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I'm having a difficult time understanding why the matrix for LTI systems is a Toeplitz matrix. I can see why $h_{n,m} = h_{n' + q,m' + q}$ given that $n' = n - q$ and $m' = m - q$, and $$\sum_{m'= -\infty}^\infty h_{n' + q,m' + q} x[m'] = y[n']$$ and $$\sum_{m= -\infty}^\infty h_{n,m} x[m] = y[n],$$ but the leap to the equivalence $h_{n,m} = h_{n + q,m + q}$ escapes me.

Peter K.
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soccermom9424
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1 Answers1

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This reference should help. If you carry out the matrix-vector multiplication, you will see they are the same.

Also note that it is only possible to express the convolution as a matrix multiplication when the filter impulse response is of finite duration, i.e. a FIR filter.

The equivalence $h_{n,m} = h_{n+q,m+q}$ comes from the "TI" part of LTI, that is, time-invariance. If you delay the input by $q$ samples, the output should be the same, just delayed in time by $q$.

Robert L.
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  • The online course I'm following is building up to the convolution now, but hasn't quite gotten there. The given explanation for equivalence is that there is a correspondence between $x[m']$ and $x[m]$ as well as $y[n']$ and $y[n]$, and thus the equivalence must hold, which was not very satisfactory reasoning for me. Furthermore, it was claimed there was a trivial change of variables to arrive at the same conclusion using the first sum, but it doesn't really seem obvious to me – soccermom9424 Dec 17 '18 at 20:59
  • Ah, that context is useful. As a side note, this online course sounds way more rigorous than the signals course I took in college! – Robert L. Dec 17 '18 at 21:09
  • The last addition to the response really helped! I can see how if the input signal is time shifted now then the Toeplitz matrix is required to see the same time-shifted response in the output. I think that's how the professor for the online course meant to go about it explaining it but your wording made a world of difference! – soccermom9424 Dec 18 '18 at 14:36