When determining $y(n) = (x * h)(n)$, what if we use $x(n) = e^{i\omega n}$?

\begin{align*}
y(n) &= \sum \limits_{k\in\Z}h(k)x(n-k) \\
y(n) &= \sum \limits_{k\in\Z}h(k)e^{i \omega (n-k)} \\ 
y(n) &= \left( \sum \limits_{k\in\Z}h(k)e^{-i \omega k} \right) e^{i \omega n} \\ 
y(n) &= H(\omega)e^{i \omega n}
\end{align*}

Since the summation part of the formula has no dependency on $n$, it relies entirely on the frequency $\omega$, thus $H(\omega)$ is the \emph{frequency response}. $H(\omega)$ is generally complex-valued. The magnitude $\left| H(\omega) \right|$ is known as the \emph{magnitude response}, or magnitude of frequency response, and the phase $\angle H(\omega)$ is the \emph{phase response}.

A system where the input signal is a complex exponential can be described as follows:

$$ e^{i \omega n} \to \fbox{H} \to H(\omega) e^{i \omega n}$$

We say that complex exponentials are eigenfunctions of LTI systems($A\vec{v}=\lambda\vec{v}$). Note that $\omega$ plots are continuous, even though the output signal is in discrete time. Generally we plot $\omega$ in $2\pi$ intervals from $-\pi$ to $\pi$.

\begin{problem}
Can the system be LTI?
$$ e^{i\frac{\pi}{4}n} \to \fbox{F} \to e^{i\frac{\pi}{3}n} $$
\end{problem}

\begin{solution}
Not linear time-invariant, since LTI systems cannot create new frequencies. They can, however, eliminate frequencies in the input.
\end{solution}


Frequency Response