Waves/Fourier Transforms

Fourier Transform

So far, you've learned how to superpose a finite number of sinusoidal waves. However, a wave in general can't be expressed as the sum of a finite number of sines and cosines. Fortunately, we have a theorem called Fourier's theorem which basically states that under certain technical assumptions, any function, f(x) is equal to an integral over sines and cosines. In other words,

$f(x)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(c_1(k)\cos (kx)+c_2(k)\sin (kx))dk$.

Now, if we're given the wave function when t=0, φ(x,0) and the velocity of each sine wave as a function of its wave number, v(k), then we can compute φ(x,t) for any t by taking the inverse Fourier transform of φ(x,0) conducting a phase shift, and then taking the Fourier transform.

Fortunately, the inverse Fourier transform is very similar to the Fourier transform itself.

$c_1(k)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)\cos (kx)dx{c}_2(k)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)\sin (kx)dx$

This tells us that, since waves which are very spread out, like the sine wave, have a narrow range of wave numbers, wave functions whose wave numbers are very spread out will only be significant at a narrow range of positions.

Fourier Transform Properties

	Signal	Fourier transform unitary, angular frequency	Fourier transform unitary, ordinary frequency	Remarks
	$g(t)\equiv$ $\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}G(\omega )e^{i\omega t}d\omega$	$G(\omega )\equiv$ $\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t)e^{-i\omega t}dt$	$G(f)\equiv$ $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t)e^{-i2\pi ft}dt$
1	$a\cdot g(t)+b\cdot h(t)$	$a\cdot G(\omega )+b\cdot H(\omega )$	$a\cdot G(f)+b\cdot H(f)$	Linearity
2	$g(t-a)$	$e^{-ia\omega }G(\omega )$	$e^{-i2\pi af}G(f)$	Shift in time domain
3	$e^{iat}g(t)$	$G(\omega -a)$	$G(f-\frac{a}{2\pi })$	Shift in frequency domain, dual of 2
4	$g(at)$	$\frac{1}{|a|}G\left(\frac{\omega }{a}\right)$	$\frac{1}{|a|}G\left(\frac{f}{a}\right)$	If $|a|$ is large, then $g(at)$ is concentrated around 0 and $\frac{1}{|a|}G\left(\frac{\omega }{a}\right)$ spreads out and flattens
5	$G(t)$	$g(-\omega )$	$g(-f)$	Duality property of the Fourier transform. Results from swapping "dummy" variables of $t$ and $\omega$.
6	$\frac{d^{n}g(t)}{d{t}^n}$	$(i\omega {)}^{n}G(\omega )$	$(i2\pi f{)}^{n}G(f)$	Generalized derivative property of the Fourier transform
7	$t^{n}g(t)$	$i^n\frac{d^{n}G(\omega )}{d{\omega }^n}$	${\left(\frac{i}{2\pi }\right)}^n\frac{d^{n}G(f)}{d{f}^n}$	This is the dual to 6
8	$(g*h)(t)$	$\sqrt{2\pi }G(\omega )H(\omega )$	$G(f)H(f)$	$g*h$ denotes the convolution of $g$ and $h$ — this rule is the convolution theorem
9	$g(t)h(t)$	$\frac{(G*H)(\omega )}{\sqrt{2\pi }}$	$(G*H)(f)$	This is the dual of 8

Fourier Transform Pairs

Time Domain	Fourier Domain
$x(t)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(j\omega )e^{j\omega t}d\omega$	$X(j\omega )=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j\omega t}dt$
$1$	$2\pi \delta (\omega )$
$-0.5+u(t)$	$\frac{1}{j\omega }$
$\delta (t)$	$1$
$\delta (t-c)$	$e^{-j\omega c}$
$u(t)$	$\pi \delta (\omega )+\frac{1}{j\omega }$
$e^{-bt}u(t)$	$\frac{1}{j\omega +b}$
$\cos {\omega }_{0}t$	$\pi [\delta (\omega +{\omega }_0)+\delta (\omega -{\omega }_0)]$
$\cos ({\omega }_{0}t+\theta )$	$\pi [e^{-j\theta }\delta (\omega +{\omega }_0)+e^{j\theta }\delta (\omega -{\omega }_0)]$
$\sin {\omega }_{0}t$	$j\pi [\delta (\omega +{\omega }_0)-\delta (\omega -{\omega }_0)]$
$\sin ({\omega }_{0}t+\theta )$	$j\pi [e^{-j\theta }\delta (\omega +{\omega }_0)-e^{j\theta }\delta (\omega -{\omega }_0)]$
$rect(\frac{t}{\tau })$	$\tau sinc\frac{\tau \omega }{2\pi }$
$\tau sinc\frac{\tau t}{2\pi }$	$2\pi p_{\tau }(\omega )$
$(1-\frac{2|t|}{\tau })p_{\tau }(t)$	$\frac{\tau }{2}sin{c}^2\frac{\tau \omega }{4\pi }$
$\frac{\tau }{2}sin{c}^2(\frac{\tau t}{4\pi })$	$2\pi (1-\frac{2|\omega |}{\tau })p_{\tau }(\omega )$

Note: $sinc(x)=\sin (x)/x$ ; $p_{\tau }(t)$ is the rectangular pulse function of width $\tau$

further reading

• Engineering Analysis/Multi-Dimensional Fourier Series

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