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, $y_{ss}[n]$?


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 $119<k\leq120119$


1 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.
  • $\begingroup$ 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. $\endgroup$ Commented Apr 25, 2023 at 13:52
  • $\begingroup$ @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. $\endgroup$
    – Juancho
    Commented Apr 25, 2023 at 15:39
  • $\begingroup$ @Juancho I see. In that case I am updating the question and adding how I think the steady state output can be computed. $\endgroup$ Commented Apr 25, 2023 at 16:04
  • $\begingroup$ @AhsonYousef : The $y_{ss}$ expression doesn't take account of $h[n]$ being FIR : Only 120 sum terms be used to calculate the output. $\endgroup$
    – Peter K.
    Commented Apr 25, 2023 at 16:13
  • 1
    $\begingroup$ @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. $\endgroup$
    – Peter K.
    Commented Apr 25, 2023 at 19:28

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