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

From LNTwww
 
(33 intermediate revisions by 4 users not shown)
Line 1: Line 1:
  
{{quiz-Header|Buchseite=Lineare zeitinvariante Systeme/Folgerungen aus dem Zuordnungssatz
+
{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem
 
}}
 
}}
  
[[File:P_ID1756__LZI_Z_3_1.png|right|Betrachtete Impulsantworten]]
+
[[File:P_ID1756__LZI_Z_3_1.png|right|frame|Considered impulse responses]]
Der Zusammenhang zwischen dem Real– und dem Imginärteil der Übertragungsfunktion realisierbarer kausaler Systeme wird durch die Hilbert–Transformation beschrieben. Hierbei gilt:
+
The relation between the real part and the imaginary part of the transfer function of realizable causal systems is described by the Hilbert transformation.  Here, the following holds:
$${\rm Im} \left\{ H(f) \right \} =  - \frac{1}{\pi }\int_{-\infty}^{
+
:$${\rm Im} \left\{ H(f) \right \} =  - \frac{1}{\pi }\int_{-\infty}^{
 
+\infty}
 
+\infty}
 
  { \frac{{\rm Re} \left\{ H(\nu) \right \}}{f - \nu}}\hspace{0.1cm}{\rm
 
  { \frac{{\rm Re} \left\{ H(\nu) \right \}}{f - \nu}}\hspace{0.1cm}{\rm
 
  d}\nu \hspace{0.05cm},$$
 
  d}\nu \hspace{0.05cm},$$
$${\rm Re} \left\{ H(f) \right \} =    \frac{1}{\pi }\int_{-\infty}^{
+
:$${\rm Re} \left\{ H(f) \right \} =    \frac{1}{\pi }\int_{-\infty}^{
 
+\infty}
 
+\infty}
 
  { \frac{{\rm Im} \left\{ H(\nu) \right \}}{f - \nu}}\hspace{0.1cm}{\rm
 
  { \frac{{\rm Im} \left\{ H(\nu) \right \}}{f - \nu}}\hspace{0.1cm}{\rm
 
  d}\nu \hspace{0.05cm}.$$
 
  d}\nu \hspace{0.05cm}.$$
  
Als gemeinsames Kurzzeichen verwendet man für diese beiden Integraltransformationen:
+
The following is used as a common abbreviation for these two integral transformations:
$${\rm Im} \left\{ H(f) \right \}  \quad
+
:$${\rm Im} \left\{ H(f) \right \}  \quad
 
\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad
 
\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad
 
{\rm Re} \left\{ H(f) \right \}\hspace{0.05cm}.$$
 
{\rm Re} \left\{ H(f) \right \}\hspace{0.05cm}.$$
  
Da sich die Hin– und die Rücktransformation lediglich durch das Vorzeichen unterscheiden, genügt eine Gleichung. Dabei gilt:
+
Since the transformation and its inverse differ only by the sign,  one equation is sufficient.  Here, the following applies:
  
* Zur Berechnung des durch den Pfeil markierten Operanden wird das positive Vorzeichen verwendet.
+
* To compute the operand marked by the arrow the positive sign is used.
  
* Dagegen ist zur Berechnung des durch den Kreis markierten Operanden das Minuszeichen zu berücksichtigen.
+
* In contrast to this,  the minus sign is taken into account for the computation of the operand marked by the circle.
  
Die Hilbert–Transformation gilt viel allgemeiner als nur für den hier beschriebenen Anwendungsfall. Zum Beispiel wird sie auch verwendet, um zu einem reellen Bandpass–Signal das dazugehörige (komplexe) analytische Signal zu ermitteln.
 
  
Bei dieser Aufgabe soll zu den in der Grafik gegebenen kausalen Impulsantworten $h(t)$ die zugehörigen Frequenzgänge $H(f)$ entsprechend der Fourierrücktransformation ermittelt werden. Zerlegt man $H(f)$ jeweils in Real– und Imaginärteil, so können daraus Hilbert–Korrespondenzen abgeleitet werden.
+
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.
  
''Hinweise:''
+
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.  
*Die Aufgabe gehört zum Kapitel  [[Lineare_zeitinvariante_Systeme/Folgerungen_aus_dem_Zuordnungssatz|Folgerungen aus dem Zuordnungssatz]].
 
*Bezug genommen wird insbesondere auf die Therieseite [[Lineare_zeitinvariante_Systeme/Folgerungen_aus_dem_Zuordnungssatz#Hilbert.E2.80.93Transformation|Hilbert-Transformation]].
 
*Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
 
  
 +
If  $H(f)$  is decomposed into real and imaginary parts respectively,  then Hilbert correspondences can be derived from it.
  
===Fragebogen===
+
 
 +
 
 +
 
 +
 
 +
 
 +
Please note:
 +
*The exercise belongs to the chapter   [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem|Conclusions from the Allocation Theorem]].
 +
*In particular, reference is made to the theory page  [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert_transform|Hilbert transform]].
 +
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Ermitteln Sie ausgehend von $h_1(t) = \alpha \cdot \delta(t)$ die Hilbert&ndash;Transformierte einer Konstanten $\alpha$. Welche Aussagen treffen zu?
+
{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="()"}
- Die Hilbert&ndash;Transformierte einer Konstanten $\alpha$ ist ebenfalls $\alpha$.
+
- The Hilbert transform of a constant &nbsp;$\alpha$&nbsp; is also &nbsp;$\alpha$.
+ Die Hilbert&ndash;Transformierte einer Konstanten $\alpha$ ist $0$.
+
+ The Hilbert transform of a constant &nbsp;$\alpha$&nbsp; is zero.
- Die Hilbert&ndash;Transformierte einer Konstanten $\alpha$ verläuft sinusförmig.
+
- The Hilbert transform of a constant &nbsp;$\alpha$&nbsp; is sinusoidal.
  
  
{Ermitteln Sie ausgehend von $h_2(t) =  \delta(t- \tau)$ die Hilbert&ndash;Transformierte einer Cosinusfunktion. Welche Aussagen treffen zu?
+
{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 $0$.
+
- 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 $h_3(t)$ die Hilbert&ndash;Transformierte der Funktion ${\rm si}(2 \pi fT) = {\rm sin}(2 \pi fT)/(2 \pi fT)$ 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 ${\rm sin}^2\hspace{-0.05cm}(\pi fT)/(\pi fT)$.
+
+ The Hilbert transform is &nbsp;${\rm sin}^2\hspace{-0.05cm}(\pi fT)/(\pi fT)$.
+ Die Hilbert Transformierte lautet ${\rm sin}( \pi fT) \cdot {\rm si}( \pi fT)$.
+
+ The Hilbert transform is &nbsp;${\rm sin}( \pi fT) \cdot {\rm si}( \pi fT)$.
  
  
{Lässt sich aus der Impulsantwort $h_4(t)$ 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.
  
  
Line 67: Line 75:
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die Fourier&ndash;Transformierte von $h_1(t) = \alpha \cdot \delta(t)$ lautet:
+
'''(1)'''&nbsp; The&nbsp; <u>second proposed solution</u>&nbsp; is correct:
$$H_1(f) = \alpha \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re} \left\{ H_1(f) \right \}  = \alpha ,
+
*The Fourier transform of&nbsp; $h_1(t) = \alpha \cdot \delta(t)$&nbsp; is:
 +
:$$H_1(f) = \alpha \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re} \left\{ H_1(f) \right \}  = \alpha ,
 
  \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}.$$
  
Richtig ist somit der <u>zweite Lösungsvorschlag</u>.
 
  
  
'''(2)'''&nbsp; Mit dem [[Signaldarstellung/Gesetzmäßigkeiten_der_Fouriertransformation#Verschiebungssatz|Verschiebungssatz]] und dem [[Signaldarstellung/Zum_Rechnen_mit_komplexen_Zahlen#Darstellung_nach_Betrag_und_Phase|Satz von Euler]] erhält man für die Impulsantwort $h_2(t)$ den Frequenzgang:
+
'''(2)'''&nbsp; The&nbsp; <u>last proposed solution</u>&nbsp; is correct:
$$H_2(f) ={\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f \tau} = \cos (2\pi
+
*The following frequency response is obtained for the impulse response&nbsp; $h_2(t)$&nbsp; considering the&nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Shifting Theorem|shifting theorem]]&nbsp; and&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation by magnitude and phase|Euler's theorem]]&nbsp;:
 +
:$$H_2(f) ={\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f \tau} = \cos (2\pi
 
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 entsprechend dem <u>letzten Lösungsvorschlag</u> 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}
-\sin (2\pi f \tau)\hspace{0.7cm}{\rm oder}\hspace{0.7cm}
+
-\sin (2\pi f \tau)\hspace{0.7cm}{\rm or}\hspace{0.7cm}
 
\cos (2\pi f \tau)  \hspace{0.3cm}
 
\cos (2\pi f \tau)  \hspace{0.3cm}
 
\bullet\hspace{-0.05cm}\!\!-\!\!\!-\!\!\!-\!-\!\hspace{-0.1cm}\rightarrow\hspace{0.3cm}
 
\bullet\hspace{-0.05cm}\!\!-\!\!\!-\!\!\!-\!-\!\hspace{-0.1cm}\rightarrow\hspace{0.3cm}
Line 91: Line 100:
  
  
'''(3)'''&nbsp; Für die rechteckförmige Impulsantwort $h_3(t)$ mit der Breite $T$ und der Höhe $1/T$ erhält man die Spektralfunktion entsprechend dem [[Signaldarstellung/Fouriertransformation_und_-rücktransformation#Das_erste_Fourierintegral|ersten Fourierintegral]]:
+
 
$$H_3(f) =    \int_{-\infty}^{
+
'''(3)'''&nbsp; <u>Both proposed solutions</u>&nbsp; are correct:
 +
*For the rectangular impulse response&nbsp; $h_3(t)$&nbsp; of width&nbsp; $T$&nbsp; and height&nbsp; $1/T$&nbsp; the spectral function is obtained according to the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_first_Fourier_integral|first Fourier integral]]:
 +
:$$H_3(f) =    \int_{-\infty}^{
 
+\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
 
  { 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
Line 107: Line 118:
 
T}}{{\rm j}\cdot 2\pi f T} \hspace{0.05cm}.$$
 
T}}{{\rm j}\cdot 2\pi f T} \hspace{0.05cm}.$$
  
Mit dem  [[Signaldarstellung/Zum_Rechnen_mit_komplexen_Zahlen#Darstellung_nach_Betrag_und_Phase|Eulerschen Satz]] kann hierfür auch geschrieben werden:
+
*Using&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation by magnitude and phase|Euler's theorem]]&nbsp; this can also be rewritten as follows:
$$H_3(f) = \frac{1-\cos (2\pi
+
:$$H_3(f) = \frac{1-\cos (2\pi
 
f T) + {\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \sin (2\pi f
 
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
 
T)}{{\rm j}\cdot 2\pi f T} =  \frac{\sin (2\pi f T)}{ 2\pi
Line 114: Line 125:
 
T}\hspace{0.05cm}.$$
 
T}\hspace{0.05cm}.$$
  
Weiter gilt mit der Umformung $1 - \cos(\alpha) = 2 \cdot \sin^2(\alpha/2)$:
+
*Furthermore, the following holds considering the transformation&nbsp; $1 - \cos(\alpha) = 2 \cdot \sin^2(\alpha/2)$:
$${\rm Re}\hspace{-0.05cm} \left\{ H_3(f) \right \}  =  {\rm si} (2\pi f T)\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si}(x)= {\rm sin}(x)/x
+
:$${\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}
 
  \hspace{0.05cm}, \hspace{0.5cm}
 
{\rm Im} \hspace{-0.05cm}\left\{ H_3(f) \right \}  =  -\frac{\sin^2 (\pi f
 
{\rm Im} \hspace{-0.05cm}\left\{ H_3(f) \right \}  =  -\frac{\sin^2 (\pi f
Line 121: Line 132:
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
Daraus folgt, dass <u>beide Lösungsalternativen</u> richtig sind.
 
  
  
'''(4)'''&nbsp; <u>Nein</u>. Die Impulsantwort $h_4(t)$ ist nicht kausal, so dass aus dem dazugehörigen Fourier&ndash;Spektrum $H_4(f)$ keine Hilbert&ndash;Korrespondenz abgeleitet werden kann.
+
'''(4)'''&nbsp; <u>No</u> is correct:
 +
*The impulse response &nbsp; $h_4(t)$&nbsp; is not causal so that no Hilbert correspondence can be derived from the associated Fourier spectrum&nbsp; $H_4(f)$&nbsp;.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Lineare zeitinvariante Systeme|^3.1 Folgerungen aus dem Zuordnungssatz^]]
+
[[Category:Linear and Time-Invariant Systems: Exercises|^3.1 Conclusions from the Assignment Theorem^]]

Latest revision as of 16:45, 9 October 2021

Considered impulse responses

The relation between the real part 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 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
$$\cos (2\pi f \tau) \hspace{0.3cm} \leftarrow\hspace{-0.05cm}\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.3cm} -\sin (2\pi f \tau)\hspace{0.7cm}{\rm or}\hspace{0.7cm} \cos (2\pi f \tau) \hspace{0.3cm} \bullet\hspace{-0.05cm}\!\!-\!\!\!-\!\!\!-\!-\!\hspace{-0.1cm}\rightarrow\hspace{0.3cm} \sin (2\pi f \tau) \hspace{0.05cm}.$$


(3)  Both proposed solutions  are correct:

  • For the rectangular impulse response  $h_3(t)$  of width  $T$  and height  $1/T$  the spectral function is obtained according to the  first Fourier integral:
$$H_3(f) = \int_{-\infty}^{ +\infty} { h_3(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f t}}\hspace{0.1cm}{\rm d}t \hspace{0.05cm} = \frac{1}{T} \cdot \int_{0}^{ T} { {\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f t}}\hspace{0.1cm}{\rm d}t = \left [\frac{1}{-{\rm j}\cdot 2\pi f T} \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f\hspace{0.05cm} t} \right ]_{0}^{T} = \frac{1-{\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f\hspace{0.05cm} T}}{{\rm j}\cdot 2\pi f T} \hspace{0.05cm}.$$
$$H_3(f) = \frac{1-\cos (2\pi f T) + {\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \sin (2\pi f T)}{{\rm j}\cdot 2\pi f T} = \frac{\sin (2\pi f T)}{ 2\pi f T} - {\rm j}\cdot \frac{1 - \cos (2\pi f T)}{ 2\pi f T}\hspace{0.05cm}.$$
  • Furthermore, the following holds considering the transformation  $1 - \cos(\alpha) = 2 \cdot \sin^2(\alpha/2)$:
$${\rm Re}\hspace{-0.05cm} \left\{ H_3(f) \right \} = {\rm si} (2\pi f T)\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si}(x)= {\rm sin}(x)/x \hspace{0.05cm}, \hspace{0.5cm} {\rm Im} \hspace{-0.05cm}\left\{ H_3(f) \right \} = -\frac{\sin^2 (\pi f T)}{ \pi f T}= - {\rm si} (\pi f T) \cdot {\rm sin} (\pi f T) \hspace{0.05cm}.$$


(4)  No is correct:

  • The impulse response   $h_4(t)$  is not causal so that no Hilbert correspondence can be derived from the associated Fourier spectrum  $H_4(f)$ .