Why are LTI filters not able to output frequencies not present at the input?

Question

I know this a very basic question but I'm just having troubles understanding why.

Thanks in advance!

Answer 1

Because for an LTI system, every input sine has an output which is a sine of the same frequency, with a potential shift and an amplification factor. In other words, input $x(t) = a\sin(\omega t+c)$ will be turned into some $y(t) = aH_\omega \sin(\omega t+d)$, where $H_\omega$ is the amplification factor for every sine with frequency $\omega $.

This come from the theory: LTI are diagonalized by complex exponentials (in a pedantic way). As a result, an absent sine at a given frequency is a zero amplitude sine (in other words, $a=0$), and the output will be a shifted sine, with zero amplitude whatever the amplification $H_\omega$ (since $aH_\omega=0$), still a zero amplitude sine.

More details:

• Are complex exponentials the only eigenfunctions of LTI systems
• Impulse Response to LTI

Answer 2

May be this will make some senses.

Let the Fourier transform of the signal $s(t)$ and the LTI system $h(t)$ be $S(\omega)$ and $H(\omega)$ respectively. Then the output spectrum $Y(\omega)$ is $$Y(\omega) = S(\omega)H(\omega).$$ Obviously, for frequencies (e.g., $\omega_0$) that are not present in $s(t)$, we have $S(\omega_0) = 0$. Therefore, $Y(\omega_0) = 0$ as well.