10/10,13,16,17 - Fourier Transform Properties: Difference between revisions
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|<math>=e^{-j\,2\pi f\,t_0}\,F[x(t)]</math> |
|<math>=e^{-j\,2\pi f\,t_0}\,F[x(t)]</math> |
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|Why isn't this <math>F[x(u)]</math> |
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===Frequency Shifting=== |
===Frequency Shifting=== |
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{| border="0" cellpadding="0" cellspacing="0" |
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===The Game (frequency domain)=== |
===The Game (frequency domain)=== |
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*You can play the game in the frequency or time domain, but not both at |
*You can play the game in the frequency or time domain, but it's not advisable to play it in both at same time |
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**Then how can you use the Fourier Transform, but can't build up to it? |
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{| border="1" cellpadding="5" cellspacing="0" |
{| border="1" cellpadding="5" cellspacing="0" |
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|Proportionality |
|Proportionality |
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|<math> \int_{-\infty}^{\infty}x(\lambda)\cdot 1\cdot e^{j\,2\,\pi f\,\lambda}\,d\lambda=X( |
|<math> \int_{-\infty}^{\infty}x(\lambda)\cdot 1\cdot e^{j\,2\,\pi f\,\lambda}\,d\lambda=X(f)</math> |
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|<math> \Longrightarrow </math> |
|<math> \Longrightarrow </math> |
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|<math> \int_{-\infty}^{\infty}x(\lambda)\cdot H(f)\cdot e^{j\,2\,\pi f\,\lambda}\,d\lambda=X( |
|<math> \int_{-\infty}^{\infty}x(\lambda)\cdot H(f)\cdot e^{j\,2\,\pi f\,\lambda}\,d\lambda=X(f)\,H(f)</math> |
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|Superposition |
|Superposition |
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*Having trouble seeing <math>F\left[x(t)*h(t)\right]=X(f)\cdot H(f)</math> |
*Having trouble seeing <math>F\left[x(t)*h(t)\right]=X(f)\cdot H(f)</math> |
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*Since we were dealing in the frequency domain, is that the reason why multiplying one side did not result in a convolution on the other? |
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===The Game (Time Domain??)=== |
===The Game (Time Domain??)=== |
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|<math>\int_{-\infty}^{\infty} h(\lambda)\cdot e^{j\,2\,\pi f_0\,(t-\lambda)}\,d\lambda</math> |
|<math>\int_{-\infty}^{\infty} h(\lambda)\cdot e^{j\,2\,\pi f_0\,(t-\lambda)}\,d\lambda</math> |
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|<math>d\lambda\,\!</math> from [[10/3,6 - The Game]] |
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|Why d lambda instead of dt? |
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|<math> \Longrightarrow </math> |
|<math> \Longrightarrow </math> |
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|<math>X(f_0)\cdot e^{j\,2\,\pi f_0\,t}\,H(f_0)</math> |
|<math>X(f_0)\cdot e^{j\,2\,\pi f_0\,t}\,H(f_0)</math> |
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|Proportionality |
|Proportionality |
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|<math> \int_{-\infty}^{\infty}X(f_0)\cdot e^{j\,2\,\pi f_0\,t}\,d f_0=x(t)</math> |
|<math> \int_{-\infty}^{\infty}X(f_0)\cdot e^{j\,2\,\pi f_0\,t}\,d f_0=x(t)</math> |
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|<math> \Longrightarrow </math> |
|<math> \Longrightarrow </math> |
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|<math>\int_{-\infty}^{\infty}X(f_0)H(f_0)\cdot e^{j\,2\,\pi f_0\,t}\,d f_0=F^{-1}\left[X(f)H(f)\right]</math> |
|<math>\int_{-\infty}^{\infty}X(f_0)H(f_0)\cdot e^{j\,2\,\pi f_0\,t}\,d f_0=F^{-1}\left[X(f)H(f)\right]</math> |
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|Superposition |
|Superposition |
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===Relation to the Fourier Series=== |
===Relation to the Fourier Series=== |
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|<math>=\sum_{n=1}^{\infty}\alpha_{-n} e^{-j\,2\pi n\,t/T}+\alpha_0+\sum_{n=1}^{\infty}\alpha_n e^{j\,2\pi n\,t/T}</math> |
|<math>=\sum_{n=1}^{\infty}\alpha_{-n} e^{-j\,2\pi n\,t/T}+\alpha_0+\sum_{n=1}^{\infty}\alpha_n e^{j\,2\pi n\,t/T}</math> |
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|Let <math>n=-m\,\!</math> and reverse the order of summation |
| Let <math>n=-m\,\!</math> and reverse the order of summation |
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|Note that <math>\alpha_{-n} e^{-j\,2\pi n\,t/T}</math> is the complex conjugate of <math>\alpha_n e^{j\,2\pi n\,t/T}</math> |
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|<math>=\alpha_0+2\Re\left[\sum_{n=1}^{\infty}\alpha_n e^{j\,2\pi n\,t/T}\right]</math> |
|<math>=\alpha_0+2\Re\left[\sum_{n=1}^{\infty}\alpha_n e^{j\,2\pi n\,t/T}\right]</math> |
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|<math>x(t)+x(t)^*=2 \,\Re \left[x(t)\right]</math> |
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|<math>=\alpha_0+\sum_{n=1}^{\infty}2\Re\left[\left|\alpha_n\right| e^{j\left(\,2\pi n\,t/T+\theta_n\right)}\right]</math> |
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|<math>=\alpha_0+\sum_{n=1}^{\infty}2\left|\alpha_n\right|\cos\left(\frac{2\pi n\,t}{T}+\theta_n\right)</math> |
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*Does the server reset every hour? |
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*How can we assume that the answer exists in the real domain? |
*How can we assume that the answer exists in the real domain? |
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===Aside: Polar coordinates=== |
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Remember from [[10/02 - Fourier Series]] that <math> \alpha_n = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j\,2\,\pi \,n\,t/T}\, dt</math> |
Remember from [[10/02 - Fourier Series]] that <math> \alpha_n = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j\,2\,\pi \,n\,t/T}\, dt</math> |
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*Rectangular coordinates: <math>a+j\,b\!</math> |
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*<math>\alpha_n=\left| \alpha_n \right|\,e^{j\theta}</math>? |
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*Polar coordinates: <math>\left| a+j\,b \right|\,e^{j\theta}</math> |
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*Rest of page |
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*<math>\theta = tan^{-1} \frac{b}{a}</math> |
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{| border="0" cellpadding="0" cellspacing="0" |
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|<math>a+j\,b\,\!</math> |
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|<math>=\sqrt{a^2+b^2}\left(\cos(\theta)+j\,\sin(\theta)\right)</math> |
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|<math>=\sqrt{a^2+b^2}\left(e^{j\,\theta}\right)</math> |
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|<math>=\sqrt{a^2+b^2}\left(e^{j\,tan^{-1} \left(\frac{b}{a}\right)}\right)</math> |
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|<math>=\left|a+j\,b\right|\,e^{j\,tan^{-1} \left(\frac{b}{a}\right)}</math> |
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===Building up to <math>F\left[u(t)\right]</math>=== |
===Building up to <math>F\left[u(t)\right]</math>=== |
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{| border="0" cellpadding="0" cellspacing="0" |
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|<math>e^{i \pi}\,\!</math> |
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|<math>= \cos \pi + i \sin \pi.\,\!</math> |
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|Euler's Identity |
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|<math>r_0(t)\,\!</math> |
|<math>r_0(t)\,\!</math> |
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|<math>=-r_0(-t)\,\!</math> |
|<math>=-r_0(-t)\,\!</math> |
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|Real odd function of t |
|Real odd function of t |
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|<math>F\left[r_0(t)\right]\,\!</math> |
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|<math>=\int_{-\infty}^{\infty}r_0(t)\,e^{-j\,2\pi\,f\,t}\,dt</math> |
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|<math>=\int_{-\infty}^{\infty}r_0(t)\,\left[ \cos (-2\pi\,f\,t) + j \sin (-2\pi\,f\,t) \right ]\,dt</math> |
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|<math>=\int_{-\infty}^{\infty}r_0(t)\,\left[ \cos (2\pi\,f\,t) - j \sin (2\pi\,f\,t) \right ]\,dt</math> |
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|<math>\cos(-x) = \cos(x)\,\!</math> & <math>\sin(-x) = -\sin(x)\,\!</math> |
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|<math>=j\int_{-\infty}^{\infty}r_0(t)\, \sin (-2\pi\,f\,t)\,dt</math> |
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|<math>r_0(t)\cdot j \sin (-2\pi\,f\,t)</math> = Real odd. Integrates out over symmetric limits. |
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|<math>=\,\!</math>Imaginary Odd function of <math>f\,\!</math> |
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|<math>r_e(t)\,\!</math> |
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|<math>=r_e(-t)\,\!</math> |
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|Real even function of t |
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|<math>F\left[r_e(t)\right]\,\!</math> |
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|<math>=\int_{-\infty}^{\infty}r_e(t)\,e^{-j\,2\pi\,f\,t}\,dt</math> |
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|<math>=\int_{-\infty}^{\infty}r_e(t)\,\left[ \cos (-2\pi\,f\,t) + j \sin (-2\pi\,f\,t) \right ]\,dt</math> |
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|<math>=\int_{-\infty}^{\infty}r_e(t)\,\left[ \cos (2\pi\,f\,t) - j \sin (2\pi\,f\,t) \right ]\,dt</math> |
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|<math>\cos(-x) = \cos(x)\,\!</math> & <math>\sin(-x) = -\sin(x)\,\!</math> |
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|<math>=\int_{-\infty}^{\infty}r_e(t)\, \cos(-2\pi\,f\,t)\,dt</math> |
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|<math>r_e(t)\cdot \cos (-2\pi\,f\,t)</math> = Real odd. Integrates out over symmetric limits. |
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|<math>=\,\!</math>Real Even function of <math>f\,\!</math> |
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===Definitions=== |
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{| border="0" cellpadding="0" cellspacing="0" |
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|<math>x(t)\,\!</math> |
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|<math>=x_e(t)+x_o(t)\,\!</math> |
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|Can't x(t) have parts that aren't even or odd? You can break any function down into a Taylor series. There are even and odd powers in the series. |
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|<math>x_e(t)\,\!</math> |
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|<math>=\frac{x(t)+x(-t)}{2}</math> |
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|<math>x_o(t)\,\!</math> |
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|<math>=\frac{x(t)-x(-t)}{2}</math> |
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|<math>u(t)\,\!</math> |
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|<math>=\frac{1+\sgn (t)}{2}</math> |
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|<math>\sgn (t) = \begin{cases} 1, & t>0 \\ 0, & t=0 \\ -1, & t<0\end{cases}</math> |
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|<math>u_e(t)\,\!</math> |
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|<math>=\frac{1}{2}</math> |
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|<math>u_o(t)\,\!</math> |
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|<math>=\frac{\sgn (t)}{2}</math> |
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===<math>F\left[u(t)\right]</math>=== |
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{| border="0" cellpadding="0" cellspacing="0" |
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|<math>F\left[\frac{1}{2}\right]</math> |
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|<math>=\int_{-\infty}^{\infty} \frac{1}{2} e^{-j\,2\,\pi\,f\,t}\,dt</math> |
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|<math>=\frac{1}{2} \delta (f)</math> |
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|<math>F\left[\frac{\sgn (t)}{2}\right]</math> |
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|<math>=\int_{-\infty}^{\infty} \frac{\sgn (t)}{2} e^{-j\,2\,\pi\,f\,t}\,dt</math> |
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|<math>=\frac{1}{2}\left[\int_{-\infty}^{0} -1\cdot e^{-j\,2\,\pi\,f\,t}\,dt+\int_{0}^{\infty} 1\cdot e^{-j\,2\,\pi\,f\,t}\,dt\right]</math> |
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|<math>=\frac{1}{2}\left[\int_{0}^{-\infty} 1\cdot e^{-j\,2\,\pi\,f\,t}\,dt+\int_{0}^{\infty} 1\cdot e^{-j\,2\,\pi\,f\,t}\,dt\right]</math> |
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|<math>=\underbrace{\frac{1}{2}\int_{0}^{\infty} -e^{j\,2\,\pi\,f\,u}\,du}_{\begin{matrix}u=-t \\ du=-dt\end{matrix}}+\underbrace{\frac{1}{2}\int_{0}^{\infty} e^{-j\,2\,\pi\,f\,u}\,du}_{\begin{matrix}u=t \\ du=dt\end{matrix}}</math> |
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|<math>=\int_{0}^{\infty} \frac{-e^{j\,2\,\pi\,f\,u} + e^{-j\,2\,\pi\,f\,u}}{2}\,du</math> |
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|<math>=\int_{0}^{\infty} -j\,\frac{e^{j\,2\,\pi\,f\,u} - e^{-j\,2\,\pi\,f\,u}}{2j}\,du</math> |
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|<math>=\int_{0}^{\infty} -j\,\sin(2\,\pi\,f\,u)\,du</math> |
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|<math>\ne \int_{0}^{\infty} \cos(2\,\pi\,f\,u)\,du</math> |
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Latest revision as of 16:47, 3 December 2008
Properties of the Fourier Transform
Linearity
Time Invariance (Delay)
Let and | ||
Why isn't this |
Frequency Shifting
Double Sideband Modulation
Differentiation in Time
Thus is a linear filter with transfer function |
The Game (frequency domain)
- You can play the game in the frequency or time domain, but it's not advisable to play it in both at same time
Input | LTI System | Output | Reason |
Given | |||
Proportionality | |||
Superposition | |||
Time Invariance | |||
Proportionality | |||
Superposition |
- Having trouble seeing
The Game (Time Domain??)
Input | LTI System | Output | Reason |
Proportionality | |||
from 10/3,6 - The Game | |||
Proportionality | |||
Superposition |
Relation to the Fourier Series
Let and reverse the order of summation | ||
Note that is the complex conjugate of | ||
- How can we assume that the answer exists in the real domain?
Aside: Polar coordinates
Remember from 10/02 - Fourier Series that
- Rectangular coordinates:
- Polar coordinates:
Building up to
Euler's Identity | ||
Real odd function of t | ||
& | ||
= Real odd. Integrates out over symmetric limits. | ||
Imaginary Odd function of | ||
Real even function of t | ||
& | ||
= Real odd. Integrates out over symmetric limits. | ||
Real Even function of |
Definitions
Can't x(t) have parts that aren't even or odd? You can break any function down into a Taylor series. There are even and odd powers in the series. | ||