MATH 414 Lecture 15

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Fourier Transforms

Given $f(t)$, we define ${\hat {f}}(\lambda )$ as

${\begin{aligned}{\mathcal {F}}[f](\lambda )={\hat {f}}(\lambda )&={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }f(t)\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t\\{\mathcal {F}}^{-1}[{\hat {f}}](t)={\frac {f(t+)+f(t-)}{2}}&={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }{\hat {f}}(\lambda )\,\mathrm {e} ^{i\,\lambda \,t}\,\mathrm {d} t\end{aligned}}$

Example

For example, $f(t)=\theta (t)\,\mathrm {e} ^{-t}$, where $\theta (t)$ is the Heaviside function.

${\begin{aligned}{\mathcal {F}}[f(t)](\lambda )&={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }\theta (t)\,\mathrm {e} ^{-t}\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t\\&={\frac {1}{\sqrt {2\pi }}}\,\int _{0}^{\infty }\mathrm {e} ^{-t-i\,\lambda \,t}\,\mathrm {d} t\\&={\frac {1}{\sqrt {2\pi }}}\,\left({\frac {1}{1+i\,\lambda }}\right)-{\frac {1}{1+i\,\lambda }}\,\lim _{t\to \infty }\mathrm {e} ^{-\left(1+i\,\lambda \right)\,t}\\&={\frac {1}{\sqrt {2\pi }}}\,{\frac {1}{1+i\,\lambda }}\end{aligned}}$

Properties

Linearity

${\mathcal {F}}$ and ${\mathcal {F}}^{-1}$ are linear transformations. That is, if $\alpha$ and $\beta$ are scalars, then

Theorem. ${\mathcal {F}}\left[\alpha \,f+\beta \,g\right]=\alpha \,{\mathcal {F}}\left[f\right]+\beta \,{\mathcal {F}}\left[g\right]$

Proof.

${\begin{aligned}{\mathcal {F}}\left[\alpha \,f+\beta \,g\right]&={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }(\alpha \,f+\beta \,g)\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t\\&=\alpha \left({\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }f(t)\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t\right)+\beta \left({\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }g(t)\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t\right)\\&=\alpha \,{\mathcal {F}}[f]+\beta \,{\mathcal {F}}[g]\end{aligned}}$

quod erat demonstrandum

Product of Powers

(Proven by Leibniz's Rule)

${\mathcal {F}}\left[t^{n}\,f(t)\right]=i^{n}\,{\frac {\mathrm {d} ^{n}{\hat {f}}}{\mathrm {d} \lambda ^{n}}}$

(and its inverse)

${\mathcal {F}}^{-1}\left[\lambda ^{n}\,{\hat {f}}(\lambda )\right]=(-i)^{n}\,{\frac {\mathrm {d} ^{n}f}{\mathrm {d} t^{n}}}$

Derivatives

(Proven with Integration by parts)

${\mathcal {F}}\left[f^{(n)}(t)\right]=\left(i\,\lambda \right)^{n}\,{\hat {f}}(\lambda )$

(and its inverse)

${\mathcal {F}}^{-1}\left[{\frac {\mathrm {d} ^{n}{\hat {f}}}{\mathrm {d} \lambda ^{n}}}\right](t)=\left(-i\,t\right)^{n}\,f(t)$

Translation / Shift

Given $f(t)$, suppose we want to find $f(t-a)$ (shift to right by $a$ units)

Theorem. ${\mathcal {F}}\left[f(t-a)\right]=\mathrm {e} ^{-i\,\lambda \,a}\,{\mathcal {F}}\left[f\right]$

Proof.

${\mathcal {F}}\left[f(t-a)\right]={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }f(t-a)\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t$

Let $\tau =t-a$, then

${\mathcal {F}}\left[f(\tau )\right]={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }f(\tau )\,\mathrm {e} ^{-i\,\lambda \,(\tau +a)}\,\mathrm {d} t=\mathrm {e} ^{-i\,\lambda \,\tau }{\hat {f}}(\tau )$

quod erat demonstrandum

Translation in spatial domain is change of phase in time domain

Scaling

${\begin{aligned}{\mathcal {F}}\left[f(b\,t)\right]&={\frac {1}{b}}\,{\hat {f}}\left({\frac {\lambda }{b}}\right)\\{\mathcal {F}}^{-1}\left[{\hat {f}}(c\,\lambda )\right]&={\frac {1}{c}}\,f\left({\frac {t}{c}}\right)\end{aligned}}$

Laplace Transform

Let $f(t)=\theta (t)\,g(t)$, where $\theta (t)$ is the Heaviside function. In other words, if $f(t)=0$ for $t<0$, then

Theorem. ${\mathcal {F}}\left[f(t)\right]={\frac {1}{\sqrt {2\pi }}}\,{\mathcal {L}}\left[f(t)\right]\left(i\,\lambda \right)$

Proof.

${\mathcal {F}}\left[f(t)\right]={\frac {1}{\sqrt {2\pi }}}\,\int _{0}^{\infty }g(t)\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t$

if we let $s=i\,\lambda$, we see that ${\mathcal {F}}[f]={\frac {1}{\sqrt {2\pi }}}\,{\mathcal {L}}\left\{f\right\}(s)={\frac {1}{\sqrt {2\pi }}}\,{\mathcal {L}}\left\{f\right\}\left(i\,\lambda \right)$

quod erat demonstrandum

Example

The fourier transform of the tent function $f(t)=\pi -\left|x\right|$ for $x\in \left[-\pi ,\pi \right]$ is ${\mathcal {F}}\left[f\right]={\sqrt {\frac {2}{\pi }}}\,\left({\frac {1-\cos {(\pi \,\lambda )}}{\lambda ^{2}}}\right)$.

find the Fourier transform of $g(t)={\begin{cases}1&x\in \left[-\pi ,0\right]\\-1&x\in \left[0,\pi \right]\\0&{\mbox{otherwise}}\end{cases}}$

Observe that $g(t)=f'(t)$, so ${\mathcal {F}}\left[g\right]=i\,\lambda \,{\mathcal {F}}\left[f\right]=i\,{\sqrt {\frac {2}{\pi }}}\,\left({\frac {1-\cos {(\pi \,\lambda )}}{\lambda }}\right)$