10/09 - Fourier Transform: Difference between revisions

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==Fourier Transform==
==Fourier Transform==
Remember from [[10/02 - Fourier Series]]
Remember from [[10/02 - Fourier Series]]
*<math> \alpha_m = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j2\pi mt/T}\, dt</math>
*<math> \alpha_n = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j\,2\,\pi \,n\,t/T}\, dt</math>
*<math>x(t) = x(t+T) = \sum_{n=-\infty}^\infty \alpha_m e^{j2\pi m/T}</math>
*<math>x(t) = x(t+T) = \sum_{n=-\infty}^\infty \alpha_n e^{j\,2\pi \,n/T}</math>


If we let <math> T \rightarrow \infty</math>
If we let <math> T \rightarrow \infty</math>
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|<math>=X(f)\,\!</math>
|<math>=X(f)\,\!</math>
|<math>=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}dt</math>
|<math>=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}dt</math>
|<math>=\left \langle x(t) \mid e^{j2\pi ft}\right \rangle</math>
|<math>=\left \langle x(t) \mid e^{j2\pi ft}\right \rangle_t</math>
|-
|-
|<math>F^{-1}[x(t)]\,\!</math>
|<math>F^{-1}[X(f)]\,\!</math>
|<math>=x(t)\,\!</math>
|<math>=x(t)\,\!</math>
|<math>=\int_{-\infty}^{\infty} X(f) e^{j2\pi ft}dt</math>
|<math>=\int_{-\infty}^{\infty} X(f) e^{j2\pi ft}df</math>
|<math>=\left \langle X(f) \mid e^{-j2\pi ft}\right \rangle</math>
|<math>=\left \langle X(f) \mid e^{-j2\pi ft}\right \rangle_f</math>
|}
|}
==Examples==
==Examples==
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>\int_{-\infty}^{\infty}e^{j2\pi ft}e^{-j2\pi f\lambda}df</math>
|<math>=\left\langle e^{j2\pi ft}\mid e^{j2\pi ft}\right\rangle_f</math>
|<math>=\delta(t-\lambda)\,\!</math>
|-
|<math>\int_{-\infty}^{\infty}e^{j2\pi tf}e^{-j2\pi tf_0}dt</math>
|<math>=\left\langle e^{j2\pi tf}\mid e^{j2\pi tf_0}\right\rangle_t</math>
|<math>=\delta(f-f_0)\,\!</math>
|}
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>F^{-1}[F[x(t)]]\,\!</math>
|<math>=\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty} x(\lambda) e^{-j2\pi f\lambda}d\lambda\right ]e^{j2\pi ft} df</math>
|<math>=\int_{-\infty}^{\infty}X(f) e^{j2\pi ft} df</math>
|<math>=x(t)\,\!</math>
|-
|
|
|<math>=\int_{-\infty}^{\infty}x(\lambda) \int_{-\infty}^{\infty}e^{j2\pi f(t-\lambda)} df d\lambda</math>
|<math>=\int_{-\infty}^{\infty}x(\lambda) \delta(t-\lambda) d\lambda</math>
|<math>=x(t)\,\!</math>
|-
|
|<math>=\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty} x(\lambda) e^{-j\omega\lambda}d\lambda\right ]e^{j\omega t} \frac{1}{2\pi}d\omega </math>
|<math>=\int_{-\infty}^{\infty}x(\lambda) \left [ \frac{1}{2\pi} \int_{-\infty}^{\infty}  e^{j(t-\omega) \lambda}d\omega\right ] d\lambda</math>
|<math>=\int_{-\infty}^{\infty}x(\lambda) \delta(t-\omega) d\lambda</math>
|<math>=x(t)\,\!</math>
|}
===Sifting property of the delta function===
The dirac delta function is defined as any function, denoted as <math> \delta(t-u)\,\!</math>, that works for all variables that makes the following equation true:
<math>x(t)=\int_{-\infty}^{\infty} x(u)\delta(t-u) du</math>
*When dealing with <math>\omega\,\!</math>, it behaves slightly different than dealing with <math>f\,\!</math>. When dealing with <math>=\int_{-\infty}^{\infty}x(\lambda) \left [ \frac{1}{2\pi} \int_{-\infty}^{\infty}  e^{j(t-\omega) \lambda}d\omega\right ] d\lambda</math>, note that the delta function is <math>\frac{1}{2\pi} \int_{-\infty}^{\infty}  e^{j(t-\omega) \lambda}d\omega</math>. The <math>\frac{1}{2\pi}</math> is tacked onto the front. Thus, when dealing with <math>\omega\,\!</math>, you will often need to multiply it by <math>2\pi\,\!</math> to cancel out the <math>\frac{1}{2\pi}</math>.
===More properties of the delta function===
<math>\delta(a\,t) = \frac{1}{\left | a \right |}</math>
<math>\delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f)</math>
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>\int_{-\infty}^{\infty}\delta(a\,t)\,dt</math>
|<math>=\int_{-\infty}^{\infty}\delta(u\,t)\,\frac{du}{\left|a\right|}</math>
|Let <math>a\,t=u</math> and <math>du=a\,dt</math>
|-
|
|<math>=\frac{1}{\left|a\right|}</math>
|}

Latest revision as of 13:49, 4 December 2008

ej2πnt/Tej2πmt/T =ej2πnt/Tej2πmt/Tdt
=ej2π(nm)t/Tdt
=T/2T/2ej2π(nm)t/Tdt Assuming the function is perodic with the period T
=Tδm,n

Fourier Transform

Remember from 10/02 - Fourier Series

  • αn=1TT/2T/2x(t)ej2πnt/Tdt
  • x(t)=x(t+T)=n=αnej2πn/T

If we let T

1T df
nT f Remember f=2πnT
T
n=1T ()df

Definitions

F[x(t)] =X(f) =x(t)ej2πftdt =x(t)ej2πftt
F1[X(f)] =x(t) =X(f)ej2πftdf =X(f)ej2πftf

Examples

ej2πftej2πfλdf =ej2πftej2πftf =δ(tλ)
ej2πtfej2πtf0dt =ej2πtfej2πtf0t =δ(ff0)
F1[F[x(t)]] =[x(λ)ej2πfλdλ]ej2πftdf =X(f)ej2πftdf =x(t)
=x(λ)ej2πf(tλ)dfdλ =x(λ)δ(tλ)dλ =x(t)
=[x(λ)ejωλdλ]ejωt12πdω =x(λ)[12πej(tω)λdω]dλ =x(λ)δ(tω)dλ =x(t)

Sifting property of the delta function

The dirac delta function is defined as any function, denoted as δ(tu), that works for all variables that makes the following equation true: x(t)=x(u)δ(tu)du

  • When dealing with ω, it behaves slightly different than dealing with f. When dealing with =x(λ)[12πej(tω)λdω]dλ, note that the delta function is 12πej(tω)λdω. The 12π is tacked onto the front. Thus, when dealing with ω, you will often need to multiply it by 2π to cancel out the 12π.

More properties of the delta function

δ(at)=1|a|

δ(ω)=δ(2πf)=12πδ(f)

δ(at)dt =δ(ut)du|a| Let at=u and du=adt
=1|a|