Exercise 3.7: Optimal Nyquist Equalization once again

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

$|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} = \ $

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

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

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

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.