Inverse of accumulator: u[n] and δ[n] − δ[n−1]

Invertibility

A system is invertible if distinct inputs give distinct outputs, so the input can be recovered from the output. For an LTI system with impulse response $h[n]$, an inverse system $h'[n]$ exists such that the cascade gives back the input:

$$h[n]*h'[n]=\delta[n]$$

 x[n] +------+  y[n] +-------+  x[n]
 ---->| h[n] |------>| h'[n] |------>
      +------+       +-------+
   overall response h*h' = delta[n]

Example: $h[n]=u[n]$ is the accumulator, $y[n]=\sum_{k=-\infty}^{n}x[k]$. Its inverse is the first difference, $h'[n]=\delta[n]-\delta[n-1]$, i.e. $w[n]=y[n]-y[n-1]$, which recovers $x[n]$. By contrast, $y[n]=0$ and $y[n]=x^2[n]$ are not invertible.

Proof that $u[n]*h'[n]=\delta[n]$

Using the shifting property of the impulse, $f[n]*\delta[n-n_0]=f[n-n_0]$:

$$\begin{aligned} h[n]*h'[n] &= u[n]*\big(\delta[n]-\delta[n-1]\big)\\ &= u[n]*\delta[n]-u[n]*\delta[n-1]\\ &= u[n]-u[n-1] \end{aligned}$$

Now evaluate $u[n]-u[n-1]$:

So it is 1 only at $n=0$:

$$u[n]-u[n-1]=\delta[n]$$

The convolution sum gives the same result: $\sum_k u[k]h'[n-k]=\sum_k u[k](\delta[n-k]-\delta[n-1-k])=u[n]-u[n-1]=\delta[n]$.

Hence the cascade of the accumulator and the first difference is the identity system, and $h'[n]=\delta[n]-\delta[n-1]$ is the inverse of $h[n]=u[n]$. Check in the z-domain: $H(z)H'(z)=\frac{1}{1-z^{-1}}\cdot(1-z^{-1})=1$ (for $|z|>1$), which is the transform of $\delta[n]$.