Aufgaben:Exercise 4.1Z: High-Pass System - Revision history

Hwang at 12:48, 17 November 2022

2022-11-17T12:48:22Z

Guenter at 13:26, 5 May 2021

2021-05-05T13:26:48Z

Guenter at 13:22, 5 May 2021

2021-05-05T13:22:09Z

Guenter at 12:46, 5 May 2021

2021-05-05T12:46:38Z

Guenter at 12:19, 5 May 2021

2021-05-05T12:19:09Z

Javier: Text replacement - "Lowpass" to "Low-Pass"

2021-05-04T07:34:30Z

Text replacement - "Lowpass" to "Low-Pass"

Javier: Javier moved page Aufgaben:Exercise 4.1Z: Highpass System to Aufgaben:Exercise 4.1Z: High-Pass System

2021-05-04T07:30:15Z

Javier moved page Exercise 4.1Z: Highpass System to Exercise 4.1Z: High-Pass System

Javier: Text replacement - "band pass" to "band-pass"

2021-05-03T15:13:21Z

Text replacement - "band pass" to "band-pass"

Javier: Text replacement - "bandpass" to "band-pass"

2021-05-03T15:12:59Z

Text replacement - "bandpass" to "band-pass"

Javier: Text replacement - "high pass" to "high-pass"

2021-05-03T15:10:52Z

Text replacement - "high pass" to "high-pass"

@@ Line 4: / Line 4: @@
 [[File:EN_Sig_Z_4_1.png|right|frame|Simplest examples for <br>"low-pass" and "high-pass"]]
-The relationships shown on the page&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals#Properties_of_BP-Signals|"Properties of Bandpass Signals"]]&nbsp; apply not only to signals and spectra, but in the same way to
+The relationships shown in the section&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals#Properties_of_BP-Signals|"Properties of Bandpass Signals"]]&nbsp; apply not only to signals and spectra, but in the same way to
 *the frequency response&nbsp; $H(f)$&nbsp; and
 *the impulse response&nbsp; $h(t)$&nbsp;  of an LTI system.
@@ Line 87: / Line 87: @@
-'''(2)'''&nbsp; On the page&nbsp; "Properties of Bandpass Signals"&nbsp; it was shown that any band-pass signal can be represented as the difference of two low-pass signals.
+'''(2)'''&nbsp; In the section&nbsp; "Properties of Bandpass Signals"&nbsp; it was shown that any band-pass signal can be represented as the difference of two low-pass signals.
 [[File:P_ID703__Sig_Z_4_1_b.png|right|frame|Red:&nbsp; Low-pass&nbsp; $\rm (TP)$,<br>Blue:&nbsp; High-pass&nbsp; $\rm (HP)$]]
@@ Line 101: / Line 101: @@
-One arrives at exactly the same result starting from the concrete circuit on the information page.&nbsp; Corresponding to a frequency-dependent voltage divider with resistors&nbsp; $R$&nbsp; and&nbsp; $1/(j\omega C)$&nbsp; applies:
+One arrives at exactly the same result starting from the concrete circuit in the information section.&nbsp; Corresponding to a frequency-dependent voltage divider with resistors&nbsp; $R$&nbsp; and&nbsp; $1/(j\omega C)$&nbsp; applies:
 :$$H_{\rm HP}(f) =  \frac{R}{R + 1/({\rm j}\cdot \omega \cdot C)} =
 \frac{{\rm j}\cdot \omega \cdot C \cdot R}{1+{\rm j} \cdot \omega

← Older revision		Revision as of 13:26, 5 May 2021
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−	{{quiz-Header\|Buchseite=~~Signal Representation~~/~~Differences and Similarities of LP and BP Signals~~	+	{{quiz-Header\|Buchseite=Signal_Representation/Differences_and_Similarities_of_Low-Pass_and_Band-Pass_Signals
	}}		}}

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Sig_Z_4_1.png|right|frame|Examples for low-pass and high-pass]]
+[[File:EN_Sig_Z_4_1.png|right|frame|Simplest examples for <br>"low-pass" and "high-pass"]]
 The relationships shown on the page&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals#Properties_of_BP-Signals|"Properties of Bandpass Signals"]]&nbsp; apply not only to signals and spectra, but in the same way to
 *the frequency response&nbsp; $H(f)$&nbsp; and
@@ Line 71: / Line 71: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp;  The correct <u> solutions are 2 and 3</u>:
-*Der Gleichsignalübertragungsfaktor ist&nbsp; $H_{\rm TP}(f = 0) = 1$.
+*The DC transmission factor is&nbsp;&nbsp; $H_{\rm TP}(f = 0) = 1$.
 *For the magnitude frequency response holds:
 :$$|H_{\rm TP}(f)| = \frac{1}{\sqrt{1 + ( f / f_{\rm G})^2}} .$$
-*Thus, the value at&nbsp; $f_{\rm G}$ is equal to $\sqrt{1/2}$. The power transfer function&nbsp; $|H_{\rm TP}(f)|^2$&nbsp; is therefore only half as large at&nbsp; $f = f_{\rm G}$&nbsp; as at&nbsp; $f = 0$, which is the reason for the designation &bdquo;3dB&ndash;cut-off frequency&rdquo; for&nbsp; $f_{\rm G}$&nbsp;.
+*Thus, the value at&nbsp; $f_{\rm G}$ is equal to $\sqrt{1/2}$.&nbsp; The power transfer function&nbsp; $|H_{\rm TP}(f)|^2$&nbsp; is therefore only half as large at&nbsp; $f = f_{\rm G}$&nbsp; as at&nbsp; $f = 0$, which is the reason for the designation "3dB cut-off frequency".
 *The phase function is generally calculated according to the following equation:
 :$$\varphi_{\rm TP}(f) = -\arctan\frac{{\rm Im}\left[H_{\rm
@@ Line 87: / Line 87: @@
-'''(2)'''&nbsp; On the page [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals#Properties_of_BP-Signals|Properties of Bandpass Signals]]it was shown that any band-pass signal can be represented as the difference of two low-pass signals.
+'''(2)'''&nbsp; On the page&nbsp; "Properties of Bandpass Signals"&nbsp; it was shown that any band-pass signal can be represented as the difference of two low-pass signals.
-[[File:P_ID703__Sig_Z_4_1_b.png|right|frame|Low-pass and high-pass spectrum]]
+[[File:P_ID703__Sig_Z_4_1_b.png|right|frame|Red:&nbsp; Low-pass&nbsp; $\rm (TP)$,<br>Blue:&nbsp; High-pass&nbsp; $\rm (HP)$]]
 The same applies to frequency responses:
 :$$H_{\rm BP}(f) = H_1(f) - H_2(f).$$
 *If we set&nbsp; $H_2(f) = H_{\rm TP}(f)$&nbsp; and consider&nbsp; $H_1(f) = 1$&nbsp; as the limiting case of a low-pass function with an infinitely large bandwidth, we get:
 :$$H_{\rm HP}(f) = 1 - H_{\rm TP}(f).$$
-*As the graph shows, the result is now a high-pass because of &nbsp; $H_1(f) = 1$&nbsp;. With the given low-pass function&nbsp; $H_{\rm TP}(f)$&nbsp; one further obtains:
+*As the graph shows, the result is now a high-pass because of &nbsp; $H_1(f) = 1$.&nbsp; With the given low-pass function&nbsp; $H_{\rm TP}(f)$&nbsp; one further obtains:
 :$$H_{\rm HP}(f) = 1 - \frac{1}{1 + {\rm j} \cdot f / f_{\rm G}}  =
 \frac{{\rm j} \cdot f / f_{\rm G}}{1 + {\rm j} \cdot f / f_{\rm
 G}} .$$
-*For&nbsp; $f = 0$&nbsp; &nbsp; $H_{\rm HP}(f = 0) \;\underline{= 0}$ is obtained. Note that it is the actual (complex) function&nbsp; $H_{\rm TP}(f)$&nbsp; and not its magnitude that is to be subtracted. Therefore, the above sketch is to be understood only qualitatively.
+*For&nbsp; $f = 0$:&nbsp; &nbsp; $H_{\rm HP}(f = 0) \;\underline{= 0}$ is obtained.&nbsp; Note that it is the actual (complex) function&nbsp; $H_{\rm TP}(f)$&nbsp; and not its magnitude that is to be subtracted.&nbsp; Therefore, the above sketch is to be understood only qualitatively.
-One arrives at exactly the same result starting from the concrete circuit on the information page. Corresponding to a frequency-dependent voltage divider with resistors&nbsp; $R$&nbsp; and&nbsp; $1/(j\omega C)$&nbsp; applies:
+One arrives at exactly the same result starting from the concrete circuit on the information page.&nbsp; Corresponding to a frequency-dependent voltage divider with resistors&nbsp; $R$&nbsp; and&nbsp; $1/(j\omega C)$&nbsp; applies:
 :$$H_{\rm HP}(f) =  \frac{R}{R + 1/({\rm j}\cdot \omega \cdot C)} =
 \frac{{\rm j}\cdot \omega \cdot C \cdot R}{1+{\rm j} \cdot \omega
@@ Line 109: / Line 109: @@
-'''(3)'''&nbsp; : The correct <u>solutions are 2 and 3</u>:
+'''(3)'''&nbsp;  The correct <u>solutions are 2 and 3</u>:
 *The magnitude function of the high-pass is:
 :$$|H_{\rm HP}(f)| = \frac{|f / f_{\rm G}|}{\sqrt{1 + ( f / f_{\rm
 G})^2}} .$$
-*At the cut-off frequency&nbsp; $f_{\rm G}$&nbsp; , the magnitudes of the high-pass and low-pass frequency response are thus equal, namely&nbsp; $0.707$ each.
+*At the cut-off frequency&nbsp; $f_{\rm G}$,&nbsp; the magnitudes of the high-pass and low-pass frequency response are thus equal, namely&nbsp; $0.707$ each.
-*On the other hand, the first statement is obviously wrong:&nbsp; According to this, the value&nbsp; $|H_{HP}(f = f_{\rm G})| = 1 – 0.707 \approx 0.293$&nbsp; should result.
+*On the other hand, the first statement is obviously wrong: &nbsp; According to this, the value&nbsp; $|H_{HP}(f = f_{\rm G})| = 1 - 0.707 \approx 0.293$&nbsp; should result.
-[[File:EN_Sig_Z_4_1_c.png|right|frame|Low-pass and high-pass phase response]]
+[[File:EN_Sig_Z_4_1_c.png|right|frame|Phase responses of <br>low-pass and high-pass ]]
 *The frequency response calculated under&nbsp; '''(2)'''&nbsp; can also be represented as follows:
 :$$H_{\rm HP}(f) = \frac{( f / f_{\rm G})^2 + {\rm j} \cdot f /
 f_{\rm G}}{{1 + ( f / f_{\rm G})^2}} .$$
-:This gives for the phase function:
+*This gives for the phase function:
 :$$\varphi_{\rm HP}(f) =-\arctan\frac{f / f_{\rm G}}{( f /
 f_{\rm G})^2}=
@@ Line 125: / Line 125: @@
 :$$\Rightarrow \hspace{0.3cm}\varphi_{\rm HP}(f)  =  {\rm arctan} (\frac{f}{f_{\rm G}}) - \frac{\pi}{2}= \varphi_{\rm TP}(f)- \frac{\pi}{2}.$$
-*At positive frequencies, therefore, the same course as with the low-pass system results, except for a shift of&nbsp; $\pi /2$&nbsp; downwards. Since the phase function is odd, there is an upward shift of&nbsp; $\pi /2$&nbsp; at negative frequencies.
+*Therefore:&nbsp; At positive frequencies, the same course as with the low-pass system results, except for a shift of&nbsp; $\pi /2$&nbsp; downwards.&nbsp; Since the phase function is odd, there is an upward shift of&nbsp; $\pi /2$&nbsp; at negative frequencies.
@@ Line 131: / Line 131: @@
 '''(4)'''&nbsp; Proposed <u>solutions 2 and 3</u> are correct:
-*Due to the linearity of the (inverse) Fourier transformation, the following applies to the time course for&nbsp; $t > 0$:
+*Due to the linearity of the (inverse) Fourier transform, the following applies to the time course for&nbsp; $t > 0$:
 :$$h_{\rm HP}(t) = h_1(t) - h_2(t)= \delta(t) - \frac{1}{\tau}\cdot
 {\rm e}^{-t / \tau} .$$
 *The Dirac function is the inverse Fourier transform of the constant frequency function „1”.
-*The second component is identical to the LP impulse response except for the sign.
+*The second component is identical to the low-pass impulse response except for the sign.
-*The Dirac function causes&nbsp; $h_{\rm HP}(t)$&nbsp; to be infinitely large at time&nbsp; $t = 0$&nbsp;.In contrast, the following applies for&nbsp; $t = \tau$:
+*The Dirac function causes&nbsp; $h_{\rm HP}(t)$&nbsp; to be infinitely large at time&nbsp; $t = 0$.&nbsp; In contrast, the following applies for&nbsp; $t = \tau$:
 :$$h_{\rm HP}(t = \tau) =  - \frac{1}{\tau}\cdot {\rm e}^{-1} =  -
-{{\rm e}\cdot \tau}^{-1}.$$
+{{\rm e}/ \tau}.$$
 *The last statement is therefore also false due to the sign.
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:EN_Sig_Z_4_1.png|right|frame|Examples for low-pass and high-pass]]
-The relationships shown on the page&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals#Properties_of_BP-Signals|Properties of Bandpass Signals]]&nbsp; apply not only to signals and spectra, but also in the same way to
+The relationships shown on the page&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals#Properties_of_BP-Signals|"Properties of Bandpass Signals"]]&nbsp; apply not only to signals and spectra, but in the same way to
 *the frequency response&nbsp; $H(f)$&nbsp; and
-*the impulse response&nbsp; $h(t)$
+*the impulse response&nbsp; $h(t)$&nbsp;  of an LTI system.
-of an LTI system. These are also related via the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse|Fourier Transform]]&nbsp;. More information on this can be found in the book&nbsp; [[Linear and timeinvariant systems]].
+These are also related via the&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse|Fourier Transform]].&nbsp; More information can be found in the book&nbsp; [[Linear_and_Time_Invariant_Systems|"Linear and Time-Invariant Systems"]].
-The circuit shown in the upper diagram is the simplest realisation of a low-pass filter:
+The circuit shown in the upper diagram is the simplest realisation of a low-pass&nbsp; (German:&nbsp; "Tiefpass" &nbsp; &rArr; &nbsp; $\text{TP}$)&nbsp; filter:
 *For very high frequencies, the capacitance&nbsp; $C$&nbsp; acts as a short circuit, so that high-frequency components are no longer included in the output signal.
 *In contrast, low-frequency signal components are only imperceptibly attenuated by the voltage divider.
-*With the 3dB&ndash;cut-off frequency&nbsp; $f_{\rm G}$&nbsp;, the following applies to the frequency response:
+*With the 3dB cut-off frequency&nbsp; $f_{\rm G}$&nbsp; the following applies to the frequency response:
 :$$H_{\rm TP}(f) = \frac{1}{1 + {\rm j} \cdot f / f_{\rm G}}  =
 |H_{\rm TP}(f)|\cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \varphi_{\rm TP}(f)} .$$
-In the second part of the equation, the frequency response&nbsp; $H_{\rm TP}(f)$&nbsp; is split according to magnitude and phase.
+In the second part of this equation, the frequency response&nbsp; $H_{\rm TP}(f)$&nbsp; is split according to magnitude and phase.
-The impulse response&nbsp; $h_{\rm TP}(t)$&nbsp; is obtained by Fourier transformation of&nbsp; $H_{\rm TP}(f)$, where the time constant&nbsp; $\tau = R \cdot C = {1}/({2\pi \cdot f_{\rm G}}) $&nbsp; is to be set.
+The impulse response&nbsp; $h_{\rm TP}(t)$&nbsp; is obtained by Fourier transform of&nbsp; $H_{\rm TP}(f)$, where the constant&nbsp; $\tau = R \cdot C = {1}/({2\pi \cdot f_{\rm G}}) $.
-For&nbsp; $t < 0$&nbsp; the impulse response is identically zero, for positive times:
+For&nbsp; $t < 0$&nbsp; the impulse response is identically zero,&nbsp; anf for positive times:
 :$$h_{\rm TP}(t) = \frac{1}{\tau} \cdot {\rm e}^{-t / \tau} .$$
 The circuit shown below describes a high-pass filter whose frequency response&nbsp; $H_{\rm HP}(f)$&nbsp; and impulse response&nbsp; $h_{\rm HP}(t)$&nbsp; are to be determined in this task. Such a high-pass can also be interpreted as a limiting case of a band-pass.
@@ Line 35: / Line 31: @@
 ''Hints:''
-*The task belongs to the chapter&nbsp; [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals|Differences and Similarities of Low-Pass and Bandpass Signals]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Differences_and_Similarities_of_Low-Pass_and_Band-Pass_Signals|Differences and Similarities of Low-Pass and Band-Pass Signals]].
@@ Line 43: / Line 39: @@
 <quiz display=simple>
-{Which of the following statements are true for the TP system??
+{Which of the following statements are true for the low-pass system??
 |type="[]"}
-- Der Gleichsignalübertragungsfaktor beträgt&nbsp; $H_{\rm TP}(f = 0) = 2$.
+- The DC transmission factor is&nbsp; $H_{\rm TP}(f = 0) = 2$.
-+ $|H_{\rm TP}(f)|$&nbsp; is smaller by&nbsp; $f = f_{\rm G}$&nbsp; um&nbsp; $\sqrt{2}$&nbsp; than at&nbsp; $f = 0$.
++ $|H_{\rm TP}(f= f_{\rm G})|$&nbsp; is smaller by&nbsp; $\sqrt{2}$&nbsp; than&nbsp; $|H_{\rm TP}(f= 0)|$.
-+ The phase function is: &nbsp; $\varphi_{\rm TP}(f) = \arctan(f/f_{\rm G})$.
++ The phase function is&nbsp; $\varphi_{\rm TP}(f) = \arctan(f/f_{\rm G})$.
-{Justify why&nbsp; $H_{\rm HP}(f) = 1 - H_{\rm TP}(f)$&nbsp; must always hold. Calculate&nbsp; $H_{\rm HP}(f)$, in particular the value at&nbsp; $f = 0$.
+{Justify why&nbsp; $H_{\rm HP}(f) = 1 - H_{\rm TP}(f)$&nbsp; must always hold.&nbsp; Calculate&nbsp; $H_{\rm HP}(f)$,&nbsp; in particular the value at&nbsp; $f = 0$.
 |type="{}"}
 $H_{\rm HP}(f = 0)\ = \ $ { 0. }
-{Which of the following statements is true?
+{Which of the following statements are true?
 |type="[]"}
-- Es gilt &nbsp;$|H_{\rm HP}(f = f_{\rm G})| = 1 - |H_{\rm TP}(f = f_{\rm G})|$ is true.
+- It holds: &nbsp;$|H_{\rm HP}(f = f_{\rm G})| = 1 - |H_{\rm TP}(f = f_{\rm G})|$.
-+ Es gilt &nbsp;$|H_{\rm HP}(f = f_{\rm G})| = |H_{\rm TP}(f = f_{\rm G})|$ is true.
++ It holds: &nbsp;$|H_{\rm HP}(f = f_{\rm G})| = |H_{\rm TP}(f = f_{\rm G})|$.
 + For positive frequencies it holds: &nbsp; $\varphi_{\rm HP}(f) = \varphi_{\rm TP}(f) - \pi/2$.
-{Calculate the impulse response&nbsp; $h_{\rm  TP}(t)$. Interpret the result. Which of the following statements are correct?
+{Calculate the impulse response&nbsp; $h_{\rm  TP}(t)$.&nbsp; Interpret the result.&nbsp; Which of the following statements are correct?
 |type="[]"}
-- &nbsp;$h_{\rm HP}(t) = \tau/t \cdot {\rm e}^{-t/\tau}$ is true.
+- It holds: &nbsp;$h_{\rm HP}(t) = \tau/t \cdot {\rm e}^{-t/\tau}$.
-+ &nbsp;$h_{\rm HP}(t) = \delta (t) - 1/\tau \cdot {\rm e}^{-t/\tau}$ is true.
++ It holds: &nbsp;$h_{\rm HP}(t) = \delta (t) - 1/\tau \cdot {\rm e}^{-t/\tau}$.
 + At time&nbsp; $t = 0$&nbsp; the impulse response is infinite.
-- At time&nbsp; $t = \tau$&nbsp; the impulse response is equal to. ${\rm e}/\tau$.
+- At time&nbsp; $t = \tau$&nbsp; the impulse response is equal to&nbsp; ${\rm e}/\tau$.

← Older revision		Revision as of 12:19, 5 May 2021
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	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:Signal Representation: Exercises\|^4.1 Differences~~/Similarities of LP~~ and ~~BP Signals~~^]]	+	[[Category:Signal Representation: Exercises\|^4.1 Differences between Low-Pass and Band-Pass^]]

← Older revision		Revision as of 07:34, 4 May 2021
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	''Hints:''		''Hints:''
−	*The task belongs to the chapter  [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals\|Differences and Similarities of ~~Lowpass~~ and Bandpass Signals]].	+	*The task belongs to the chapter  [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals\|Differences and Similarities of Low-Pass and Bandpass Signals]].

← Older revision		Revision as of 15:13, 3 May 2021
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	:$$h_{\rm TP}(t) = \frac{1}{\tau} \cdot {\rm e}^{-t / \tau} .$$		:$$h_{\rm TP}(t) = \frac{1}{\tau} \cdot {\rm e}^{-t / \tau} .$$

−	The circuit shown below describes a high-pass filter whose frequency response  $H_{\rm HP}(f)$  and impulse response  $h_{\rm HP}(t)$  are to be determined in this task. Such a high-pass can also be interpreted as a limiting case of a band pass.	+	The circuit shown below describes a high-pass filter whose frequency response  $H_{\rm HP}(f)$  and impulse response  $h_{\rm HP}(t)$  are to be determined in this task. Such a high-pass can also be interpreted as a limiting case of a band-pass.

← Older revision		Revision as of 15:12, 3 May 2021
Line 91:		Line 91:


−	'''(2)'''  On the page [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals#Properties_of_BP-Signals\|Properties of Bandpass Signals]]it was shown that any ~~bandpass~~ signal can be represented as the difference of two low-pass signals.	+	'''(2)'''  On the page [[Signal_Representation/Differences_and_Similarities_of_LP_and_BP_Signals#Properties_of_BP-Signals\|Properties of Bandpass Signals]]it was shown that any band-pass signal can be represented as the difference of two low-pass signals.

	[[File:P_ID703__Sig_Z_4_1_b.png\|right\|frame\|Low-pass and high-pass spectrum]]		[[File:P_ID703__Sig_Z_4_1_b.png\|right\|frame\|Low-pass and high-pass spectrum]]