I am having problems to formulate the signal d[n] that is given in the image below. What I am not getting here is that the result imho should be a scalar but no matter what I do I always end up multiplying a vector, the filter F_2(z), with a scalar.
Given:
\begin{align*}
H(z) &= 3 + 2z^{-1} \\
F_1(z) = F_2(z) &= z^{-1} + 0.1z^{-2}
\end{align*}
The equation for d[n] should look something like this:
\begin{align*}
d[n] &= \mathbf{f_2}^T \cdot \big( \mathbf{h}^T \mathbf{w}[n] + \nu[n] \big)
\end{align*}
But as you can see, the result wouldn't be a scalar.
What am I missing here?
Hint: It appears that we might have to rearrange the filtes F1, F2 somehow. But I don't know how that can help me.
Best Answer
In your upper path, the transfere function of \$H(z)\$ uses the actual input sample at time \$n_0\$ and the next sample at time \$n_{1}\$ Also, \$F_{2}(Z)\$ processes the delayed samples of \$H(z)\$. Therefore we need to define the input as a Matrix in form of delayed input vectors for \$H(z)\$
\$w[n]={\begin{bmatrix} w_0 & w_{-1}\\ w_{-1} & w_{-2}\\ w_{-2} & w_{-3} \end{bmatrix}}\$
For \$n_0\$ we get following expression: \$d[n]=\mathbf{f_2}\cdot(\mathbf{w}[n] \cdot \mathbf{h^T}+ \mathbf{v}[n])\$
And this gives: \$d[n]=\begin{bmatrix} 0 & 1 & 0.1 \end{bmatrix}\cdot(\begin{bmatrix} w_0 & w_{-1}\\ w_{-1} & w_{-2}\\ w_{-2} & w_{-3} \end{bmatrix}\cdot\begin{bmatrix} 3\\ 2 \end{bmatrix} + \begin{bmatrix} v_0\\ v_{-1}\\ v_{-2} \end{bmatrix})\$
which outputs a scalar for \$d[n]\$