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> \ | *<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 \ | *<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>=\left \langle x(t) \mid e^{j2\pi ft}\right \rangle_t</math> | |<math>=\left \langle x(t) \mid e^{j2\pi ft}\right \rangle_t</math> | ||
|- | |- | ||
|<math>F^{-1}[ | |<math>F^{-1}[X(f)]\,\!</math> | ||
|<math>=x(t)\,\!</math> | |<math>=x(t)\,\!</math> | ||
|<math>=\int_{-\infty}^{\infty} X(f) e^{j2\pi ft}df</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_f</math> | |<math>=\left \langle X(f) \mid e^{-j2\pi ft}\right \rangle_f</math> | ||
|} | |} | ||
==Examples== | ==Examples== | ||
{| border="0" cellpadding="0" cellspacing="0" | {| border="0" cellpadding="0" cellspacing="0" | ||
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<math>x(t)=\int_{-\infty}^{\infty} x(u)\delta(t-u) du</math> | <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>. | *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
Assuming the function is perodic with the period T | ||
Fourier Transform
Remember from 10/02 - Fourier Series
If we let
Remember | ||
Definitions
Examples
Sifting property of the delta function
The dirac delta function is defined as any function, denoted as , that works for all variables that makes the following equation true:
- When dealing with , it behaves slightly different than dealing with . When dealing with , note that the delta function is . The is tacked onto the front. Thus, when dealing with , you will often need to multiply it by to cancel out the .
More properties of the delta function
Let and | ||