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Below is the excerpt from Discrete Time Signal Processing by Alan Oppenheim.

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I don't get how $(\delta[n+1] - \delta[n]) * \delta[n-1])$ becomes $\delta[n] - \delta[n-1]$. The convolution sum operator "*" 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?

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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 $n_0=1$.

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The effet of $\delta[n-n_0]$ (as a filter) is a shift: $x[n]*\delta[n-n_0]=x[n-n_0]$. For instance, $\delta[n]$ does nothing (a $0$ shift), $\delta[n-1]$ shifts by one to the right, $\delta[n+1]$ shifts by one to the left.

So $\delta[n+1]*\delta[n-1]$ shifts by one to the left, then by one to the right. Finally, it performs a zero shift: $\delta[n+1]*\delta[n-1]=\delta[n]$. And $\delta[n]*\delta[n-1]$ does a zero shift, followed by a shift to the right: $\delta[n]*\delta[n-1]=\delta[n-1]$.

Hence, by linearity,

$$(\delta[n+1]-\delta[n])*\delta[n-1]=\delta[n]-\delta[n-1]\,.$$

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