What does it mean to find the steady state output in this case?

Question

I have a given $h[n] = 0.96^n$ for $n \in \{0,1,...,119\}$ and a test signal $x[n] = \cos(0.05\pi n)$ for $n \in \{0,1,...,N_x -1\}$ where $N_x = 120,000$.

Now, I can analytically find the output $y[n]$ using the convolution sum:

$$y[n] = \sum_{k=0}^{N_x + N_h - 1} x[k] h[n-k] = \sum_{k=0}^{120119} \cos(0.05\pi k) \cdot 0.96^{(n-k)}$$

and then expanding the $cosine$ into complex exponentials and so on. But what does it mean exactly to find the steady state output, $y_{ss}[n]$?

Edit:

Expression for what $y_{ss}$ should be based on @juancho's comment. $$y_{ss}[n] = \sum_{k= n-120}^{120119} \cos(0.05\pi k) \cdot 0.96^{(n-k)}$$ for $119<k\leq120119$

Answer 1

1. Your systems is a linear time invariant (LTI) system
2. An important property of LTI systems is that the output to a sine wave is also a sine wave of the same frequency just with a different magnitude and phase.
3. So we can write the output at some points as $$y_{ss}[n] = A\cdot \cos(0.05\pi n + \varphi), n >= N_0$$ That's the steady state output
4. It typically takes some amount of time/samples until the output reaches steady state simply because the input is not a pure sine wave but a gated sine wave and $x[n] = 0, n <0$. As long as there are initial zeros in the convolution sum, the output is not steady state yet.