Aufgaben:Exercise 5.9: Minimization of the MSE - Revision history

Guenter at 14:35, 23 February 2022

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Guenter at 14:16, 23 February 2022

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Guenter at 14:14, 23 February 2022

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Hwang at 09:27, 1 February 2022

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Guenter at 10:42, 12 January 2022

2022-01-12T10:42:23Z

Guenter: Guenter moved page Aufgaben:Aufgabe 5.9: Minimierung des MQF to Aufgaben:Exercise 5.9: Minimization of the MSE

2022-01-12T09:17:59Z

Guenter moved page Aufgabe 5.9: Minimierung des MQF to Exercise 5.9: Minimization of the MSE

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2021-05-28T14:44:54Z

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Javier: Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

2021-03-23T12:43:42Z

Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

Javier: Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

2020-07-09T09:46:16Z

Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

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 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Only the last solution</u> is correct:
+'''(1)'''&nbsp; <u>Only the last solution</u>&nbsp; is correct:
-*The task  &nbsp; &rArr; &nbsp; "Minimization of the mean square error" already points to the&ndash;Kolmogorow filter.
+*The task&nbsp; "Minimization of the mean square error"&nbsp; already points to the Wiener&ndash;Kolmogorow filter.
-*The matched filter, on the other hand, is used to concentrate the signal energy and thereby maximize the S/N ratio for a given time.
+*The matched filter,&nbsp; on the other hand,&nbsp; is used to concentrate the signal energy and thereby maximize the S/N ratio for a given time.
-'''(2)'''&nbsp; For the optimal frequency response, according to Wiener and Kolmogorov, the following applies in general:
+'''(2)'''&nbsp; For the optimal frequency response,&nbsp; according to Wiener and Kolmogorov,&nbsp; the following applies in general:
 :$$H(f) = H_{\rm WF} (f) = \frac{1}{{1 + {\it \Phi} _n (f)/{\it \Phi} _s (f)}}.$$
-*With the given power-spectral density, it is also possible to write for this:
+*With the given power-spectral density,&nbsp; it is also possible to write for this:
 :$$H(f) = \frac{1}{{1 + {N_0 }/({{2{\it \Phi} _0 })}\cdot \left[ {1 + ( {f/f_0 } )^2 } \right]}} = \frac{1}{{1 + {1}/{Q}\cdot \left[ {1 + ( {f/f_0 } )^2 } \right]}}.$$
@@ Line 86: / Line 86: @@
-'''(3)'''&nbsp; For the filter calculated in subtask&nbsp; '''(2)''',&nbsp; taking symmetry into account, the following holds:
+'''(3)'''&nbsp; For the filter calculated in subtask&nbsp; '''(2)''',&nbsp; taking symmetry into account,&nbsp; the following holds:
 :$${\rm{MSE = }}\int_{-\infty}^{+\infty}  H(f) \cdot {\it \Phi} _n (f) \,\, {\rm{d}}f = \int_{0}^{+\infty}  \frac{N_0}{{1 + {N_0 }/({{2{\it \Phi} _0 })}\cdot \left[ {1 + ( {f/f_0 } )^2 } \right]}} \,\, {\rm{d}}f .$$
-*For this, with &nbsp;$Q = 2 \cdot {\it \Phi}_0/N_0$&nbsp; and &nbsp;$a^2 = Q + 1$,&nbsp; we can also write:<br />
+*For this,&nbsp; with &nbsp;$Q = 2 \cdot {\it \Phi}_0/N_0$&nbsp; and &nbsp;$a^2 = Q + 1$,&nbsp; we can also write:<br />
 :$${\rm{MSE = }}\int_0^\infty  {\frac{{2{\it \Phi} _0 }}{{ Q+1 + ( {f/f_0 })^2 }}} \,\, {\rm{d}}f = 2{\it \Phi} _0  \cdot f_0 \int_0^\infty  {\frac{1}{a^2  + x^2 }}\,\, {\rm{d}}x.$$
-*With the integral given, this leads to the result:
+*This leads to the result with the integral given:
 :$${\rm{MSE}} = \frac{{2{\it \Phi} _0 f_0 }}{{\sqrt {1 + Q} }}\left( {\arctan ( \infty  ) - \arctan ( 0 )} \right) = \frac{{{\it \Phi} _0 f_0 {\rm{\pi }}}}{{\sqrt {1 + Q} }}.$$
-*Normalizing MSE to the useful power&nbsp; $P_s$, we obtain for&nbsp; $Q=3$:
+*Normalizing MSE to the power&nbsp; $P_s$&nbsp; of the&nbsp; signal $s(t)$,&nbsp; we obtain for&nbsp; $Q=3$:
 :$$\frac{\rm{MSE}}{P_s}  = \frac{1}{{\sqrt {1 + Q} }} \hspace{0.15cm}\underline { = 0.5}.$$
@@ Line 101: / Line 101: @@
 '''(4)'''&nbsp; From the calculation in subtask&nbsp; '''(3)''',&nbsp; the condition&nbsp; $Q \ge 99$ &nbsp; &#8658; &nbsp; $Q_{\rm min} \hspace{0.15cm}\underline{= 99}$ follows directly for&nbsp; ${\rm MSE}/P_s \ge 0.1.$
-*The larger&nbsp; $Q$&nbsp; is, the smaller the mean square error becomes.
+*The larger&nbsp; $Q$,&nbsp;  the smaller the mean square error  becomes.
-'''(5)'''&nbsp; <u>Only the second solution</u> is correct:
+'''(5)'''&nbsp; <u>Only the second solution</u>&nbsp; is correct:
 *A frequency response of the same shape as the Wiener&ndash;Kolmogorov filter &nbsp; &#8658; &nbsp; $H(f) = K \cdot H_{\rm WF}(f)$&nbsp; with&nbsp; $K \ne 1$&nbsp; always leads to large distortions.
-*This can be illustrated, for example, by the noise-free case&nbsp; $(Q \to \infty)$.&nbsp; In this case,&nbsp; $d(t) = K \cdot s(t)$&nbsp; and the optimization task would be extremely poorly solved despite good conditions.
+*This can be illustrated by the noise-free case&nbsp; $(Q \to \infty)$:&nbsp; $d(t) = K \cdot s(t)$&nbsp; and the optimization task would be extremely poorly solved despite good conditions.
-*From the equation
+*One might incorrectly conclude from the equation
 :$${\rm{MSE}} = \int_{ - \infty }^{ + \infty } {H_{\rm WF} (f)}  \cdot \it{\Phi} _n (f)\,\,{\rm{d}}f$$
-:one might incorrectly conclude that a filter&nbsp; $H(f) = 2 \cdot H_{\rm WF}(f)$&nbsp; only doubles the mean squared error.
+:that a filter&nbsp; $H(f) = 2 \cdot H_{\rm WF}(f)$&nbsp; only doubles the mean squared error.
-*However, this is not the case, since&nbsp; $H(f)$&nbsp; is then no longer a Wiener filter and the above equation is also no longer applicable.
+*However,&nbsp;  this is not the case,&nbsp; since&nbsp; $H(f)$&nbsp; is then no longer a Wiener filter and the above equation is no longer applicable.
-[[File:P_ID651__Sto_A_5_9_e.png|right|frame|Power-spectral density with the Wiener filter]]
+The second statement is true,&nbsp; as can be seen from the accompanying diagram.
-The second statement is true, as can be seen from the accompanying diagram.
+*The dots mark the frequency response&nbsp; $H_{\rm WF}(f)$&nbsp; of the filter for&nbsp; $Q = 3$&nbsp; and for&nbsp; $Q = 10$, resp.
 *For larger&nbsp; $Q (= 10)$,&nbsp; high components are attenuated less than for lower&nbsp; $Q (= 3)$.
-*Therefore, in the case of&nbsp; $Q = 10$,&nbsp; the filter output signal also contains more higher frequency components, which are due to the noise&nbsp;   $n(t)$.&nbsp;
+*Therefore,&nbsp; in the case of&nbsp; $Q = 10$,&nbsp; the filter output signal contains more higher frequency components,&nbsp; which are due to the noise&nbsp;   $n(t)$.&nbsp;
 {{ML-Fuß}}

@@ Line 41: / Line 41: @@
-{Determine the frequency response&nbsp; $H(f)$&nbsp; of the optimum filter for this purpose.&nbsp; <br>What values result with&nbsp; $Q = 3$&nbsp; at&nbsp; $f = 0$&nbsp; and&nbsp; $f = 2f_0$?
+{Determine the frequency response&nbsp; $H(f)=H_{\rm WF} (f)$&nbsp; of the optimum filter for this purpose.&nbsp; <br>What values result with&nbsp; $Q = 3$&nbsp; at&nbsp; $f = 0$&nbsp; and&nbsp; $f = 2f_0$?
 |type="{}"}
 $H(f = 0) \ = \ $ { 0.75 3% }
@@ Line 47: / Line 47: @@
-{Let&nbsp; $Q = 3$.&nbsp; Calculate the mean square error&nbsp; $(\rm MSE)$&nbsp; with respect to&nbsp; $P_s$&nbsp; or the best possible filter.
+{Let&nbsp; $Q = 3$.&nbsp; Calculate the mean square error&nbsp; $(\rm MSE)$&nbsp; with respect to&nbsp; $P_s$&nbsp; of the best possible filter.
 |type="{}"}
 ${\rm MSE}/P_s \ =  \ $ { 0.5 3% }

@@ Line 4: / Line 4: @@
 [[File:P_ID652__Sto_A_5_9.png|right|frame|Power-spectral densities <br>with the Wiener filter ]]
-Given a stochastic useful signal&nbsp; $s(t)$ of which only the power-spectral density (PSD) is known:
+Given a stochastic signal&nbsp; $s(t)$ of which only the power-spectral density&nbsp; $\rm (PSD)$&nbsp; is known:
 :$${\it \Phi} _s (f) = \frac{\it{\Phi} _{\rm 0} }{1 + ( {f/f_0 } )^2 }.$$
 This power-spectral density&nbsp; ${\it \Phi} _s (f)$&nbsp; is shown in blue in the accompanying diagram.
@@ Line 10: / Line 10: @@
 *The average power of&nbsp; $s(t)$&nbsp; is obtained by integration over the power-spectral density:
 :$$P_s  = \int_{ - \infty }^{ + \infty } {{\it \Phi} _s (f)}\, {\rm d} f = {\it \Phi} _0  \cdot f_0  \cdot {\rm{\pi }}.$$
-*Additively superimposed on the useful signal&nbsp; $s(t)$&nbsp; is white noise&nbsp; $n(t)$&nbsp; with noise power density&nbsp; ${\it \Phi}_n(f) = N_0/2.$
+*Additively superimposed on this signal&nbsp; $s(t)$&nbsp; is white noise&nbsp; $n(t)$&nbsp; with noise power density&nbsp; ${\it \Phi}_n(f) = N_0/2.$
-*As an abbreviation, we use&nbsp; $Q = 2 \cdot {\it \Phi}_0/N_0$, where&nbsp; $Q$&nbsp; could be interpreted as "quality".
+*As an abbreviation,&nbsp; we use&nbsp; $Q = 2 \cdot {\it \Phi}_0/N_0$,&nbsp; where&nbsp; $Q$&nbsp; could be interpreted as&nbsp; "quality".
 *Note that&nbsp; $Q$&nbsp; does not represent a signal&ndash;to&ndash;noise power ratio.
-In this exercise, we want to determine the frequency response&nbsp; $H(f)$&nbsp; of a filter that minimizes the mean square error&nbsp; $\rm (MSE)$&nbsp; between the useful signal&nbsp; $s(t)$&nbsp; and the filter output signal&nbsp; $d(t)$:&nbsp;
+In this exercise,&nbsp; we want to determine the frequency response&nbsp; $H(f)$&nbsp; of a filter that minimizes the mean square error&nbsp; $\rm (MSE)$&nbsp; between the useful signal&nbsp; $s(t)$&nbsp; and the filter output signal&nbsp; $d(t)$:&nbsp;
 :$${\rm{MSE}} = \mathop {\lim }\limits_{T_{\rm M}  \to \infty } \frac{1}{T_{\rm M} }\int_{ - T_{\rm M} /2}^{T_{\rm M} /2} {\left| {d(t) - s(t)} \right|^2 \, {\rm{d}}t.}$$
@@ Line 21: / Line 21: @@
 Notes:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Wiener–Kolmogorow–Filter|Wiener–Kolmogorow Filter]].
+*For the optimal frequency response,&nbsp; according to Wiener and Kolmogorov,&nbsp; the following applies in general:
 *The following indefinite integral is given for solving:
 :$$\int {\frac{1}{a^2  + x^2 }} \, {\rm{d}}x ={1}/{a} \cdot \arctan \left( {{x}/{a}} \right).$$
@@ Line 38: / Line 36: @@
 {Which of the following statements are true?
 |type="[]"}
-- $H(f)$&nbsp; is a Gaussian low pass.
+- $H(f)$&nbsp; is a Gaussian low-pass.
 - $H(f)$&nbsp; represents a matched filter.
 + $H(f)$&nbsp; represents a Wiener&ndash;Kolmogorov filter.
-{Determine the frequency response&nbsp; $H(f)$&nbsp; of the optimum filter for this purpose.&nbsp; What values result with&nbsp; $Q = 3$&nbsp; at&nbsp; $f = 0$&nbsp; and&nbsp; $f = 2f_0$?
+{Determine the frequency response&nbsp; $H(f)$&nbsp; of the optimum filter for this purpose.&nbsp; <br>What values result with&nbsp; $Q = 3$&nbsp; at&nbsp; $f = 0$&nbsp; and&nbsp; $f = 2f_0$?
 |type="{}"}
 $H(f = 0) \ = \ $ { 0.75 3% }
@@ Line 54: / Line 52: @@
-{How large must the "quality factor"&nbsp; $Q$&nbsp; be chosen at least, so that for the quotient the value&nbsp; ${\rm MSE}/P_s = 0.1$&nbsp; can be reached?
+{How must the&nbsp; "quality factor"&nbsp; $Q$&nbsp; be chosen at least,&nbsp; so that for the quotient the value&nbsp; ${\rm MSE}/P_s = 0.1$&nbsp; can be reached?
 |type="{}"}
 $Q_\text{min} \ =  \ $ { 99 3% }

← Older revision		Revision as of 10:42, 12 January 2022
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−	[[Category:Theory of Stochastic Signals: Exercises\|^5.5 ~~Wiener–Kolmogorow–Filter~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^5.5 Wiener-Kolmogorov Filter^]]

@@ Line 11: / Line 11: @@
 :$$P_s  = \int_{ - \infty }^{ + \infty } {{\it \Phi} _s (f)}\, {\rm d} f = {\it \Phi} _0  \cdot f_0  \cdot {\rm{\pi }}.$$
 *Additiv überlagert ist dem Nutzsignal&nbsp; $s(t)$&nbsp; Weißes Rauschen&nbsp; $n(t)$&nbsp; mit der Rauschleistungsdichte&nbsp; ${\it \Phi}_n(f) = N_0/2.$
-*Als Abkürzung verwenden wir&nbsp; $Q = 2 \cdot {\it \Phi}_0/N_0$, wobei&nbsp; $Q$&nbsp; als "Qualität&rdquo; interpretiert werden könnte.
+*Als Abkürzung verwenden wir&nbsp; $Q = 2 \cdot {\it \Phi}_0/N_0$, wobei&nbsp; $Q$&nbsp; als "Qualität" interpretiert werden könnte.
 *Zu beachten ist, dass&nbsp; $Q$&nbsp; kein Signal&ndash;zu&ndash;Rauschleistungsverhältnis darstellt.
@@ Line 54: / Line 54: @@
-{Wie groß muss der "Qualitätsfaktor&rdquo;&nbsp; $Q$&nbsp; mindestens gewählt werden, damit für den Quotienten der Wert&nbsp; ${\rm MQF}/P_s = 0.1$&nbsp; erreicht werden kann?
+{Wie groß muss der "Qualitätsfaktor"&nbsp; $Q$&nbsp; mindestens gewählt werden, damit für den Quotienten der Wert&nbsp; ${\rm MQF}/P_s = 0.1$&nbsp; erreicht werden kann?
 |type="{}"}
 $Q_\text{min} \ =  \ $ { 99 3% }
@@ Line 71: / Line 71: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; Richtig ist <u>nur der letzte Lösungsvorschlag</u>:
-*Die Aufgabenstellung  &nbsp; &rArr; &nbsp; "Minimierung des mittleren quadratischen Fehlers&rdquo; weist bereits auf das Filter nach Wiener&ndash;Kolmogorow hin.
+*Die Aufgabenstellung  &nbsp; &rArr; &nbsp; "Minimierung des mittleren quadratischen Fehlers" weist bereits auf das Filter nach Wiener&ndash;Kolmogorow hin.
 *Das Matched&ndash;Filter verwendet man dagegen, um die Signalenergie zu bündeln und dadurch für einen vorgegebenen Zeitpunkt das S/N&ndash;Verhältnis zu maximieren.

← Older revision		Revision as of 12:43, 23 March 2021
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−	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^5.5 Wiener–Kolmogorow–Filter^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^5.5 Wiener–Kolmogorow–Filter^]]

@@ Line 25: / Line 25: @@
 ''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Stochastische_Signaltheorie/Wiener–Kolmogorow–Filter|Wiener–Kolmogorow–Filter]].
+*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Wiener–Kolmogorow–Filter|Wiener–Kolmogorow–Filter]].
 *Zur Lösung vorgegeben wird das folgende unbestimmte Integral: