Exercise 3.3: Noise at Channel Equalization

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

$\sigma_d/s_0 \ = \ $

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

${\ \rm dB}$

$\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) \ = \ $

$\ \cdot 10^0$

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

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

$\ \cdot 10^6$

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.