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

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

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.
• Is there a way to find the number of samples until the output reaches the steady state? I cannot figure that out directly from the sum, it seems. Apr 25 at 13:52
• @AhsonYousef your filter is a FIR filter, so after 120 samples you reach steady state. This is not valid for IIR (recursive) filters where, in theory, you never reach steady state. Apr 25 at 15:39
• @Juancho I see. In that case I am updating the question and adding how I think the steady state output can be computed. Apr 25 at 16:04
• @AhsonYousef : The $y_{ss}$ expression doesn't take account of $h[n]$ being FIR : Only 120 sum terms be used to calculate the output.
– Peter K.
Apr 25 at 16:13
• @AhsonYousef Yes: the sum should only be over 120 elements of the FIR filter. The value of $n$ will range over $N_x + H_h - 1$ values, but each value of $n$ only requires a sum over 120 elements.
– Peter K.
Apr 25 at 19:28