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\noindent
For $m = 2$ note $i \dfrac{}{\omega} \left(
\frac{1}{i\omega - a} \right) = \frac{1}{(i\omega
- a)^2}$ and recall (8.12) $\mathcal{F}(tf(t)) =
i\tilde{f}'(\omega)$, so
\[ \mathcal{F}^{-1} \left( \frac{1}{(i\omega -
a)^2} \right) = \begin{cases}
    te^{at} & t > 0 \\
    0 & t < 0
\end{cases} \]
By induction,
\[ \mathcal{F}^{-1} \left( \frac{1}{(i\omega -
a)^m} \right) = \begin{cases}
    \frac{t^{m - 1}}{(m - 1)!} e^{at} & t > 0 \\
    0 & t < 0
\end{cases} \tag{8.34} \]
Thus the response function takes the form
\[ R(t) = \sum_j \sum_m \Gamma_{jm} \frac{t^{m -
1}}{(m - 1)!} e^{c_j t} \quad t > 0
\tag{8.35} \]
We can solve (8.31) in Green's function form
(8.30) or directly invert $\tilde{R}(\omega)
\tilde{J}(\omega)$ for polynomial
$\tilde{J}(\omega)$.

\begin{example*}[Damped oscillator]
    Solve
    \[ \mathcal{L} y \equiv y'' + 2p y' + (p^2 +
    q^2) y = f(t) \]
    with damping $p > 0$ and homogeneous initial
    conditions $y(0) = y'(0) = 0$. Fourier
    Transform is
    \[ (i\omega)^2 \tilde{y} + 2ip\omega \tilde{y}
    + (p^2 + q^2) \tilde{y} = \tilde{f} \]
    \[ \tilde{y} = \frac{\tilde{f}}{-\omega^2 +
    2ip\omega + p^2 + q^2} \equiv \tilde{R}
    \tilde{f} \]
    Inverting with convolution theorem (8.17)
    \[ y(t) = \int_0^t r(t - \tau) f(\tau) \dd
    \tau \]
    with response
    \[ R(t - \tau) = \frac{1}{2\pi}
    \int_{-\infty}^\infty \frac{e^{i\omega(t -
    \tau)} \dd \omega}{p^2 + q^2 + 2ip\omega -
    \omega^2} \]
\end{example*}

\noindent
Exercise: Show $\mathcal{L} R(t - \tau) = \delta(t
- \tau)$ using (8.23). That is, the response
function for $R(t - \tau)$ is the Green's function
(see Example sheet 3, Q4).

\subsection{Discrete Fourier Transforms}

\subsubsection*{Discrete sampling \& the Nyquist
frequency}

Sample a signal $h(t)$ at equal times $t_n =
n\Delta$ with time-sampling $\Delta$, and values
\[ h_n = h(n\Delta), \quad n = \dots, -2, 1, 0, 1,
2, \dots \tag{8.36} \]
i.e. with sampling frequency $\frac{1}{\Delta}$
($\omega_s = 2\pi f_s = \frac{2\pi}{\Delta}$). \\
\begin{hiddenflashcard}[Nyquist-frequency]
\prompt{What is the Nyquist frequency?} \\
\[ \cloze{f_c = \frac{1}{2\Delta}} \]
\end{hiddenflashcard}
The \emph{Nyquist frequency} $f_c =
\frac{1}{2\Delta}$ (8.37) is the highest frequency
actually sampled at $\Delta$. Suppose we have a
signal with given frequency $f$.
\begin{align*}
    g_f(t)
    &= A \cos (2\pi ft + \phi) \\
    &= \Re(A e^{2\pi i ft + \phi}) \\
    &= \half (A e^{i\phi} e^{2\pi i ft} + A
    e^{-i\phi} e^{-2\pi i ft}) \tag{8.38}
\end{align*}
(i.e. for real complex Fourier Series, the sum of
positive frequencies $f$ and negative frequency
$-f$ modes). \\
What happens if we sample at Nyquist $f = f_c$?
\begin{align*}
    g_{f_c}(t_n)
    &= A \cos (2\pi \left(
    \frac{1}{2\Delta}n\Delta + \phi \right) \\
    &= A \cos \pi n \cos \phi + A \sin n\pi \sin
    \phi \\
    &= A' \cos (2\pi f_c t_n) \tag{8.39}
\end{align*}
with $A' = A \cos \phi$. So phase / amplitude
information is lost (no distinction) and we can
identify $f_c \leftrightarrow -f_c$ i.e. (8.38)
and (8.39) are \emph{aliased} together. \\
What happens if we sample above $f > f_c$?
Exercise: Take $f = f_c + \delta f > f_c$ and show
that ($\delta f < f_c$)
\begin{align*}
    g_f(t_n)
    &= A \cos (2\pi (f_c + \delta f)t_n + \phi) \\
    &= A \cos (2\pi (f_c - \delta f)t_n - \phi)
    \tag{8.40}
\end{align*}
So the effect is to \emph{alias} a ``ghost
signal'' to frequency $f_c - \delta f$ (actually -
$-(f_c - \delta f)$).

\subsubsection*{Sampling Theorem}

A signal $g(t)$ is \emph{bandwidth limited} if it
contains no frequencies above
$\omega_{\text{max}} = 2\pi f_{\text{max}}$, i.e.
$\tilde{g}(\omega) = 0$ for $|\omega| >
\omega_{\text{max}}$. So
\begin{align*}
    g(t)
    &= \frac{1}{2\pi} \int_{-\infty}^\infty
    \tilde{g}(\omega) e^{i\omega t} \dd \omega \\
    &= \frac{1}{2\pi}
    \int_{-\omega_{\text{max}}}^{\omega_{\text{max}}}
    \tilde{g}(\omega) e^{i\omega t} \dd \omega
    \tag{8.41}
\end{align*}
Set sampling to satisfy Nyquist condition
\[ \Delta = \frac{1}{2f_{\text{max}}} \]
then
\[ g_n \equiv g(t_n) = \frac{1}{2\pi}
\int_{-\omega_{\text{max}}}^{\omega_{\text{max}}}
\tilde{g}(\omega) e^{i\pi
n\omega/\omega_{\text{max}}} \dd \omega \]
which is complex Fourier series coefficient (1.13)
$c_n \times \frac{\omega_{\text{max}}}{\pi}$ ($x
\to \omega$). The Fourier Series represents a
periodic function (period $2\omega_{\text{max}}$)
\[ \tilde{g}_{\text{per}}(\omega) =
\frac{\pi}{\omega_{\text{max}}} \sum_{n =
-\infty}^\infty g_n e^{-i\pi n\omega /
\omega_{\text{max}}} \tag{8.42} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/0ba745fa69d211ed.png}
\end{center}
The actual Fourier Transform $\tilde{g}(\omega)$
is found by multiplying by a ``top hat''
\[ \tilde{h}(\omega) = \begin{cases}
    1 & |\omega| \le \omega_{\text{max}} \\
    0 & \text{otherwise}
\end{cases} \]
i.e.
\[ \tilde{g}(\omega) =
\tilde{g}_{\text{per}}(\omega) \tilde{h}(\omega)
\tag{8.43} \]
which is an \emph{exact} relation. Inverting with
(8.42):
\begin{align*}
    g(t)
    &= \frac{1}{2\pi} \int_{-\infty}^\infty
    \tilde{g}_{\text{per}}(\omega)
    \tilde{h}(\omega) e^{i\omega t} \dd \omega \\
    &= \frac{1}{2\omega_{\text{max}}} \sum_{n =
    -\infty}^\infty g_n
    \int_{-\omega_{\text{max}}}^{\omega_{\text{max}}}
    \exp \left( i\omega \left( t -
    \frac{n\pi}{\omega_{\text{max}}} \right)
    \right) \dd \omega \\
    &= \sum_{n = -\infty}^\infty g_n
    \frac{\sin(\omega_{\text{max}} t - \pi
    n)}{\omega_{\text{max}} t - \pi n} \tag{8.44}
\end{align*}
So $g(t)$ can be exactly represented after
sampling at discrete times $t_n$ (sampling
theorem).