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

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'''(2)'''&nbsp; The <u>last proposed solution</u> is correct:
 
'''(2)'''&nbsp; The <u>last proposed solution</u> is correct:
*Mit dem&nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Shifting Theorem|shifting theorem]]&nbsp; and dem&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation by magnitude and phase|Euler's theorem]]&nbsp; erhält man für die Impulsantwort&nbsp; $h_2(t)$&nbsp; den Frequenzgang:
+
*The following frequency response is obtained for the impulse response&nbsp; $h_2(t)$&nbsp; considering the&nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Shifting Theorem|shifting theorem]]&nbsp; and the&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation by magnitude and phase|Euler's theorem]]&nbsp;:
 
:$$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
 
f \tau) - {\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \sin (2\pi
 
f \tau) - {\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \sin (2\pi
 
f \tau)\hspace{0.05cm}.$$
 
f \tau)\hspace{0.05cm}.$$
  
*Daraus ergibt sich  die Hilbert&ndash;Korrespondenz
+
*This results in the Hilbert correspondence
 
:$$\cos (2\pi f \tau) \hspace{0.3cm}
 
:$$\cos (2\pi f \tau) \hspace{0.3cm}
 
\leftarrow\hspace{-0.05cm}\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.3cm}
 
\leftarrow\hspace{-0.05cm}\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.3cm}
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'''(3)'''&nbsp; Richtig sind <u>beide Lösungsvorschläge</u>:
+
'''(3)'''&nbsp; <u>Both proposed solutions</u> are correct:
*Für die rechteckförmige Impulsantwort&nbsp; $h_3(t)$&nbsp; mit Breite&nbsp; $T$&nbsp; und Höhe&nbsp; $1/T$&nbsp; erhält man die Spektralfunktion gemäß dem&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Das_erste_Fourierintegral|ersten Fourierintegral]]:
+
*For the rectangular impulse response&nbsp; $h_3(t)$&nbsp; of width&nbsp; $T$&nbsp; and height&nbsp; $1/T$&nbsp; erhält man die Spektralfunktion gemäß dem&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Das_erste_Fourierintegral|ersten Fourierintegral]]:
 
:$$H_3(f) =    \int_{-\infty}^{
 
:$$H_3(f) =    \int_{-\infty}^{
 
+\infty}
 
+\infty}

Revision as of 09:24, 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:

$$H_2(f) ={\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f \tau} = \cos (2\pi f \tau) - {\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \sin (2\pi f \tau)\hspace{0.05cm}.$$
  • This results in the Hilbert correspondence
$$\cos (2\pi f \tau) \hspace{0.3cm} \leftarrow\hspace{-0.05cm}\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.3cm} -\sin (2\pi f \tau)\hspace{0.7cm}{\rm oder}\hspace{0.7cm} \cos (2\pi f \tau) \hspace{0.3cm} \bullet\hspace{-0.05cm}\!\!-\!\!\!-\!\!\!-\!-\!\hspace{-0.1cm}\rightarrow\hspace{0.3cm} \sin (2\pi f \tau) \hspace{0.05cm}.$$


(3)  Both proposed solutions are correct:

  • For the rectangular impulse response  $h_3(t)$  of width  $T$  and height  $1/T$  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.