Aufgaben:Exercise 3.3Z: Optimization of a Coaxial Cable System

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Normalized system parameters for different cut-off frequencies

We consider a redundancy-free binary transmission system with the following specifications:

• The transmission pulses are NRZ rectangular and have energy  $E_{\rm B} = s_0^2 \cdot T$.

• The channel is a coaxial cable with characteristic cable attenuation  $a_* = 40 \, {\rm dB}$.

• AWGN noise with (one-sided) noise power density  $N_0 = 0.0001 \cdot E_{\rm B}$  is present.

• The receiver frequency response  $H_{\rm E}(f)$  includes an ideal channel equalizer  $H_{\rm K}^{\rm -1}(f)$  and a Gaussian low-pass filter  $H_{\rm G}(f)$  with cutoff frequency  $f_{\rm G}$  for noise power limitation.

The table shows the eye opening   $\ddot{o}(T_{\rm D})$   as well as the detection noise rms value   $\sigma_{\rm d}$   – each normalized to the transmitted amplitude  $s_0$  – for different cutoff frequencies  $f_{\rm G}$.  The cutoff frequency is to be chosen such that the worst-case error probability is as small as possible,  with the following definition:

$$p_{\rm U} = {\rm Q} \left( \frac{\ddot{o}(T_{\rm D})/2}{ \sigma_d} \right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}\right)$$

• This quantity represents an upper bound for the mean error probability  $p_{\rm S}$:    $p_{\rm S} \le p_{\rm U}$.

• For  $f_{\rm G} \cdot T ≥ 0.4$,  a lower bound can also be given:   $p_{\rm S} \ge p_{\rm U}/4$.

Notes:

• The exercise belongs to the chapter  "Consideration of Channel Distortion and Equalization".

• Use the interaction module  "Complementary Gaussian Error Functions"  for numerical evaluation of the Q-function..

Questions

1

Within the given grid,  determine the optimal cutoff frequency with respect to the  "worst-case error probability"  criterion.

$f_\text{G, opt} \cdot T \ = \ $

2

What values does this give for the  "worst-case signal-to-noise ratio"  and the worst-case error probability?

$f_\text{G} = \text{G, opt:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ ${\ \rm dB}$

$\hspace{4.07cm}p_{\rm U} \ = \ $ $\ \rm \%$

3

To what value would we need to reduce the noise power density  $N_0$  (with respect to signal energy)  so that  $p_{\rm U}$  is not greater than  $10^{\rm -6}$?

$N_0/E_{\rm B} \ = \ $ $\ \cdot 10^{\rm -5}$

4

For the assumptions made in  (3),  give a lower and an upper bound for the  "average error probability"   $p_{\rm S}$. 

$p_\text{ S, min}\hspace{0.02cm} \ = \ $ $\ \cdot 10^{\rm -6}$

$p_\text{ S, max} \ = \ $ $\ \cdot 10^{\rm -6}$

Solution

Solution

(1)  For the optimization it is sufficient to maximize the quotient $\ddot{o}(T_{\rm D})/\sigma_d$:

• This is maximized from the values given in the table for the cutoff frequency  $f_{\rm G, opt} \cdot T \underline {= 0.4}$  with $0.735/0.197 \approx 3.73$.
• As a comparison:   For  $f_{\rm G} \cdot T = 0.3$  the result is $0.192/0.094 \approx 2.04$ due to the smaller eye opening and for  $f_{\rm G} \cdot T = 0.5$  the quotient is also smaller than for the optimum: $1.159/0.379 \approx 3.05$.
• An even larger cutoff frequency leads to a very large noise rms value without simultaneously increasing the vertical eye opening in the same way.

(2)  Using the result from (1), we further obtain:

$$\rho_{\rm U} = \left ( {3.73}/{2} \right )^2 \approx 3.48 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline { = 5.41\,{\rm dB}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q}\left ( {3.73}/{2} \right) \hspace{0.15cm}\underline {\approx 0.031} \hspace{0.05cm}.$$

(3) 

• With the given $10 \cdot {\rm lg} \, E_{\rm B}/N_0 = 40 \ \rm dB$, i.e. $E_{\rm B}/N_0 = 10^4$, the worst-case signal-to-noise ratio has been found to be $10 \cdot {\rm lg} \, \rho_{\rm U} \approx 5.41 \, {\rm dB}$.
• However, for the worst-case error probability $p_{\rm U} = 10^{\rm -6}$, $10 \cdot {\rm lg} \, \rho_{\rm U} > 13.55 \, {\rm dB}$ must be obtained.
• This is achieved by increasing the quotient $E_{\rm B}/N_0$ accordingly:

$$10 \cdot {\rm lg}\hspace{0.1cm}{E_{\rm B}}/{N_0} = 40\,{\rm dB} \hspace{0.1cm}+\hspace{0.1cm}13.55\,{\rm dB} \hspace{0.1cm}-\hspace{0.1cm}5.41\,{\rm dB}= 48.14\,{\rm dB}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {E_{\rm B}}/{N_0} = 10^{4.814}\approx 65163 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} {N_0}/{E_{\rm B}}\hspace{0.15cm}\underline { = 1.53 \cdot 10^{-5}} \hspace{0.05cm}.$$

(4) 

• The upper bound for $p_{\rm S}$ is equal to the worst-case error probability $p_{\rm U} = \underline {10^{\rm -6}}$.
• The lower bound is $\underline {0.25 \cdot 10^{\rm -6}}$, which is smaller by a factor of 4.