Aufgaben:Exercise 1.3Z: Exponentially Decreasing Impulse Response - Revision history

Oezer at 17:17, 18 July 2021

2021-07-18T17:17:46Z

Guenter at 10:34, 12 July 2021

2021-07-12T10:34:34Z

Guenter at 10:26, 12 July 2021

2021-07-12T10:26:34Z

Oezer at 16:08, 1 July 2021

2021-07-01T16:08:27Z

Oezer at 12:17, 1 July 2021

2021-07-01T12:17:04Z

Oezer at 12:16, 1 July 2021

2021-07-01T12:16:47Z

Oezer at 12:16, 1 July 2021

2021-07-01T12:16:08Z

Oezer at 12:09, 1 July 2021

2021-07-01T12:09:49Z

Oezer at 12:05, 1 July 2021

2021-07-01T12:05:40Z

Oezer at 12:04, 1 July 2021

2021-07-01T12:04:54Z

@@ Line 93: / Line 93: @@
 *In contrast, a high-pass filter would have to satisfy the following condition:
 :$$H(f = 0) = \int_{-\infty}^{+\infty}h(t) \hspace{0.15cm}{\rm d}t = 0.$$
-*$H(f)$&nbsp; is a complex function. The phase response is&nbsp; (see&nbsp; [[Aufgaben:Exercise_1.1Z:_Low-Pass_1st_and_2nd_Order|Exercise 1.1Z]]):
+*$H(f)$&nbsp; is a complex function. The phase response is&nbsp; (see&nbsp; [[Aufgaben:Exercise_1.1Z:_Low-Pass_Filter_of_1st_and_2nd_Order|Exercise 1.1Z]]):
 :$$b(f) = \arctan {f}/{f_{\rm G}}.$$
 *For the frequency&nbsp; $f = f_{\rm G}$&nbsp; one obtains&nbsp; $b(f = f_{\rm G}) = π/4 = 45^\circ$.

@@ Line 20: / Line 20: @@
 ''Please note:''
-*The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain|System Description in Time Domain]]..
+*The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain|System Description in Time Domain]].
 *The following definite integral is given:
 :$$\int_{ 0 }^{ \infty } \frac{1}{1+x^2}  \hspace{0.1cm}{\rm d}x = {\pi}/{2} .$$
@@ Line 57: / Line 57: @@
 </quiz>
-===Sample solution===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; The frequency response&nbsp; $H(f)$&nbsp; is the Fourier transform of&nbsp; $h(t)$:
@@ Line 78: / Line 78: @@
 *Using the given definite integral with the result&nbsp; $π/2$&nbsp; the following is obtained:
 :$$h(t=0)= \frac{1}{2 T} \hspace{0.15cm}\underline{= {\rm 500\cdot 1/s}}.$$
-*The result shows that the impulse response at&nbsp; $t = 0$&nbsp; is equal to the mean value of the left-hand&ndash; and right-hand limits.
+*The result shows that the impulse response at&nbsp; $t = 0$&nbsp; is equal to the mean value of the left-hand and right-hand limits.
-'''(3)'''&nbsp; The amplitude response in this task or in general with the 3dB-cutoff frequency&nbsp; $f_{\rm G}$ is:
+'''(3)'''&nbsp; The amplitude response in this task or in general with the 3dB cut-off frequency&nbsp; $f_{\rm G}$ is:
 :$$|H(f)|  =  \frac{1}{\sqrt{1+(2\pi fT)^2}} = \frac{1}{\sqrt{1+(f/f_{\rm G})^2}}.$$
-*By comparing coefficients one obtains:
+*By comparing the coefficients one obtains:
 :$$f_{\rm G} = \frac{1}{2\pi T} \hspace{0.15cm}\underline{= {\rm 159 \hspace{0.1cm} Hz}}.$$
-'''(4)'''&nbsp; <u>The first approach</u> is correct:
+'''(4)'''&nbsp; <u>The first statement</u> is correct:
-*Due to&nbsp; $h(t) = 0$&nbsp; &nbsp;for&nbsp; $t < 0$&nbsp; the system is indeed causal. It is a <u>low-pass filter of first order</u>.
+*Due to&nbsp; $h(t) = 0$&nbsp; &nbsp;for&nbsp; $t < 0$:&nbsp; The system is indeed causal.&nbsp; It is a <u>low-pass filter of first order</u>.
 *In contrast, a high-pass filter would have to satisfy the following condition:
 :$$H(f = 0) = \int_{-\infty}^{+\infty}h(t) \hspace{0.15cm}{\rm d}t = 0.$$
-*$H(f)$&nbsp; is a complex function. The phase response is&nbsp; (see&nbsp; [[Aufgaben:Exercise_1.1Z:_Low-Pass_1st_and_2nd_Order|exercise 1.1Z]]):
+*$H(f)$&nbsp; is a complex function. The phase response is&nbsp; (see&nbsp; [[Aufgaben:Exercise_1.1Z:_Low-Pass_1st_and_2nd_Order|Exercise 1.1Z]]):
 :$$b(f) = \arctan {f}/{f_{\rm G}}.$$
 *For the frequency&nbsp; $f = f_{\rm G}$&nbsp; one obtains&nbsp; $b(f = f_{\rm G}) = π/4 = 45^\circ$.

@@ Line 4: / Line 4: @@
 [[File:P_ID819__LZI_Z_1_3.png |right|frame|Decreasing impulse response]]
 The impulse response&nbsp; $h(t)$&nbsp; of an LTI system, which
-*is identically zero for all times&nbsp; $t < 0$&nbsp;,
+*is identically zero for all times&nbsp; $t < 0$,
-*changes abruptly at time&nbsp; $t > 0$&nbsp;, and
+*changes abruptly at time&nbsp; $t > 0$,&nbsp; and
 *decreases for&nbsp; $t > 0$&nbsp; according to an exponential function:
 :$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T},$$ was measured.
-Let the parameter be&nbsp; $T = 1 \hspace{0.05cm} \rm ms$. In the subtask&nbsp; '''(3)''' &nbsp; the 3dB–cutoff frequency&nbsp; $f_{\rm G}$&nbsp; is to be determined, which is (implicitly) defined as follows:
+Let the parameter be&nbsp; $T = 1 \hspace{0.15cm} \rm ms$.&nbsp; In the subtask&nbsp; '''(3)''' &nbsp; the 3dB cut-off frequency&nbsp; $f_{\rm G}$&nbsp; is to be determined, which is (implicitly) defined as follows:
 :$$|H(f = f_{\rm G})| = {1}/{\sqrt{2}} \cdot|H(f = 0)| .$$
@@ Line 20: / Line 20: @@
 ''Please note:''
-*The task belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain|System Description in Time Domain]].
+*The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain|System Description in Time Domain]]..
 *The following definite integral is given:
 :$$\int_{ 0 }^{ \infty } \frac{1}{1+x^2}  \hspace{0.1cm}{\rm d}x = {\pi}/{2} .$$
@@ Line 32: / Line 32: @@
 <quiz display=simple>
-{Compute the frequency response&nbsp; $H(f)$. What value is obtained for&nbsp; $f = 0$?
+{Compute the frequency response&nbsp; $H(f)$.&nbsp; What value is obtained for&nbsp; $f = 0$?
 |type="{}"}
 $H(f = 0) \ = \ $ { 1 3% }
@@ Line 42: / Line 42: @@
-{Compute the 3dB–cutoff frequency&nbsp; $f_{\rm G}$.
+{Compute the 3dB cut-off frequency&nbsp; $f_{\rm G}$.
 |type="{}"}
 $f_{\rm G}  \ =\ $ { 159 3% } &nbsp;$\rm Hz$

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain}}
-[[File:P_ID819__LZI_Z_1_3.png |right|frame|Dropping impulse response]]
+[[File:P_ID819__LZI_Z_1_3.png |right|frame|Decreasing impulse response]]
 The impulse response&nbsp; $h(t)$&nbsp; of an LTI system, which
 *is identically zero for all times&nbsp; $t < 0$&nbsp;,

@@ Line 7: / Line 7: @@
 *changes abruptly at time&nbsp; $t > 0$&nbsp;, and
 *decreases for&nbsp; $t > 0$&nbsp; according to an exponential function:
-:$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T}$$, was measured.
+:$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T},$$ was measured.
 Let the parameter be&nbsp; $T = 1 \hspace{0.05cm} \rm ms$. In the subtask&nbsp; '''(3)''' &nbsp; the 3dB–cutoff frequency&nbsp; $f_{\rm G}$&nbsp; is to be determined, which is (implicitly) defined as follows:
 :$$|H(f = f_{\rm G})| = {1}/{\sqrt{2}} \cdot|H(f = 0)| .$$

@@ Line 8: / Line 8: @@
 *für&nbsp; $t > 0$&nbsp; entsprechend einer Exponentialfunktion abfällt:
 :$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T}.$$
-Der Parameter sei&nbsp; $T = 1 \hspace{0.05cm} \rm ms$. In der Teilaufgabe&nbsp; '''(3)''' ist nach der&nbsp; 3dB–Grenzfrequenz&nbsp; $f_{\rm G}$&nbsp; gefragt, die wie folgt (implizit) definiert ist:
+Let the parameter be&nbsp; $T = 1 \hspace{0.05cm} \rm ms$. In the subtask&nbsp; '''(3)''' &nbsp; the 3dB–cutoff frequency&nbsp; $f_{\rm G}$&nbsp; is to be determined, which is (implicitly) defined as follows:
 :$$|H(f = f_{\rm G})| = {1}/{\sqrt{2}} \cdot|H(f = 0)| .$$

@@ Line 90: / Line 90: @@
 '''(4)'''&nbsp; <u>The first approach</u> is correct:
-*Wegen&nbsp; $h(t) = 0$&nbsp; &nbsp;für&nbsp; $t < 0$&nbsp; ist das System tatsächlich kausal. Es handelt sich um einen <u>Tiefpass erster Ordnung</u>.
+*Due to&nbsp; $h(t) = 0$&nbsp; &nbsp;for&nbsp; $t < 0$&nbsp; the system is indeed causal. It is a <u>low-pass filter of first order</u>.
-*Dagegen müsste ein Hochpass folgende Bedingung erfüllen:
+*In contrast, a high-pass filter would have to satisfy the following condition:
 :$$H(f = 0) = \int_{-\infty}^{+\infty}h(t) \hspace{0.15cm}{\rm d}t = 0.$$
-*$H(f)$&nbsp; ist eine komplexe Funktion. Der Phasengang lautet&nbsp; (siehe&nbsp; [[Aufgaben:Aufgabe_1.1Z:_Tiefpass_1._und_2._Ordnung|Aufgabe 1.1Z]]):
+*$H(f)$&nbsp; is a complex function. The phase response is&nbsp; (see&nbsp; [[Aufgaben:Exercise_1.1Z:_Low-Pass_1st_and_2nd_Order|Exercise 1.1Z]]):
 :$$b(f) = \arctan {f}/{f_{\rm G}}.$$
-*Für die Frequenz&nbsp; $f = f_{\rm G}$&nbsp; erhält man&nbsp; $b(f = f_{\rm G}) = π/4 = 45^\circ$.
+*For the frequency&nbsp; $f = f_{\rm G}$&nbsp; one obtains&nbsp; $b(f = f_{\rm G}) = π/4 = 45^\circ$.
-*Liegt am Eingang ein Cosinussignal der Frequenz&nbsp; $f = f_{\rm G}$&nbsp; an, so ergibt sich für das Ausgangssignal:
+*If a cosine signal of frequency&nbsp; $f = f_{\rm G}$&nbsp; is applied ot the input, the output signal is given by:
 :$$y(t) = K \cdot \cos( 2 \pi f_{\rm G} t - 45^{\circ}).$$
-*Dieses Signal ist zwar eine harmonische Schwingung, aber kein Cosinussignal.
+*This signal is a harmonic oscillation but not a cosine signal.
 {{ML-Fuß}}