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  • $y(n)$ = output signal

  • $x(n)$ = input signal

  • $\mathbf H$ = system response as a toeplitz matrix

$$\mathbf H = \begin{bmatrix}h(0)&&&\\h(1)&h(0)&&\\h(2)&h(1)&h(0)&\\\vdots&&&\ddots\end{bmatrix}$$

I understand that $\mathbf H$, with its transposable (orthonormal) properties, makes it easy achieve $x(n)$ when we only know $y(n)$ during deconvolution. But how did we just assume $\mathbf H$ to be of this shape. Why not some other matrix shape when doing deconvolution? Is there a certain reason for this?

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For any one still interested: I personally think it is because of the assumption that the system is causal + time invariant; output is only dependent on past inputs. Therefore you can see that that its lower triangular. The toeplitz structure is because of the fact the system is assumed to be time invariant. If you think about what causality and time invariance means then you can imagine this structure.

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To apply de-convolution, you're assuming that $y = h * x + \epsilon$ is a reasonable model, where $*$ denotes convolution; i.e. you convolve some filter $h$ with a signal $x$ and add some noise $\epsilon$.

Now, assume you have $n$ samples of $y$. Then, write those samples of $y$ in terms of samples of $x$. When you write the convolution as a matrix, you get precisely $y=H x $ as you've written above.

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