Aufgaben:Exercise 3.7: Optimal Nyquist Equalization once again

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Transversal filter
frequency response

We assume the following for this exercise:

• binary bipolar NRZ rectangular pulses

$$|H_{\rm S}(f)|= {\rm sinc}(f T) \hspace{0.05cm},$$

• coaxial cable with characteristic cable attenuation  $a_* = 9.2 \ {\rm Np} \ (\approx 80 \ \rm dB)$:

$$|H_{\rm K}(f)|= {\rm e}^{ -9.2\cdot \sqrt{2 \cdot |f| \cdot T} }\hspace{0.05cm},$$

• optimal Nyquist equalizer consisting of matched filter and transversal filter:

$$H_{\rm E}(f) = H_{\rm MF}(f) \cdot H_{\rm TF}(f)$$

$$\hspace{0.8cm}{\rm where}\hspace{0.2cm}H_{\rm MF}(f) = H_{\rm S}^{\star}(f) \cdot H_{\rm K}^{\star}(f)\hspace{0.05cm},\hspace{0.2cm}H_{\rm TF}(f) =\frac{1}{\sum\limits_{\kappa = -\infty}^{+\infty} |H_{\rm SK}(f -{\kappa}/{T})|^2}\hspace{0.05cm}.$$

Here,  $H_{\rm SK}(f) = H_{\rm S}(f) \cdot H_{\rm K}(f)$  denotes the product of transmitter and channel frequency response.

Because of Nyquist equalization,  the eye is maximally open.  For the error probability holds:

$$p_{\rm S} \left ( = p_{\rm U} \right ) = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right )\hspace{0.05cm}.$$

The normalized noise power at the decision is given by the following equations:

$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot\int_{-\infty}^{+\infty} |H_{\rm E}(f)|^2 \,{\rm d} f\hspace{0.5cm} = \hspace{0.5cm}\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot\int_{-1/(2T)}^{+1/(2T)} H_{\rm TF}(f) \,{\rm d} f \hspace{0.5cm}= \hspace{0.5cm}T\cdot \int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f\hspace{0.05cm}.$$

The validity of this equation follows from the periodicity of the transversal filter frequency response  $H_{\rm TF}(f)$.

• In the graph,  the normalized noise power can be seen as the area highlighted in red.

• As an approximation,  the normalized noise power can be calculated by the triangular area shown in blue in the graph.

Notes:

• The exercise belongs to the chapter  "Linear Nyquist Equalization".

• To determine the error probability you can use the interactive calculation module  "Complementary Gaussian Error Functions". 

Questions

1 Calculate the magnitude of the transmitter channel frequency response for the frequencies  $f = 0$,  $f = 1/(2T)= f_{\rm Nyq}$  and  $f = 1/T = 2 \cdot f_{\rm Nyq}$.

$|H_{\rm SK} (f = 0)| \hspace{0.8cm} = \ $

$|H_{\rm SK} (f = f_{\rm Nyq})| \hspace{0.2cm} = \ $ $\ \cdot 10^{-5}$

$|H_{\rm SK} (f = 1/T)| \hspace{0.25cm} = \ $

2 Calculate the maximum value of  $H_{\rm TF}(f)$  at frequency  $f = f_{\rm Nyq}$.

$|H_{\rm TF} (f = f_{\rm Nyq})| \hspace{0.2cm} = \ $ $\ \cdot 10^8$

3 Calculate the normalized noise power according to the triangular approximation.

$\sigma_{d, \ \rm norm}^2 \hspace{0.2cm} = \ $ $\ \cdot 10^7$

4 What is the symbol error probability with  $s_0^2 \cdot T/N_0 = 10^8$?

$p_{\rm S} \hspace{0.2cm} = \ $ $\ \%$

Solution

Solution

(1)  In general,  for all frequencies  $f \ge 0$:  

$$|H_{\rm SK}(f)|= {\rm sinc}(f T) \cdot {\rm e}^{ -9.2\cdot \sqrt{2 \cdot |f| \cdot T} }\hspace{0.05cm}.$$

• This gives the special cases we are looking for:

$$f= 0 \text{:} \ \hspace{0.1cm}|H_{\rm SK}(f = 0)|= {\rm sinc}(0) \cdot {\rm e}^0 \hspace{0.15cm}\underline {= 1}\hspace{0.05cm},$$

$$ f= f_{\rm Nyq}\text{:} \ \hspace{0.1cm}|H_{\rm SK}(f = \frac{1}{2T})|= {\rm sinc}({1}/{2}) \cdot {\rm e}^{-9.2}\hspace{0.15cm}\underline { \approx 6.43 \cdot 10^{-5}}\hspace{0.05cm},$$

$$ f= {1}/{T} \text{:}\ \hspace{0.1cm}|H_{\rm SK}(f = \frac{1}{T})|= {\rm sinc}({1}) \cdot {\rm e}^{...}\hspace{0.15cm}\underline { = 0} \hspace{0.05cm}.$$

(2)  The graph shows that  $H_{\rm TF}(f)$  becomes maximal at  $f = f_{\rm Nyq}$. 

• It follows with the given equation that

$${\sum\limits_{\kappa = -\infty}^{+\infty} |H_{\rm SK}(f -\frac{\kappa}{T})|^2}$$

is minimal at the Nyquist frequency. 

• However,  for $f = f_{\rm Nyq}$,  of the infinite sum, only the terms with  $\kappa = 0$  and  $\kappa = 1$  contribute relevantly to the result.
• From this follows further with the result from subtask  (1):

$${\rm Max} \left [ H_{\rm TF}(f) \right ] \ = \ H_{\rm TF}(f = f_{\rm Nyq})={1}/{2 \cdot |H_{\rm SK}(f = f_{\rm Nyq}) |^2} = \ \frac{1}{2 \cdot (6.43 \cdot 10^{-5})^2}=\frac{10^{10}}{82.69} \hspace{0.15cm}\underline {\approx 1.21 \cdot 10^{8}}\hspace{0.05cm}.$$

(3)  Approximating the integral over  $H_{\rm TF}(f)$  by the triangular area plotted in the graph,  we obtain:

$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot\int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f \approx T \cdot\frac{1}{2}\cdot 1.21 \cdot 10^{8}\cdot (0.64 -0.36)\hspace{0.15cm}\underline {\approx 1.7\cdot 10^{7}} \hspace{0.05cm}.$$

(4)  According to the given equation,  we obtain for the (mean) symbol error probability:

$$p_{\rm S} = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right) = {\rm Q}\left ( \sqrt{\frac{10^{8}}{1.7\cdot 10^{7}}} \right ) \approx {\rm Q}(2.42)\hspace{0.3cm}\Rightarrow\hspace{0.3cm} p_{\rm S} \hspace{0.15cm}\underline {\approx 0.8 \%} \hspace{0.05cm}.$$

Since a binary Nyquist system is present,  the worst–case probability  $p_{\rm U}$  is just as large.