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

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	*Since the imaginary part is odd only the real part has to be integrated over.		*Since the imaginary part is odd only the real part has to be integrated over.
	*Using the symmetry property one obtains:		*Using the symmetry property one obtains:
	:$$h(t=0)=2 \cdot \int_{ 0 }^{ \infty } \frac{1}{1+(2\pi fT)^2} \hspace{0.1cm}{\~~rmd~~}f = \frac{1}{\pi T} \cdot \int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x .$$		:$$h(t=0)=2 \cdot \int_{ 0 }^{ \infty } \frac{1}{1+(2\pi fT)^2} \hspace{0.1cm}{\rm d}f = \frac{1}{\pi T} \cdot \int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x .$$
	*Using the given definite integral with the result  $π/2$  the following is obtained:		*Using the given definite integral with the result  $π/2$  the following is obtained:
	:$$h(t=0)= \frac{1}{2 T} \hspace{0.15cm}\underline{= {\rm 500\cdot 1/s}}.$$		:$$h(t=0)= \frac{1}{2 T} \hspace{0.15cm}\underline{= {\rm 500\cdot 1/s}}.$$
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^1.2 System Description in Time Domain^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^1.2 System Description in Time Domain^]]
	[[de:Aufgaben:1.~~3Z_Exponentiell_abfallende_Impulsantwort~~]]		[[de:Aufgaben:Aufgabe 1.3Z: Exponentiell abfallende Impulsantwort]]

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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The frequency response  $H(f)$  is the Fourier transform of  $h(t)$:		'''(1)'''  The frequency response  $H(f)$  is the Fourier transform of  $h(t)$:
	:$$H(f) = \int_{-\infty}^{+\infty}h(t) \cdot {\rm e}^{\hspace{0.05cm}{-\rm j}2\pi ft}\hspace{0.15cm} {\rm d}t = \frac{1}{T} \cdot \int_{0}^{+\infty} {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T}) t}\hspace{0.15cm}		:$$H(f) = \int_{-\infty}^{+\infty}h(t) \cdot {\rm e}^{\hspace{0.05cm}{-\rm j}2\pi ft}\hspace{0.15cm} {\rm d}t = \frac{1}{T} \cdot \int_{0}^{+\infty} {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T}) t}\hspace{0.15cm}{\rm d}t.$$
	{\rm d}t.$$
	*Integration leads to the result:		*Integration leads to the result:
	:$$H(f) = \left[ \frac{-1/T}{{\rm j}2\pi f+{1}/{T}} \cdot {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T})		:$$H(f) = \left[ \frac{-1/T}{{\rm j}2\pi f+{1}/{T}} \cdot {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T})t}\right]_{0}^{\infty}= \frac{1}{1+{\rm j} \cdot 2\pi fT}.$$
	t}\right]_{0}^{\infty}= \frac{1}{1+{\rm j} \cdot 2\pi fT}.$$
	*At frequency  $f = 0$  the frequency response has the value  $H(f = 0) \; \underline{= 1}$.		*At frequency  $f = 0$  the frequency response has the value  $H(f = 0) \; \underline{= 1}$.

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	*Since the imaginary part is odd only the real part has to be integrated over.		*Since the imaginary part is odd only the real part has to be integrated over.
	*Using the symmetry property one obtains:		*Using the symmetry property one obtains:
	:$$h(t=0)=2 \cdot \int_{ 0 }^{ \infty } \frac{1}{1+(2\pi fT)^2} \hspace{0.1cm}{\rm		:$$h(t=0)=2 \cdot \int_{ 0 }^{ \infty } \frac{1}{1+(2\pi fT)^2} \hspace{0.1cm}{\rmd}f = \frac{1}{\pi T} \cdot \int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x .$$
	d}f = \frac{1}{\pi T} \cdot \int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x .$$
	*Using the given definite integral with the result  $π/2$  the following is obtained:		*Using the given definite integral with the result  $π/2$  the following is obtained:
	:$$h(t=0)= \frac{1}{2 T} \hspace{0.15cm}\underline{= {\rm 500\cdot 1/s}}.$$		:$$h(t=0)= \frac{1}{2 T} \hspace{0.15cm}\underline{= {\rm 500\cdot 1/s}}.$$

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^1.2 System Description in Time Domain^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^1.2 System Description in Time Domain^]]
			[[de:Aufgaben:1.3Z_Exponentiell_abfallende_Impulsantwort]]

Oezer at 17:17, 18 July 2021

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

Guenter at 10:34, 12 July 2021

2021-07-12T10:34:34Z

Guenter at 10:26, 12 July 2021

2021-07-12T10:26:34Z

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	[[File:P_ID819__LZI_Z_1_3.png \|right\|frame\|Decreasing impulse response]]		[[File:P_ID819__LZI_Z_1_3.png \|right\|frame\|Decreasing impulse response]]
	The impulse response  $h(t)$  of an LTI system, which		The impulse response  $h(t)$  of an LTI system, which
	*is identically zero for all times  $t < 0$~~ ~~,		*is identically zero for all times  $t < 0$,
	*changes abruptly at time  $t > 0$ , and		*changes abruptly at time  $t > 0$,  and
	*decreases for  $t > 0$  according to an exponential function:		*decreases for  $t > 0$  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  $T = 1 \hspace{0.~~05cm~~} \rm ms$. In the subtask  '''(3)'''   the ~~3dB–cutoff~~ frequency  $f_{\rm G}$  is to be determined, which is (implicitly) defined as follows:		Let the parameter be  $T = 1 \hspace{0.15cm} \rm ms$.  In the subtask  '''(3)'''   the 3dB cut-off frequency  $f_{\rm G}$  is to be determined, which is (implicitly) defined as follows:
	:$$\|H(f = f_{\rm G})\| = {1}/{\sqrt{2}} \cdot\|H(f = 0)\| .$$		:$$\|H(f = f_{\rm G})\| = {1}/{\sqrt{2}} \cdot\|H(f = 0)\| .$$

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	''Please note:''		''Please note:''
	*The ~~task~~ belongs to the chapter  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain\|System Description in Time Domain]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain\|System Description in Time Domain]]..
	*The following definite integral is given:		*The following definite integral is given:
	:$$\int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x = {\pi}/{2} .$$		:$$\int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x = {\pi}/{2} .$$
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	<quiz display=simple>		<quiz display=simple>
	{Compute the frequency response  $H(f)$. What value is obtained for  $f = 0$?		{Compute the frequency response  $H(f)$.  What value is obtained for  $f = 0$?
	\|type="{}"}		\|type="{}"}
	$H(f = 0) \ = \ $ { 1 3% }		$H(f = 0) \ = \ $ { 1 3% }
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	{Compute the ~~3dB–cutoff~~ frequency  $f_{\rm G}$.		{Compute the 3dB cut-off frequency  $f_{\rm G}$.
	\|type="{}"}		\|type="{}"}
	$f_{\rm G} \ =\ $ { 159 3% }  $\rm Hz$		$f_{\rm G} \ =\ $ { 159 3% }  $\rm Hz$

Oezer at 16:08, 1 July 2021

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	{{quiz-Header\|Buchseite=Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain}}		{{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  $h(t)$  of an LTI system, which		The impulse response  $h(t)$  of an LTI system, which
	*is identically zero for all times  $t < 0$ ,		*is identically zero for all times  $t < 0$ ,

Oezer at 12:17, 1 July 2021

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

Oezer at 12:16, 1 July 2021

2021-07-01T12:16:47Z

← Older revision		Revision as of 14:16, 1 July 2021
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	*changes abruptly at time  $t > 0$ , and		*changes abruptly at time  $t > 0$ , and
	*decreases for  $t > 0$  according to an exponential function:		*decreases for  $t > 0$  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  $T = 1 \hspace{0.05cm} \rm ms$. In the subtask  '''(3)'''   the 3dB–cutoff frequency  $f_{\rm G}$  is to be determined, which is (implicitly) defined as follows:		Let the parameter be  $T = 1 \hspace{0.05cm} \rm ms$. In the subtask  '''(3)'''   the 3dB–cutoff frequency  $f_{\rm G}$  is to be determined, which is (implicitly) defined as follows:
	:$$\|H(f = f_{\rm G})\| = {1}/{\sqrt{2}} \cdot\|H(f = 0)\| .$$		:$$\|H(f = f_{\rm G})\| = {1}/{\sqrt{2}} \cdot\|H(f = 0)\| .$$

Oezer at 12:16, 1 July 2021

2021-07-01T12:16:08Z

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	[[File:P_ID819__LZI_Z_1_3.png \|right\|frame\|Dropping impulse response]]		[[File:P_ID819__LZI_Z_1_3.png \|right\|frame\|Dropping impulse response]]
	~~Gemessen wurde die Impulsantwort~~  $h(t)$  ~~eines LZI–Systems~~,		The impulse response  $h(t)$  of an LTI system, which
	*~~die für alle Zeiten~~  $t < 0$  ~~identisch Null ist~~,		*is identically zero for all times  $t < 0$ ,
	*~~sich zur Zeit~~  $t > 0$  ~~sprungartig verändert~~, ~~und~~		*changes abruptly at time  $t > 0$ , and
	*~~für~~  $t > 0$  ~~entsprechend einer Exponentialfunktion abfällt~~:		*decreases for  $t > 0$  according to an exponential function:
	:$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T}.$$		:$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T}$$ was measured.
	Let the parameter be  $T = 1 \hspace{0.05cm} \rm ms$. In the subtask  '''(3)'''   the 3dB–cutoff frequency  $f_{\rm G}$  is to be determined, which is (implicitly) defined as follows:		Let the parameter be  $T = 1 \hspace{0.05cm} \rm ms$. In the subtask  '''(3)'''   the 3dB–cutoff frequency  $f_{\rm G}$  is to be determined, which is (implicitly) defined as follows:
	:$$\|H(f = f_{\rm G})\| = {1}/{\sqrt{2}} \cdot\|H(f = 0)\| .$$		:$$\|H(f = f_{\rm G})\| = {1}/{\sqrt{2}} \cdot\|H(f = 0)\| .$$

← Older revision		Revision as of 12:34, 12 July 2021
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	''Please note:''		''Please note:''
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain\|System Description in Time Domain]]..		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain\|System Description in Time Domain]].
	*The following definite integral is given:		*The following definite integral is given:
	:$$\int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x = {\pi}/{2} .$$		:$$\int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x = {\pi}/{2} .$$
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	</quiz>		</quiz>

	===~~Sample solution~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The frequency response  $H(f)$  is the Fourier transform of  $h(t)$:		'''(1)'''  The frequency response  $H(f)$  is the Fourier transform of  $h(t)$:
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	*Using the given definite integral with the result  $π/2$  the following is obtained:		*Using the given definite integral with the result  $π/2$  the following is obtained:
	:$$h(t=0)= \frac{1}{2 T} \hspace{0.15cm}\underline{= {\rm 500\cdot 1/s}}.$$		:$$h(t=0)= \frac{1}{2 T} \hspace{0.15cm}\underline{= {\rm 500\cdot 1/s}}.$$
	*The result shows that the impulse response at  $t = 0$  is equal to the mean value of the left-hand~~–~~ and right-hand limits.		*The result shows that the impulse response at  $t = 0$  is equal to the mean value of the left-hand and right-hand limits.



	'''(3)'''  The amplitude response in this task or in general with the 3dB-~~cutoff~~ frequency  $f_{\rm G}$ is:		'''(3)'''  The amplitude response in this task or in general with the 3dB cut-off frequency  $f_{\rm G}$ is:
	:$$\|H(f)\| = \frac{1}{\sqrt{1+(2\pi fT)^2}} = \frac{1}{\sqrt{1+(f/f_{\rm G})^2}}.$$		:$$\|H(f)\| = \frac{1}{\sqrt{1+(2\pi fT)^2}} = \frac{1}{\sqrt{1+(f/f_{\rm G})^2}}.$$
	*By comparing coefficients one obtains:		*By comparing the coefficients one obtains:
	:$$f_{\rm G} = \frac{1}{2\pi T} \hspace{0.15cm}\underline{= {\rm 159 \hspace{0.1cm} Hz}}.$$		:$$f_{\rm G} = \frac{1}{2\pi T} \hspace{0.15cm}\underline{= {\rm 159 \hspace{0.1cm} Hz}}.$$



	'''(4)'''  <u>The first ~~approach~~</u> is correct:		'''(4)'''  <u>The first statement</u> is correct:
	*Due to  $h(t) = 0$   for  $t < 0$  ~~the~~ system is indeed causal. It is a <u>low-pass filter of first order</u>.		*Due to  $h(t) = 0$   for  $t < 0$:  The system is indeed causal.  It is a <u>low-pass filter of first order</u>.
	*In contrast, a high-pass filter would have to satisfy the following condition:		*In contrast, a high-pass filter would have to satisfy the following condition:
	:$$H(f = 0) = \int_{-\infty}^{+\infty}h(t) \hspace{0.15cm}{\rm d}t = 0.$$		:$$H(f = 0) = \int_{-\infty}^{+\infty}h(t) \hspace{0.15cm}{\rm d}t = 0.$$
	*$H(f)$  is a complex function. The phase response is  (see  [[Aufgaben:Exercise_1.1Z:_Low-Pass_1st_and_2nd_Order\|~~exercise~~ 1.1Z]]):		*$H(f)$  is a complex function. The phase response is  (see  [[Aufgaben:Exercise_1.1Z:_Low-Pass_1st_and_2nd_Order\|Exercise 1.1Z]]):
	:$$b(f) = \arctan {f}/{f_{\rm G}}.$$		:$$b(f) = \arctan {f}/{f_{\rm G}}.$$
	*For the frequency  $f = f_{\rm G}$  one obtains  $b(f = f_{\rm G}) = π/4 = 45^\circ$.		*For the frequency  $f = f_{\rm G}$  one obtains  $b(f = f_{\rm G}) = π/4 = 45^\circ$.