ \sin (\alpha - \beta)\big]$$
the above equation can be transformed:
$$w(t ) = \frac{2\hspace{0.05cm}{\rm V}}{\pi f_2 t}\cdot \big [\sin (5\pi f_2 t) + \sin (-3\pi f_2 t)\big ] = 10\hspace{0.05cm}{\rm V} \cdot \frac{\sin (5\pi f_2 t)}{5\pi f_2 t}- 6\hspace{0.05cm}{\rm V} \cdot \frac{\sin (3\pi f_2 t)}{3\pi f_2 t}.$$
  • This shows that both signals are actually identical   ⇒   Proposed solution 1:
$$w(t) = 10 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 5 \pi f_2 t) - 6 \hspace{0.05cm}{\rm V} \cdot {\rm si} ( 3 \pi f_2 t) = d(t).$$