# replace input signal with $\delta$ will have the impulse response of the function?

Is my impulse response right? By definition,the impulse response is the output when the input is a impulse signal,so

$$y[n]=\sum\limits ^{n}_{k=-\infty}\frac{1}{2^{n-k}}\ x[k]$$,the impulse response of $$y[n]=\sum\limits ^{n}_{k=-\infty}\frac{1}{2^{n-k}}\ \delta[k]$$ ,Is my thinking right?