Exercise 3.6Z:Optimum Nyquist Equalizer for Exponential Pulse

Two-sided exponential input pulse

As in Exercise 3.6 we consider again the optimal Nyquist equalizer.

The input pulse $g_x(t)$ is a two-sided exponential function:

$$g_x(t) = {\rm e }^{ - |t|/T}\hspace{0.05cm}.$$

Through a transversal filter of $N$–th order with the impulse response

$$h_{\rm TF}(t) = \sum_{\lambda = -N}^{+N} k_\lambda \cdot \delta(t - \lambda \cdot T)$$

it is possible that the output pulse $g_y(t)$ has zero crossings at $t/T = ±1, \ \text{...} \ , \ t/T = ±N$,
while $g_y(t = 0) = 1$.

However, in the general case, the precursors and trailers with $| \nu | > N$ lead to intersymbol interference.

Note: The exercise belongs to the chapter "Linear Nyquist Equalization".

Questions

$g_x(0)\ = \ $

$g_x(1)\ = \ $

$g_x(2)\ = \ $

$k_0 \ = \ $

$k_1 \ = \ $

$g_2 \ = \ $

$g_3\ = \ $

	For the given input pulse $g_x(t)$, no improvement is possible with a second-order transversal filter.
	The first statement is independent of the input pulse $g_x(t)$.
	For the given input pulse, a further improvement is obtained with an "infinite" transversal filter.

Solution

(1) The five first samples of the input pulse at distance $T$ are:

$$g_x(0)\hspace{0.25cm}\underline{ = 1},\hspace{0.2cm}g_x(1) \hspace{0.25cm}\underline{= 0.368},\hspace{0.25cm}g_x(2) \hspace{0.25cm}\underline{= 0.135},\hspace{0.2cm}g_x(3) = 0.050,\hspace{0.2cm}g_x(4) {= 0.018} \hspace{0.05cm}.$$

(2) According to "solution to Exercise 3.6", we arrive at the following system of equations:

$$2t = T \hspace{-0.1cm}:\hspace{0.2cm}g_1 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} k_0 \cdot g_x(1) +k_1 \cdot [g_x(0)+g_x(2)]= 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \frac{k_1}{k_0} = - \frac{g_x(1)}{g_x(0)+g_x(2)} \hspace{0.05cm},$$

$$t = 0 \hspace{-0.1cm}:\hspace{0.2cm}g_0 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} k_0 \cdot g_x(0) + k_1 \cdot 2 \cdot g_x(1) = 1\hspace{0.3cm}\Rightarrow \hspace{0.3cm} k_1 = \frac{1-k_0}{0.736} \hspace{0.05cm}.$$

This leads to the result:

$$k_0 - 0.324 \cdot 0.736 \cdot k_0 = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} k_0 \hspace{0.15cm}\underline {= 1.313}, \hspace{0.2cm}k_1\hspace{0.15cm}\underline { = -0.425} \hspace{0.05cm}.$$

Input pulse (top), output pulse for $N = 1$ (bottom)

(3) For time $t = 2T$ holds:

$$g_2 \ = \ k_0 \cdot g_x(2) +k_1 \cdot [g_x(1)+g_x(3)]$$

$$\Rightarrow\hspace{0.3cm} g_2 \ = \ 1.313 \cdot 0.050 -0.425 \cdot [0.135+0.018]\hspace{0.15cm}\underline {\approx 0} \hspace{0.05cm}.$$

Similarly, the output pulse at time $t = 3T$ is also zero:

$$g_3 \ = \ k_0 \cdot g_x(3) +k_1 \cdot [g_x(2)+g_x(4)$$

$$\Rightarrow\hspace{0.3cm}g_3 \ = \ 1.313 \cdot 0.135 -0.425 \cdot [0.368+0.050]\hspace{0.15cm}\underline {\approx 0} \hspace{0.05cm}.$$

The figure shows that for this exponentially decaying pulse, the first-order transversal filter provides complete equalization.

Outside the interval $-T < t < T$, $g_y(t)$ is identically zero.

Inside it results in a triangular shape.

(4) Only the first statement is correct:

Since already with a first-order delay filter all precursors and trailers are compensated, also with a second-order filter and also for $N → ∞$ no further improvements result.

However, this result applies exclusively to the (bilaterally) exponentially decaying input pulse.

For almost any other pulse shape, the larger $N$ is, the better the result.