Difference between revisions of "Aufgaben:Exercise 1.6Z: Interpretation of the Frequency Response"

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{{quiz-Header|Buchseite=Lineare zeitinvariante Systeme/Einige systemtheoretische Tiefpassfunktionen}}
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{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory}}
  
[[File:P_ID862__LZI_Z_1_6.png|right|Impulsantwort und Eingangssignale]]  
+
[[File:P_ID862__LZI_Z_1_6.png|right|frame|Impulse response and input signals]]  
Mit dieser Aufgabe soll der Einfluss eines Tiefpasses $H(f)$ auf cosinusförmige Signale der Form
+
The task is meant to investigate the influence of a low-pass filter  $H(f)$  on cosinusoidal signals of the form
$$x_i(t) = A_x \cdot {\rm cos}(2\pi  f_i  t )$$
+
:$$x_i(t) = A_x \cdot {\rm cos}(2\pi  f_i  t ).$$  
veranschaulicht werden. In der Grafik sehen Sie die Signale $x_i(t)$, wobei der Index $i$ die Frequenz in $\rm kHz$ angibt. So beschreibt $x_2(t)$ ein $2 \ \rm  kHz$–Signal.
+
In the graph you can see the signals  $x_i(t)$ where the index $i$  indicates the frequency in $\rm kHz$ . So,  $x_2(t)$  describes a  $2 \hspace{0.09cm} \rm  kHz$–signal.
  
Die Signalamplitude beträgt jeweils $A_x =\ \rm  V$. Das Gleichsignal $x_0(t)$ ist als Grenzfall eines Cosinussignals mit der Frequenz $f_0 =0$ zu interpretieren.
+
The signal amplitude in each case is $A_x = 1 \hspace{0.05cm} \rm  V$. The direct (DC) signal  $x_0(t)$  is to be interpreted as a limiting case of a cosine signal with frequeny  $f_0 =0$.
  
Die obere Skizze zeigt die rechteckige Impulsantwort $h(t)$ des Tiefpasses. Der dazugehörige Frequenzgang lautet:
+
The upper sketch shows the rectangular impulse response  $h(t)$  of the low-pass filter. Its frequency response is:
$$H(f) = {\rm si}(\pi {f}/{ {\rm \Delta}f}) .$$
+
:$$H(f) = {\rm si}(\pi {f}/{ {\rm \Delta}f}) .$$
Aufgrund der Linearität und der Tatsache, dass $H(f)$ reell und gerade ist, sind die Ausgangssignale ebenfalls cosinusförmig:
+
Due to linearity and the fact that $H(f)$  is real and even the output signals are also cosine-shaped:
$$y_i(t) = A_i \cdot {\rm cos}(2\pi  f_i  t ) .$$
+
:$$y_i(t) = A_i \cdot {\rm cos}(2\pi  f_i  t ) .$$
Gesucht werden die Signalamplituden $A_i$ am Ausgang für die verschiedenen Eingangsfrequenzen $f_i$, wobei die Lösung ausschließlich im Zeitbereich gefunden werden soll. Dieser etwas umständliche Lösungsweg soll dazu dienen, den Zusammenhang zwischen dem Zeit– und dem Frequenzbereich deutlich zu machen.
+
*The signal amplitudes $A_i$  at the output for different frequencies  $f_i$ are searched-for and the solution is to be found in the time domain only.  
  
''Hinweise:''
+
*This somewhat circuitous solution is intended to make the basic relationship between the time and frequency domains clear.
*Die Aufgabe gehört bezieht sich auf das Kapitel  [[Lineare_zeitinvariante_Systeme/Einige_systemtheoretische_Tiefpassfunktionen|Einige systemtheoretische Tiefpassfunktionen]].
 
*Entgegen der sonst üblichen Definition der Amplitude können die $„A_i”$ durchaus negativ sein. Dies entspricht dann der Funktion „Minus-Cosinus”.
 
*Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
 
  
  
  
===Fragebogen===
+
 
 +
 
 +
 
 +
 
 +
 
 +
 
 +
''Please note:''
 +
*The exercise belongs to the chapter   [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory|Some Low-Pass Functions in Systems Theory]].
 +
*Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. Then, this corresponds to the function "minus-cosine".
 +
 +
 
 +
 
 +
 
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welcher Tiefpass liegt hier vor?
+
{Which low-pass filter is at hand here?
 
|type="[]"}
 
|type="[]"}
- Idealer Tiefpass.
+
- Ideal low-pass filter,
+ Spalttiefpass.
+
+ slit low-pass filter,
- Gaußtiefpass.
+
- Gaussian low-pass filter.
  
  
{Geben Sie die äquivalente Bandbreite von $H(f)$ an.
+
{State the equivalent bandwidth of&nbsp; $H(f)$&nbsp;.
 
|type="{}"}
 
|type="{}"}
$\Delta f  \ =$ { 2 3% } $\ \rm kHz}
+
$\Delta f  \ =\ $ { 2 3% } $\ \rm kHz$
  
  
{Berechnen Sie allgemein die Amplitude $A_i$ in Abhängigkeit von $x_i(t)$ und $h(t)$. Welche der folgenden Punkte sind bei der Berechnung zu berücksichtigen?
+
{In general, compute the amplitude&nbsp; $A_i$&nbsp; as a function of&nbsp; $x_i(t)$&nbsp; and&nbsp; $h(t)$. Which of the following should be considered in the calculations?
 
|type="[]"}
 
|type="[]"}
+ Beim Cosinussignal gilt $A_i = y_i(t = 0)$.
+
+ For the cosine signal, &nbsp;$A_i = y_i(t = 0)$ holds.
- Es gilt $y_i(t) = x_i(t) · h(t)$.
+
- The following holds: &nbsp;$y_i(t) = x_i(t) · h(t)$.
+ Es gilt $y_i(t) = x_i(t) ∗ h(t)$.
+
+ The following holds: &nbsp;$y_i(t) = x_i(t) ∗ h(t)$.
  
  
{Welche der folgenden Ergebnisse treffen für $A_0, A_2$ und $A_4$ zu?
+
{Which of the following results are true for&nbsp; $A_0, A_2$&nbsp; and&nbsp; $A_4$&nbsp;? &nbsp; The following still holds: &nbsp; $A_i = y_i(t = 0)$.
 
|type="[]"}
 
|type="[]"}
 
- $A_0 = 0$.
 
- $A_0 = 0$.
+ $A_0 = 1 \ \rm V $.
+
+ $A_0 = 1 \hspace{0.05cm} \rm V $.
 
+ $A_2 = 0$.
 
+ $A_2 = 0$.
- $A_2 = 1 \ \rm V $.
+
- $A_2 = 1 \hspace{0.05cm} \rm V $.
 
+ $A_4 = 0$.
 
+ $A_4 = 0$.
- $A_4 =1 \ \rm V $.
+
- $A_4 =1 \hspace{0.05cm} \rm V $.
  
  
{Berechnen Sie die Amplituden $A_1$ und $A_3$ für ein $1 \ \rm kHz$- bzw. $3 \ \rm kHz$-Signal. Interpretieren Sie die Ergebnisse anhand der Spektralfunktion.
+
{Compute the amplitudes&nbsp; $A_1$&nbsp; and&nbsp; $A_3$&nbsp; for a&nbsp; $1 \ \rm kHz$&ndash; and&nbsp; $3 \ \rm kHz$&ndash;signal. <br>Interpret the results using the spectral functions.
 
|type="{}"}
 
|type="{}"}
$A_1 =$ { 0.637 5%  } V
+
$A_1 \ = \ $ { 0.637 5%  } $\ \rm V$
$A_3 =$ { -0.215--0.205  } V
+
$A_3 \ = \ $ { -0.215--0.205  } $\ \rm V$
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
:'''a)''' Es handelt sich um einen $\rm \underline{Spalttiefpass}$.
+
'''(1)'''&nbsp; <u>Approach 2</u> is correct: &nbsp; It is a <u>slit low-pass filter</u>.
 +
 
 +
 
 +
'''(2)'''&nbsp; The (equivalent) time duration of the impulse response is&nbsp; $Δt = 0.5 \ \rm ms$. &nbsp; The equivalent bandwidth is equal to the reciprocal:
 +
:$$Δf = 1/Δt \ \rm \underline{= \ 2 \ kHz}.$$
  
  
:'''b)''' Die (äquivalente) Zeitdauer der Impulsantwort ist $Δt =$ 0.5 ms. Die äquivalente Bandbreite ist gleich dem Kehrwert $Δf = 1/Δt \ \rm \underline{= \ 2 kHz}$.
+
'''(3)'''&nbsp; <u>Approaches 1 and 3</u> are correct:
 +
*The amplitude is&nbsp; $A_i = y_i(t = 0)$ since&nbsp; $y_i(t)$&nbsp; is cosine-shaped. The output signal is calculated by convolution for this purpose:
 +
:$$A_i = y_i (t=0) = \int\limits_{ - \infty }^{ + \infty } {x_i ( \tau  )}  \cdot h ( {0 - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$
 +
*Considering the symmetry and the time limitation of&nbsp; $h(t)$ the following result is obtained:
 +
:$$A_i = \frac{A_x}{\Delta t} \cdot \int\limits_{ - \Delta t /2 }^{ + \Delta t /2 } {\rm cos}(2\pi  f_i  \tau )\hspace{0.1cm}{\rm d}\tau.$$
  
  
:'''c)''' Da $y_i(t)$ cosinusförmig ist, ist die Amplitude gleich dem Signalwert bei $t =$ 0. Das Ausgangssignal soll hier über die Faltung berechnet werden:
 
$$A_i = y_i (t=0) = \int\limits_{ - \infty }^{ + \infty } {x_i ( \tau  )}  \cdot h ( {0 - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$
 
:Berücksichtigt man die Symmetrie und die zeitliche Begrenzung von $h(t)$, so kommt man zum Ergebnis:
 
$$A_i = \frac{A_x}{\Delta t} \cdot \int\limits_{ - \Delta t /2 }^{ + \Delta t /2 } {\rm cos}(2\pi  f_i  \tau )\hspace{0.1cm}{\rm d}\tau.$$
 
:Richtig sind also die $\rm \underline{\ Lösungsvorschläge \ 1 \ und \ 3}$.
 
  
 +
'''(4)'''&nbsp; <u>Approaches 2, 3 and 5</u> are correct:
 +
*For the direct (DC) signal  &nbsp;$x_0(t) = A_x$&nbsp;, set &nbsp;$f_i = 0$&nbsp; and one obtains &nbsp;$A_0 = A_x \ \rm \underline{ = \ 1 \hspace{0.05cm} V}$.
 +
*In contrast to this, for the cosine frequencies &nbsp;$f_2 = 2 \ \rm kHz$&nbsp; and &nbsp;$f_4 = 4 \ \rm kHz$&nbsp; the integral vanishes in each case because then it is integrated over one and two periods, respectively: &nbsp; $A_2 \ \rm \underline{ = \hspace{0.05cm} 0}$&nbsp; und&nbsp; $A_4 \hspace{0.05cm} \rm \underline{ = \ 0}$.
 +
*In the frequency domain, the cases which are dealt with here correspond to:
 +
:$$H(f=0) = 1, \hspace{0.3cm}H(f=\Delta f) = 0, \hspace{0.3cm}H(f=2\Delta f) = 0.$$
  
:'''d)''' Beim Gleichsignal $x_0(t) = A_x$ ist $f_i =$ 0 zu setzen und man erhält $A_0 = A_x \ \rm \underline{ = \ 1 \ V}$.
 
:Dagegen verschwindet bei den Cosinusfrequenzen $f_2 =$ 2 kHz und $f_4 =$ 4 kHz jeweils das Integral, da dann genau über eine bzw. zwei Periodendauern zu integrieren ist: $A_2 \ \rm \underline{ = \ 0}$ und $A_4 \ \rm \underline{ = \ 0}$.
 
:Im Frequenzbereich entsprechen die hier behandelten Fälle:
 
$$H(f=0) = 1, \hspace{0.3cm}H(f=\Delta f) = 0, \hspace{0.3cm}H(f=2\Delta f) = 0.$$
 
  
  
:'''e)''' Das Ergebnis von c) lautet unter Berücksichtigung der Symmetrie für $f_i = f_1$:
+
'''(5)'''&nbsp; The result of subtask&nbsp; '''(3)'''&nbsp; – considering the symmetry – is for&nbsp; $f_i = f_1$:
$$A_1= \frac{2A_x}{\Delta t} \cdot \int\limits_{ 0 }^{  \Delta t /2 } {\rm cos}(2\pi  f_1  \tau )\hspace{0.1cm}{\rm
+
:$$A_1= \frac{2A_x}{\Delta t} \cdot \int\limits_{ 0 }^{  \Delta t /2 } {\rm cos}(2\pi  f_1  \tau )\hspace{0.1cm}{\rm
 
  d}\tau =  \frac{2A_x}{2\pi  f_1 \cdot \Delta t} \cdot {\rm sin}(2\pi  f_1  \frac{\Delta t}{2}
 
  d}\tau =  \frac{2A_x}{2\pi  f_1 \cdot \Delta t} \cdot {\rm sin}(2\pi  f_1  \frac{\Delta t}{2}
 
  )= A_x \cdot {\rm si}(\pi  f_1  \Delta t ).$$
 
  )= A_x \cdot {\rm si}(\pi  f_1  \Delta t ).$$
:Mit $f_1 · Δt =$ 0.5 lautet somit das Ergebnis:
+
*Taking &nbsp; $f_1 · Δt = 0.5$&nbsp;into account the result is:
$$A_1 = A_x \cdot {\rm si}(\frac{\pi}{2} ) = \frac{2A_x}{\pi} \hspace{0.15cm}\underline{= 0.637\,{\rm V}}.$$
+
:$$A_1 = A_x \cdot {\rm si}(\frac{\pi}{2} ) = \frac{2A_x}{\pi} \hspace{0.15cm}\underline{= 0.637\,{\rm V}}.$$
:Entsprechend erhält man mit $f_3 · Δt =$ 1.5:
+
*Correspondingly, the following is obtained using&nbsp; $f_3 · Δt = 1.5$:
$$A_3 = A_x \cdot {\rm si}(\frac{3\pi}{2} ) = -\frac{2A_x}{3\pi} = -\frac{A_1}{3}\hspace{0.15cm}\underline{= -0.212\,{\rm V}}.$$
+
:$$A_3 = A_x \cdot {\rm si}({3\pi}/{2} ) = -\frac{2A_x}{3\pi} = -{A_1}/{3}\hspace{0.15cm}\underline{= -0.212\,{\rm V}}.$$
:Genau zu den gleichen Ergebnissen aber deutlich schneller kommt man durch die Anwendung der Gleichung $A_i = A_x · H(f = f_i)$.
+
*The exact same results are obtained but much faster by applying the equation:
 +
:$$A_i = A_x · H(f = f_i).$$
  
:Bereits aus den Grafiken auf der Angabenseite erkennt man, dass das Integral über $x_1(t)$ im markierten Bereich positiv und das Integral über $x_3(t)$ negativ ist. Es ist allerdings anzumerken, dass man im Allgemeinen als Amplitude meist den Betrag bezeichnet (siehe Hinweis auf der Angabenseite).
+
*From the graphs on the information page it is already obvious that the integral over&nbsp; $x_1(t)$&nbsp; is positive in the marked area and the integral over&nbsp; $x_3(t)$&nbsp; is negative.  
 +
*However, it should be noted that in general amplitude is usually referred to as the magnitude (See notice on the information page).
  
 
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[[Category:Aufgaben zu Lineare zeitinvariante Systeme|^1.3 Einige systemtheoretische Tiefpassfunktionen^]]
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[[Category:Linear and Time-Invariant Systems: Exercises|^1.3 Some Low-Pass Functions in Systems Theory^]]

Latest revision as of 18:33, 7 September 2021

Impulse response and input signals

The task is meant to investigate the influence of a low-pass filter  $H(f)$  on cosinusoidal signals of the form

$$x_i(t) = A_x \cdot {\rm cos}(2\pi f_i t ).$$

In the graph you can see the signals  $x_i(t)$ where the index $i$  indicates the frequency in $\rm kHz$ . So,  $x_2(t)$  describes a  $2 \hspace{0.09cm} \rm kHz$–signal.

The signal amplitude in each case is $A_x = 1 \hspace{0.05cm} \rm V$. The direct (DC) signal  $x_0(t)$  is to be interpreted as a limiting case of a cosine signal with frequeny  $f_0 =0$.

The upper sketch shows the rectangular impulse response  $h(t)$  of the low-pass filter. Its frequency response is:

$$H(f) = {\rm si}(\pi {f}/{ {\rm \Delta}f}) .$$

Due to linearity and the fact that $H(f)$  is real and even the output signals are also cosine-shaped:

$$y_i(t) = A_i \cdot {\rm cos}(2\pi f_i t ) .$$
  • The signal amplitudes $A_i$  at the output for different frequencies  $f_i$ are searched-for and the solution is to be found in the time domain only.
  • This somewhat circuitous solution is intended to make the basic relationship between the time and frequency domains clear.





Please note:

  • The exercise belongs to the chapter  Some Low-Pass Functions in Systems Theory.
  • Contrary to the usual definition of the amplitude, the "$A_i$" may well be negative. Then, this corresponds to the function "minus-cosine".




Questions

1

Which low-pass filter is at hand here?

Ideal low-pass filter,
slit low-pass filter,
Gaussian low-pass filter.

2

State the equivalent bandwidth of  $H(f)$ .

$\Delta f \ =\ $

$\ \rm kHz$

3

In general, compute the amplitude  $A_i$  as a function of  $x_i(t)$  and  $h(t)$. Which of the following should be considered in the calculations?

For the cosine signal,  $A_i = y_i(t = 0)$ holds.
The following holds:  $y_i(t) = x_i(t) · h(t)$.
The following holds:  $y_i(t) = x_i(t) ∗ h(t)$.

4

Which of the following results are true for  $A_0, A_2$  and  $A_4$ ?   The following still holds:   $A_i = y_i(t = 0)$.

$A_0 = 0$.
$A_0 = 1 \hspace{0.05cm} \rm V $.
$A_2 = 0$.
$A_2 = 1 \hspace{0.05cm} \rm V $.
$A_4 = 0$.
$A_4 =1 \hspace{0.05cm} \rm V $.

5

Compute the amplitudes  $A_1$  and  $A_3$  for a  $1 \ \rm kHz$– and  $3 \ \rm kHz$–signal.
Interpret the results using the spectral functions.

$A_1 \ = \ $

$\ \rm V$
$A_3 \ = \ $

$\ \rm V$


Solution

(1)  Approach 2 is correct:   It is a slit low-pass filter.


(2)  The (equivalent) time duration of the impulse response is  $Δt = 0.5 \ \rm ms$.   The equivalent bandwidth is equal to the reciprocal:

$$Δf = 1/Δt \ \rm \underline{= \ 2 \ kHz}.$$


(3)  Approaches 1 and 3 are correct:

  • The amplitude is  $A_i = y_i(t = 0)$ since  $y_i(t)$  is cosine-shaped. The output signal is calculated by convolution for this purpose:
$$A_i = y_i (t=0) = \int\limits_{ - \infty }^{ + \infty } {x_i ( \tau )} \cdot h ( {0 - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$
  • Considering the symmetry and the time limitation of  $h(t)$ the following result is obtained:
$$A_i = \frac{A_x}{\Delta t} \cdot \int\limits_{ - \Delta t /2 }^{ + \Delta t /2 } {\rm cos}(2\pi f_i \tau )\hspace{0.1cm}{\rm d}\tau.$$


(4)  Approaches 2, 3 and 5 are correct:

  • For the direct (DC) signal  $x_0(t) = A_x$ , set  $f_i = 0$  and one obtains  $A_0 = A_x \ \rm \underline{ = \ 1 \hspace{0.05cm} V}$.
  • In contrast to this, for the cosine frequencies  $f_2 = 2 \ \rm kHz$  and  $f_4 = 4 \ \rm kHz$  the integral vanishes in each case because then it is integrated over one and two periods, respectively:   $A_2 \ \rm \underline{ = \hspace{0.05cm} 0}$  und  $A_4 \hspace{0.05cm} \rm \underline{ = \ 0}$.
  • In the frequency domain, the cases which are dealt with here correspond to:
$$H(f=0) = 1, \hspace{0.3cm}H(f=\Delta f) = 0, \hspace{0.3cm}H(f=2\Delta f) = 0.$$


(5)  The result of subtask  (3)  – considering the symmetry – is for  $f_i = f_1$:

$$A_1= \frac{2A_x}{\Delta t} \cdot \int\limits_{ 0 }^{ \Delta t /2 } {\rm cos}(2\pi f_1 \tau )\hspace{0.1cm}{\rm d}\tau = \frac{2A_x}{2\pi f_1 \cdot \Delta t} \cdot {\rm sin}(2\pi f_1 \frac{\Delta t}{2} )= A_x \cdot {\rm si}(\pi f_1 \Delta t ).$$
  • Taking   $f_1 · Δt = 0.5$ into account the result is:
$$A_1 = A_x \cdot {\rm si}(\frac{\pi}{2} ) = \frac{2A_x}{\pi} \hspace{0.15cm}\underline{= 0.637\,{\rm V}}.$$
  • Correspondingly, the following is obtained using  $f_3 · Δt = 1.5$:
$$A_3 = A_x \cdot {\rm si}({3\pi}/{2} ) = -\frac{2A_x}{3\pi} = -{A_1}/{3}\hspace{0.15cm}\underline{= -0.212\,{\rm V}}.$$
  • The exact same results are obtained – but much faster – by applying the equation:
$$A_i = A_x · H(f = f_i).$$
  • From the graphs on the information page it is already obvious that the integral over  $x_1(t)$  is positive in the marked area and the integral over  $x_3(t)$  is negative.
  • However, it should be noted that in general amplitude is usually referred to as the magnitude (See notice on the information page).