Lets look at the frequency domain behavior of LTI systems. To do this, we pass in a pure frequency by letting $x(n) = e^{i\omega n}$, or $x(t) = e^{i\omega t}$. We know that we can write a signal as a convolution sum, so lets start with this and pass in a pure frequency:

\begin{align*}
y(n) &= \sum \limits_{k\in\Z} f(k) x(n-k) \\
y(n) &= \sum \limits_{k\in\Z} f(k) e^{i\omega(n-k)} \\
y(n) &= \sum \limits_{k\in\Z} f(k) e^{i\omega n}e^{i\omega k} \\
y(n) &= \lr{\sum \limits_{k\in\Z} f(k)e^{i\omega k}} e^{i\omega n} \\
y(n) &= F(\omega) e^{i\omega n} \\
\end{align*}

The same frequency came out of the signal, only scaled by the frequency response of the system.

\begin{nicebox}
$$ F(\omega) = \sum \limits_{k \in \Z} f(k) e^{-i\omega k} $$
\end{nicebox}


The meaning of $F(\omega)$, the frequency response, has to do with frequency content. This tells us how much the system "likes" $\omega$, by suppressing or amplifying it. This motivates the term "filter", since certain frequencies are removed, kept, or amplified. In general, $F(\omega) \in \C$, since it is a linear combination of complex exponentials.

\begin{example}
Given a the impulse response in the graph below,
\begin{enumerate}
\item Express $y(n)$ in terms of $x$
\item Determine expression for $F(\omega)$
\item Plot $\abs{F(\omega)}$
\item Plot $ \angle F(\omega)$
\end{enumerate}

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\begin{align*}
f(n) &= \frac{1}{2}\delta(n) + \frac{1}{2}\delta(n-1) \\
y(n) &= \frac{1}{2}x(n) + \frac{1}{2}x(n-1)
\end{align*}

What about the frequency response? Well, we know that $F(\omega) = \sum \limits_{\ell \in \Z} f(\ell)e^{-i\omega\ell}$ Think about how many non-zero terms there are:

\begin{align*}
F(\omega) &= \sum \limits_{\ell \in \Z} f(\ell)e^{-i\omega\ell} \\
F(\omega) &= \sum \limits_{\ell = 0}^1 f(\ell)e^{-i\omega\ell} \quad \mbox{only looking at } \ell \in \{0,1\}\\
F(\omega) &= h(0)e^{0} + h(1)e^{-i\omega}\\
F(\omega) &= \frac{1}{2} + \frac{1}{2}e^{-i\omega}
\end{align*}

Note that we could have used another method to find $F(\omega)$

\begin{align*}
y(n) &= \frac{1}{2}x(n) + \frac{1}{2}x(n-1) \\
y(n) &= \frac{1}{2}e^{i\omega n} + \frac{1}{2}e^{i\omega (n-1)} \quad \mbox{let } x(n) = e^{i\omega n} \\
y(n) &= \lr{\frac{1}{2} + \frac{1}{2}e^{-i\omega}}e^{i\omega n} \\
F(\omega)e^{i\omega n} &= \lr{\frac{1}{2} + \frac{1}{2}e^{-i\omega}}e^{i\omega n} \quad \mbox{since } y(n) = F(\omega)e^{i\omega n} \\
F(\omega) &= \frac{1}{2} + \frac{1}{2}e^{-i\omega}
\end{align*}

We can simplify this expression for $F(\omega)$ further by realizing the symmetry in the coefficients. We can factor out half of the power:

\begin{align*}
F(\omega) &= \frac{1}{2} + \frac{1}{2}e^{-i\omega} \\
F(\omega) &= e^{-i\omega/2} \left[ \frac{1}{2}e^{i\omega/2} + \frac{1}{2}e^{-i\omega/2} \right] \quad \mbox{factor out } e^{-i\omega/2} \\
F(\omega) &= e^{-i\omega/2} \frac{\left[ e^{i\omega/2} + e^{-i\omega/2} \right]}{2} \\
F(\omega) &= e^{-i\omega/2} \cos(\omega/2) \\
F(\omega) &= \cos(\omega/2) e^{-i\omega/2}
\end{align*}

Since $\cos(\omega/2)$ is not always positive, we need to take a slight detour. $F(\omega + 2\pi) = F(\omega)$ in \emph{discrete time}.

\end{example}

\begin{proof}

\begin{align*}
F(\omega + 2\pi) &= \sum \limits_{\ell \in \Z} f(\ell)e^{-i(\omega + 2\pi)\ell} \\
F(\omega + 2\pi) &= \sum \limits_{\ell \in \Z} f(\ell)e^{-i\omega} e^{-i 2\pi\ell} \\
F(\omega + 2\pi) &= \sum \limits_{\ell \in \Z} f(\ell)e^{-i\omega} \quad \mbox{since } \ell \in \Z \to e^{-i 2\pi\ell} = 1 \\
F(\omega + 2\pi) &= F(\omega)
\end{align*}

\end{proof}

So we only plot in some contiguous interval of length $2\pi$. So we have $F(\omega) = \cos(\omega/2)e^{-i\omega/2}$ This gives us $\left| F(\omega) \right| = \left| \cos(\omega/2) \right| = \cos(\omega/2) $ for $\left| \omega \right| \leq \pi$.

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To determine the phase, we can recognize a pattern by looking at the frequency response in terms of its magnitude and phase:

\begin{align*}
F(\omega) &= \abs{F(\omega)}e^{i\angle F(\omega)} 
\end{align*}

if $F(\omega) = \cos(\omega/2) e^{-i\omega/2}$, then by pattern recognition, $\abs{F(\omega)} = \cos(\omega/2)$, and $\angle F(\omega) = -\omega/2$. In this case, we have established that the cosine is positive on $(-\pi,\pi)$, but if it were to have been negative, we would have to shift the phase by $\pm \pi$.



Frequency-Domain Behavior