3 - Non-periodic Functions

Look carefully at the signs in your exponential for the Fourier transform ($e^{-j2\pi ft}$) and its inverse ($e^{j2\pi ft}$). It is correct in the integral form, but not in the bra-ket notation that follows... -Brandon

Lets look at what happens if our signals are not periodic. We can achieve this by setting our period T to infinity such that $\underset{T\to \mathrm{\infty }}{\lim }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}(1/T\underset{-T/2}{\overset{T/2}{\int }}x(t^{\text{'}})e^{-j2\pi n{t}^{\text{'}}/T}d{t}^{\text{'}})e^{j2\pi nt/T},$
where
$1/T\underset{-T/2}{\overset{T/2}{\int }}x(t^{\text{'}})e^{-j2\pi n{t}^{\text{'}}/T}d{t}^{\text{'}}={\alpha }_n$
first we need to remove the restiction x(t) = x(t + T) by following these steps.
1/T $⟹df$
n/T $⟹df$
$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}1/T⟹\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}()df$
${\alpha }_n⟹X(f)$
this leads us to the equation
$x(t)=\underset{T\to \mathrm{\infty }}{\lim }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}(1/T\underset{-T/2}{\overset{T/2}{\int }}x(t^{\text{'}})e^{-j2\pi n{t}^{\text{'}}/T}d{t}^{\text{'}})e^{j2\pi nt/T},$
and if we replace n/T with f and take the integral with respect to f we get
$x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})e^{-j2\pi f{t}^{\text{'}}}d{t}^{\text{'}})e^{j2\pi ft}df,$
where $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})e^{-j2\pi f{t}^{\text{'}}}d{t}^{\text{'}}=X(f)$
simplifying the equation to
$x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)e^{j2\pi ft}df=<X(f)|e^{j2\pi ft}>$ = Inverse Fourier Transform
and
$X(f)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j2\pi ft}dt=<x(t)|e^{j2\pi ft}>=$Fourier Transform
Now using the x(t) equation and rearranging it gives us $x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})e^{-j2\pi f{t}^{\text{'}}}d{t}^{\text{'}})e^{j2\pi ft}df=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f(t-t^{\text{'}})}df)d{t}^{\text{'}}$
where
$e^{j2\pi f(t-t^{\text{'}})}=\delta (t-t^{\text{'}})$
Similarly for X(f)
$X(f)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f^{\text{'}})e^{j2\pi f^{\text{'}}t}d{f}^{\text{'}})e^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f^{\text{'}})(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi t(f^{\text{'}}-f)}dt)d{f}^{\text{'}}$
where
$e^{j2\pi t(f^{\text{'}}-f)}=\delta (f^{\text{'}}-f)=\delta (f-f^{\text{'}})$

This works out nicely for us in both the time and frequency domain because this give us the inpulse function for both where they are non-zero only when t = t' or f = f' depending on which equation you use