Aufgaben:Exercise 3.3: Noise at Channel Equalization

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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.

• In both cases, a Gaussian low-pass filter is used to limit noise power

$$H_{\rm G}(f) = {\rm exp}(- \pi \cdot \frac{f^2}{(2f_{\rm G})^2})$$

with normalized cutoff frequency  $f_{\rm G} \cdot T = 0.35$  is used, so that both systems also have the same eye opening with  $\ddot{o}(T_{\rm D} = 0) = 0.478 \cdot s_0$. 

• 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 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:

$$\sigma_d^2 = {N_0}/{2} \cdot \int_{-\infty}^{+\infty} |H_{\rm E}(f)|^2 \,{\rm d} f = \frac{N_0 \cdot f_{\rm G}}{\sqrt{2} \cdot \alpha^2} \hspace{0.05cm}.$$

• 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 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  $f_{\rm G} \cdot T = 0.35)$ is:

$${\it \Phi}_{d{\rm N}}(f) = {N_0}/{2} \cdot \frac{|H_{\rm G }(f)|^2}{|H_{\rm K}(f)|^2} = {N_0}/{2} \cdot {\rm exp}\left [18.4 \cdot \sqrt{2 f T} - 2\pi \cdot \frac{(f \cdot T)^2}{(2 \cdot 0.35)^2} \right ] \hspace{0.05cm}.$$

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.

For the system  $\rm B$,  the worst-case error probability

$$p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}} \right) \hspace{0.2cm}{\rm with} \hspace{0.2cm} \rho_{\rm U} = \frac{[\ddot{o}(T_{\rm D})/2]^2}{ \sigma_d^2}$$

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  Consideration of Channel Distortion and Equalization.
• Use the  Complementary Gaussian Error Functions interaction module for numerical evaluation of the Q function.

Questions

1

What (normalized) noise rms value occurs in system  $\rm B$? 

$\sigma_d/s_0 \ = \ $

2

What noise rms value occurs for system  $\rm A$  when it leads to exactly the same (worst-case) error probability as system  $\rm B$? 

$\sigma_d/s_0 \ = \ $

3

By what attenuation factor  $\alpha$  is system  $\rm A$  equivalent to system  $\rm B$  in terms of (worst-case) error probability?

$20 \cdot {\rm lg} \ \alpha \ = \ $ ${\ \rm dB}$

4

What is the noise power density referenced to  $N_0/2$   $($at  $f = 0)$  before the decision for system  $\rm A$  and system  $\rm B$?

$\text{System A:}\hspace{0.4cm} {\it \Phi}_{d \rm N} (f = 0)/(N_0/2) \ = \ $ $\ \cdot 10^6$

$\text{System B:}\hspace{0.42cm} {\it \Phi}_{d \rm N} (f = 0)/(N_0/2) \ = \ $ $\ \cdot 10^0$

5

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?

$f_{\rm max} \cdot T\ = \ $

6

By what factor is the noise power density at frequency  $f_{\rm max}$  greater than at  $f = 0$?

${\it \Phi}_{d \rm N}(f_{\rm max})/{\it \Phi}_{d \rm N}(0)\ = \ $ $\ \cdot 10^6$

Solution

Solution

(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:

$$\sqrt{\rho_{\rm U}} = \frac{\ddot{o}(T_{\rm D})/2}{ \sigma_d}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \sigma_d = \frac{0.478 \cdot s_0/2}{ \sqrt{30.2}} \hspace{0.15cm}\underline { \approx 0.044 \cdot s_0 }\hspace{0.05cm}.$$

(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   ⇒   $\sigma_d/s_0 \underline{= 0.044}.$

(3)  According to the specification section:

$$\alpha^2 = \frac{N_0 \cdot f_{\rm G}}{\sqrt{2} \cdot \sigma_d^2} = \frac{10^{-9} \cdot s_0^2 \cdot T \cdot f_{\rm G}}{\sqrt{2} \cdot \sigma_d^2} = 10^{-9} \cdot \frac{ f_{\rm G} \cdot T}{\sqrt{2} \cdot (\sigma_d/s_0)^2}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \alpha^2 = 10^{-9} \cdot \frac{ 0.35}{\sqrt{2} \cdot 0.044^2} \approx 1.28 \cdot 10^{-7} \hspace{0.05cm}.$$

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

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

(4)  For system  $\rm B$,  because $H_{\rm E}(f = 0) = 1$, the normalized value is equal to $1$, that means, it is ${\it \Phi}_{d \rm N}(f = 0) = N_0/2$.

In contrast, for system  $\rm A$,  this value is larger by $1/\alpha^2$ due to the components of the frequency-independent cable attenuation $\alpha$:

$${\rm System}\hspace{0.15cm}{\rm A:}\hspace{0.1cm}\frac{{\it \Phi}_{d{\rm N}}(f = 0)}{N_0/2} = \frac{1}{\alpha^2} \hspace{0.15cm}\underline {\approx 7.8 \cdot 10^{6}} \hspace{0.05cm}, \hspace{1.05cm}{\rm System\hspace{0.15cm}B}: \frac{{\it \Phi}_{d \rm N}(f = 0)}{N_0/2} \, \underline {= 1}.$$

(5)  ${\it \Phi}_{d \rm N}(f)$ is maximal if the exponent

$$18.4 \cdot \sqrt{2 f T} - 2\pi \cdot \frac{(f \cdot T)^2}{0.49}$$

has the maximum value. Thus, with $x = f \cdot T$, the optimization function is:

$$y(x) = 26.022 \cdot \sqrt{x} - 12.823 \cdot x^2 \approx 26 \cdot \sqrt{x} - 13 \cdot x^2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{{\rm d}y}{{\rm d}x} = \frac{26} {2\cdot \sqrt{x}} - 13 \cdot 2 \cdot x = 0$$

$$\Rightarrow \hspace{0.3cm} \frac{1} { \sqrt{x}} = 2 \cdot x \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\frac{1} { x} = 4 \cdot x^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} x^3 = 0.25 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} x \approx 0.63 \hspace{0.05cm}.$$

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

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

$$y(x_{\rm max}) \approx 26 \cdot \sqrt{0.63} - 13 \cdot 0.63^2 \hspace{0.15cm}\underline {\approx 15.477}.$$

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 $f = 0$.

• 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.