- 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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A definitive feature of LTI systems is that they cannot generate any new frequencies which is not already present in their input $x(t)$. 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 $Y(\omega) = H(\omega)X(\omega)$ only when the system is LTI. (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, you can see the the frequency region of support (i.e. those set, $R_y$, of frequencies for which $Y(\omega)$ is non-zero), $R_y$, for the output $Y(\omega)$ is the intersection of that of the regions of support $R_x$ and $R_h$ for $X(\omega)$ and $H(\omega)$: $$R_y = R_x \cap R_h$$ 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 supports of $X(\omega)$ and $Y(\omega)$., 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 an 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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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.
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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