Given a discrete-time white noise model: $$y_k = \sum_{i=0}^{\infty} \alpha^{-i}a_{k-i} + \eta_k$$ where $E(|a_k|^2)=2$, $|\alpha|>1$, $\eta_k$ are uncorrelated zero mean random variables with $E(|\eta_k|^2) = 2N_0$.

  1. Find the Z-domain transfer function for this channel model.

  2. Find the coefficients of a Zero Forcing linear equalizer for this system, that operates on the samples $y_k$ and removes all the intersymbol interference.

  3. For what values of $\alpha$ the channel becomes free of intersymbol interference?

My attempt:

  1. For part 1, $F(z) = z + \alpha^{-1}z^{-1} + \alpha^{-2}z^{-2} + \ldots$. However I don't know how to do the rest.

1 Answer 1


The following are hints as this is reads as a homework question. Please note that these are not ground breaking hints and can be found in a good text or summary on zero forcing somewhere I'm sure.

Hint 1

The first step would be to look at the model and be able to pick out what the channel is (you have channel convolved with signal plus noise). It seems you have identified the channel as, $h_k = \alpha^{-k}u_k$, where $u_k$ is the unit step function, and the z-transform can be found by directly applying the formula.

\begin{align} H(z) &= \sum_{k=-\infty}^{\infty}h_k z^{-k} \\ &= \sum_{k=-\infty}^{\infty} \big(\alpha^{-k}u_k \big)z^{-k} \\ &= \sum_{k=0}^{\infty} \alpha^{-k} z^{-k} \\ &= \alpha^0z^0+\alpha^{-1}z^{-1}+\alpha^{-2}z^{-2}+\cdots \end{align}

Just be careful when simplifying...

Hint 2

Zero forcing (ZF) means that the equalizer is trying to completely eliminate intersymbol interference (ISI), and ZF does achieve this in the noise free case. For the noise free case, in the z-domain your model becomes:

$$ Y(z) = H(z)A(z) $$

The ZF equalizer should be $W_{\text{ZF}}(z)=\frac{1}{H(z)}$ so that the equalized signal will have eliminated the ISI caused by the channel.

Hint 3

This should become clear when you consider what happens to each symbol as it goes through the channel. Take a single symbol being sent over the channel at $k=0$. The received noise free version is $a_0+\alpha^{-1}a_{-1}+\cdots$, or in words this is, $\text{Original symbol } + \text{ Previous symbol } + \cdots$. Can you pick an $\alpha$ to ensure the only non-zero value is the original symbol?

  • $\begingroup$ So for the first part and second part, I will need to determine $H(z)$ and then $W(z) = \frac{1}{H(z)}$. So I wonder how can I simply $H(z)$. I see that the McLaurin series for $\frac{1}{1-x}$ having a similar form, can I apply that one? $\endgroup$
    – thnghh
    Dec 21, 2020 at 12:53
  • $\begingroup$ Now it’s a math problem, go find the series formula $\endgroup$
    – Engineer
    Dec 21, 2020 at 13:02
  • $\begingroup$ So I figure it out for part 1 and 2, it should be $\frac{az}{az-1}$ and $\frac{az-1}{az}$. For part 3, intuitively $\alpha$ should be $\infty$, but it doesn't make much sense to me, as a coefficient cannot be $\infty$. $\endgroup$
    – thnghh
    Dec 21, 2020 at 13:49
  • $\begingroup$ Why can't $\alpha=\infty$? I saw the only constraint you have on $\alpha$ is that its magnitude is greater than $1$, and the magnitude of $\infty$ is greater than $1$. If you're thinking about in practice, then I suppose even if you setup a system to have no ISI there would still be some very small amount but it might be so small that it doesn't matter. $\endgroup$
    – Engineer
    Dec 21, 2020 at 13:54
  • $\begingroup$ Yes, because I'm confused about the problem in practice, but anyway it's homework so I think it's possible. Thank you Mr. Engineer. $\endgroup$
    – thnghh
    Dec 21, 2020 at 15:39

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