Impulse response of forward difference cascaded with one sample delay Below is the excerpt from Discrete Time Signal Processing by Alan Oppenheim.


I don't get how $(\delta[n+1] - \delta[n]) * \delta[n-1])$ becomes $\delta[n] - \delta[n-1]$. The convolution sum operator &quot;*&quot; here is defined as below.
$$y[n] = \sum_{k=-\infty}^{\infty} x[k]h[n-k] $$
How do I substitute the definitions of forward difference and one sample delay in this and arrive at the impulse response the author has given?
For any sequence $x[n]$ you have
$$x[n]\star\delta[n-n_0]=x[n-n_0]\tag{1}$$
because
$$x[n]\star\delta[n-n_0]=\sum_{k=-\infty}^{\infty}x[k]\delta[n-n_0-k]\tag{2}$$
Note that $\delta[n-n_0-k]$ is only non-zero for $k=n-n_0$, hence the result $(1)$.
Now use $x[n]=\delta[n+1]-\delta[n]$ and $n_0=1$.