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

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

### $F\left[u(t)\right]$

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