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  • Why $Y (\omega) = X(\omega)H(\omega)$ implies that an LTI system cannot generate any new frequencies?
  • Why if a system generates new frequencies, then it is not LTI?
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One of the definitive features of LTI systems is that they cannot generate any new frequencies which is not already present in their inputs. Please note that in this context a frequency refers to signals of the type $x(t)=e^{j\Omega_0 t}$ or $\cos(\Omega_0 t)$ which are of infinite duration, and are also referred to as eigenfunctions of LTI systems (specifically for the complex exponential only) and whose CT Fourier transforms are expressed by impulse functions in frequency domain as $X(\Omega) =2\pi \delta (\Omega - \Omega_0)$ or $X(\Omega)=\pi \delta (\Omega - \Omega_0) + \pi \delta (\Omega + \Omega_0)$ repectively.

One way to see why this is so, comes by observing the CTFT, $Y(\omega)$, of the output $y(t)$, which is given by the well known relation $Y(\omega) = H(\omega)X(\omega)$ only when the system is LTI (and also stable as a matter of fact so that $H(e^{j \omega})$ exists).

(i.e. $$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau \longleftrightarrow Y(\omega)=X(\omega)H(\omega),$$ holds only when the impulse response $h(t)$ exists and it will exist only when the system is LTI.)

From a little thought, guided by a simple graphical plot, and using the multiplication property above, one can see that the frequency region of support $R_y$ (set of frequencies for which $Y(\omega)$ is non-zero), of the output $Y(\omega)$ is given by the intersection of the regions of support $R_x$ and $R_h$ of the inputs $X(\omega)$ and frequency response $H(\omega)$ of the LTI system: $$R_y = R_x \cap R_h$$

And from set algebra we know that if $A = B \cap C$ then $A \subset B$ and $A \subset C$ . That is, an intersection is always less or equivalent to what are being intersected. Therefore, the region of support for $Y(\omega)$ will be less than or at most equal to the support of $X(\omega)$. Hence no new frequencies will be observed at the output.

Since this property is a necessary condition for being an LTI system, any system that fails to posses it, therefore, cannot be LTI.

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You can make a simple algebraic argument, given the premise that you provided. If:

$$ Y(\omega) = X(\omega) H(\omega) $$

where $X(\omega)$ is the spectrum of the input signal and $H(\omega$) is the frequency response of the system, then it's obvious that if there is some $\omega$ in the input signal for which $X(\omega) = 0$, then $Y(\omega) = 0$ as well; there is no factor $H(\omega)$ that you could multiply by to yield a nonzero value.

With that said, establishing the truth of the premise I started with above for LTI systems does take some work. However, if we assume it to be true, then the fact that an LTI system can't introduce any new frequency components to its output follows directly.

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  • $\begingroup$ the proof would be to show that for any sufficiently well-behaved signal, the Fourier Transform is invertible and both the FT and its inverse are linear. Every signal with a frequency is sufficiently well-behaved. $\endgroup$ – Marcus Müller Jun 21 '16 at 8:06
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Why $Y(ω)=X(ω)H(ω)$ implies that an LTI system cannot generate any new frequencies?

If a certain frequency $\omega_\text{abs}$is not present in our input, $X(\omega_\text{abs}) = 0$. Because 0 obeys the multiplicative identity $\forall x\in \mathbb{R},~ 0 \cdot x = 0$, $Y(\omega_\text{abs}) = 0$. Thus the frequency $\omega_\text{abs}$ is not present in the output signal.

Why if a system generates new frequencies, then it is not LTI?

Let's say our input is $x(t) = \cos(t)$. Then if we assume that our system can generate new frequencies, it is possible to obtain the output $y(t) = \cos(2\cdot t)$. Because we can not find constants $c_1, c_2$ such that $y(t) = c_1 \cos(t - c_2)$, our system is not LTI.

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  • $\begingroup$ Doesn't for checking LTI just c1 used, and not also c2? $\endgroup$ – USER Jun 21 '16 at 5:03
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    $\begingroup$ i would say that the first point, which is essentially that you can't get something non-zero by multiplying zero times anything, that is the concise answer. $\endgroup$ – robert bristow-johnson Jun 21 '16 at 5:24
  • $\begingroup$ c1 is used for linearity, c2 is used for time-shifting. We could have an LTI system that delays everything by 1 time unit. $\endgroup$ – Scott Jun 21 '16 at 13:26
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An LTI system is diagonalized by pure frequencies. Sines/cosines are eigenvectors of the linear system. In other words, any single non-zero sine or cosine (or a complex cisoid) input has a sine or cosine output of the same frequency exactly (but the output amplitude may vanish).

The only thing that may change is their amplitude or their phase. Hence, if you have no sine with a given frequency in the input, you get nothing (zero) with that frequency at the output.

The second question is answered by contraposition or regula falsi: if $A\implies B$ is true, so is $\overline{B}\implies \overline{A}$. If a system is LTI, it does not generate new frequencies. If a system generates new frequencies, it is not LTI.

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