So, I have a problem where the input and output of a system are given :

\begin{align} x(t) &= \sin(10t) \\ y(t) &=5\cos(10t+\frac{\pi}{6}) \end{align}

So, I need to determine whether this system is LTI.

First, when I saw this problem, I checked for zero input-zero output condition of LTI. By substituting $t=0$, $x(t)=0 , y(t) \neq 0$. So, I guessed it was not LTI. But the answer is given as LTI.

Can someone give some insights? Specifically, is Zero input-Zero Output for LTI always satisfied?

I require just some hints, not complete answers. Particularly the answer to whether LTI systems always obey zero input/zero output condition?

  • 1
    $\begingroup$ I have explained it in the second paragraph. Put t=0, sin(10t)=0 , but 5cos(pi/6)≠0 $\endgroup$
    – Naveen
    Mar 20 at 15:36
  • $\begingroup$ Thank-you!!!!!! $\endgroup$
    – Peter K.
    Mar 20 at 16:09

1 Answer 1


Note that for an LTI system the input can be zero at a certain time, and the output at that time can be non-zero. You can only conclude that a system is not LTI if you observe a non-zero output for zero input, i.e., $y(t)\neq 0$ if $x(t)=0$ for all $t$.

Another important thing to note is that from a single input-output pair you generally cannot determine whether a system is LTI or not. What you can say is that the given input-output pair could be from an LTI system because the sinusoidal input causes a sinusoidal output with the same frequency. The scaling and phase shift is what an LTI system can do to the input signal. However, there is no way to tell whether the system is really an LTI system. From the given input and output we simply cannot rule out that the system could be LTI.


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