Aufgaben:Exercise 3.3: Noise at Channel Equalization - Revision history

Guenter at 12:17, 17 June 2022

2022-06-17T12:17:54Z

Guenter at 11:55, 17 June 2022

2022-06-17T11:55:00Z

Guenter at 11:10, 17 June 2022

2022-06-17T11:10:34Z

Guenter at 11:01, 17 June 2022

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Guenter at 16:18, 14 June 2022

2022-06-14T16:18:33Z

Hwang at 15:55, 16 May 2022

2022-05-16T15:55:48Z

Guenter at 14:59, 7 May 2022

2022-05-07T14:59:28Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.3: Rauschen bei Kanalentzerrung to Aufgaben:Exercise 3.3: Noise at Channel Equalization

2022-05-07T14:57:10Z

Guenter moved page Aufgabe 3.3: Rauschen bei Kanalentzerrung to Exercise 3.3: Noise at Channel Equalization

Hwang at 13:12, 2 May 2022

2022-05-02T13:12:40Z

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Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:52:38Z

Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

@@ Line 77: / Line 77: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; From $10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$ follows $\rho_{\rm U} = 10^{\rm 1.48} &asymp; 30.2$ and continue with the given equation:
+'''(1)'''&nbsp; From&nbsp; $10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$&nbsp; follows&nbsp; $\rho_{\rm U} = 10^{\rm 1.48} &asymp; 30.2$&nbsp; and with the given equation:
 :$$\sqrt{\rho_{\rm U}} = \frac{\ddot{o}(T_{\rm D})/2}{ \sigma_d}\hspace{0.3cm}\Rightarrow
 \hspace{0.3cm} \sigma_d = \frac{0.478 \cdot s_0/2}{ \sqrt{30.2}}
@@ Line 83: / Line 83: @@
-'''(2)'''&nbsp; With the same error probability $p_{\rm U}$ (and thus the same $\rho_{\rm U}$), $\sigma_d$ must have exactly the same value as calculated in subtask '''(1)''', since the eye opening also remains the same &nbsp; &#8658; &nbsp; $\sigma_d/s_0 \underline{= 0.044}.$
+'''(2)'''&nbsp; With the same worst-case error probability&nbsp; $p_{\rm U}$&nbsp; (and thus the same&nbsp; $\rho_{\rm U}$),&nbsp; $\sigma_d$ must have exactly the same value as calculated in subtask&nbsp; '''(1)''',&nbsp; since the eye opening also remains the same &nbsp; &#8658; &nbsp; $\sigma_d/s_0 \underline{= 0.044}.$
@@ Line 95: / Line 95: @@
   \hspace{0.05cm}.$$
-Expressed in ${\rm dB}$, one thus obtains
+Expressed in&nbsp; ${\rm dB}$,&nbsp; one thus obtains
 :$$20 \cdot {\rm lg}\hspace{0.1cm}\alpha = 10 \cdot {\rm lg}\hspace{0.1cm}\alpha^2 =
    -70\,{\rm dB}\hspace{0.1cm}+\hspace{0.1cm}10 \cdot {\rm lg}\hspace{0.1cm}1.28\hspace{0.15cm}\underline { =
@@ Line 101: / Line 101: @@
-'''(4)'''&nbsp; For system &nbsp;$\rm B$,&nbsp; because $H_{\rm E}(f = 0) = 1$, the normalized value is equal to $1$, that means, it is ${\it \Phi}_{d \rm N}(f = 0) = N_0/2$.
+'''(4)'''&nbsp; For system &nbsp;$\rm B$,&nbsp; because of&nbsp; $H_{\rm E}(f = 0) = 1$,&nbsp; the normalized value is equal&nbsp; to $1$,&nbsp; that means,&nbsp; it is&nbsp; ${\it \Phi}_{d \rm N}(f = 0) = N_0/2$.
-In contrast, for system &nbsp;$\rm A$,&nbsp; this value is larger by $1/\alpha^2$ due to the components of the frequency-independent cable attenuation $\alpha$:
+In contrast,&nbsp; for system &nbsp;$\rm A$,&nbsp; this value is larger&nbsp; by the factor&nbsp; $1/\alpha^2$&nbsp; due to the components of the frequency-independent cable attenuation&nbsp; $\alpha$:
 :$${\rm System}\hspace{0.15cm}{\rm A:}\hspace{0.1cm}\frac{{\it \Phi}_{d{\rm N}}(f = 0)}{N_0/2}  = \frac{1}{\alpha^2} \hspace{0.15cm}\underline {\approx 7.8 \cdot 10^{6}} \hspace{0.05cm}, \hspace{1.05cm}{\rm System\hspace{0.15cm}B}: \frac{{\it \Phi}_{d \rm N}(f = 0)}{N_0/2} \, \underline {= 1}.$$
-'''(5)'''&nbsp; ${\it \Phi}_{d \rm N}(f)$ is maximal if the exponent
+'''(5)'''&nbsp; ${\it \Phi}_{d \rm N}(f)$&nbsp; is maximal if the exponent
 :$$18.4 \cdot \sqrt{2  f  T} - 2\pi \cdot \frac{(f \cdot T)^2}{0.49}$$
-has the maximum value. Thus, with $x = f \cdot T$, the optimization function is:
+has the maximum value.&nbsp; Thus,&nbsp; with $x = f \cdot T$,&nbsp; the optimization function is:
 :$$y(x) = 26.022 \cdot  \sqrt{x} - 12.823 \cdot x^2 \approx 26 \cdot
 \sqrt{x} - 13 \cdot x^2 \hspace{0.3cm}
@@ Line 121: / Line 121: @@
 \hspace{0.05cm}.$$
-This gives $f_{\rm max} \cdot T\hspace{0.15cm}\underline {\approx 0.63}$.
+This gives&nbsp; $f_{\rm max} \cdot T\hspace{0.15cm}\underline {\approx 0.63}$.
-'''(6)'''&nbsp; With $x_{\rm max} = 0.63$ we get the function value
+'''(6)'''&nbsp; With&nbsp; $x_{\rm max} = 0.63$&nbsp; we get the function value
 :$$y(x_{\rm max})  \approx 26 \cdot \sqrt{0.63} - 13 \cdot 0.63^2
 \hspace{0.15cm}\underline {\approx 15.477}.$$
-[[File:EN_Dig_A_3_3f.png|frame|right|Noise component $d_{\rm N}(t)$]]
+[[File:EN_Dig_A_3_3f.png|frame|right|Noise component&nbsp; $d_{\rm N}(t)$&nbsp; of the detection signal]]
 It follows:
-*The noise power density at the (normalized) frequency $f \cdot T \approx 0.63$ is larger than at the frequency $e^{\rm 15.5} \underline{\approx 5.4 \cdot 10^6}$ by a factor of $f = 0$.
+*The noise PSD at the&nbsp; (normalized)&nbsp; frequency&nbsp; $f \cdot T \approx 0.63$&nbsp; is larger than at the frequency&nbsp; $f = 0$&nbsp;  by a factor of&nbsp; $e^{\rm 15.5} \ \underline{\approx 5.4 \cdot 10^6}$.
-*Thus, periodic components with period $T_0 \approx 1.6 \cdot T$ predominate in the noise component $d_{\rm N}(t)$.
+*Thus,&nbsp; periodic components with period&nbsp; $T_0 \approx 1.6 \cdot T$&nbsp; predominate in the noise component&nbsp; $d_{\rm N}(t)$.
 *The graph shows a simulation and confirms this result.

@@ Line 61: / Line 61: @@
 $20 \cdot {\rm lg} \ \alpha \ = \ $ { -70.967--66.833 } ${\ \rm dB}$
-{What is the noise power-spectral density normalized to &nbsp;$N_0/2$&nbsp; &nbsp;$($at &nbsp;$f = 0)$&nbsp; before the decision for system &nbsp;$\rm A$&nbsp; and system &nbsp;$\rm B$?
+{What is the noise power-spectral density&nbsp; $($at &nbsp;$f = 0)$&nbsp; normalized to &nbsp;$N_0/2$&nbsp;  before the decision for system &nbsp;$\rm A$&nbsp; and system &nbsp;$\rm B$?
 |type="{}"}
 $\text{System A:}\hspace{0.4cm} {\it \Phi}_{d \rm N} (f = 0)/(N_0/2) \ = \ $ { 7.8 3% } $\ \cdot 10^6$
 $\text{System B:}\hspace{0.42cm} {\it \Phi}_{d \rm N} (f = 0)/(N_0/2)  \ = \ $ { 1 3% } $\ \cdot 10^0$
-{For the rest of the exercise, we will only consider system &nbsp;$\rm B$. At what frequency &nbsp;$f_{\rm max}$&nbsp; does &nbsp;${\it \Phi}_{d \rm N}(f)$&nbsp; have its maximum?
+{For the rest of the exercise,&nbsp; we will only consider system &nbsp;$\rm B$.&nbsp; At what frequency &nbsp;$f_{\rm max}$&nbsp; does &nbsp;${\it \Phi}_{d \rm N}(f)$&nbsp; have its maximum?
 |type="{}"}
 $f_{\rm max} \cdot T\ = \ ${ 0.63 3% }
-{By what factor is the noise power density at frequency &nbsp;$f_{\rm max}$&nbsp; greater than at &nbsp;$f = 0$?
+{By what factor is the noise power-spectral density at frequency &nbsp;$f_{\rm max}$&nbsp; greater than at &nbsp;$f = 0$?
 |type="{}"}
 ${\it \Phi}_{d \rm N}(f_{\rm max})/{\it \Phi}_{d \rm N}(0)\ = \ $ { 5.4 3% } $\ \cdot 10^6$

@@ Line 21: / Line 21: @@
 \hspace{0.05cm}\sqrt{2 f T}}\hspace{0.05cm}.$$
-* Thus, the equation for the noise power density before the decision $($with &nbsp;$f_{\rm G} \cdot T = 0.35)$ is:
+* Thus,&nbsp; the equation for the noise power density&nbsp; $\rm (PSD)$&nbsp; before the decision is&nbsp; $($with &nbsp;$f_{\rm G} \cdot T = 0.35)$:
 :$${\it \Phi}_{d{\rm N}}(f) = {N_0}/{2} \cdot \frac{|H_{\rm G
 }(f)|^2}{|H_{\rm K}(f)|^2} = {N_0}/{2} \cdot {\rm exp}\left
@@ Line 27: / Line 27: @@
 \right ] \hspace{0.05cm}.$$
-This function curve &nbsp;$\rm B$&nbsp; is shown in red in the above graph. The noise power density for system &nbsp;$\rm A$&nbsp; is drawn in blue.
+*The noise PSD of system &nbsp;$\rm B$&nbsp; is shown in red in the above graph and the noise PSD of system &nbsp;$\rm A$&nbsp; is drawn in blue.
-For the system &nbsp;$\rm B$,&nbsp; the worst-case error probability
+*For the system &nbsp;$\rm B$,&nbsp; the worst-case error probability
 :$$p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}
-   \right) \hspace{0.2cm}{\rm with} \hspace{0.2cm} \rho_{\rm U} = \frac{[\ddot{o}(T_{\rm D})/2]^2}{ \sigma_d^2}$$
+   \right) \hspace{0.4cm}{\rm with} \hspace{0.4cm} \rho_{\rm U} = \frac{[\ddot{o}(T_{\rm D})/2]^2}{ \sigma_d^2}$$
-was determined. The measurement resulted in &nbsp;$p_{\rm U} = 4 \cdot 10^{\rm -8}$, which corresponds to the signal-to-noise ratio &nbsp;$10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$.&nbsp;
+:was determined.&nbsp; The measurement resulted in &nbsp;$p_{\rm U} = 4 \cdot 10^{\rm -8}$,&nbsp; which corresponds to the signal-to-noise ratio &nbsp;$10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$.&nbsp;
+* Use the &nbsp;[[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|"Complementary Gaussian Error Functions"]] interaction module for numerical evaluation of the Q&ndash;function.
@@ Line 50: / Line 49: @@
 <quiz display=simple>
-{What (normalized) noise rms value occurs in system &nbsp;$\rm B$?&nbsp;
+{What&nbsp; (normalized)&nbsp; noise rms value occurs in system &nbsp;$\rm B$?&nbsp;
 |type="{}"}
 $\sigma_d/s_0 \ = \ $ { 0.044 3% }
-{What noise rms value occurs for system &nbsp;$\rm A$&nbsp; when it leads to exactly the same (worst-case) error probability as system &nbsp;$\rm B$?&nbsp;
+{What noise rms value occurs for system &nbsp;$\rm A$&nbsp; when it leads to exactly the same&nbsp; "worst-case error probability"&nbsp; as system &nbsp;$\rm B$?&nbsp;
 |type="{}"}
 $\sigma_d/s_0 \ = \ $ { 0.044 3% }
-{By what attenuation factor &nbsp;$\alpha$&nbsp; is system &nbsp;$\rm A$&nbsp; equivalent to system &nbsp;$\rm B$&nbsp; in terms of (worst-case) error probability?
+{By what attenuation factor &nbsp;$\alpha$&nbsp; is system &nbsp;$\rm A$&nbsp; equivalent to system &nbsp;$\rm B$&nbsp; in terms of worst-case error probability?
 |type="{}"}
 $20 \cdot {\rm lg} \ \alpha \ = \ $ { -70.967--66.833 } ${\ \rm dB}$
-{What is the noise power density referenced to &nbsp;$N_0/2$&nbsp; &nbsp;$($at &nbsp;$f = 0)$&nbsp; before the decision for system &nbsp;$\rm A$&nbsp; and system &nbsp;$\rm B$?
+{What is the noise power-spectral density normalized to &nbsp;$N_0/2$&nbsp; &nbsp;$($at &nbsp;$f = 0)$&nbsp; before the decision for system &nbsp;$\rm A$&nbsp; and system &nbsp;$\rm B$?
 |type="{}"}
 $\text{System A:}\hspace{0.4cm} {\it \Phi}_{d \rm N} (f = 0)/(N_0/2) \ = \ $ { 7.8 3% } $\ \cdot 10^6$

@@ Line 4: / Line 4: @@
 [[File:P_ID1407__Dig_A_3_3.png |right|frame|Noise PSD before the decision]]
-We consider two different system variants, both of which use NRZ rectangular transmission pulses and are affected by AWGN noise.
+We consider two different system variants,&nbsp; both using NRZ rectangular transmission pulses and are affected by AWGN noise.
-*In both cases, a Gaussian low-pass filter is used to limit noise power
+*To limit the noise power,&nbsp; in both cases a Gaussian low-pass filter
 :$$H_{\rm G}(f) = {\rm exp}(- \pi \cdot
 \frac{f^2}{(2f_{\rm G})^2})$$
 :with normalized cutoff frequency &nbsp;$f_{\rm G} \cdot T = 0.35$&nbsp; is used, so that both systems also have the same eye opening with &nbsp;$\ddot{o}(T_{\rm D} = 0) = 0.478 \cdot s_0$.&nbsp;
-*The transmitted energy &nbsp;$E_{\rm B} = s_0^2 \cdot T$&nbsp; spent per bit is larger than the noise power density &nbsp;$N_0$ &nbsp; &#8658; &nbsp; $10\cdot {\rm lg} \, E_{\rm B}/N_0 = 90 \, {\rm dB}$ by a factor of &nbsp;$10^9$&nbsp;.
+*The transmission energy &nbsp;$E_{\rm B} = s_0^2 \cdot T$&nbsp; spent per bit is larger than the noise power density &nbsp;$N_0$ &nbsp;  by a factor of &nbsp;$10^9$&nbsp; &#8658; &nbsp; $10\cdot {\rm lg} \, E_{\rm B}/N_0 = 90 \, {\rm dB}$.
+* The channel frequency response of system &nbsp;$\rm A$&nbsp; is frequency-independent: &nbsp; $H_{\rm K}(f) = \alpha$.&nbsp; Accordingly,&nbsp;  for the receiver filter &nbsp;$H_{\rm E}(f) = H_{\rm G}(f)/\alpha$&nbsp; must be assumed,&nbsp; so that the following applies to the detection noise power:
 :$$\sigma_d^2 = {N_0}/{2} \cdot \int_{-\infty}^{+\infty}
 |H_{\rm E}(f)|^2 \,{\rm d} f = \frac{N_0 \cdot f_{\rm G}}{\sqrt{2}
 \cdot \alpha^2} \hspace{0.05cm}.$$
 * In contrast, system &nbsp;$\rm B$&nbsp; assumes a coaxial cable with characteristic attenuation (at half the bit rate) &nbsp;$a_* = 80 \, {\rm dB}$&nbsp; $($or &nbsp;$9.2 \, {\rm Np})$&nbsp; so that the magnitude frequency response is:
 :$$|H_{\rm K}(f)| = {\rm e}^{- 9.2 \hspace{0.05cm} \cdot
 \hspace{0.05cm}\sqrt{2 f T}}\hspace{0.05cm}.$$
 * Thus, the equation for the noise power density before the decision $($with &nbsp;$f_{\rm G} \cdot T = 0.35)$ is:
 :$${\it \Phi}_{d{\rm N}}(f) = {N_0}/{2} \cdot \frac{|H_{\rm G

@@ Line 129: / Line 129: @@
 :$$y(x_{\rm max})  \approx 26 \cdot \sqrt{0.63} - 13 \cdot 0.63^2
 \hspace{0.15cm}\underline {\approx 15.477}.$$
-[[File:P_ID1408__Dig_A_3_3f.png|frame|right|Noise component $d_{\rm N}(t)$]]
+[[File:EN_Dig_A_3_3f.png|frame|right|Noise component $d_{\rm N}(t)$]]
 It follows:
 *The noise power density at the (normalized) frequency $f \cdot T \approx 0.63$ is larger than at the frequency $e^{\rm 15.5} \underline{\approx 5.4 \cdot 10^6}$ by a factor of $f = 0$.

@@ Line 41: / Line 41: @@
 ''Notes:''
-*The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization|Consideration of Channel Distortion and Equalization]].
+*The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization|"Consideration of Channel Distortion and Equalization"]].
-* Use the &nbsp;[[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|Complementary Gaussian Error Functions]] interaction module for numerical evaluation of the Q function.
+* Use the &nbsp;[[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|"Complementary Gaussian Error Functions"]] interaction module for numerical evaluation of the Q function.

← Older revision		Revision as of 14:59, 7 May 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^3.3 ~~Kanalverzerrungen und Entzerrung~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^3.3 Channel Distortion and Equalization^]]

← Older revision	Revision as of 14:57, 7 May 2022
(No difference)

← Older revision		Revision as of 12:52, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^3.3 Kanalverzerrungen und Entzerrung^]]	+	[[Category:Digital Signal Transmission: Exercises\|^3.3 Kanalverzerrungen und Entzerrung^]]