View source for 10/10,13,16,17 - Fourier Transform Properties

==Properties of the Fourier Transform==
===Linearity===
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>F\left [ a\,x(t)+b\,x(t)\right ]</math>
|<math>=\int_{-\infty}^{\infty}\left[ a\,x(t)+b\,x(t)\right]e^{-j\,2\pi f\,t}\,dt</math>
|-
|
|<math>=a\int_{-\infty}^{\infty}\,x(t)\,e^{-j\,2\pi f\,t}\,dt + b\int_{-\infty}^{\infty} \,x(t)\,e^{-j\,2\pi f\,t}\,dt</math>
|-
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|<math>=a\,F[x(t)]+b\,F[x(t)]</math>
|}
===Time Invariance (Delay)===
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>F[x(t-t_0)]\,\!</math>
|<math>=\int_{-\infty}^{\infty} x(t-t_0)\,e^{-j\,2\pi f\,t}\,dt</math>
|Let <math>u=t-t_0\,\!</math> and <math> du=dt\,\!</math>
|-
|
|<math>=\int_{-\infty}^{\infty} x(u)\,e^{-j\,2\pi f\,(u+t_0)}\,du</math>
|-
|
|<math>=e^{-j\,2\pi f\,t_0}\int_{-\infty}^{\infty} x(u)\,e^{-j\,2\pi f\,u}\,du</math>
|-
|
|<math>=e^{-j\,2\pi f\,t_0}\,F[x(t)]</math>
|Why isn't this <math>F[x(u)]</math>
|}

===Frequency Shifting===
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>F\left[e^{j\,2\pi f\,t}x(t)\right]</math>
|<math>=\int_{-\infty}^{\infty}\left[e^{j\,2\pi f_0\,t}x(t)\right]e^{-j\,2\pi f\,t}\,dt</math>
|-
|
|<math>=\int_{-\infty}^{\infty}x(t)\,e^{-j\,2\pi (f-f_0)\,t}\,dt</math>
|-
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|<math>=X(f-f_0)\,\!</math>
|}

===Double Sideband Modulation===
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>F[cos(2\pi f_0\,t)\cdot x(t)]</math>
|<math>=\int_{-\infty}^{\infty}\frac{e^{j\,2\pi f_0\,t} + e^{-j\,2\pi f_0\,t}}{2}x(t)\,e^{-j\,2\pi f\,t}\,dt</math>
|-
|
|<math>=\frac{1}{2}\int_{-\infty}^{\infty}x(t)\left[e^{-j\,2\pi (f-f_0)\,t} + e^{-j\,2\pi (f+f_0)\,t}\right]\,dt</math>
|-
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|<math>=\frac{1}{2}X(f-f_0)+\frac{1}{2}X(f+f_0)</math>
|}

===Differentiation in Time===
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>x(t)\,\!</math>
|<math>=F^{-1}\left[X(f)\right]</math>
|-
|<math>F\left[\frac{dx}{dt}\right]</math>
|<math>=F\left[\frac{d}{dt}F^{-1}\left[X(f)\right]\right]</math>
|-
|
|<math>=F\left[\frac{d}{dt}\int_{-\infty}^{\infty}X(f)\,e^{j\,2\pi f\,t}\,df\right]</math>
|-
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|<math>=F\left[\int_{-\infty}^{\infty}j\,2\pi fX(f)e^{j\,2\pi f\,t}\,df\right]</math>
|-
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|<math>=F\left[j\,2\pi f\,F^{-1}[X(f)]\right]</math>
|-
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|<math>=j\,2\pi f\,X(f)</math>
|Thus <math>\frac{dx}{dt}</math> is a linear filter with transfer function <math>j\,2\pi f</math>
|}

===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
{| border="1" cellpadding="5" cellspacing="0"
|-
|Input
|LTI System
|Output
|Reason
|-
|<math> \delta (t) \,\!</math>
|<math> \Longrightarrow </math>
|<math> h(t) \,\!</math>
|Given
|-
|<math> \delta (t)\,e^{-j\,2\,\pi f\,t}</math>
|<math> \Longrightarrow </math>
|<math> h(t)\,e^{-j\,2\,\pi f\,t}</math>
|Proportionality
|-
|<math> \int_{-\infty}^{\infty}\delta (t)\,e^{-j\,2\,\pi f\,t}\,dt=F[\delta(t)]=1</math>
|<math> \Longrightarrow </math>
|<math> \int_{-\infty}^{\infty}h(t)\,e^{-j\,2\,\pi f\,t}\,dt=F[h(t)]=H(f)</math>
|Superposition 
|-
|<math> \int_{-\infty}^{\infty}\delta (t-\lambda)\,e^{-j\,2\,\pi f\,t}\,dt = F[\delta(t-\lambda)]=1\cdot e^{-j\,2\,\pi f\,\lambda}</math>
|<math> \Longrightarrow </math>
|<math> H(f)\cdot e^{-j\,2\,\pi f\,\lambda}</math>
|Time Invariance
|-
|<math> x(\lambda)\cdot 1\cdot e^{-j\,2\,\pi f\,\lambda}</math>
|<math> \Longrightarrow </math>
|<math> x(\lambda)\cdot H(f)\cdot e^{-j\,2\,\pi f\,\lambda}</math>
|Proportionality
|-
|<math> \int_{-\infty}^{\infty}x(\lambda)\cdot 1\cdot e^{j\,2\,\pi f\,\lambda}\,d\lambda=X(f)</math>
|<math> \Longrightarrow </math>
|<math> \int_{-\infty}^{\infty}x(\lambda)\cdot H(f)\cdot e^{j\,2\,\pi f\,\lambda}\,d\lambda=X(f)\,H(f)</math>
|Superposition
|}
*Having trouble seeing <math>F\left[x(t)*h(t)\right]=X(f)\cdot H(f)</math>

===The Game (Time Domain??)===
{| border="1" cellpadding="5" cellspacing="0"
|-
|Input
|LTI System
|Output
|Reason
|-
|<math> 1\cdot e^{j\,2\,\pi f_0\,t}</math>
|<math> \Longrightarrow </math>
|<math> h(t)* e^{j\,2\,\pi f_0\,t}=e^{j\,2\,\pi f_0\,t}*h(t)</math>
|Proportionality
|-
|
|
|<math>\int_{-\infty}^{\infty} h(\lambda)\cdot e^{j\,2\,\pi f_0\,(t-\lambda)}\,d\lambda</math>
|<math>d\lambda\,\!</math> from [[10/3,6 - The Game]]
|-
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|
|<math>e^{j\,2\,\pi f_0\,\lambda}\int_{-\infty}^{\infty} h(\lambda)\cdot e^{-j\,2\,\pi f_0\,\lambda}\,d\lambda</math>
|
|-
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|<math>e^{j\,2\,\pi f_0\,t}\,H(f_0)</math>
|
|-
|<math> X(f_0)\cdot e^{j\,2\,\pi f_0\,t}</math>
|<math> \Longrightarrow </math>
|<math>X(f_0)\cdot e^{j\,2\,\pi f_0\,t}\,H(f_0)</math>
|Proportionality
|-
|<math> \int_{-\infty}^{\infty}X(f_0)\cdot e^{j\,2\,\pi f_0\,t}\,d f_0=x(t)</math>
|<math> \Longrightarrow </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>
|Superposition
|}

===Relation to the Fourier Series===
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>x(t)\,\!</math>
|<math>=x(t+T)\,\!</math>
|-
|
|<math>=\sum_{n=-\infty}^{\infty}\alpha_n e^{j\,2\pi n\,t/T}</math>
|-
|
|<math>=\underbrace{\sum_{m=-\infty}^{-1}\alpha_m e^{j\,2\pi m\,t/T}}_{Negative frequencies}+\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>
| Let <math>n=-m\,\!</math> and reverse the order of summation
|-
|
|
|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>
|-
|
|<math>=\alpha_0+2\Re\left[\sum_{n=1}^{\infty}\alpha_n e^{j\,2\pi n\,t/T}\right]</math>
|<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>
|-
|
|<math>=\alpha_0+\sum_{n=1}^{\infty}2\left|\alpha_n\right|\cos\left(\frac{2\pi n\,t}{T}+\theta_n\right)</math>
|}
*How can we assume that the answer exists in the real domain?

===Aside: Polar coordinates===
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>
*Rectangular coordinates: <math>a+j\,b\!</math>
*Polar coordinates: <math>\left| a+j\,b \right|\,e^{j\theta}</math>
*<math>\theta = tan^{-1} \frac{b}{a}</math>

{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>a+j\,b\,\!</math>
|<math>=\sqrt{a^2+b^2}\left(\cos(\theta)+j\,\sin(\theta)\right)</math>
|-
|
|<math>=\sqrt{a^2+b^2}\left(e^{j\,\theta}\right)</math>
|-
|
|<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>
|}

===Building up to <math>F\left[u(t)\right]</math>===
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>e^{i \pi}\,\!</math>
|<math>= \cos \pi + i \sin \pi.\,\!</math>
|Euler's Identity
|-
|<math>r_0(t)\,\!</math>
|<math>=-r_0(-t)\,\!</math>
|Real odd function of t
|-
|<math>F\left[r_0(t)\right]\,\!</math>
|<math>=\int_{-\infty}^{\infty}r_0(t)\,e^{-j\,2\pi\,f\,t}\,dt</math>
|-
|
|<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>
|<math>\cos(-x) = \cos(x)\,\!</math> & <math>\sin(-x) = -\sin(x)\,\!</math>
|-
|
|<math>=j\int_{-\infty}^{\infty}r_0(t)\, \sin (-2\pi\,f\,t)\,dt</math>
|<math>r_0(t)\cdot j \sin (-2\pi\,f\,t)</math> = Real odd. Integrates out over symmetric limits.
|-
|
|<math>=\,\!</math>Imaginary Odd function of <math>f\,\!</math>
|-
|<math>r_e(t)\,\!</math>
|<math>=r_e(-t)\,\!</math>
|Real even function of t
|-
|<math>F\left[r_e(t)\right]\,\!</math>
|<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>
|<math>\cos(-x) = \cos(x)\,\!</math> & <math>\sin(-x) = -\sin(x)\,\!</math>
|-
|
|<math>=\int_{-\infty}^{\infty}r_e(t)\, \cos(-2\pi\,f\,t)\,dt</math>
|<math>r_e(t)\cdot \cos (-2\pi\,f\,t)</math> = Real odd. Integrates out over symmetric limits.
|-
|
|<math>=\,\!</math>Real Even function of <math>f\,\!</math>
|}

===Definitions===
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>x(t)\,\!</math>
|<math>=x_e(t)+x_o(t)\,\!</math>
|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.
|-
|<math>x_e(t)\,\!</math>
|<math>=\frac{x(t)+x(-t)}{2}</math>
|-
|<math>x_o(t)\,\!</math>
|<math>=\frac{x(t)-x(-t)}{2}</math>
|-
|<math>u(t)\,\!</math>
|<math>=\frac{1+\sgn (t)}{2}</math>
|<math>\sgn (t) = \begin{cases} 1, & t>0 \\ 0, & t=0 \\ -1, & t<0\end{cases}</math>
|-
|<math>u_e(t)\,\!</math>
|<math>=\frac{1}{2}</math>
|-
|<math>u_o(t)\,\!</math>
|<math>=\frac{\sgn (t)}{2}</math>
|}

===<math>F\left[u(t)\right]</math>===
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>F\left[\frac{1}{2}\right]</math>
|<math>=\int_{-\infty}^{\infty} \frac{1}{2} e^{-j\,2\,\pi\,f\,t}\,dt</math>
|-
|
|<math>=\frac{1}{2} \delta (f)</math>
|-
|<math>F\left[\frac{\sgn (t)}{2}\right]</math>
|<math>=\int_{-\infty}^{\infty} \frac{\sgn (t)}{2} e^{-j\,2\,\pi\,f\,t}\,dt</math>
|-
|
|<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>
|-
|
|<math>\ne \int_{0}^{\infty} \cos(2\,\pi\,f\,u)\,du</math>
|}