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

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	[[Category:Digital Signal Transmission: Exercises\|^3.3 Channel Distortion and Equalization^]]		[[Category:Digital Signal Transmission: Exercises\|^3.3 Channel Distortion and Equalization^]]
	[[de:Aufgaben:3.~~3_Rauschen_bei_Kanalentzerrung~~]]		[[de:Aufgaben:Aufgabe 3.3: Rauschen bei Kanalentzerrung]]

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	'''(3)'''  According to the specification section:		'''(3)'''  According to the specification section:
	:$$\alpha^2 = \frac{N_0 \cdot f_{\rm G}}{\sqrt{2} \cdot \sigma_d^2}		:$$\alpha^2 = \frac{N_0 \cdot f_{\rm G}}{\sqrt{2} \cdot \sigma_d^2}= \frac{10^{-9} \cdot s_0^2 \cdot T \cdot f_{\rm G}}{\sqrt{2}\cdot \sigma_d^2} = 10^{-9} \cdot \frac{ f_{\rm G} \cdotT}{\sqrt{2} \cdot (\sigma_d/s_0)^2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \alpha^2 = 10^{-9} \cdot \frac{0.35}{\sqrt{2} \cdot 0.044^2} \approx 1.28 \cdot 10^{-7}\hspace{0.05cm}.$$
	= \frac{10^{-9} \cdot s_0^2 \cdot T \cdot f_{\rm G}}{\sqrt{2}
	\cdot \sigma_d^2} = 10^{-9} \cdot \frac{ f_{\rm G} \~~cdot~~
	T}{\sqrt{2} \cdot (\sigma_d/s_0)^2}\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} \alpha^2 = 10^{-9} \cdot \frac{
	0.35}{\sqrt{2} \cdot 0.044^2} \approx 1.28 \cdot 10^{-7}
	\hspace{0.05cm}.$$

	Expressed in  ${\rm dB}$,  one thus obtains		Expressed in  ${\rm dB}$,  one thus obtains

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	has the maximum value.  Thus,  with $x = f \cdot T$,  the optimization function is:		has the maximum value.  Thus,  with $x = f \cdot T$,  the optimization function is:
	:$$y(x) = 26.022 \cdot \sqrt{x} - 12.823 \cdot x^2 \approx 26 \cdot\sqrt{x} - 13 \cdot x^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \frac{{\rm d}y}{{\rm d}x} = \frac{26}{2\cdot \sqrt{x}} - 13 \cdot 2 \cdot x = 0$$		:$$y(x) = 26.022 \cdot \sqrt{x} - 12.823 \cdot x^2 \approx 26 \cdot\sqrt{x} - 13 \cdot x^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \frac{{\rm d}y}{{\rm d}x} = \frac{26}{2\cdot \sqrt{x}} - 13 \cdot 2 \cdot x = 0$$
	:$$\Rightarrow \hspace{0.3cm} \frac{1} { \sqrt{x}} = 2 \cdot x\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\frac{1} { x} = 4 \~~cdotx~~^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} x^3 = 0.25\hspace{0.3cm}\Rightarrow \hspace{0.3cm} x \approx 0.63\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} \frac{1} { \sqrt{x}} = 2 \cdot x\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\frac{1} { x} = 4 \cdot x^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} x^3 = 0.25\hspace{0.3cm}\Rightarrow \hspace{0.3cm} x \approx 0.63\hspace{0.05cm}.$$

	This gives  $f_{\rm max} \cdot T\hspace{0.15cm}\underline {\approx 0.63}$.		This gives  $f_{\rm max} \cdot T\hspace{0.15cm}\underline {\approx 0.63}$.

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	We consider two different system variants,  both using NRZ rectangular transmission pulses and are affected by AWGN noise.		We consider two different system variants,  both using NRZ rectangular transmission pulses and are affected by AWGN noise.
	*To limit the noise power,  in both cases a Gaussian low-pass filter		*To limit the noise power,  in both cases a Gaussian low-pass filter
	:$$H_{\rm G}(f) = {\rm exp}(- \pi \cdot		:$$H_{\rm G}(f) = {\rm exp}(- \pi \cdot\frac{f^2}{(2f_{\rm G})^2})$$
	\frac{f^2}{(2f_{\rm G})^2})$$

	:with normalized cutoff frequency  $f_{\rm G} \cdot T = 0.35$  is used, so that both systems also have the same eye opening with  $\ddot{o}(T_{\rm D} = 0) = 0.478 \cdot s_0$.		:with normalized cutoff frequency  $f_{\rm G} \cdot T = 0.35$  is used, so that both systems also have the same eye opening with  $\ddot{o}(T_{\rm D} = 0) = 0.478 \cdot s_0$.
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	* The channel frequency response of system  $\rm A$  is frequency-independent:   $H_{\rm K}(f) = \alpha$.  Accordingly,  for the receiver filter  $H_{\rm E}(f) = H_{\rm G}(f)/\alpha$  must be assumed,  so that the following applies to the detection noise power:		* The channel frequency response of system  $\rm A$  is frequency-independent:   $H_{\rm K}(f) = \alpha$.  Accordingly,  for the receiver filter  $H_{\rm E}(f) = H_{\rm G}(f)/\alpha$  must be assumed,  so that the following applies to the detection noise power:
	:$$\sigma_d^2 = {N_0}/{2} \cdot \int_{-\infty}^{+\infty}		:$$\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}.$$
	\|H_{\rm E}(f)\|^2 \,{\rm d} f = \frac{N_0 \cdot f_{\rm G}}{\sqrt{2}
	\cdot \alpha^2} \hspace{0.05cm}.$$

	* In contrast,  system  $\rm B$  assumes a coaxial cable with characteristic attenuation  $($at half the bit rate$)$  $a_* = 80 \, {\rm dB}$  $($or  $9.2 \, {\rm Np})$  so that the channel magnitude frequency response is:		* In contrast,  system  $\rm B$  assumes a coaxial cable with characteristic attenuation  $($at half the bit rate$)$  $a_* = 80 \, {\rm dB}$  $($or  $9.2 \, {\rm Np})$  so that the channel magnitude frequency response is:
	:$$\|H_{\rm K}(f)\| = {\rm e}^{- 9.2 \hspace{0.05cm} \cdot		:$$\|H_{\rm K}(f)\| = {\rm e}^{- 9.2 \hspace{0.05cm} \cdot\hspace{0.05cm}\sqrt{2 f T}}\hspace{0.05cm}.$$
	\hspace{0.05cm}\sqrt{2 f T}}\hspace{0.05cm}.$$

	* Thus,  the equation for the noise power density  $\rm (PSD)$  before the decision is  $($with  $f_{\rm G} \cdot T = 0.35)$:		* Thus,  the equation for the noise power density  $\rm (PSD)$  before the decision is  $($with  $f_{\rm G} \cdot T = 0.35)$:
	:$${\it \Phi}_{d{\rm N}}(f) = {N_0}/{2} \cdot \frac{\|H_{\rm G		:$${\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[18.4 \cdot \sqrt{2 f T} - 2\pi \cdot \frac{(f \cdot T)^2}{(2 \cdot 0.35)^2}\right ] \hspace{0.05cm}.$$
	}(f)\|^2}{\|H_{\rm K}(f)\|^2} = {N_0}/{2} \cdot {\rm exp}\left
	[18.4 \cdot \sqrt{2 f T} - 2\pi \cdot \frac{(f \cdot T)^2}{(2 \cdot 0.35)^2}
	\right ] \hspace{0.05cm}.$$

	*The noise PSD of system  $\rm B$  is shown in red in the above graph and the noise PSD of system  $\rm A$  is drawn in blue.		*The noise PSD of system  $\rm B$  is shown in red in the above graph and the noise PSD of system  $\rm A$  is drawn in blue.

	*For the system  $\rm B$,  the worst-case error probability		*For the system  $\rm B$,  the worst-case error probability
	:$$p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}		:$$p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}\right) \hspace{0.4cm}{\rm with} \hspace{0.4cm} \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  $p_{\rm U} = 4 \cdot 10^{\rm -8}$,  which corresponds to the signal-to-noise ratio  $10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$.		:was determined.  The measurement resulted in  $p_{\rm U} = 4 \cdot 10^{\rm -8}$,  which corresponds to the signal-to-noise ratio  $10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$.
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From  $10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$  follows  $\rho_{\rm U} = 10^{\rm 1.48} ≈ 30.2$  and with the given equation:		'''(1)'''  From  $10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$  follows  $\rho_{\rm U} = 10^{\rm 1.48} ≈ 30.2$  and with the given equation:
	:$$\sqrt{\rho_{\rm U}} = \frac{\ddot{o}(T_{\rm D})/2}{ \sigma_d}\hspace{0.3cm}\Rightarrow		:$$\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}}\hspace{0.15cm}\underline { \approx 0.044 \cdot s_0 }\hspace{0.05cm}.$$
	\hspace{0.3cm} \sigma_d = \frac{0.478 \cdot s_0/2}{ \sqrt{30.2}}
	\hspace{0.15cm}\underline { \approx 0.044 \cdot s_0 }\hspace{0.05cm}.$$


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	Expressed in  ${\rm dB}$,  one thus obtains		Expressed in  ${\rm dB}$,  one thus obtains
	:$$20 \cdot {\rm lg}\hspace{0.1cm}\alpha = 10 \cdot {\rm lg}\hspace{0.1cm}\alpha^2 =		:$$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 { =-68.9\,{\rm dB}} \hspace{0.05cm}.$$
	-70\,{\rm dB}\hspace{0.1cm}+\hspace{0.1cm}10 \cdot {\rm lg}\hspace{0.1cm}1.28\hspace{0.15cm}\underline { =
	-68.9\,{\rm dB}} \hspace{0.05cm}.$$


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	has the maximum value.  Thus,  with $x = f \cdot T$,  the optimization function is:		has the maximum value.  Thus,  with $x = f \cdot T$,  the optimization function is:
	:$$y(x) = 26.022 \cdot \sqrt{x} - 12.823 \cdot x^2 \approx 26 \cdot		:$$y(x) = 26.022 \cdot \sqrt{x} - 12.823 \cdot x^2 \approx 26 \cdot\sqrt{x} - 13 \cdot x^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \frac{{\rm d}y}{{\rm d}x} = \frac{26}{2\cdot \sqrt{x}} - 13 \cdot 2 \cdot x = 0$$
	\sqrt{x} - 13 \cdot x^2 \hspace{0.3cm}		:$$\Rightarrow \hspace{0.3cm} \frac{1} { \sqrt{x}} = 2 \cdot x\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\frac{1} { x} = 4 \cdotx^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} x^3 = 0.25\hspace{0.3cm}\Rightarrow \hspace{0.3cm} x \approx 0.63\hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} \frac{{\rm d}y}{{\rm d}x} = \frac{26}
	{2\cdot \sqrt{x}} - 13 \cdot 2 \cdot x = 0$$
	:$$\Rightarrow \hspace{0.3cm} \frac{1} { \sqrt{x}} = 2 \cdot x
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\frac{1} { x} = 4 \~~cdot~~
	x^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} x^3 = 0.25
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} x \approx 0.63
	\hspace{0.05cm}.$$

	This gives  $f_{\rm max} \cdot T\hspace{0.15cm}\underline {\approx 0.63}$.		This gives  $f_{\rm max} \cdot T\hspace{0.15cm}\underline {\approx 0.63}$.
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	'''(6)'''  With  $x_{\rm max} = 0.63$  we get the function value		'''(6)'''  With  $x_{\rm max} = 0.63$  we get the function value

	:$$y(x_{\rm max}) \approx 26 \cdot \sqrt{0.63} - 13 \cdot 0.63^2		:$$y(x_{\rm max}) \approx 26 \cdot \sqrt{0.63} - 13 \cdot 0.63^2\hspace{0.15cm}\underline {\approx 15.477}.$$
	\hspace{0.15cm}\underline {\approx 15.477}.$$
	[[File:EN_Dig_A_3_3f.png\|frame\|right\|Noise component  $d_{\rm N}(t)$  of the detection signal]]		[[File:EN_Dig_A_3_3f.png\|frame\|right\|Noise component  $d_{\rm N}(t)$  of the detection signal]]
	It follows:		It follows:

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	[[Category:Digital Signal Transmission: Exercises\|^3.3 Channel Distortion and Equalization^]]		[[Category:Digital Signal Transmission: Exercises\|^3.3 Channel Distortion and Equalization^]]
			[[de:Aufgaben:3.3_Rauschen_bei_Kanalentzerrung]]

Guenter at 12:17, 17 June 2022

2022-06-17T12:17:54Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From $10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$ follows $\rho_{\rm U} = 10^{\rm 1.48} ≈ 30.2$ and ~~continue~~ with the given equation:		'''(1)'''  From  $10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$  follows  $\rho_{\rm U} = 10^{\rm 1.48} ≈ 30.2$  and with the given equation:
	:$$\sqrt{\rho_{\rm U}} = \frac{\ddot{o}(T_{\rm D})/2}{ \sigma_d}\hspace{0.3cm}\Rightarrow		:$$\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}}		\hspace{0.3cm} \sigma_d = \frac{0.478 \cdot s_0/2}{ \sqrt{30.2}}
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	'''(2)'''  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   ⇒   $\sigma_d/s_0 \underline{= 0.044}.$		'''(2)'''  With the same worst-case 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   ⇒   $\sigma_d/s_0 \underline{= 0.044}.$


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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	Expressed in ${\rm dB}$, one thus obtains		Expressed in  ${\rm dB}$,  one thus obtains
	:$$20 \cdot {\rm lg}\hspace{0.1cm}\alpha = 10 \cdot {\rm lg}\hspace{0.1cm}\alpha^2 =		:$$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 { =		-70\,{\rm dB}\hspace{0.1cm}+\hspace{0.1cm}10 \cdot {\rm lg}\hspace{0.1cm}1.28\hspace{0.15cm}\underline { =
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	'''(4)'''  For system  $\rm B$,  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)'''  For system  $\rm B$,  because of  $H_{\rm E}(f = 0) = 1$,  the normalized value is equal  to $1$,  that means,  it is  ${\it \Phi}_{d \rm N}(f = 0) = N_0/2$.

	In contrast, for system  $\rm A$,  this value is larger by $1/\alpha^2$ due to the components of the frequency-independent cable attenuation $\alpha$:		In contrast,  for system  $\rm A$,  this value is larger  by the factor  $1/\alpha^2$  due to the components of the frequency-independent cable attenuation  $\alpha$:
	:$${\rm System}\hspace{0.15cm}{\rm A:}\hspace{0.1cm}\frac{{\it \Phi}_{d{\rm N}}(f = 0)}{N_0/2} = \frac{1}{\alpha^2} \hspace{0.15cm}\underline {\approx 7.8 \cdot 10^{6}} \hspace{0.05cm}, \hspace{1.05cm}{\rm System\hspace{0.15cm}B}: \frac{{\it \Phi}_{d \rm N}(f = 0)}{N_0/2} \, \underline {= 1}.$$		:$${\rm System}\hspace{0.15cm}{\rm A:}\hspace{0.1cm}\frac{{\it \Phi}_{d{\rm N}}(f = 0)}{N_0/2} = \frac{1}{\alpha^2} \hspace{0.15cm}\underline {\approx 7.8 \cdot 10^{6}} \hspace{0.05cm}, \hspace{1.05cm}{\rm System\hspace{0.15cm}B}: \frac{{\it \Phi}_{d \rm N}(f = 0)}{N_0/2} \, \underline {= 1}.$$


	'''(5)'''  ${\it \Phi}_{d \rm N}(f)$ is maximal if the exponent		'''(5)'''  ${\it \Phi}_{d \rm N}(f)$  is maximal if the exponent
	:$$18.4 \cdot \sqrt{2 f T} - 2\pi \cdot \frac{(f \cdot T)^2}{0.49}$$		:$$18.4 \cdot \sqrt{2 f T} - 2\pi \cdot \frac{(f \cdot T)^2}{0.49}$$

	has the maximum value. Thus, with $x = f \cdot T$, the optimization function is:		has the maximum value.  Thus,  with $x = f \cdot T$,  the optimization function is:
	:$$y(x) = 26.022 \cdot \sqrt{x} - 12.823 \cdot x^2 \approx 26 \cdot		:$$y(x) = 26.022 \cdot \sqrt{x} - 12.823 \cdot x^2 \approx 26 \cdot
	\sqrt{x} - 13 \cdot x^2 \hspace{0.3cm}		\sqrt{x} - 13 \cdot x^2 \hspace{0.3cm}
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	This gives $f_{\rm max} \cdot T\hspace{0.15cm}\underline {\approx 0.63}$.		This gives  $f_{\rm max} \cdot T\hspace{0.15cm}\underline {\approx 0.63}$.


	'''(6)'''  With $x_{\rm max} = 0.63$ we get the function value		'''(6)'''  With  $x_{\rm max} = 0.63$  we get the function value

	:$$y(x_{\rm max}) \approx 26 \cdot \sqrt{0.63} - 13 \cdot 0.63^2		:$$y(x_{\rm max}) \approx 26 \cdot \sqrt{0.63} - 13 \cdot 0.63^2
	\hspace{0.15cm}\underline {\approx 15.477}.$$		\hspace{0.15cm}\underline {\approx 15.477}.$$
	[[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  $d_{\rm N}(t)$  of the detection signal]]
	It follows:		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  (normalized)  frequency  $f \cdot T \approx 0.63$  is larger than at the frequency  $f = 0$  by a factor of  $e^{\rm 15.5} \ \underline{\approx 5.4 \cdot 10^6}$.

	*Thus, periodic components with period $T_0 \approx 1.6 \cdot T$ predominate in the noise component $d_{\rm N}(t)$.		*Thus,  periodic components with period  $T_0 \approx 1.6 \cdot T$  predominate in the noise component  $d_{\rm N}(t)$.

	*The graph shows a simulation and confirms this result.		*The graph shows a simulation and confirms this result.

Guenter at 11:55, 17 June 2022

2022-06-17T11:55:00Z

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	$20 \cdot {\rm lg} \ \alpha \ = \ $ { -70.967--66.833 } ${\ \rm dB}$		$20 \cdot {\rm lg} \ \alpha \ = \ $ { -70.967--66.833 } ${\ \rm dB}$

	{What is the noise power-spectral density ~~normalized to~~  $~~N_0/2~~$   ~~$($at~~  $~~f = 0)~~$  before the decision for system  $\rm A$  and system  $\rm B$?		{What is the noise power-spectral density  $($at  $f = 0)$  normalized to  $N_0/2$  before the decision for system  $\rm A$  and system  $\rm B$?
	\|type="{}"}		\|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 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% } $\ \cdot 10^0$		$\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  $\rm B$. At what frequency  $f_{\rm max}$  does  ${\it \Phi}_{d \rm N}(f)$  have its maximum?		{For the rest of the exercise,  we will only consider system  $\rm B$.  At what frequency  $f_{\rm max}$  does  ${\it \Phi}_{d \rm N}(f)$  have its maximum?
	\|type="{}"}		\|type="{}"}
	$f_{\rm max} \cdot T\ = \ ${ 0.63 3% }		$f_{\rm max} \cdot T\ = \ ${ 0.63 3% }

	{By what factor is the noise power density at frequency  $f_{\rm max}$  greater than at  $f = 0$?		{By what factor is the noise power-spectral density at frequency  $f_{\rm max}$  greater than at  $f = 0$?
	\|type="{}"}		\|type="{}"}
	${\it \Phi}_{d \rm N}(f_{\rm max})/{\it \Phi}_{d \rm N}(0)\ = \ $ { 5.4 3% } $\ \cdot 10^6$		${\it \Phi}_{d \rm N}(f_{\rm max})/{\it \Phi}_{d \rm N}(0)\ = \ $ { 5.4 3% } $\ \cdot 10^6$

Guenter at 11:10, 17 June 2022

2022-06-17T11:10:34Z

← Older revision		Revision as of 13:10, 17 June 2022
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	\hspace{0.05cm}\sqrt{2 f T}}\hspace{0.05cm}.$$		\hspace{0.05cm}\sqrt{2 f T}}\hspace{0.05cm}.$$

	* Thus, the equation for the noise power density before the decision $($with  $f_{\rm G} \cdot T = 0.35)$ is:		* Thus,  the equation for the noise power density  $\rm (PSD)$  before the decision is  $($with  $f_{\rm G} \cdot T = 0.35)$:
	:$${\it \Phi}_{d{\rm N}}(f) = {N_0}/{2} \cdot \frac{\|H_{\rm G		:$${\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		}(f)\|^2}{\|H_{\rm K}(f)\|^2} = {N_0}/{2} \cdot {\rm exp}\left
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	\right ] \hspace{0.05cm}.$$		\right ] \hspace{0.05cm}.$$

	~~This function curve~~  $\rm B$  is shown in red in the above graph~~. The~~ noise ~~power density for~~ system  $\rm A$  is drawn in blue.		*The noise PSD of system  $\rm B$  is shown in red in the above graph and the noise PSD of system  $\rm A$  is drawn in blue.

	For the system  $\rm B$,  the worst-case error probability		*For the system  $\rm B$,  the worst-case error probability
	:$$p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}		:$$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  $p_{\rm U} = 4 \cdot 10^{\rm -8}$, which corresponds to the signal-to-noise ratio  $10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$.		:was determined.  The measurement resulted in  $p_{\rm U} = 4 \cdot 10^{\rm -8}$,  which corresponds to the signal-to-noise ratio  $10 \cdot {\rm lg} \, \rho_{\rm U} = 14.8 \, {\rm dB}$.




			Notes:
			*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization\|"Consideration of Channel Distortion and Equalization"]].

			* Use the  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]] interaction module for numerical evaluation of the Q–function.
	~~''Notes:''~~
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization\|"Consideration of Channel Distortion and Equalization"]].
	* Use the  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]] interaction module for numerical evaluation of the Q function.


Line 50:		Line 49:

	<quiz display=simple>		<quiz display=simple>
	{What (normalized) noise rms value occurs in system  $\rm B$?		{What  (normalized)  noise rms value occurs in system  $\rm B$?
	\|type="{}"}		\|type="{}"}
	$\sigma_d/s_0 \ = \ $ { 0.044 3% }		$\sigma_d/s_0 \ = \ $ { 0.044 3% }

	{What noise rms value occurs for system  $\rm A$  when it leads to exactly the same (worst-case) error probability as system  $\rm B$?		{What noise rms value occurs for system  $\rm A$  when it leads to exactly the same  "worst-case error probability"  as system  $\rm B$?
	\|type="{}"}		\|type="{}"}
	$\sigma_d/s_0 \ = \ $ { 0.044 3% }		$\sigma_d/s_0 \ = \ $ { 0.044 3% }

	{By what attenuation factor  $\alpha$  is system  $\rm A$  equivalent to system  $\rm B$  in terms of (worst-case) error probability?		{By what attenuation factor  $\alpha$  is system  $\rm A$  equivalent to system  $\rm B$  in terms of worst-case error probability?
	\|type="{}"}		\|type="{}"}
	$20 \cdot {\rm lg} \ \alpha \ = \ $ { -70.967--66.833 } ${\ \rm dB}$		$20 \cdot {\rm lg} \ \alpha \ = \ $ { -70.967--66.833 } ${\ \rm dB}$

	{What is the noise power density ~~referenced~~ to  $N_0/2$   $($at  $f = 0)$  before the decision for system  $\rm A$  and system  $\rm B$?		{What is the noise power-spectral density normalized to  $N_0/2$   $($at  $f = 0)$  before the decision for system  $\rm A$  and system  $\rm B$?
	\|type="{}"}		\|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 A:}\hspace{0.4cm} {\it \Phi}_{d \rm N} (f = 0)/(N_0/2) \ = \ $ { 7.8 3% } $\ \cdot 10^6$

Guenter at 11:01, 17 June 2022

2022-06-17T11:01:10Z

← Older revision		Revision as of 13:01, 17 June 2022
Line 4:		Line 4:

	[[File:P_ID1407__Dig_A_3_3.png \|right\|frame\|Noise PSD before the decision]]		[[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,  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,  in both cases a Gaussian low-pass filter
	:$$H_{\rm G}(f) = {\rm exp}(- \pi \cdot		:$$H_{\rm G}(f) = {\rm exp}(- \pi \cdot
	\frac{f^2}{(2f_{\rm G})^2})$$		\frac{f^2}{(2f_{\rm G})^2})$$

	:with normalized cutoff frequency  $f_{\rm G} \cdot T = 0.35$  is used, so that both systems also have the same eye opening with  $\ddot{o}(T_{\rm D} = 0) = 0.478 \cdot s_0$.		:with normalized cutoff frequency  $f_{\rm G} \cdot T = 0.35$  is used, so that both systems also have the same eye opening with  $\ddot{o}(T_{\rm D} = 0) = 0.478 \cdot s_0$.
	*The ~~transmitted~~ energy  $E_{\rm B} = s_0^2 \cdot T$  spent per bit is larger than the noise power density  $N_0$   ⇒   $10\cdot {\rm lg} \, E_{\rm B}/N_0 = 90 \, {\rm dB}$ ~~by a factor of  $10^9$ ~~.		*The transmission energy  $E_{\rm B} = s_0^2 \cdot T$  spent per bit is larger than the noise power density  $N_0$   by a factor of  $10^9$  ⇒   $10\cdot {\rm lg} \, E_{\rm B}/N_0 = 90 \, {\rm dB}$.

			* The channel frequency response of system  $\rm A$  is frequency-independent:   $H_{\rm K}(f) = \alpha$.  Accordingly,  for the receiver filter  $H_{\rm E}(f) = H_{\rm G}(f)/\alpha$  must be assumed,  so that the following applies to the detection noise power:
	~~The two systems differ as follows:~~
	* The channel frequency response of system  $\rm A$  is frequency independent:   $H_{\rm K}(f) = \alpha$. Accordingly,  $H_{\rm E}(f) = H_{\rm G}(f)/\alpha$  must be assumed ~~for the receiver filter~~, so that the following applies to the detection noise power:
	:$$\sigma_d^2 = {N_0}/{2} \cdot \int_{-\infty}^{+\infty}		:$$\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}		\|H_{\rm E}(f)\|^2 \,{\rm d} f = \frac{N_0 \cdot f_{\rm G}}{\sqrt{2}
	\cdot \alpha^2} \hspace{0.05cm}.$$		\cdot \alpha^2} \hspace{0.05cm}.$$
	* In contrast, system  $\rm B$  assumes a coaxial cable with characteristic attenuation (at half the bit rate)  $a_* = 80 \, {\rm dB}$  $($or  $9.2 \, {\rm Np})$  so that the magnitude frequency response is:
			* In contrast,  system  $\rm B$  assumes a coaxial cable with characteristic attenuation  $($at half the bit rate$)$  $a_* = 80 \, {\rm dB}$  $($or  $9.2 \, {\rm Np})$  so that the channel magnitude frequency response is:
	:$$\|H_{\rm K}(f)\| = {\rm e}^{- 9.2 \hspace{0.05cm} \cdot		:$$\|H_{\rm K}(f)\| = {\rm e}^{- 9.2 \hspace{0.05cm} \cdot
	\hspace{0.05cm}\sqrt{2 f T}}\hspace{0.05cm}.$$		\hspace{0.05cm}\sqrt{2 f T}}\hspace{0.05cm}.$$

	* Thus, the equation for the noise power density before the decision $($with  $f_{\rm G} \cdot T = 0.35)$ is:		* Thus, the equation for the noise power density before the decision $($with  $f_{\rm G} \cdot T = 0.35)$ is:
	:$${\it \Phi}_{d{\rm N}}(f) = {N_0}/{2} \cdot \frac{\|H_{\rm G		:$${\it \Phi}_{d{\rm N}}(f) = {N_0}/{2} \cdot \frac{\|H_{\rm G

Guenter at 16:18, 14 June 2022

2022-06-14T16:18:33Z

← Older revision		Revision as of 18:18, 14 June 2022
Line 129:		Line 129:
	:$$y(x_{\rm max}) \approx 26 \cdot \sqrt{0.63} - 13 \cdot 0.63^2		:$$y(x_{\rm max}) \approx 26 \cdot \sqrt{0.63} - 13 \cdot 0.63^2
	\hspace{0.15cm}\underline {\approx 15.477}.$$		\hspace{0.15cm}\underline {\approx 15.477}.$$
	[[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:		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 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 $f = 0$.