I'm currently stuck on a problem. I have a DT LTI system $\mathcal{T}$ that depicts the first backward difference:
$$y[n] = \mathcal{T} \{ {x[n] \} } = x[n] - x[n-1]$$
Its impulse response is therefore: $h[n] = \delta[n] - \delta[n-1]$, where $\delta[n]$ is the DT unit impulse.
I want to compute the impulse response $h'$ of the system $\mathcal{T}'$ that, when cascaded with $\mathcal{T}$, recovers my input:
$$\mathcal{T}'\{\mathcal{T}\{x[n]\}\} = x[n]$$
I know the answer is that $h'[n]$ is the DT unit step $u[n]$, but I can't quite figure out why.
Here is my idea so far:
- I know that the impulse response of the cascaded system is: $$h_{casc}[n] = \mathcal{T}'\{\mathcal{T}\{\delta[n]\}\} = \delta[n]$$
- I know that this impulse response is the convolution of the impulse responses of $\mathcal{T}'$ and $\mathcal{T}$, so: $$\delta[n] = h[n] \ast h'[n] = h'[n] \ast (\delta[n] - \delta[n-1]) = h'[n] - h'[n-1]$$
- So for $n=0$ I have: $$h'[0] - h'[-1] = 1$$
- And for $n \ne 0$ I have: $$h'[n] - h'[n-1] = 0 \Leftrightarrow h'[n] = h'[n-1]$$
Still, this isn't enough to solve for $h'$ if I am not mistaken, as for example the following different signals would satisfy this:
- $... 4 \ 4 \ 4 \ 4 \ 4 \ 5 \ 5 \ 5 \ 5 \ 5 \ldots$, Here: $h'[-1] = 4$, $h'[0] = 5$
- $... 2 \ 2 \ 2 \ 2 \ 2 \ 3 \ 3 \ 3 \ 3 \ 3 \ldots$, Here: $h'[-1] = 2$, $h'[0] = 3$
Can anyone give me a hint what I am missing here or perhaps point me to an error or in the right direction?