Given an input-output pair of a LTI system \begin{gather*} x[ n] \ =\ 2\delta [ n+2] -\delta [ n+1] +\delta [ n-1]\\ y[ n] \ =\ 4 \delta [ n+2] +\ 4\delta [ n+1] -\delta [ n-1] \end{gather*}

My primary question is, is a single input-output pair enough to find the impulse response of system? Secondary, given some input-output pair(s), can it be the case that such input-output pair(s) is/are impossible for an LTI system?

I know a property from class for discrete time convolution that if two signals with duration or support of L and M are convolved, the output has a length L+M-1. Does this always hold, or can the output signal still have a finite convolution if one of the signals being convolved has an infinite duration. Keeping this property in mind, my $\displaystyle x[ n] \ $and$\displaystyle \ y[ n]$ have support of 4. If $\displaystyle h[ n] \ $is my impulse response, shouldn't its support or duration be 1?

And such $\displaystyle h[ n] \ $should be non zero at n=0 after some inspection.

That being said, there exists no $\displaystyle h[ n] \ $ with duration 1 as can be seen.

I haven't been taught the Z-transform method, but I used matlab to compute the inverse Z transform to find $\displaystyle h[ n] \ $

enter image description here

It looks like $\displaystyle h[ n] \ $ has infinite support, also what is r4 root that symsum is summing? Is it the real root of the denominator.

Overall I can't reconcile if an LTI can have given input-output pair, and if it does, shouldn't it have finite support as per the theorem talking about support of convolution of two signals is L+M-1. If it indeed has finite support, then it turns out it doesn't exist.

I am really stuck and I dont have sources that can guide me.


1 Answer 1


These are interesting questions. First, not all input signals are useful for characterizing an LTI system by observing the corresponding response. If the input signal has a spectral zero, i.e., if its discrete-time Fourier transform (DTFT) is zero for some frequency, then the system is not excited at that frequency, and, consequently, we can't obtain a complete description of the system because we don't know how the system reacts to a sinusoidal component at that frequency.

Assuming that the input signal has no spectral zeros, we can determine the impulse response of an LTI system by observing its response. Note, however, that we must assume that the system is LTI. If it isn't, and if we don't know anything else about the system, then a single input-output pair won't help to characterize the system.

Given that the input signal has no spectral zeros and that we know that the system is LTI, any input-output pair will allow us to completely characterize the system.

Coming to your example, your argument about the support of the impulse response seems convincing. The argument goes as follows: the input and output sequences have the same lengths, hence the impulse response must have length $1$, which is just a scaling. Since the output is not a scaled version of the input, the system can't be LTI. However, this line of thought is wrong.

Note that the response $y[n]$ could be infinitely long, but only have finitely many non-zero values. A simple example of such a case is a system with the (infinitely long) impulse response


and with the input signal


Considering the convolution sum $y[n]=(x\star h)[n]$, we note that all terms of $y[n]$ cancel out apart from $y[0]=1$.

Your example is similar in the sense that there exists a causal and stable LTI system with an infinitely long impulse response generating the given response for the given input sequence. Its impulse response is the inverse $\mathcal{Z}$-transform of the transfer function


Note that in order for $(3)$ to make sense we need to check if $x[n]$ has no spectral zeros, which is indeed the case.

The figure below shows the first $50$ values of the impulse response corresponding to the transfer function $(3)$. Given the input sequence in your example, an LTI system with this impulse response will generate the given finite length output, even though its impulse response is infinitely long.

enter image description here

  • $\begingroup$ Wow! exactly the answer which I was looking for. Thanks alot for this! $\endgroup$
    – Anmoldeep
    Commented Sep 30, 2023 at 15:23

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