From the textbook Signals and Systems by Oppenheimer ...

If we define $h_k[n]$ to be the impulse response to $\delta[n-k]$, then for the system to be time-invariant, we must have

$$h_k[n] = h_0[n-k]$$

I don't understand how the last equation follows. By definition of time-invariance, if the input is shifted by $k$ units, the output is shifted by $k$ units. From the given definition of $h_k[n]$, $h_0[n]$ is the impulse response to $\delta[n]$. Shouldn't the time-invariance property read

$$h_k[n] = h_0[n]$$

EDIT: The text goes on to add:

Since $\delta[n-k]$ is the time-shifted version of $\delta[n]$, $h_k[n]$ is the time-shifted version of $h_0[n]$

But saying it is the time-shifted version does not mean equivalance. What gives?


1 Answer 1


Time-invariance means that if $y[n]$ is the response to $x[n]$, then $y[n-k]$ is the response to $x[n-k]$. So if $h_0[n]$ is the response to the impulse $\delta[n]$, then - if the system is time-invariant - its response to $\delta[n-k]$ must be $h_0[n-k]$.

If $h_k[n]=h_0[n]$ were true, then the response to any input $\delta[n-k]$ would be $h_0[n]$, regardless of $k$, which basically means that the system output would be independent of the input signal.

  • $\begingroup$ My mistake was in equating $h_k[n]$ and $h_0[n]$. Thanks. $\endgroup$
    – curryage
    May 19, 2014 at 13:35

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