Difference between revisions of "Aufgaben:Exercise 3.1Z: Hilbert Transform"

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<quiz display=simple>
 
<quiz display=simple>
{Determine the Hilbert transform of a constant &nbsp;$\alpha$ starting from &nbsp;$h_1(t) = \alpha \cdot \delta(t)$&nbsp;. <br>Which statements are true?
+
{Determine the Hilbert transform of a constant &nbsp;$\alpha$ beginning with &nbsp;$h_1(t) = \alpha \cdot \delta(t)$&nbsp;. <br>Which statements are true?
 
|type="()"}
 
|type="()"}
 
- The Hilbert transform of a constant &nbsp;$\alpha$&nbsp; is also &nbsp;$\alpha$.
 
- The Hilbert transform of a constant &nbsp;$\alpha$&nbsp; is also &nbsp;$\alpha$.
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{Determine the the Hilbert transform of a cosine function starting from &nbsp;$h_2(t) =  \delta(t- \tau)$&nbsp;. <br>Which statements are true?
+
{Determine the Hilbert transform of a cosine function beginning with &nbsp;$h_2(t) =  \delta(t- \tau)$&nbsp;. <br>Which statements are true?
 
|type="()"}
 
|type="()"}
- Die Hilbert&ndash;Transformierte von einem Cosinus ist eine Konstante.
+
- The Hilbert transform of a cosine is a constant.
- Die Hilbert&ndash;Transformierte einer Cosinusfunktion ist Null.
+
- The Hilbert transform of a cosine function is zero.
+ Die Hilbert&ndash;Transformierte von einem Cosinus verläuft sinusförmig.
+
+ The Hilbert transform of a cosine is sinusoidal.
  
  
{Ermitteln Sie ausgehend vom rechteckförmigen &nbsp;$h_3(t)$&nbsp; die Hilbert&ndash;Transformierte der Funktion &nbsp;${\rm si}(2 \pi fT) = {\rm sin}(2 \pi fT)/(2 \pi fT)$. <br>Welche Aussagen treffen zu?
+
{Determine the Hilbert transform of the function &nbsp;${\rm si}(2 \pi fT) = {\rm sin}(2 \pi fT)/(2 \pi fT)$ beginning with the rectangular &nbsp;$h_3(t)$&nbsp;. <br>Which statements are true?
 
|type="[]"}
 
|type="[]"}
+ Die Hilbert Transformierte lautet &nbsp;${\rm sin}^2\hspace{-0.05cm}(\pi fT)/(\pi fT)$.
+
+ The Hilbert transform is &nbsp;${\rm sin}^2\hspace{-0.05cm}(\pi fT)/(\pi fT)$.
+ Die Hilbert Transformierte lautet &nbsp;${\rm sin}( \pi fT) \cdot {\rm si}( \pi fT)$.
+
+ The Hilbert transform is &nbsp;${\rm sin}( \pi fT) \cdot {\rm si}( \pi fT)$.
  
  
{Lässt sich aus der Impulsantwort &nbsp;$h_4(t)$&nbsp; eine Hilbert&ndash;Korrespondenz ableiten?
+
{Can a Hilbert correspondence be derived from the impulse response &nbsp;$h_4(t)$&nbsp;?
 
|type="()"}
 
|type="()"}
- Ja.
+
- Yes.
+ Nein.
+
+ No.
  
  
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===Solution===
 
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig ist der <u>zweite Lösungsvorschlag</u>:
+
'''(1)'''&nbsp; The <u>second proposed solution</u> is correct:
*Die Fourier&ndash;Transformierte von&nbsp; $h_1(t) = \alpha \cdot \delta(t)$&nbsp; lautet:
+
*The Fourier transform of&nbsp; $h_1(t) = \alpha \cdot \delta(t)$&nbsp; is:
 
:$$H_1(f) = \alpha \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re} \left\{ H_1(f) \right \}  = \alpha ,
 
:$$H_1(f) = \alpha \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re} \left\{ H_1(f) \right \}  = \alpha ,
 
  \hspace{0.2cm}{\rm Im} \left\{ H_1(f) \right \}  = 0\hspace{0.05cm}.$$
 
  \hspace{0.2cm}{\rm Im} \left\{ H_1(f) \right \}  = 0\hspace{0.05cm}.$$
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'''(2)'''&nbsp; Richtig ist  der <u>letzte Lösungsvorschlag</u>:
+
'''(2)'''&nbsp; The <u>last proposed solution</u> is correct:
 
*Mit dem&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Verschiebungssatz|Verschiebungssatz]]&nbsp; und dem&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Darstellung_nach_Betrag_und_Phase|Satz von Euler]]&nbsp; erhält man für die Impulsantwort&nbsp; $h_2(t)$&nbsp; den Frequenzgang:
 
*Mit dem&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Verschiebungssatz|Verschiebungssatz]]&nbsp; und dem&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Darstellung_nach_Betrag_und_Phase|Satz von Euler]]&nbsp; erhält man für die Impulsantwort&nbsp; $h_2(t)$&nbsp; den Frequenzgang:
 
:$$H_2(f) ={\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f \tau} = \cos (2\pi
 
:$$H_2(f) ={\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f \tau} = \cos (2\pi

Revision as of 09:13, 24 September 2021

Considered impulse responses

The relation between the real and the imaginary part of the transfer function of realizable causal systems is described by the Hilbert transformation. Here, the following holds:

$${\rm Im} \left\{ H(f) \right \} = - \frac{1}{\pi }\int_{-\infty}^{ +\infty} { \frac{{\rm Re} \left\{ H(\nu) \right \}}{f - \nu}}\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm},$$
$${\rm Re} \left\{ H(f) \right \} = \frac{1}{\pi }\int_{-\infty}^{ +\infty} { \frac{{\rm Im} \left\{ H(\nu) \right \}}{f - \nu}}\hspace{0.1cm}{\rm d}\nu \hspace{0.05cm}.$$

The following is used as a common abbreviation for these two integral transformations:

$${\rm Im} \left\{ H(f) \right \} \quad \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad {\rm Re} \left\{ H(f) \right \}\hspace{0.05cm}.$$

Since the transformation and its inverse differ only by the sign, one equation is sufficient. Here, the following applies:

  • To compute the operand marked by the arrow the positive sign is used.
  • In contrast to this, the minus sign is taken into account for the computation of the operand marked by the circle.


The Hilbert transformation pertains much more generally than only to the case of application described here. For example, it is also used to determine the (complex) analytical signal corresponding to a real band-pass signal.

In this exercise, the corresponding frequency responses  $H(f)$  are to be determined for the causal impulse responses  $h(t)$  given in the diagram according to the inverse Fourier transformation.

If  $H(f)$  is decomposed into in real and imaginary parts respectively, then Hilbert correspondences can be derived from it.





Please note:


Questions

1

Determine the Hilbert transform of a constant  $\alpha$ beginning with  $h_1(t) = \alpha \cdot \delta(t)$ .
Which statements are true?

The Hilbert transform of a constant  $\alpha$  is also  $\alpha$.
The Hilbert transform of a constant  $\alpha$  is zero.
The Hilbert transform of a constant  $\alpha$  is sinusoidal.

2

Determine the Hilbert transform of a cosine function beginning with  $h_2(t) = \delta(t- \tau)$ .
Which statements are true?

The Hilbert transform of a cosine is a constant.
The Hilbert transform of a cosine function is zero.
The Hilbert transform of a cosine is sinusoidal.

3

Determine the Hilbert transform of the function  ${\rm si}(2 \pi fT) = {\rm sin}(2 \pi fT)/(2 \pi fT)$ beginning with the rectangular  $h_3(t)$ .
Which statements are true?

The Hilbert transform is  ${\rm sin}^2\hspace{-0.05cm}(\pi fT)/(\pi fT)$.
The Hilbert transform is  ${\rm sin}( \pi fT) \cdot {\rm si}( \pi fT)$.

4

Can a Hilbert correspondence be derived from the impulse response  $h_4(t)$ ?

Yes.
No.


Solution

(1)  The second proposed solution is correct:

  • The Fourier transform of  $h_1(t) = \alpha \cdot \delta(t)$  is:
$$H_1(f) = \alpha \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re} \left\{ H_1(f) \right \} = \alpha , \hspace{0.2cm}{\rm Im} \left\{ H_1(f) \right \} = 0\hspace{0.05cm}.$$


(2)  The last proposed solution is correct:

$$H_2(f) ={\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f \tau} = \cos (2\pi f \tau) - {\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \sin (2\pi f \tau)\hspace{0.05cm}.$$
  • Daraus ergibt sich die Hilbert–Korrespondenz
$$\cos (2\pi f \tau) \hspace{0.3cm} \leftarrow\hspace{-0.05cm}\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.3cm} -\sin (2\pi f \tau)\hspace{0.7cm}{\rm oder}\hspace{0.7cm} \cos (2\pi f \tau) \hspace{0.3cm} \bullet\hspace{-0.05cm}\!\!-\!\!\!-\!\!\!-\!-\!\hspace{-0.1cm}\rightarrow\hspace{0.3cm} \sin (2\pi f \tau) \hspace{0.05cm}.$$


(3)  Richtig sind beide Lösungsvorschläge:

  • Für die rechteckförmige Impulsantwort  $h_3(t)$  mit Breite  $T$  und Höhe  $1/T$  erhält man die Spektralfunktion gemäß dem  ersten Fourierintegral:
$$H_3(f) = \int_{-\infty}^{ +\infty} { h_3(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f t}}\hspace{0.1cm}{\rm d}t \hspace{0.05cm} = \frac{1}{T} \cdot \int_{0}^{ T} { {\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f t}}\hspace{0.1cm}{\rm d}t = \left [\frac{1}{-{\rm j}\cdot 2\pi f T} \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f\hspace{0.05cm} t} \right ]_{0}^{T} = \frac{1-{\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f\hspace{0.05cm} T}}{{\rm j}\cdot 2\pi f T} \hspace{0.05cm}.$$
$$H_3(f) = \frac{1-\cos (2\pi f T) + {\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \sin (2\pi f T)}{{\rm j}\cdot 2\pi f T} = \frac{\sin (2\pi f T)}{ 2\pi f T} - {\rm j}\cdot \frac{1 - \cos (2\pi f T)}{ 2\pi f T}\hspace{0.05cm}.$$
  • Weiter gilt mit der Umformung  $1 - \cos(\alpha) = 2 \cdot \sin^2(\alpha/2)$:
$${\rm Re}\hspace{-0.05cm} \left\{ H_3(f) \right \} = {\rm si} (2\pi f T)\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si}(x)= {\rm sin}(x)/x \hspace{0.05cm}, \hspace{0.5cm} {\rm Im} \hspace{-0.05cm}\left\{ H_3(f) \right \} = -\frac{\sin^2 (\pi f T)}{ \pi f T}= - {\rm si} (\pi f T) \cdot {\rm sin} (\pi f T) \hspace{0.05cm}.$$


(4)  Richtig ist Nein:

  • Die Impulsantwort  $h_4(t)$  ist nicht kausal, so dass aus dem dazugehörigen Fourier–Spektrum  $H_4(f)$  keine Hilbert–Korrespondenz abgeleitet werden kann.