0
$\begingroup$

Determine $g(t)$ for minimal symbol error probability, given $h(t)$, $K=1$ and Minimum distance receiver.

$h(t)= \begin{cases} 0, & \text{$t<0$}\\ 2, & \text{$t\in(0\;, 10^{-3})$}\\ 1, & \text{$t\in(10^{-3}\;,2\cdot 10^{-3})$}\\ 0, & \text{$t>2\cdot 10^{-3}$} \end{cases} $

This is ISI-free if $h(t)*g(t)|_{t=kT}= \begin{cases} 1, & \text{if $k=0$}\\ 0, & \text{if $k\neq0$} \end{cases} $

Using a matched filter as $g(t)=ch^*(-t)$ the convolution becomes $$(g*h)(t)=\int_{-\infty}^\infty{h(\tau)g(-\tau)d\tau }=c\int_{-\infty}^\infty h(\tau)h^*(\tau)d\tau=1$$

How do I calculate the $\int_{-\infty}^\infty h(\tau)h^*(\tau)d\tau=1$ integral or the pulse energy?
From what I understand it is the same as the pulse energy (which I don't know how to calculate either).

From here I understand how to determine $g(t)$ and it will be the $c$-scaled and time inversed variant of $h(t)$, but I don't really understand how to calculate the pulse energy.

Is $E_h=\int_{-\infty}^{\infty}|h(\tau)|^2d\tau$$\rightarrow$ $E_h=\int_{0}^{10^{-3}}2^2d\tau+\int_{10^{-3}}^{2*10^{-3}}1^2d\tau$ the correct way?

$\endgroup$
2
  • $\begingroup$ A key problem is that your convolution integral is not taking into account the time instant $t$ where the convolution is evaluated. $\endgroup$
    – cjferes
    Mar 8 at 19:01
  • $\begingroup$ Can you explain how you mean? $\endgroup$
    – Ehter
    Mar 8 at 19:20
1
$\begingroup$

Your computation using the energy of $h(t)$, $E_h$, to find the value of $c$, is correct. But note that you are only forcing the zero ISI condition for $k=0$. You still need to check the zero ISI criterion whenever $k\neq0$, but there is no mention of the value of $T$, and we can't derive what happens in those cases until we know the value of $T$.

For a general approach, I'd recommend going to the basics and not taking shortcuts. One key problem in your definition is that your convolution integral is not taking into account the time instant $t$, and therefore, you only compute one value of the convolution (for $t=0$), instead of all the values that are relevant (namely, $kT$ for all $k$).


The convolution of $h(t)$ and $g(t)$ is defined as $$p(t) = (g* h)(t) = \int_{-\infty}^{\infty}g(\tau)h(t-\tau)\,d\tau = \int_{-\infty}^{\infty}h(\tau)g(t-\tau)\,d\tau$$

and we need to satisfy the Nyquist first criterion for zero ISI, $$p(t)\Big|_{t=kT}=(g* h)(kT)=\left\{\begin{array}{rl}1,&k=0\\0,&k\neq0\end{array}\right.$$

Lastly, the matched filter is defined as $g(t)=ch^*(T-t)$ in its causal version, and $g(t)=ch^*(-t)$ in its non-causal version. For simplicity, let's use the noncausal version: $$\begin{align} p(t)& = (g* h)(t) = c\int_{-\infty}^{\infty}h(\tau)h^*(\tau-t)\,d\tau\\ \end{align}$$

In the particular case of $t=0$ (equivalently, $k=0$), we have $$\begin{align} p(0)& = (g* h)(0) = 1\\ &=c\int_{-\infty}^{\infty}h(\tau)h^*(\tau)\,d\tau= c\int_{-\infty}^{\infty}|h(\tau)|^2\,d\tau\\ &= c\Bigg(\int_{0}^{10^{-3}}|2|^2\,d\tau+\int_{10^{-3}}^{2\cdot10^{-3}}|1|^2\,d\tau\Bigg)=c\big(4\cdot10^{-3}+10^{-3}\big)=5c\cdot10^{-3}\\ &\Rightarrow c=10^3/5 = 200 \end{align}$$

Now, for verifying that the system satisfies the zero ISI criterion, we need the value of $T$. But you can follow a similar procedure.

$\endgroup$
2
  • $\begingroup$ thank you very much for this explanation! $\endgroup$
    – Ehter
    Mar 9 at 8:43
  • $\begingroup$ I forgot to add that $T$ might be a design choice, in the sense of finding the values of $T$ for which the pulse shaping filter + matched filter system satisfies the zero ISI criterion. For that, an option would be to look for $t_0$, the values of t such that the convolution is equal to zero, and set $T=t_0$. $\endgroup$
    – cjferes
    Mar 10 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.