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. $\endgroup$ Apr 25 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
    Apr 25 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. $\endgroup$ Apr 25 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.
    Apr 25 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.
    Apr 25 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.