10/10,13,16,17 - Fourier Transform Properties

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Properties of the Fourier Transform

Linearity

F\left [ a\,x(t)+b\,x(t)\right ] =\int_{-\infty}^{\infty}\left[ a\,x(t)+b\,x(t)\right]e^{-j\,2\pi f\,t}\,dt
=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
=a\,F[x(t)]+b\,F[x(t)]

Time Invariance (Delay)

F[x(t-t_0)]\,\! =\int_{-\infty}^{\infty} x(t-t_0)\,e^{-j\,2\pi f\,t}\,dt Let u=t-t_0\,\! and  du=dt\,\!
=\int_{-\infty}^{\infty} x(u)\,e^{-j\,2\pi f\,(u+t_0)}\,du
=e^{-j\,2\pi f\,t_0}\int_{-\infty}^{\infty} x(u)\,e^{-j\,2\pi f\,u}\,du
=e^{-j\,2\pi f\,t_0}\,F[x(t)] Why isn't this F[x(u)]

Frequency Shifting

F\left[e^{j\,2\pi f\,t}x(t)\right] =\int_{-\infty}^{\infty}\left[e^{j\,2\pi f_0\,t}x(t)\right]e^{-j\,2\pi f\,t}\,dt
=\int_{-\infty}^{\infty}x(t)\,e^{-j\,2\pi (f-f_0)\,t}\,dt
=X(f-f_0)\,\!

Double Sideband Modulation

F[cos(2\pi f_0\,t)\cdot x(t)] =\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
=\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
=\frac{1}{2}X(f-f_0)+\frac{1}{2}X(f+f_0)

Differentiation in Time

x(t)\,\! =F^{-1}\left[X(f)\right]
F\left[\frac{dx}{dt}\right] =F\left[\frac{d}{dt}F^{-1}\left[X(f)\right]\right]
=F\left[\frac{d}{dt}\int_{-\infty}^{\infty}X(f)\,e^{j\,2\pi f\,t}\,df\right]
=F\left[\int_{-\infty}^{\infty}j\,2\pi fX(f)e^{j\,2\pi f\,t}\,df\right]
=F\left[j\,2\pi f\,F^{-1}[X(f)]\right]
=j\,2\pi f\,X(f) Thus \frac{dx}{dt} is a linear filter with transfer function j\,2\pi f

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
 \delta (t) \,\!  \Longrightarrow  h(t) \,\! Given
 \delta (t)\,e^{-j\,2\,\pi f\,t}  \Longrightarrow  h(t)\,e^{-j\,2\,\pi f\,t} Proportionality
 \int_{-\infty}^{\infty}\delta (t)\,e^{-j\,2\,\pi f\,t}\,dt=F[\delta(t)]=1  \Longrightarrow  \int_{-\infty}^{\infty}h(t)\,e^{-j\,2\,\pi f\,t}\,dt=F[h(t)]=H(f) Superposition
 \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}  \Longrightarrow  H(f)\cdot e^{-j\,2\,\pi f\,\lambda} Time Invariance
 x(\lambda)\cdot 1\cdot e^{-j\,2\,\pi f\,\lambda}  \Longrightarrow  x(\lambda)\cdot H(f)\cdot e^{-j\,2\,\pi f\,\lambda} Proportionality
 \int_{-\infty}^{\infty}x(\lambda)\cdot 1\cdot e^{j\,2\,\pi f\,\lambda}\,d\lambda=X(f)  \Longrightarrow  \int_{-\infty}^{\infty}x(\lambda)\cdot H(f)\cdot e^{j\,2\,\pi f\,\lambda}\,d\lambda=X(f)\,H(f) Superposition
  • Having trouble seeing F\left[x(t)*h(t)\right]=X(f)\cdot H(f)

The Game (Time Domain??)

Input LTI System Output Reason
 1\cdot e^{j\,2\,\pi f_0\,t}  \Longrightarrow  h(t)* e^{j\,2\,\pi f_0\,t}=e^{j\,2\,\pi f_0\,t}*h(t) Proportionality
\int_{-\infty}^{\infty} h(\lambda)\cdot e^{j\,2\,\pi f_0\,(t-\lambda)}\,d\lambda d\lambda\,\! from 10/3,6 - The Game
e^{j\,2\,\pi f_0\,\lambda}\int_{-\infty}^{\infty} h(\lambda)\cdot e^{-j\,2\,\pi f_0\,\lambda}\,d\lambda
e^{j\,2\,\pi f_0\,t}\,H(f_0)
 X(f_0)\cdot e^{j\,2\,\pi f_0\,t}  \Longrightarrow X(f_0)\cdot e^{j\,2\,\pi f_0\,t}\,H(f_0) Proportionality
 \int_{-\infty}^{\infty}X(f_0)\cdot e^{j\,2\,\pi f_0\,t}\,d f_0=x(t)  \Longrightarrow \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] Superposition

Relation to the Fourier Series

x(t)\,\! =x(t+T)\,\!
=\sum_{n=-\infty}^{\infty}\alpha_n e^{j\,2\pi n\,t/T}
=\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}
=\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} Let n=-m\,\! and reverse the order of summation
Note that \alpha_{-n} e^{-j\,2\pi n\,t/T} is the complex conjugate of \alpha_n e^{j\,2\pi n\,t/T}
=\alpha_0+2\Re\left[\sum_{n=1}^{\infty}\alpha_n e^{j\,2\pi n\,t/T}\right] x(t)+x(t)^*=2 \,\Re \left[x(t)\right]
=\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]
=\alpha_0+\sum_{n=1}^{\infty}2\left|\alpha_n\right|\cos\left(\frac{2\pi n\,t}{T}+\theta_n\right)
  • How can we assume that the answer exists in the real domain?

Aside: Polar coordinates

Remember from 10/02 - Fourier Series that  \alpha_n = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j\,2\,\pi \,n\,t/T}\, dt

  • Rectangular coordinates: a+j\,b\!
  • Polar coordinates: \left| a+j\,b \right|\,e^{j\theta}
  • \theta = tan^{-1} \frac{b}{a}
a+j\,b\,\! =\sqrt{a^2+b^2}\left(\cos(\theta)+j\,\sin(\theta)\right)
=\sqrt{a^2+b^2}\left(e^{j\,\theta}\right)
=\sqrt{a^2+b^2}\left(e^{j\,tan^{-1} \left(\frac{b}{a}\right)}\right)
=\left|a+j\,b\right|\,e^{j\,tan^{-1} \left(\frac{b}{a}\right)}

Building up to F\left[u(t)\right]

e^{i \pi}\,\! = \cos \pi + i \sin \pi.\,\! Euler's Identity
r_0(t)\,\! =-r_0(-t)\,\! Real odd function of t
F\left[r_0(t)\right]\,\! =\int_{-\infty}^{\infty}r_0(t)\,e^{-j\,2\pi\,f\,t}\,dt
=\int_{-\infty}^{\infty}r_0(t)\,\left[ \cos (-2\pi\,f\,t) + j \sin (-2\pi\,f\,t) \right ]\,dt
=\int_{-\infty}^{\infty}r_0(t)\,\left[ \cos (2\pi\,f\,t) - j \sin (2\pi\,f\,t) \right ]\,dt \cos(-x) = \cos(x)\,\! & \sin(-x) = -\sin(x)\,\!
=j\int_{-\infty}^{\infty}r_0(t)\, \sin (-2\pi\,f\,t)\,dt r_0(t)\cdot j \sin (-2\pi\,f\,t) = Real odd. Integrates out over symmetric limits.
=\,\!Imaginary Odd function of f\,\!
r_e(t)\,\! =r_e(-t)\,\! Real even function of t
F\left[r_e(t)\right]\,\! =\int_{-\infty}^{\infty}r_e(t)\,e^{-j\,2\pi\,f\,t}\,dt
=\int_{-\infty}^{\infty}r_e(t)\,\left[ \cos (-2\pi\,f\,t) + j \sin (-2\pi\,f\,t) \right ]\,dt
=\int_{-\infty}^{\infty}r_e(t)\,\left[ \cos (2\pi\,f\,t) - j \sin (2\pi\,f\,t) \right ]\,dt \cos(-x) = \cos(x)\,\! & \sin(-x) = -\sin(x)\,\!
=\int_{-\infty}^{\infty}r_e(t)\, \cos(-2\pi\,f\,t)\,dt r_e(t)\cdot \cos (-2\pi\,f\,t) = Real odd. Integrates out over symmetric limits.
=\,\!Real Even function of f\,\!

Definitions

x(t)\,\! =x_e(t)+x_o(t)\,\! 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.
x_e(t)\,\! =\frac{x(t)+x(-t)}{2}
x_o(t)\,\! =\frac{x(t)-x(-t)}{2}
u(t)\,\! =\frac{1+\sgn (t)}{2} \sgn (t) = \begin{cases} 1, & t>0 \\ 0, & t=0 \\ -1, & t<0\end{cases}
u_e(t)\,\! =\frac{1}{2}
u_o(t)\,\! =\frac{\sgn (t)}{2}

F\left[u(t)\right]

F\left[\frac{1}{2}\right] =\int_{-\infty}^{\infty} \frac{1}{2} e^{-j\,2\,\pi\,f\,t}\,dt
=\frac{1}{2} \delta (f)
F\left[\frac{\sgn (t)}{2}\right] =\int_{-\infty}^{\infty} \frac{\sgn (t)}{2} e^{-j\,2\,\pi\,f\,t}\,dt
=\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]
=\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]
=\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}}
=\int_{0}^{\infty} \frac{-e^{j\,2\,\pi\,f\,u} + e^{-j\,2\,\pi\,f\,u}}{2}\,du
=\int_{0}^{\infty} -j\,\frac{e^{j\,2\,\pi\,f\,u} - e^{-j\,2\,\pi\,f\,u}}{2j}\,du
=\int_{0}^{\infty} -j\,\sin(2\,\pi\,f\,u)\,du
\ne \int_{0}^{\infty} \cos(2\,\pi\,f\,u)\,du