Aufgaben:Exercise 1.5Z: Sinc-shaped Impulse Response - Revision history

Oezer at 12:54, 7 September 2021

2021-09-07T12:54:00Z

2021-09-07T12:52:36Z

2021-09-07T12:51:43Z

2021-09-07T12:50:02Z

2021-09-07T12:46:26Z

2021-09-06T20:19:32Z

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2021-09-06T20:07:46Z

@@ Line 57: / Line 57: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; A comparison with the equations on the page&nbsp; [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Ideal_low-pass_filter_–_Rectangular-in-frequency|ideal low-pass filter]] or applying the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_second_Fourier_integral |inverse Fourier transformation]]&nbsp; shows that&nbsp; $H(f)$&nbsp; is an ideal low-pass filter:
+'''(1)'''&nbsp; A comparison with the equations on the page&nbsp; [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Ideal_low-pass_filter_–_Rectangular-in-frequency|Ideal low-pass filter]] or applying the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_second_Fourier_integral |inverse Fourier transformation]]&nbsp; shows that&nbsp; $H(f)$&nbsp; is an ideal low-pass filter:
 :$$H(f) = \left\{ \begin{array}{c} \hspace{0.25cm}K  \\  K/2 \\ 0 \\  \end{array} \right.\quad \quad \begin{array}{*{10}c}   {\rm{f\ddot{u}r}}  \\ {\rm{f\ddot{u}r}}
 \\   {\rm{f\ddot{u}r}}  \\ \end{array}\begin{array}{*{20}c}

@@ Line 57: / Line 57: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; A comparison with the equations on the page&nbsp; [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Ideal_low-pass_filter_–_Rectangular-in-frequency
+'''(1)'''&nbsp; A comparison with the equations on the page&nbsp; [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Ideal_low-pass_filter_–_Rectangular-in-frequency|ideal low-pass filter]] or applying the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_second_Fourier_integral |inverse Fourier transformation]]&nbsp; shows that&nbsp; $H(f)$&nbsp; is an ideal low-pass filter:
 :$$H(f) = \left\{ \begin{array}{c} \hspace{0.25cm}K  \\  K/2 \\ 0 \\  \end{array} \right.\quad \quad \begin{array}{*{10}c}   {\rm{f\ddot{u}r}}  \\ {\rm{f\ddot{u}r}}
 \\   {\rm{f\ddot{u}r}}  \\ \end{array}\begin{array}{*{20}c}

@@ Line 58: / Line 58: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; A comparison with the equations on the page&nbsp; [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Ideal_low-pass_filter_–_Rectangular-in-frequency
-|Ideal low-pass filter]] or applying the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_second_Fourier_integral |inverse Fourier transformation]]&nbsp; shows that&nbsp; $H(f)$&nbsp; is an ideal low-pass filter:
+|ideal low-pass filter]] or applying the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_second_Fourier_integral |inverse Fourier transformation]]&nbsp; shows that&nbsp; $H(f)$&nbsp; is an ideal low-pass filter:
 :$$H(f) = \left\{ \begin{array}{c} \hspace{0.25cm}K  \\  K/2 \\ 0 \\  \end{array} \right.\quad \quad \begin{array}{*{10}c}   {\rm{f\ddot{u}r}}  \\ {\rm{f\ddot{u}r}}
 \\   {\rm{f\ddot{u}r}}  \\ \end{array}\begin{array}{*{20}c}

@@ Line 57: / Line 57: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; A comparison with the equations on the page&nbsp; [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Idealer_Tiefpass_.E2.80.93_K.C3.BCpfm.C3.BCller.E2.80.93Tiefpass|Ideal low-pass filter]] or applying the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_second_Fourier_integral |inverse Fourier transformation]]&nbsp; shows that&nbsp; $H(f)$&nbsp; is an ideal low-pass filter:
+'''(1)'''&nbsp; A comparison with the equations on the page&nbsp; [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Ideal_low-pass_filter_–_Rectangular-in-frequency
 :$$H(f) = \left\{ \begin{array}{c} \hspace{0.25cm}K  \\  K/2 \\ 0 \\  \end{array} \right.\quad \quad \begin{array}{*{10}c}   {\rm{f\ddot{u}r}}  \\ {\rm{f\ddot{u}r}}
 \\   {\rm{f\ddot{u}r}}  \\ \end{array}\begin{array}{*{20}c}

@@ Line 57: / Line 57: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; A comparison with the equations on the page&nbsp; [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Idealer_Tiefpass_.E2.80.93_K.C3.BCpfm.C3.BCller.E2.80.93Tiefpass|Ideal low-pass filter]] or applying the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Das_zweite_Fourierintegral|Fourierrücktransformation]]&nbsp; shows that&nbsp; $H(f)$&nbsp; is an ideal low-pass filter:
+'''(1)'''&nbsp; A comparison with the equations on the page&nbsp; [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Idealer_Tiefpass_.E2.80.93_K.C3.BCpfm.C3.BCller.E2.80.93Tiefpass|Ideal low-pass filter]] or applying the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_second_Fourier_integral |inverse Fourier transformation]]&nbsp; shows that&nbsp; $H(f)$&nbsp; is an ideal low-pass filter:
 :$$H(f) = \left\{ \begin{array}{c} \hspace{0.25cm}K  \\  K/2 \\ 0 \\  \end{array} \right.\quad \quad \begin{array}{*{10}c}   {\rm{f\ddot{u}r}}  \\ {\rm{f\ddot{u}r}}
 \\   {\rm{f\ddot{u}r}}  \\ \end{array}\begin{array}{*{20}c}

@@ Line 90: / Line 90: @@
-'''(3)'''&nbsp; The frequency response has the value&nbsp; $K = 0.5$  at&nbsp; $f = f_0 = 100 \ \rm Hz$&nbsp; according to the calculations for the subtask&nbsp; '''(1)'''&nbsp;. Therefore,
+'''(3)'''&nbsp; The frequency response has the value&nbsp; $K = 0.5$ at&nbsp; $f = f_0 = 100 \ \rm Hz$&nbsp; according to the calculations for subtask&nbsp; '''(1)'''&nbsp;. Therefore,
 :$$A_y = A_x/2 = 2\ \rm  V$$ is obtained.
-*The same result is obtained by convolution according to the above equation..
+*The same result is obtained by convolution according to the above equation.
 *For&nbsp; $a = 2f_0/Δf = 0.2$&nbsp; the integral is equal to&nbsp; $π/2$&nbsp; and one obtains
 :$$y(t = 0 ) = {A_y } = \frac{A_x}{\pi} \cdot \frac{\pi}{2} = \frac{A_x}{2} \hspace{0.15cm}\underline{= 2\,{\rm V}}.$$

@@ Line 84: / Line 84: @@
   {\rm cos}(2\pi \cdot  f_0
 \cdot \tau ) \hspace{0.1cm}{\rm d}\tau.$$
-*With the substitution&nbsp; $u = π · Δf · τ$&nbsp; this can also be formulated as follows:
+*With the substitution&nbsp; $u = π · Δf · τ$&nbsp;, this can also be formulated as follows:
 :$$y(t = 0 ) = \frac{A_x }{\pi} \cdot \int_{ 0 }^{ \infty } \frac{\sin(u) \cdot \cos(a \cdot u)}{u}   \hspace{0.15cm}{\rm d}u .$$
 *Here, the constant is&nbsp; $a = 2f_0/Δf = 2$. With this value, the given integral yields zero: &nbsp; $y(t = 0 ) = {A_y } = 0.$

@@ Line 99: / Line 99: @@
 '''(4)'''&nbsp; The transition from the band-pass to the band-stop is exactly at&nbsp; $f = 0.5   \ \rm kHz$&nbsp; and for this singular location the following holds:
 :$$H(f = f_0) = K/2.$$
-*Somit ist die Amplitude des Ausgangssignals nur halb so groß wie in der Teilaufgabe&nbsp; '''(3)'''&nbsp;  berechnet, nämlich&nbsp; $A_y \;  \underline{= 1  \, \rm V}$.
+*Thus, the amplitude of the output signal is only half as large as calculated in subtask&nbsp; '''(3)'''&nbsp;, namely&nbsp; $A_y \;  \underline{= 1  \, \rm V}$.
-*Zum gleichen Ergebnis kommt man mit &nbsp;$a = 2f_0/Δf = 1$&nbsp; über die Faltung.
+*The same result is obtained with&nbsp;$a = 2f_0/Δf = 1$&nbsp; by convolution.
 {{ML-Fuß}}