ECE 307
Fourier Transform
Z. Aliyazicioglu Electrical & Computer Engineering Dept. Cal Poly Pomona
Fourier Transform
The Fourier transform (FT) is the extension of the Fourier series to nonperiodic signals. The Fourier transform of a signal exist if satisfies the following condition.
∞ 2 ∫ xt() dt< ∞ −∞
∞ The Fourier transform X()ω = ∫ xte ()− jtω dt −∞
The inverse Fourier transform (IFT) of X(ω) is x(t)and given by
1 ∞ x()tXed= ∫ (ω ) jtω ω 2π −∞
1 Fourier Transform
Also, The Fourier transform can be defined in terms of frequency of Hertz as
∞ X()fxtedt= ∫ () − jft2π −∞
and corresponding inverse Fourier transform is
∞ x()tXfedf= ∫ () jft2π −∞
Fourier Transform
Example: Determine the Fourier transform of a rectangular pulse shown in the following figure x(t)
h a /2 ωωaa − jj − jtω h Xhedtee()ω ==− 22 ∫ − jω t −a /2 ωa -a/2 a/2 sin( ) 2haω ==sin( ) ha 2 ω 2 ωa 2 ωa = hasinc 2π
2 Fourier Transform
Example: To find in frequency domain,
a /2 22ππfa fa − jj − jft2π h Xf()==∫ he dt e22 − e −a /2 − jf2π h = 1, hfasin(π ) ==sin(πfa ) ha a = 1 ππffa ω = hasinc fa X()ω = 2sinc () 2π
>> h=1; >> a=1; >> f=-3.5:0.01:3.5; >> w=2*pi*f; >> x=h*a*sinc(w*a/(2*pi)); >> plot (w,x) >> title ('X(\omega)') >> xlabel('\omega'); >>
Fourier Transform
h = 1, a = 2 ω2 X()ω = 2sinc 2π
>> h=1; >> a=1; >> f=-3.5:0.01:3.5; >> w=2*pi*f; >> x=abs(h*a*sinc(w*a/(2*pi))); >> subplot (2,1,1) >> plot (w,x) >> title ('|X(\omega)|') >> xlabel('\omega') >> xp=phase(h*a*sinc(w*a/(2*pi))); >> subplot (2,1,2) >> plot (w,xp) >> title ('phase X(\omega)') >> xlabel('\omega')
3 Fourier Transform
Example
Determine the Fourier transform of the Delta function δ(t)
∞ Xtedte()ωδ===∫ ()−−jtωω j0 1 −∞
X(ω) 1
ω
Fourier Transform
Properties of the Fourier Transform We summarize several important properties of the Fourier Transform as follows.
1. Linearity (Superposition)
xt()⇔ X (ω ) If xt11()⇔ X (ω ) and 22
Then, ax11() t+⇔+ ax 2 2 () t aX 1 1 (ω ) aX 2 2 (ω )
Proof: ∞∞∞ axt()+= axte ()−−jtωω dta xte () jt dta + xte () − jt ω dt ∫∫∫[]11 2 2 1 1 2 2 −∞ −∞ −∞
=+aX11()ωω aX 2 2 ()
4 Fourier Transform
Properties of the Fourier Transform
2. Time Shifting
If xt()⇔ X (ω )
− jωt0 Then, xt()−⇔ t0 X ()ω e
Proof: Let τ =−tt0 then tt= τ + 0 and dt= dτ
∞∞ jt() xt()−= t e− jtω dt x ()τ e−+ωτ 0 dτ ∫∫0 −∞ −∞ ∞ jt = exed− ω 0 ∫ ()τ − jωτ τ −∞ jt = eX− ω 0 ()ω
Fourier Transform
Let yt()=− xt ( t0 )
j tjt YXeXee()ωω== ()−−ω 00 () ωjX∠ ()ω ω
= Xe()ω jX(())∠−ωω t0
Ye()ωωjY∠ ()ω = Xe () j(())∠−Xtωω0
Therefore, the amplitude spectrum of the time shifted signal is the same as the amplitude spectrum of the original signal, and the phase spectrum of the time-shifted signal is the sum of the phase spectrum of the original signal and a linear phase term.
5 Fourier Transform
Example: Determine the Fourier transform of the following time shifted rectangular pulse.
x(t) h a ωa − jω Xhae()ω = sinc 2 2π t
0 a >> h=1; >> a=1; >> f=-3.5:0.01:3.5; >> w=2*pi*f; >> x=abs(h*a*sinc(w*a/(2*pi)).*exp(- j*w*1/2)); >> subplot (2,1,1) >> plot (w,x) >> title ('|X(\omega)|') >> xlabel('\omega') >> xp=phase(h*a*sinc(w*a/(2*pi)).*exp(- j*w.*1/2)); >> subplot (2,1,2) >> plot (w,xp) >> xlabel('\omega') >> title ('phaseX(\omega)')
Fourier Transform
3. Time Scaling
If xt()⇔ X (ω ) then 1 ω xat()⇔ X () aa
Proof: Let τ = at then ta= τ / and dt= (1/ a ) dτ
If , a>0 then If , a<0 then
∞∞ω − j τ ∞∞ω − jtω 1 − j τ 1 x()at e dt= x ()τ ea dτ xate()− jtω dt= x ()ττ ea d ∫∫ ∫∫a −∞ −∞ a −∞ −∞ ∞ ω 1 ω 11− j τ ω = X() ==∫ xe()ττa d X ( ) aa aaa−∞
6 Fourier Transform
Example. if , xt () ⇔ X ( ω ) then find the Fourier transform of the following signals 1 −ω xt(2)−⇔ X ( ) a. 22 b. xt(/5)⇔ 5 X (5)ω 1 −ω c. xt(5(−− 2)) ⇔ X ( ) e− jω 2 55
Example: Find the Fourier transform of the following signal.
ω xt11()=∏ () t ⇔ X (ω ) = sinc a. 2π 11ω ω xt() (5) t X ( ) X ( ) sinc b. 221=∏ ⇔ω = = 555 10π c. ω xt331()=∏ ( t /5) ⇔ X (ωω ) = 5 X (5) = 5sinc 0.4π
Fourier Transform
4. Duality (Symmetry) If xt()⇔ X (ω ) then
Xt()⇔− 2π x (ω ) or X()txf⇔ (− )
Proof: Since t and ω are arbitrary variables in the inverse Fourier transform
1 ∞ x()tXed= ∫ (ω ) jtω ω 2π −∞ we can replace ω with t and t with - ω to get
∞ 1 − jtω Therefore, xXtedt()−=ω ∫ () 2π −∞ ∞ F{}Xt()==−∫ Xte ()− jtω dt 2π x (ω ) −∞
7 Fourier Transform
Similarly, if we can replace f with t and t with -f in the inverse Fourier transform
∞ x()tXfedf= ∫ () jft2π −∞
to get ∞ x()−=fXtedf∫ ()− jft2π −∞ Therefore, F{Xt()} = x (− f )
Fourier Transform
Example: xt()=⇔δ () t X (ω ) = 1 Applying symmetry property,
xt()=⇔ 1 X (ω ) = 2πδ ( −= ω ) 2 πδ ( ω ) (δ () ω is even function)
or xt()=⇔ 1 Xf () =−=δ ( f )δ () f
Example: ta ω xt()=⇔= rect X (ω ) a sinc a 2π ta −ω ω xt()=⇔== a sinc X (ωπ ) 2 rect 2 π rect 2π aa
a Let c = then ac= 2π 2π ωω1 x() t=⇔== a sinc() ct X (ωπ ) 2 rect rect 22πcc π c
8 Fourier Transform
Time Reversal If xt()⇔ X (ω ) then
xt()−⇔− X (ω )
Proof: Let −=t τ . Then t = −τ and dt= − dτ
∞∞ ∫∫xtedt()−=−=−−−−jtωωτ x ()τ e j() dτω X ( ) −∞ −∞
Fourier Transform
Frequency Shifting If xt()⇔ X (ω ) then
− jtωc xte()⇔− X (ω ωc )
Proof:
∞∞ xte()jtωωωcc e− jtω dt==− xte ()−− j() t dt X (ω ω ) ∫∫ c −∞ −∞
9 Fourier Transform
Example: Determine the Fourier transform of cos ω c t and sinωct
11 xt()==+ cosω t ejtωωcc e− jt ⇔=−++ X (ωπδωωδωω )[] ( ) ( ) ccc22
or 11 1 x()tteeXfffff==+ cosωδδjtωωcc− jt ⇔=−++ ()[] ( ) ( ) ccc22 2 X(f) 1/2
-fc fc f The phase spectrum is zero everywhere.
Fourier Transform
11jtωωcc− jt xt()== sinωc t e − e ⇔=−−−+ X (ωπδωωδωω ) j [] (cc ) ( ) 22jj 11 − j x()ttee== sinωδδjtωωcc −− jt ⇔= Xfffff ()[] ( −−+ ) ( ) c 22jj 2cc
|X(f)| 1/2
-fc fc f θ(f) π/2
fc
-fc f -π/2
10 Fourier Transform
7. Modulation If xt()⇔ X (ω ) then 1 xt( )cos(ωccc t )⇔−++[] X (ωω ) X ( ωω ) 2
Proof:
∞∞ 1 jt jt xt()cos(ω te )−−jtωω dt=+ xt () eωωcc e e jt dt ∫∫c −∞ −∞ 2 ∞∞ 1 −−jt()ωω −+ jt () ωω =+ xte()cc dt xte () dt 2 ∫∫ −∞ −∞ 1 =−++[]XX()()ωωcc ωω 2
Fourier Transform
8. Time Differentiation: If xt()⇔ X (ω ) then General case
n dx() t dxt() n ⇔ jXω ()ω n ⇔ ()()jXω ω dt dt
Proof: Taking the derivative of the inverse Fourier transform
1 ∞ x()tXed= ∫ (ω ) jtω ω 2π −∞ we obtain dx() t 1 ∞ = ∫ jωXed()ωωjtω dt 2π −∞
Therefore dx() t ⇔ jXω ()ω dt
11 Fourier Transform
9. Time Differentiation: If xt()⇔ X (ω ) then General case
dXn ()ω dX() jω txtnn()⇔ j tx() t⇔ j dω n dω Proof: Taking derivative of Fourier Transform
∞ − jtω X()ω = ∫ xte () dt with respect to ω, we obtain −∞
dX()ω ∞ =−∫ ()()jtxte− jtω dt dω −∞ dX() jω Therefore tx() t⇔ j dω
Fourier Transform
10 Conjugate If xt()⇔ X (ω ) then
xt**()⇔− X (ω )
Proof: * ∞∞ *()*−−−jtωω j t ∫∫xte() dt= xte () dt=− X (ω ) −∞ −∞
If x(t) is real x * () txt = () so that XX()ω = * (−ω )
12 Fourier Transform
11. Convolution
If xt()⇔ X (ω ) , ht () ⇔ H ( ω ) , and yt () ⇔ Y (ω ) ∞ yt()==− ht ()* xt ()∫ h (τ ) xt (ττ ) d −∞ YHX()ω = ()()ωω Proof:
∞∞ − jtω Yhxtdedt()ωτττ=−∫∫ ()( ) −∞ −∞ Interchanging the order of integration, we obtain
∞∞ ∞∞ −−jjωτ ωτ Yhxtedtd()ω =− ()τττ ( )− jtω YhXedXhed()ω ==∫∫ ()()τω τ () ω () τ τ ∫∫−∞ −∞ −∞ −∞ = XH()()ωω
Fourier Transform
12. Multiplication
xt()⇔ X (ω ) If 11 , and xt 22 () ⇔ X (ω )
11∞ x ()tx () t⇔=− X (ωω )* X ( ) X ( vX ) ( ω vdv ) 12 1 2∫ 1 2 22ππ−∞
or
∞ x ()tx () t⇔=− X ()* f X () f X ( vX ) ( f vdv ) 12 1 2∫ 1 2 −∞
13 Fourier Transform
13. Parseval’s Theorem
If xt11()⇔ X (ω ) , then total normalized(based on one ohms resistor) energy E of and x(t) is given by
∞∞∞ 2221 ExtdtXd==∫∫∫() (ωω ) = Xfdf () −∞2π −∞ −∞ Proof
∞∞ ∞∞ 2 ** 1 − jtω x()tdtxtxtdtxtXeddt== () () () (ωω ) ∫∫ ∫∫2 −∞ −∞ −∞ π −∞
Interchanging the order of integration, we obtain
Fourier Transform
Proof (cont)
∞∞∞ 2 1 * − jtω x()tdt= X (ω ) xtedtd () ω ∫∫∫2 −∞π −∞ −∞ 1 ∞ = ∫ XXd* ()()ωωω 2π −∞ ∞ 1 2 = ∫ Xd()ωω 2π −∞
14