HW

• Signals and Systems

An Introduction to the Fourier Transform

A Fourier Transform is a representation of a function using a large number of sinusoids added together to create it.

For example, a square wave could be represented by: $x_{\mathrm{s}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{r}\mathrm{e}}(t)=\frac{4}{\pi }\sum _{k=1}^{\mathrm{\infty }}\frac{\sin ((2k-1)2\pi ft)}{(2k-1)}$

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 $\underset{T_{\zeta }\to \mathrm{\infty }}{\lim }\gamma (t)=\beta (t)$
The Fourier Series representation of $\gamma (t)$ is
$\gamma (t)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_k{e}^{j2\pi fkt}$ where $f=\frac{1}{T_{\zeta }}$
and ${\alpha }_k=\frac{1}{T_{\zeta }}\underset{-\frac{T_{\zeta }}{2}}{\overset{\frac{T_{\zeta }}{2}}{\int }}\gamma (t)e^{-j2\pi kt}dt$
${\alpha }_k$ can now be rewritten as ${\alpha }_k=\frac{1}{T_{\zeta }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\beta (t)e^{-j2\pi kt}dt$
From our initial identity then, we can write ${\alpha }_k$ as ${\alpha }_k=\frac{1}{T_{\zeta }}\mathrm{B}(kf)$
and $\gamma (t)$ becomes $\gamma (t)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{T_{\zeta }}\mathrm{B}(kf)e^{j2\pi fkt}$
Now remember that $\beta (t)=\underset{T_{\zeta }\to \mathrm{\infty }}{\lim }\gamma (t)$ and $\frac{1}{T_{\zeta }}=f.$
Which means that $\beta (t)=\underset{f\to 0}{\lim }\gamma (t)=\underset{f\to 0}{\lim }\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}f\mathrm{B}(kf)e^{j2\pi fkt}$
Which is just to say that $\beta (t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mathrm{B}(f)e^{j2\pi fkt}df$

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

Some Useful Fourier Transform Pairs

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

${ℱ}[c_1\alpha (t)+c_2\beta (t)]$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(c_1\alpha (t)+c_2\beta (t))e^{-j2\pi ft}dt$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{c}_1\alpha (t)e^{-j2\pi ft}dt+\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{c}_2\beta (t)e^{-j2\pi ft}dt$
	$=c_1\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\alpha (t)e^{-j2\pi ft}dt+c_2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\beta (t)e^{-j2\pi ft}dt=c_1\mathrm{A}(f)+c_2\mathrm{B}(f)$

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

A Second Approach to Fourier Transforms

• Fourier Transforms