Output of Zero-forcing equalizer doesn't match theoretical calculations

Question

Let's say I have a sequence $x[n]$ (signal) and perform convolution with another sequence $h[n]$ (channel), so: $$ y[n] = h[n]*x[n] = \sum_{k}{h[k]x[n-k]} \leftrightarrow Y(z) = H(z)X(z) $$

I'm looking for $C(z)$ such that: $$ C(z)= \frac{1}{H(z)} $$

So that I can compensate the channel: $$ c[n]*y[n] = \sum_{k}{c[k]y[n-k]} \leftrightarrow C(z)Y(z) = \frac{1}{H(z)}Y(z) = \frac{1}{H(z)}H(z)X(z) = X(z) $$

Let's assume I know the channel $h[n]$ and I use that to obtain $c[n]$ using Zero-Forcing equalization.

My question is, if I do:

x = [1 2 3 4 5 6 7];
h = [1 0.9 0.3];
c = zeroforcing(h); % c=[-4.7619, 4.2857, -1.4286]

y = filter(h,1,x);
x_d = filter(c,1,y);

Shouldn't x_d be a delayed version of x ? But I don't seem to get that in my simulations.

(The way I obtain c is converting h to a toeplitz matrix and doing $c[n]=H^{-1}\delta [n]$ as explained here:

$$ \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} h_2 & h_1 & 0 \\ h_3 & h_2 & h_1 \\ 0 & h_3 & h_2 \end{bmatrix}^{-1} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0.9 & 1 & 0 \\ 0.3 & 0.9 & 1 \\ 0 & 0.3 & 0.9 \end{bmatrix}^{-1} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $$

Answer 1

It seems like there might be something weird going on with that blog, but I only quickly looked. Here is the idea for the ZF equalizer:

1. Figure out the convolution matrix.

convMatrix = convmtx(h, length(x));

This will be a $9 \times 7$ matrix because the input vector $\mathbf{x}$ is length 7 and the output convolution will be of length $7+3-1=9$. It is not a square matrix but we can still get an inverse by computing the Moore-Penrose inverse (https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse), which in MATLAB can be done by calling:

invConvMatrix = pinv(convMatrix);.

2. Choose your delay. invConvMatrix is a $7 \times 9$ matrix and we are multiplying with an all zero vector except for a single one at some delay $\mathbf{\delta}$. Lets choose the delay to be 5, so we have: $\mathbf{\delta}=[0, 0, 0, 0, 1, 0, 0, 0, 0]^T$.

3. Compute the ZF equalizer:

zfEqualizer = invConvMatrix * d;

4. Generate the received signal, and generate signal after equalization

y = conv(h, x); z = conv(zfEqualizer, y);

Now lets take a look at the resulting signals, and specifically notice where the equalized signal starts, right at the delay of 5 that we choose earlier. I hope this helps.