HW

• Signals and Systems

An Introduction to the Fourier Transform

Unfortunately, the Fourier Transform isn't a Transformer. If it was, you would have seen it in the movie that came out lately.
One way to explain a Fourier Transform is to say it's a bunch of sinusoids added to create a just about any function you want. Another way to describe it is to say it's a way of representing a function in the frequency domain instead of the time domain.
For example, a square wave could be represented by: $x_{\mathrm {square} }(t)={\frac {4}{\pi }}\sum _{k=1}^{\infty }{\sin {\left((2k-1)2\pi ft\right)} \over (2k-1)}$

That's alot of numbers seemingly out of the blue, at first observance. I bet your questions are:

1. What is a Fourier Transform?
2.How do I get one?
3.If I've got one, what can I do with it?

X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt

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Suppose that we have some function, say $\beta (t)$, that is nonperiodic and finite in duration.
This means that $\beta (t)=0$ for some $T_{\alpha }<\left|t\right|$

Now let's make a periodic function $\gamma (t)$ by repeating $\beta (t)$ with a fundamental period $T_{\zeta }$. Note that $\lim _{T_{\zeta }\to \infty }\gamma (t)=\beta (t)$
The Fourier Series representation of $\gamma (t)$ is
$\gamma (t)=\sum _{k=-\infty }^{\infty }\alpha _{k}e^{j2\pi fkt}$ where $f={1 \over T_{\zeta }}$
and $\alpha _{k}={1 \over T_{\zeta }}\int _{-{T_{\zeta } \over 2}}^{T_{\zeta } \over 2}\gamma (t)e^{-j2\pi kt}\,dt$
$\alpha _{k}$ can now be rewritten as $\alpha _{k}={1 \over T_{\zeta }}\int _{-\infty }^{\infty }\beta (t)e^{-j2\pi kt}\,dt$
From our initial identity then, we can write $\alpha _{k}$ as $\alpha _{k}={1 \over T_{\zeta }}\mathrm {B} (kf)$
and $\gamma (t)$ becomes $\gamma (t)=\sum _{k=-\infty }^{\infty }{1 \over T_{\zeta }}\mathrm {B} (kf)e^{j2\pi fkt}$
Now remember that $\beta (t)=\lim _{T_{\zeta }\to \infty }\gamma (t)$ and ${1 \over {T_{\zeta }}}=f.$
Which means that $\beta (t)=\lim _{f\to 0}\gamma (t)=\lim _{f\to 0}\sum _{k=-\infty }^{\infty }f\mathrm {B} (kf)e^{j2\pi fkt}$
Which is just to say that $\beta (t)=\int _{-\infty }^{\infty }\mathrm {B} (f)e^{j2\pi fkt}\,df$

So we have that ${\mathcal {F}}[\beta (t)]=\mathrm {B} (f)=\int _{-\infty }^{\infty }\beta (t)e^{-j2\pi ft}\,dt$
Further ${\mathcal {F}}^{-1}[\mathrm {B} (f)]=\beta (t)=\int _{-\infty }^{\infty }\mathrm {B} (f)e^{j2\pi fkt}\,df$

Some Useful Fourier Transform Pairs

${\mathcal {F}}[\alpha (t)]={\frac {1}{\mid \alpha \mid }}f({\frac {\omega }{\alpha }})$

${\mathcal {F}}[c_{1}\alpha (t)+c_{2}\beta (t)]$	$=\int _{-\infty }^{\infty }(c_{1}\alpha (t)+c_{2}\beta (t))e^{-j2\pi ft}\,dt$
	$=\int _{-\infty }^{\infty }c_{1}\alpha (t)e^{-j2\pi ft}\,dt+\int _{-\infty }^{\infty }c_{2}\beta (t)e^{-j2\pi ft}\,dt$
	$=c_{1}\int _{-\infty }^{\infty }\alpha (t)e^{-j2\pi ft}\,dt+c_{2}\int _{-\infty }^{\infty }\beta (t)e^{-j2\pi ft}\,dt=c_{1}\mathrm {A} (f)+c_{2}\mathrm {B} (f)$

${\mathcal {F}}[\alpha (t-\gamma )]=e^{-j2\pi f\gamma }\mathrm {A} (f)$
${\mathcal {F}}[\alpha (t)*\beta (t)]=\mathrm {A} (f)\mathrm {B} (f)$
${\mathcal {F}}[\alpha (t)\beta (t)]=\mathrm {A} (f)*\mathrm {B} (f)$
Some other usefull pairs can be found here: Fourier Transforms

A Second Approach to Fourier Transforms

• Fourier Transforms