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.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau = t-a} , then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \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{align}}

Laplace Transform

Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(t) = \theta(t) \, g(t)} , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta(t)} is the Heaviside function. In other words, if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(t) = 0} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t < 0} , then

Theorem. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{F} \left[ f(t) \right] = \frac{1}{\sqrt{2\pi}} \, \mathcal{L} \left[ f(t) \right] \left( i \, \lambda \right)}

Proof.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s = i \, \lambda} , we see that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(t) = \pi - \left| x \right|} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \left[ -\pi, \pi \right]} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{F}\left[ f \right] = \sqrt{ \frac{2}{\pi} } \, \left( \frac{1-\cos{(\pi \, \lambda)}}{\lambda^2} \right)} .

find the Fourier transform of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(t) = f'(t)} , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)}