Let $x[n]=a\cos(\omega_{0}n)$, if we pass it towards an LTI system, we should get as an output: $$ y[n]=a|X(e^{j\omega_{0}})|\cos(\omega_{0}n+\phi_{X}(\omega_{0}))+\operatorname{transients} $$ My question is why are there transients in the response and where do they come from? I know that the steady-state value is a scaled and shifted version of the input at a frequency $\omega_{0}$
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3 Answers
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You are mixing your analysis techniques.
If $x[n]=a\cos(\omega_{0}n)$, then you cannot use the $z$ transform for your analysis; you need to use Fourier analysis. Because the signal exists for all time there is no transient.
If you're using the usual single-sided $z$ transform, then the signal must have a beginning (although it can exist into infinity for all positive values of time). Then the signal is $x[n]=u(n) a\cos(\omega_{0}n)$, where $$u(n) = \begin{cases}0 & n < 0 \\ 1 & n \ge 0\end{cases}.$$
Then there will be a transient, because the sine wave starts up at time $n = 0$.
answered Mar 23, 2022 at 20:48
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$\begingroup$ I came to realize that you are right about the second case, it explains the issue I faced $\endgroup$ Mar 23, 2022 at 21:25
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"LTI" means "the output needs to scale with the input": So if you double $a$, $y$ must also double.
Since that needs to work for any $a$, especially 0, this means that $\text{transients}$ must be zero.
Makes sense – "passing a cosine into a system" means you're passing in a cosine – not a cosine and some other transient function.
answered Mar 23, 2022 at 20:34
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$\begingroup$ Thank you for your answer, I forgot for one moment what an LTI system should do. $\endgroup$ Mar 23, 2022 at 21:26
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In DSP (discrete) LTI systems typically have delay elements (memory storage locations). Assuming all the delay elements initially contain zero-valued samples, a system's transient response is the system's output sequence that occurs until all the delay elements are filled with valid data samples.
