# Why do transients exist when we input a sine signal to an LTI system?

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 $$\omega_{0}$$

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 $$n = 0$$.

• I came to realize that you are right about the second case, it explains the issue I faced Mar 23 at 21:25

"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.

• Thank you for your answer, I forgot for one moment what an LTI system should do. Mar 23 at 21:26

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.