Aufgaben:Exercise 3.1Z: Hilbert Transform - Revision history

Guenter at 15:45, 9 October 2021

2021-10-09T15:45:35Z

Guenter at 15:42, 9 October 2021

2021-10-09T15:42:27Z

Guenter at 15:41, 9 October 2021

2021-10-09T15:41:10Z

Oezer at 20:09, 24 September 2021

2021-09-24T20:09:09Z

Oezer at 20:06, 24 September 2021

2021-09-24T20:06:46Z

Oezer at 20:06, 24 September 2021

2021-09-24T20:06:30Z

Oezer at 20:05, 24 September 2021

2021-09-24T20:05:47Z

Oezer at 20:03, 24 September 2021

2021-09-24T20:03:13Z

Oezer at 08:34, 24 September 2021

2021-09-24T08:34:11Z

Oezer at 08:31, 24 September 2021

2021-09-24T08:31:08Z

@@ Line 77: / Line 77: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The <u>second proposed solution</u> is correct:
+'''(1)'''&nbsp; The&nbsp; <u>second proposed solution</u>&nbsp; is correct:
 *The Fourier transform of&nbsp; $h_1(t) = \alpha \cdot \delta(t)$&nbsp; is:
 :$$H_1(f) = \alpha \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re} \left\{ H_1(f) \right \}  = \alpha ,
@@ Line 84: / Line 84: @@
-'''(2)'''&nbsp; The <u>last proposed solution</u> is correct:
+'''(2)'''&nbsp; The&nbsp; <u>last proposed solution</u>&nbsp; is correct:
 *The following frequency response is obtained for the impulse response&nbsp; $h_2(t)$&nbsp; considering the&nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Shifting Theorem|shifting theorem]]&nbsp; and&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation by magnitude and phase|Euler's theorem]]&nbsp;:
 :$$H_2(f) ={\rm e}^{-{\rm j}\hspace{0.05cm} 2\pi f \tau} = \cos (2\pi
@@ Line 101: / Line 101: @@
-'''(3)'''&nbsp; <u>Both proposed solutions</u> are correct:
+'''(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}^{

@@ Line 39: / Line 39: @@
 Please note:
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem|Conclusions from the Allocation Theorem]].
-*In particular, reference is made to the theory page &nbsp;[[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation|Hilbert transformation]].
+*In particular, reference is made to the theory page &nbsp;[[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert_transform|Hilbert transform]].

@@ Line 4: / Line 4: @@
 [[File:P_ID1756__LZI_Z_3_1.png|right|frame|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:
+The relation between the real part and the imaginary part of the transfer function of realizable causal systems is described by the Hilbert transformation.&nbsp; Here, the following holds:
 :$${\rm Im} \left\{ H(f) \right \} =   - \frac{1}{\pi }\int_{-\infty}^{
 +\infty}
@@ Line 19: / Line 19: @@
 {\rm Re} \left\{ H(f) \right \}\hspace{0.05cm}.$$
-Since the transformation and its inverse differ only by the sign, one equation is sufficient. Here, the following applies:
+Since the transformation and its inverse differ only by the sign,&nbsp; one equation is sufficient.&nbsp; 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.
+* In contrast to this,&nbsp; 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.
+The Hilbert transformation pertains much more generally than only to the case of application described here.&nbsp; For example,&nbsp; it is also used to determine the (complex) analytical signal corresponding to a real band-pass signal.
-In this exercise, the corresponding frequency responses &nbsp;$H(f)$&nbsp; are to be determined for the causal impulse responses &nbsp;$h(t)$&nbsp; given in the diagram according to the inverse Fourier transformation.
+In this exercise,&nbsp; the corresponding frequency responses &nbsp;$H(f)$&nbsp; are to be determined for the causal impulse responses &nbsp;$h(t)$&nbsp; given in the diagram according to the inverse Fourier transformation.
-If &nbsp;$H(f)$&nbsp; is decomposed into in real and imaginary parts respectively, then Hilbert correspondences can be derived from it.
+If &nbsp;$H(f)$&nbsp; is decomposed into real and imaginary parts respectively,&nbsp; then Hilbert correspondences can be derived from it.
@@ Line 37: / Line 37: @@
 Please note:
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem|Conclusions from the Allocation Theorem]].
 *In particular, reference is made to the theory page &nbsp;[[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation|Hilbert transformation]].

@@ Line 88: / Line 88: @@
 '''(2)'''&nbsp; The <u>last proposed solution</u> is correct:
-*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;:
+*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

@@ Line 121: / Line 121: @@
 T}}{{\rm j}\cdot 2\pi f T} \hspace{0.05cm}.$$
-*Using &nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation by magnitude and phase|Euler's theorem]]&nbsp; this can also be rewritten as follows:
+*Using&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation by magnitude and phase|Euler's theorem]]&nbsp; this can also be rewritten as follows:
 :$$H_3(f) = \frac{1-\cos (2\pi
 f T) + {\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \sin (2\pi f

@@ Line 96: / Line 96: @@
 :$$\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}
+-\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}

@@ Line 105: / Line 105: @@
 '''(3)'''&nbsp; <u>Both proposed solutions</u> are correct:
-*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 the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_first_Fourier_integral|first Fourier integral]]:
+*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]]:
 :$$H_3(f) =    \int_{-\infty}^{
 +\infty}
@@ Line 121: / Line 121: @@
 T}}{{\rm j}\cdot 2\pi f T} \hspace{0.05cm}.$$
-*Mit dem&nbsp;  [[Signal_Representation/Calculating_With_Complex_Numbers#Representation by magnitude and phase|Euler's theorem]]&nbsp; kann hierfür auch geschrieben werden:
+*Using the&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation by magnitude and phase|Euler's theorem]]&nbsp; this can also be rewritten as follows:
 :$$H_3(f) = \frac{1-\cos (2\pi
 f T) + {\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \sin (2\pi f
@@ Line 128: / Line 128: @@
 T}\hspace{0.05cm}.$$
-*Weiter gilt mit der Umformung&nbsp; $1 - \cos(\alpha) = 2 \cdot \sin^2(\alpha/2)$:
+*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
   \hspace{0.05cm}, \hspace{0.5cm}

@@ Line 138: / Line 138: @@
 '''(4)'''&nbsp; <u>No</u> is correct:
-*The impulse response &nbsp; $h_4(t)$&nbsp; is not causal so that aus dem dazugehörigen Fourier&ndash;Spektrum&nbsp; $H_4(f)$&nbsp; keine Hilbert correspondence abgeleitet werden kann.
+*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}}