Class lecture notes October 5 - HW3

Max Woesner

Homework #3 - Class lecture notes October 5

The following notes are my interpretation of the material covered in class on October 5, 2009

Nonperiodic Signals

In real life, most systems have a finite time period and can be fairly easily evaluated as periodic. However, we do want to be able to evaluate nonperiodic signals for cases when this is not possible.
A nonperiodic signal can be thought of as periodic signal with an infinite period. To deal with such signals we can take the limit as the period $T$ goes to infinity, or
$\underset{T\to \mathrm{\infty }}{\lim }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}(\frac{1}{T}\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t^{\text{'}})e^{\frac{-j2\pi n{t}^{\text{'}}}{T}}d{t}^{\text{'}})e^{\frac{j2\pi nt}{T}}$, where $\frac{1}{T}\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t^{\text{'}})e^{\frac{-j2\pi n{t}^{\text{'}}}{T}}d{t}^{\text{'}}$ is the ${\alpha }_n$ term.
We want to remove the restriction $x(t)=x(t+T)$, which we can do as follows.

$\frac{1}{T}⟶df$

$\frac{n}{T}⟶f$

$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{T}⟶\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}()df$

${\alpha }_n⟶X(f)$

So $x(t)=\underset{T\to \mathrm{\infty }}{\lim }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{T}(\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t^{\text{'}})e^{\frac{-j2\pi n{t}^{\text{'}}}{T}}d{t}^{\text{'}})e^{\frac{j2\pi nt}{T}}$
Using $\frac{n}{T}=f$ and the information above, we can rewrite the equation.
$x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})e^{-j2\pi f{t}^{\text{'}}}d{t}^{\text{'}})e^{j2\pi ft}df$, where $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})e^{-j2\pi f{t}^{\text{'}}}d{t}^{\text{'}}=X(f)$
So $$\begin{gathered} x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)e^{+j2\pi ft}df= \\ <X(f)|e^{j2\pi ft}> \end{gathered}$$ This is the inverse Fourier transform.
Also, $$\begin{gathered} X(f)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j2\pi ft}dt= \\ <x(t)|e^{j2\pi ft}> \end{gathered}$$ This is the Fourier transform.
So $x(t)={{ℱ}}^{-1}[X(f)]$ and $X(f)={ℱ}[x(t)]$
From above, $x(t)=\underset{-{\mathrm{\infty }}_f}{\overset{\mathrm{\infty }}{\int }}(\underset{-{\mathrm{\infty }}_{t^{\text{'}}}}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})e^{-j2\pi f{t}^{\text{'}}}d{t}^{\text{'}})e^{j2\pi ft}df$, so
$x(t)=\underset{-{\mathrm{\infty }}_{t^{\text{'}}}}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})(\underset{-{\mathrm{\infty }}_{\delta (t-t^{\text{'}})}}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f(t-t^{\text{'}})}df)d{t}^{\text{'}}$, where $$\begin{gathered} \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f(t-t^{\text{'}})}df= \\ <e^{j2\pi ft}|e^{j2\pi f{t}^{\text{'}}}> \end{gathered}$$