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

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
$\lim _{T\to \infty }\sum _{n=-\infty }^{\infty }({\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t^{'})e^{\frac {-j2\pi nt^{'}}{T}}dt^{'})e^{\frac {j2\pi nt}{T}}\!$, where ${\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t^{'})e^{\frac {-j2\pi nt^{'}}{T}}dt^{'}\!$ is the $\alpha _{n}\!$ term.
We want to remove the restriction $x(t)=x(t+T)\!$, which we can do as follows.

${\frac {1}{T}}\longrightarrow df\!$

${\frac {n}{T}}\longrightarrow f\!$

$\sum _{n=-\infty }^{\infty }{\frac {1}{T}}\longrightarrow \int _{-\infty }^{\infty }(\qquad )df\!$

$\alpha _{n}\longrightarrow X(f)\!$

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