# Property for time invariance of system in terms of its impulse response

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

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.

• My mistake was in equating $h_k[n]$ and $h_0[n]$. Thanks. Commented May 19, 2014 at 13:35