Aufgaben:Exercise 3.6: Transversal Filter of the Optimal Nyquist Equalizer

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Transversal filter of the
Optimal Nyquist Equalizer

At the input of the symmetric second order transversal filter  $(N = 2)$  shown in the diagram there is a triangular pulse (normalized to $1$):

$$g_x(t) = \left\{ \begin{array}{c} 1 - {|\hspace{0.05cm}t\hspace{0.05cm}|}/{(2T)} \\ \\ 0 \\ \end{array} \right. \begin{array}{*{1}c} {\rm{for}}\\ \\ {\rm{for}} \\ \end{array} \begin{array}{*{20}c}|\hspace{0.05cm}t\hspace{0.05cm}| \le 2\hspace{0.05cm}T, \\ \\ |\hspace{0.05cm}t\hspace{0.05cm}| \ge 2\hspace{0.05cm}T. \\ \end{array}$$

If all filter coefficients  $k_0$,  $k_1$  and  $k_2$  are nonzero, then the following holds for the pulse at the output:

$$g_y(t) \ = k_0 \cdot g_x(t) + k_1 \cdot \big[ g_x(t-T)+ g_x(t+T) \big] + k_2 \cdot \big[ g_x(t-2T)+ g_x(t+2T) \big]\hspace{0.05cm}.$$

By appropriate choice of filter coefficients  $k_0$,  $k_1$  and  $k_2$,  the output pulse can satisfy the following conditions:

$$g_0 = g_y(t = 0) = 1,\hspace{0.2cm}g_1 = g_y(t = \pm T) = 0,\hspace{0.2cm}g_2 = g_y(t = \pm 2 T) = 0 \hspace{0.05cm}.$$

• A first order filter  $(N = 1)$  is obtained from the above arrangement and equation by setting the coefficient  $k_2 = 0$. 
• Then, by appropriate choice of  $k_0$  and  $k_1$,   $g_0 = 1$  and  $g_1 = 0$  can be obtained. However, in this case,  $g_2 ≠ 0$  will always be  $g_2 ≠ 0$. 

Note:

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

Questions

1

What are the optimal coefficients for the first order filter   ⇒   $k_2 = 0$?

$k_0\ = \ $

$k_1\ = \ $

2

What are the output values at time points  $t = 2T$  and  $t = 3T$?

$g_2\ = \ $

$g_3\ = \ $

3

What are the optimal coefficients for the second order filter  $(N = 2)$?

$k_0\ = \ $

$k_1\ = \ $

$k_2\ = \ $

4

What are the output values at time points  $t = 3T$  and  $t = 4T$?

$g_3\ = \ $

$g_4\ = \ $

Solution

Solution

(1)  The input pulse $g_x(t)$ is given by the following samples at multiples of $T$:

$$g_x(t = 0) = 1,\hspace{0.2cm}g_x(t = \pm T) = 0.5,\hspace{0.2cm}g_x(t = \pm 2 T) = ... = 0 \hspace{0.05cm}.$$

• Thus, the following system of equations can be set up:

$$t = 0\hspace{-0.1cm}:\hspace{0.2cm}g_0 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} k_0 \cdot 1.0 + k_1 \cdot 2 \cdot 0.5 = 1\hspace{0.05cm},$$

$$t = T\hspace{-0.1cm}:\hspace{0.2cm}g_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} k_0 \cdot 0.5 + k_1 \cdot 1.0 = 0 \hspace{0.05cm}.$$

Output pulse for  $N = 1$

• From these equations, it follows that $k_0 \ \underline {= \ 2}$ and $k_1 \ \underline {= \ –1}$.

(2)  The values $g_0 = 1$  and $g_1 = 0$  have already been used as a basis for the optimization and are therefore undisputed.

• At time $t = 2T$, the output results, where $k_{-1} = k_1 = -1$:

$$g_2 = g_y(t = 2 T) = g_x(t = T) \cdot k_{-1}\hspace{0.15cm}\underline { = -0.5 = g_{-2}} \hspace{0.05cm}.$$

• Since all input values are zero at times $2T$, $3T$ and $4T$, $g_3 = g_y(t = 3T) \underline {= \ 0}$.
• This gives the output pulse $g_y(t)$ as shown in the sketch.

Output pulse for  $N = 2$

(3)  For a second order filter, the system of equations is:

$$t = 2T\hspace{-0.1cm}:\hspace{0.2cm}g_2 = k_1 \cdot 0.5 + k_2 \cdot 1.0 = 0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} k_2 = - 0.5 \cdot k_1\hspace{0.05cm},$$

$$t = T\hspace{-0.1cm}:\hspace{0.2cm}g_1= k_0 \cdot 0.5 +k_1 \cdot 1.0 + k_2 \cdot 0.5 = 0\hspace{0.05cm},$$

$$\hspace{1.6cm}\Rightarrow \hspace{0.3cm} k_1 = - {2}/{3} \cdot k_0\hspace{0.05cm},$$

$$t = 0\hspace{-0.1cm}:\hspace{0.2cm}g_0 = k_0 \cdot 1.0 + k_1 \cdot 0.5 + k_1 \cdot 0.5 = 1\hspace{0.05cm},$$

$$\hspace{1.6cm}\Rightarrow \hspace{0.3cm} k_0 = 3 \hspace{0.05cm}.$$

• Thus, the optimal coefficients are

$$k_0 \ \underline {= \ 3},k_1 \ \underline {= \ –2}, k_2 \ \underline {= \ 1}.$$

(4)  Proceeding in the same way as in subtask (2), we obtain  $g_4 \ \underline {= \ 0}$  as well as

$$g_3 = g_y(t = 3 T) = g_x(t = T) \cdot k_{-2} = 0.5 \cdot 1 \hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$

• However, the two graphs also show that for the triangular shape here, the optimal Nyquist equalization does not improve anything.
• The eye is just closed in all cases:

$$N = 0\hspace{-0.1cm}:\hspace{0.2cm} \ddot{o}/2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} g_0 - 2 \cdot g_1 = 1- 2 \cdot 0.5 = 0 \hspace{0.05cm}, $$

$$N = 1\hspace{-0.1cm}:\hspace{0.2cm} \ddot{o}/2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} g_0 - 2 \cdot |g_2 | = 1- 2 \cdot 0.5 = 0 \hspace{0.05cm}, $$

$$N = 2\hspace{-0.1cm}:\hspace{0.2cm} \ddot{o}/2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} g_0 - 2 \cdot g_3 = 1- 2 \cdot 0.5 = 0 \hspace{0.05cm}.$$