Aufgaben:Exercise 5.1Z: Cosine Square Noise Limitation - Revision history

Bene: Text replacement - "rms value" to "standard deviation"

2022-02-17T11:12:00Z

Text replacement - "rms value" to "standard deviation"

Guenter at 17:18, 9 February 2022

2022-02-09T17:18:57Z

Guenter at 17:08, 9 February 2022

2022-02-09T17:08:32Z

Guenter at 17:07, 9 February 2022

2022-02-09T17:07:21Z

Guenter at 17:06, 9 February 2022

2022-02-09T17:06:27Z

Guenter: Guenter moved page Aufgaben:Exercise 5.1Z: cos² Noise Limitation to Aufgaben:Exercise 5.1Z: Cosine Square Noise Limitation

2022-02-09T16:57:07Z

Guenter moved page Exercise 5.1Z: cos² Noise Limitation to Exercise 5.1Z: Cosine Square Noise Limitation

Hwang at 15:30, 13 January 2022

2022-01-13T15:30:16Z

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Guenter at 10:28, 12 January 2022

2022-01-12T10:28:47Z

Guenter: Guenter moved page Aufgaben:Aufgabe 5.1Z: cos² -Rauschbegrenzung to Aufgaben:Exercise 5.1Z: cos² Noise Limitation

2022-01-12T08:54:25Z

Guenter moved page Aufgabe 5.1Z: cos² -Rauschbegrenzung to Exercise 5.1Z: cos² Noise Limitation

Javier: Text replacement - "”" to """

2021-05-28T14:39:15Z

Text replacement - "”" to """

@@ Line 38: / Line 38: @@
 <quiz display=simple>
-{What is the rms value of the input signal&nbsp; $x(t)$?
+{What is the standard deviation of the input signal&nbsp; $x(t)$?
 |type="{}"}
 $\sigma_x \ = \ $  { 1 3% } $\ \rm &micro; V$
@@ Line 53: / Line 53: @@
-{Calculate the rms value of the output signal&nbsp; $y(t)$&nbsp; for&nbsp;  $f_0 = B_x/2$.
+{Calculate the standard deviation of the output signal&nbsp; $y(t)$&nbsp; for&nbsp;  $f_0 = B_x/2$.
 |type="{}"}
 $\sigma_y \ =  \ $  { 0.433 3% } $\ \rm &micro; V$
-{Calculate the rms value of&nbsp; $y(t)$&nbsp; under the condition&nbsp; $f_0 = 2 \cdot B_x$.
+{Calculate the standard deviation of&nbsp; $y(t)$&nbsp; under the condition&nbsp; $f_0 = 2 \cdot B_x$.
 |type="{}"}
 $\sigma_y \ =  \ $ { 0.731 3% } $\ \rm &micro; V$

@@ Line 22: / Line 22: @@
 Notes:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Stochastic_System_Theory|Stochastic System Theory]].
 *Reference is also made to the chapters&nbsp; [[Theory_of_Stochastic_Signals/Gaußverteilte_Zufallsgrößen|Gaussian Distributed Random Variables]]&nbsp; and&nbsp; [[Theory_of_Stochastic_Signals/Power-Spectral_Density|Power-Spectral Density]].
@@ Line 77: / Line 73: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The variance (power) &nbsp; &#8658; &nbsp; rms squared of the signal&nbsp; $x(t)$&nbsp; is
+'''(1)'''&nbsp; The variance&nbsp; of the signal&nbsp; $x(t)$&nbsp; is
 :$$\sigma _x ^2  = \frac{N_0 }{2} \cdot 2B_x  = N_0  \cdot B_x  = 10^{ - 12} \;{\rm{V}}^2
 \hspace{0.3cm}\Rightarrow\hspace{0.3cm}
@@ Line 83: / Line 79: @@
-'''(2)'''&nbsp; According to the chapter "Gaussian Distributed Random Variables" and the approximation given here&nbsp; $($for large&nbsp; $x)$,&nbsp; we obtain:
+'''(2)'''&nbsp; According to the chapter&nbsp; "Gaussian Distributed Random Variables"&nbsp; and the approximation given here&nbsp; $($for large&nbsp; $x)$,&nbsp; we obtain:
 :$$\Pr \left( {\left| {x(t)} \right| > 5\;{\rm{&micro; V}}} \right) = 2 \cdot {\rm Q}(5) = \frac{2}{{\sqrt {2{\rm{\pi }}}  \cdot 5}} \cdot {\rm{e}}^{ - 12.5}\hspace{0.15cm} \underline{ \approx 0.6 \cdot 10^{ - 6}} .$$
-'''(3)'''&nbsp; The input signal&nbsp; $x(t)$&nbsp; is mean-free&nbsp; $(m_x = 0)$.
+'''(3)'''&nbsp; The input signal&nbsp; $x(t)$&nbsp; is mean-free &nbsp;&rArr;  &nbsp; $m_x = 0$.
-*Otherwise&nbsp; ${\it Φ}_x(f)$&nbsp; would still have to contain a Dirac function at&nbsp; $f= 0$.&nbsp;
+*Otherwise&nbsp; ${\it Φ}_x(f)$&nbsp; would still have to contain a Dirac delta function at&nbsp; $f= 0$.&nbsp;
 *The mean is not changed by the linear filter &nbsp; &#8658; &nbsp; $m_y\hspace{0.05cm}\underline{ = 0}$.
-'''(4)'''&nbsp; For the power density spectrum of the output signal generally applies:
+'''(4)'''&nbsp; For the power-spectral density of the output signal generally applies:
 :$${\it \Phi}_y (f) = {N_0 }/{2} \cdot \left| {H( f )} \right|^2 .$$
-*Thus, the variance&nbsp; $\sigma _y^2$&nbsp; can be calculated.&nbsp; Taking advantage of the symmetry, we obtain:
+*Thus,&nbsp; the variance&nbsp; $\sigma _y^2$&nbsp; can be calculated.&nbsp; Taking advantage of the symmetry,&nbsp; we obtain:
 :$$\sigma _y ^2  = {N_0 }/{2} \cdot \int_{ - \infty }^{ + \infty } {\left| {H( f )} \right|^2 \hspace{0.1cm}{\rm{d}}f} =  N_0  \cdot \int_0^{f_0 } {\cos ^4 } \left( {\frac{{{\rm{\pi }}f}}{2f_0 }} \right)\hspace{0.1cm}{\rm{d}}f .$$
-*The definite integral is given.&nbsp; For each of the three solution terms, the value of the lower bound is zero.&nbsp; It follows that:
+*The definite integral is given.&nbsp; For each of the three solution terms,&nbsp; the value of the lower bound is zero.&nbsp; It follows that:
 :$$\sigma _y ^2  = {N_0}/{2} \cdot \left( {\frac{3}{8} \cdot f_0  + \frac{f_0 }{{2{\rm{\pi }}}} \cdot \sin ( {\rm{\pi }} ) + \frac{f_0 }{{16{\rm{\pi }}}} \cdot \sin ( {{\rm{2\pi }}} )} \right) = \frac{3}{8} \cdot N_0  \cdot f_0 $$
 :$$\Rightarrow \hspace{0.3cm} f_0 = B_x/2\text{:}\hspace{0.2cm}\sigma _y ^2  = \frac{3}{16} \cdot N_0  \cdot B_x  = \frac{3}{16} \cdot \sigma _x ^2  = 0.1875 \cdot 10^{ - 12} \;{\rm{V}}^2  \hspace{0.2cm}\Rightarrow \hspace{0.2cm}\sigma _y \hspace{0.15cm}\underline{ = 0.433\;{\rm{&micro; V}}}{\rm{.}}$$
@@ Line 105: / Line 101: @@
-'''(5)'''&nbsp; Now the input PDS for&nbsp; $|f| > B_x$&nbsp; has no components.
+'''(5)'''&nbsp; Now the input PSD for&nbsp; $|f| > B_x$&nbsp; has no components.
 *Therefore holds:
 :$$\sigma _y ^2  = N_0\cdot \int_0^{B_x } {\cos ^4 \left( {\frac{{{\rm{\pi }}f}}{2f_0 }} \right)\hspace{0.1cm}{\rm{d}}f = N_0  \cdot \int_0^{f_0 /2} {\cos ^4 } \left( {\frac{{{\rm{\pi }}f}}{2f_0 }} \right)\hspace{0.1cm}{\rm{d}}f.}$$
 *The numerical evaluation yields for this:
-:$$\sigma _y ^2  = N_0 \left( {\frac{3}{8} \cdot B_x  + \frac{B_x }{{2{\rm{\pi }}}} \cdot \sin ( {\frac{{\rm{\pi }}}{2}} ) + \frac{B_x }{{{\rm{16\pi }}}} \cdot \sin ( {\rm{\pi }} )} \right) = N_0  \cdot B_x \left( {\frac{3}{8} + \frac{1}{{2{\rm{\pi }}}}} \right) = 0.534\cdot \sigma _x ^2  \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\sigma _y \hspace{0.15cm}\underline{  = 0.731\;{\rm{&micro; V}}}{\rm{.}}$$
+:$$\sigma _y ^2  = N_0 \left( {\frac{3}{8} \cdot B_x  + \frac{B_x }{{2{\rm{\pi }}}} \cdot \sin ( {\frac{{\rm{\pi }}}{2}} ) + \frac{B_x }{{{\rm{16\pi }}}} \cdot \sin ( {\rm{\pi }} )} \right) = N_0  \cdot B_x \left( {\frac{3}{8} + \frac{1}{{2{\rm{\pi }}}}} \right) = 0.534\cdot \sigma _x ^2  \hspace{0.6cm}\Rightarrow \hspace{0.6cm}\sigma _y \hspace{0.15cm}\underline{  = 0.731\;{\rm{&micro; V}}}{\rm{.}}$$
-'''(6)'''&nbsp; Analogous to the sample solution of subtask&nbsp; '''(2)'''&nbsp; holds:
+'''(6)'''&nbsp; Analogous to the solution of subtask&nbsp; '''(2)'''&nbsp; holds:
 :$$\Pr \left( {\left| {y\left( t \right)} \right| > 5\;{\rm{&micro; V}}} \right) = 2 \cdot {\rm Q}\left( {\frac{{5\;{\rm{&micro; V}}}}{{0.731\;{\rm{&micro; V}}}}} \right) = 2 \cdot {\rm Q}( {6.84} ).$$
-*With the given approximation, this probability has the following value:
+*With the given approximation,&nbsp; this probability has the following value:
 :$$\Pr \left( {\left| {y\left( t \right)} \right| > 5\;{\rm{&micro; V}}} \right) \hspace{0.15cm} \underline{ \approx 8 \cdot 10^{ - 12}}.$$
 {{ML-Fuß}}

@@ Line 28: / Line 28: @@
 ''Notes:''
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Stochastic_System_Theory|Stochastic System Theory]].
-*Reference is also made to the chapters&nbsp; [[Theory_of_Stochastic_Signals/Gaußverteilte_Zufallsgrößen|Gaussian Distributed Random Variables]]&nbsp; and&nbsp; [[Theory_of_Stochastic_Signals/Theory_of_Stochastic_Signals/Power-Spectral_Density|Power-Spectral Density]].
+*Reference is also made to the chapters&nbsp; [[Theory_of_Stochastic_Signals/Gaußverteilte_Zufallsgrößen|Gaussian Distributed Random Variables]]&nbsp; and&nbsp; [[Theory_of_Stochastic_Signals/Power-Spectral_Density|Power-Spectral Density]].
 *Use the following equations if necessary:

@@ Line 28: / Line 28: @@
 ''Notes:''
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Stochastic_System_Theory|Stochastic System Theory]].
-*Reference is also made to the chapters&nbsp; [[Theory_of_Stochastic_Signals/Gaußverteilte_Zufallsgrößen|Gaussian Distributed Random Variables]]&nbsp; and&nbsp; [[Theory_of_Stochastic_Signals/Leistungsdichtespektrum_(LDS)|Power-Spectral Density]].
+*Reference is also made to the chapters&nbsp; [[Theory_of_Stochastic_Signals/Gaußverteilte_Zufallsgrößen|Gaussian Distributed Random Variables]]&nbsp; and&nbsp; [[Theory_of_Stochastic_Signals/Theory_of_Stochastic_Signals/Power-Spectral_Density|Power-Spectral Density]].
 *Use the following equations if necessary:

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID491__Sto_Z_5_1.png|right|frame|Top:&nbsp; Input PDS&nbsp; ${\it Φ}_x(f)$, <br>Bottom:&nbsp; Filter frequency response $H(f)$]]
+[[File:P_ID491__Sto_Z_5_1.png|right|frame|Top:&nbsp; Input PSD&nbsp; ${\it Φ}_x(f)$, Bottom:&nbsp; Frequency response&nbsp; $H(f)$]]
-We consider bandlimited white noise&nbsp; $x(t)$&nbsp; with the power density spectrum&nbsp; ${\it Φ}_x(f)$&nbsp; sketched above. This is constant equal to&nbsp; $N_0/2$&nbsp; in the range&nbsp; $|f| \le B_x$&nbsp;  and zero outside.
+We consider bandlimited white noise&nbsp; $x(t)$&nbsp; with the power-spectral density&nbsp; ${\it Φ}_x(f)$&nbsp; sketched above.&nbsp; This is constant equal to&nbsp; $N_0/2$&nbsp; in the range&nbsp; $|f| \le B_x$&nbsp;  and zero outside.
 Assume the following numerical values:
-*Noise power density&nbsp; $N_0 = 10^{-16} \ \rm V^2/Hz$,
+*Noise power-spectral density&nbsp; $N_0 = 10^{-16} \ \rm V^2/Hz$,
-*(one-sided) noise bandwidth&nbsp; $B_x = 10 \ \rm kHz$.
+*(one-sided)&nbsp; noise bandwidth&nbsp; $B_x = 10 \ \rm kHz$.
 This signal is applied to the input of a low-pass filter with frequency response
-$$H(f) = \left\{ {\begin{array}{*{20}c}   {\cos ^2 \left( {\frac{{{\rm{\pi }}f}}{2f_0 }} \right)} & {\rm{f\ddot{u}r}\quad \left| \it f \right| \le \it f_{\rm 0} ,}  \\   0 & {{\rm{else}}}  \\\end{array}} \right.$$
+:$$H(f) = \left\{ {\begin{array}{*{20}c}   {\cos ^2 \left( {\frac{{{\rm{\pi }}f}}{2f_0 }} \right)} & {\rm{f\ddot{u}r}\quad \left| \it f \right| \le \it f_{\rm 0} ,}  \\   0 & {{\rm{else}}}  \\\end{array}} \right.$$
-&nbsp; Here&nbsp; $f_0$&nbsp; denotes the absolute filter bandwidth, which can vary between&nbsp; $B_x/2$&nbsp; and&nbsp; $2B_x$.&nbsp;
+*Here, $f_0$&nbsp; denotes the absolute filter bandwidth,&nbsp; which can vary between&nbsp; $B_x/2$&nbsp; and&nbsp; $2B_x$.&nbsp;
-The filter output signal is denoted by&nbsp; $y(t)$.&nbsp;
+*The filter output signal is denoted by&nbsp; $y(t)$.&nbsp;
@@ Line 28: / Line 28: @@
 ''Notes:''
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Stochastic_System_Theory|Stochastic System Theory]].
-*Reference is also made to the chapters&nbsp; [[Theory_of_Stochastic_Signals/Gaußverteilte_Zufallsgrößen|Gaussian Distributed Random Variables]]&nbsp; and&nbsp; [[Theory_of_Stochastic_Signals/Leistungsdichtespektrum_(LDS)|Power Density Spectrum]].
+*Reference is also made to the chapters&nbsp; [[Theory_of_Stochastic_Signals/Gaußverteilte_Zufallsgrößen|Gaussian Distributed Random Variables]]&nbsp; and&nbsp; [[Theory_of_Stochastic_Signals/Leistungsdichtespektrum_(LDS)|Power-Spectral Density]].
 *Use the following equations if necessary:
 :$${\rm Q}(x) \approx \frac{1}{{\sqrt {2{\rm{\pi }}}  \cdot x}} \cdot {\rm{e}}^{ - x^2 /2} \hspace{0.15cm}(
-\text{für große }x),$$
+\text{for large }x),$$
 :$$\int {\rm{cos}}^{\rm{2}}( {ax} )\hspace{0.1cm}{\rm{d}}x = \frac{1}{2} \cdot x + \frac{1}{4a} \cdot \sin ( {2ax} ),$$
 :$$\int {\cos ^4 } ( {ax} )\hspace{0.1cm}{\rm{d}}x = \frac{3}{8} \cdot x + \frac{1}{4a} \cdot \sin ( {2ax} ) + \frac{1}{32a} \cdot \sin ( {4ax} ).$$

← Older revision	Revision as of 16:57, 9 February 2022
(No difference)

← Older revision		Revision as of 10:28, 12 January 2022
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−	[[Category:Theory of Stochastic Signals: Exercises\|^5.1 ~~Stochastische Systemtheorie~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^5.1 Stochastic Systems Theory^]]

← Older revision	Revision as of 08:54, 12 January 2022
(No difference)

← Older revision		Revision as of 14:39, 28 May 2021
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−	'''(2)'''  Entsprechend dem Kapitel "Gaußverteilte Zufallsgrößen~~”~~ und der hier angegebenen Näherung  $($für große  $x)$  erhält man:	+	'''(2)'''  Entsprechend dem Kapitel "Gaußverteilte Zufallsgrößen" und der hier angegebenen Näherung  $($für große  $x)$  erhält man:
	:$$\Pr \left( {\left\| {x(t)} \right\| > 5\;{\rm{µ V}}} \right) = 2 \cdot {\rm Q}(5) = \frac{2}{{\sqrt {2{\rm{\pi }}} \cdot 5}} \cdot {\rm{e}}^{ - 12.5}\hspace{0.15cm} \underline{ \approx 0.6 \cdot 10^{ - 6}} .$$		:$$\Pr \left( {\left\| {x(t)} \right\| > 5\;{\rm{µ V}}} \right) = 2 \cdot {\rm Q}(5) = \frac{2}{{\sqrt {2{\rm{\pi }}} \cdot 5}} \cdot {\rm{e}}^{ - 12.5}\hspace{0.15cm} \underline{ \approx 0.6 \cdot 10^{ - 6}} .$$