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![4
Fourier Transform
Determine the Fourier transform of the Delta function δ(t)
Example
0
( ) ( ) 1
j t j
X t e dt e
ω ω
ω δ
∞
− −
−∞
= = =
∫
1
X(ω)
ω
Fourier Transform
Properties of the Fourier Transform
We summarize several important properties of the Fourier Transform as follows.
1. Linearity (Superposition)
1 1
( ) ( )
x t X ω
⇔ 2 2
( ) ( )
x t X ω
⇔
1 1 2 2 1 1 2 2
( ) ( ) ( ) ( )
a x t a x t a X a X
ω ω
+ ⇔ +
Then,
If and
Proof:
[ ]
1 1 2 2 1 1 2 2
1 1 2 2
( ) ( ) ( ) ( )
( ) ( )
j t j t j t
a x t a x t e dt a x t e dt a x t e dt
a X a X
ω ω ω
ω ω
∞ ∞ ∞
− − −
−∞ −∞ −∞
+ = +
= +
∫ ∫ ∫](https://image.slidesharecdn.com/fouriertransform-220125153301/85/Fourier-transform-4-320.jpg)
![10
Fourier Transform
Determine the Fourier transform of and
Example: cos ct
ω sin ct
ω
[ ]
1 1
( ) cos ( ) ( ) ( )
2 2
c c
j t j t
c c c
x t t e e X
ω ω
ω ω π δ ω ω δ ω ω
−
= = + ⇔ = − + +
[ ]
1 1 1
( ) cos ( ) ( ) ( )
2 2 2
c c
j t j t
c c c
x t t e e X f f f f f
ω ω
ω δ δ
−
= = + ⇔ = − + +
or
f
1/2
fc
-fc
The phase spectrum is zero everywhere.
X(f)
Fourier Transform
[ ]
1 1
( ) sin ( ) ( ) ( )
2 2
c c
j t j t
c c c
x t t e e X j
j j
ω ω
ω ω π δ ω ω δ ω ω
−
= = − ⇔ = − − − +
[ ]
1 1
( ) sin ( ) ( ) ( )
2 2 2
c c
j t j t
c c c
j
x t t e e X f f f f f
j j
ω ω
ω δ δ
− −
= = − ⇔ = − − +
f
π/2
-fc
fc
-π/2
f
1/2
fc
-fc
|X(f)|
θ(f)](https://image.slidesharecdn.com/fouriertransform-220125153301/85/Fourier-transform-10-320.jpg)
![11
Fourier Transform
7. Modulation
If then
Proof:
( ) ( )
x t X ω
⇔
[ ]
1
( )cos( ) ( ) ( )
2
c c c
x t t X X
ω ω ω ω ω
⇔ − + +
[ ]
( ) ( )
1
( )cos( ) ( )
2
1
( ) ( )
2
1
( ) ( )
2
c c
c c
j t j t
j t j t
c
j t j t
c c
x t t e dt x t e e e dt
x t e dt x t e dt
X X
ω ω
ω ω
ω ω ω ω
ω
ω ω ω ω
∞ ∞
− −
−∞ −∞
∞ ∞
− − − +
−∞ −∞
= +
= +
= − + +
∫ ∫
∫ ∫
Fourier Transform
8. Time Differentiation:
If then
Proof:
( ) ( )
x t X ω
⇔
( )
( )
dx t
j X
dt
ω ω
⇔
( )
( ) ( )
n
n
n
d x t
j X
dt
ω ω
⇔
General case
Taking the derivative of the inverse Fourier transform
1
( ) ( )
2
j t
x t X e d
ω
ω ω
π
∞
−∞
= ∫
( ) 1
( )
2
j t
dx t
j X e d
dt
ω
ω ω ω
π
∞
−∞
= ∫
( )
( )
dx t
j X
dt
ω ω
⇔
we obtain
Therefore](https://image.slidesharecdn.com/fouriertransform-220125153301/85/Fourier-transform-11-320.jpg)
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Fourier transform
- 1. 1 Fourier Transform Z. Aliyazicioglu Electrical& Computer Engineering Dept. Cal Poly Pomona ECE 307 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. The Fourier transform The inverse Fourier transform (IFT) of X(ω) is x(t)and given by 2 ( ) x t dt ∞ −∞ < ∞ ∫ ( ) ( ) j t X x t e dt ω ω ∞ − −∞ = ∫ 1 ( ) ( ) 2 j t x t X e d ω ω ω π ∞ −∞ = ∫
- 2. 2 Fourier Transform Also, TheFourier transform can be defined in terms of frequency of Hertz as and corresponding inverse Fourier transform is 2 ( ) ( ) j ft X f x t e dt π ∞ − −∞ = ∫ 2 ( ) ( ) j ft x t X f e df π ∞ −∞ = ∫ Fourier Transform Determine the Fourier transform of a rectangular pulse shown in the following figure Example: -a/2 a/2 h t x(t) / 2 2 2 / 2 ( ) sin( ) 2 2 sin( ) 2 2 sinc 2 a a a j j j t a h X he dt e e j a h a ha a a ha ω ω ω ω ω ω ω ω ω ω π − − − = = − − = = = ∫
- 3. 3 Fourier Transform Example: Tofind in frequency domain, ( ) / 2 2 2 2 2 2 / 2 ( ) 2 sin( ) sin( ) sinc a fa fa j j j ft a h X f he dt e e j f h fa fa ha f fa ha fa π π π π π π π π − − − = = − − = = = ∫ >> 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'); >> 1, 1 ( ) 2sinc 2 h a X ω ω π = = = Fourier Transform 1, 2 2 ( ) 2sinc 2 h a X ω ω π = = = >> 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')
- 4. 4 Fourier Transform Determine theFourier transform of the Delta function δ(t) Example 0 ( ) ( ) 1 j t j X t e dt e ω ω ω δ ∞ − − −∞ = = = ∫ 1 X(ω) ω Fourier Transform Properties of the Fourier Transform We summarize several important properties of the Fourier Transform as follows. 1. Linearity (Superposition) 1 1 ( ) ( ) x t X ω ⇔ 2 2 ( ) ( ) x t X ω ⇔ 1 1 2 2 1 1 2 2 ( ) ( ) ( ) ( ) a x t a x t a X a X ω ω + ⇔ + Then, If and Proof: [ ] 1 1 2 2 1 1 2 2 1 1 2 2 ( ) ( ) ( ) ( ) ( ) ( ) j t j t j t a x t a x t e dt a x t e dt a x t e dt a X a X ω ω ω ω ω ∞ ∞ ∞ − − − −∞ −∞ −∞ + = + = + ∫ ∫ ∫
- 5. 5 Fourier Transform Properties ofthe Fourier Transform 2. Time Shifting Then, If Proof: ( ) ( ) x t X ω ⇔ 0 0 ( ) ( ) j t x t t X e ω ω − − ⇔ 0 t t τ = − 0 t t τ = + dt dτ = 0 0 0 ( ) 0 ( ) ( ) ( ) ( ) j t j t j t j j t x t t e dt x e d e x e d e X ω τ ω ω ωτ ω τ τ τ τ ω ∞ ∞ − + − −∞ −∞ ∞ − − −∞ − − = = = ∫ ∫ ∫ Let then and Fourier Transform Let 0 ( ) ( ) y t x t t = − 0 0 0 ( ) ( ( ) ) ( ) ( ) ( ) ( ) j t j t j X j X t Y X e X e e X e ω ω ω ω ω ω ω ω ω − − ∠ ∠ − = = = 0 ( ( ) ) ( ) ( ) ( ) j X t j Y Y e X e ω ω ω ω ω ∠ − ∠ = 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.
- 6. 6 Fourier Transform Example: Determinethe Fourier transform of the following time shifted rectangular pulse. 0 a h t x(t) 2 ( ) sinc 2 a j a X ha e ω ω ω π − = >> 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 then Proof: If , a>0 then ( ) ( ) x t X ω ⇔ If 1 ( ) ( ) x at X a a ω ⇔ at τ = / t a τ = (1/ ) dt a dτ = Let then and 1 ( ) ( ) 1 ( ) j j t a x at e dt x e d a X a a ω τ ω τ τ ω ∞ ∞ − − −∞ −∞ = = ∫ ∫ If , a<0 then 1 ( ) ( ) 1 1 ( ) ( ) j j t a j a x at e dt x e d a x e d X a a a ω τ ω ω τ τ τ ω τ τ ∞ ∞ − − −∞ −∞ ∞ − −∞ = = = ∫ ∫ ∫
- 7. 7 Fourier Transform Example. if, then find the Fourier transform of the following signals 1 ( 2 ) ( ) 2 2 x t X ω − − ⇔ ( /5) 5 (5 ) x t X ω ⇔ 2 1 ( 5( 2)) ( ) 5 5 j x t X e ω ω − − − − ⇔ ( ) ( ) x t X ω ⇔ a. b. c. Example: Find the Fourier transform of the following signal. 1 1 ( ) ( ) ( ) sinc 2 x t t X ω ω π = ∏ ⇔ = 2 2 1 1 1 ( ) (5 ) ( ) ( ) sinc 5 5 5 10 x t t X X ω ω ω π = ∏ ⇔ = = 3 3 1 ( ) ( /5) ( ) 5 (5 ) 5sinc 0.4 x t t X X ω ω ω π = ∏ ⇔ = = a. b. c. Fourier Transform 4. Duality (Symmetry) If then ( ) ( ) x t X ω ⇔ ( ) 2 ( ) X t x π ω ⇔ − ( ) ( ) X t x f ⇔ − or Proof: Since t and ω are arbitrary variables in the inverse Fourier transform 1 ( ) ( ) 2 j t x t X e d ω ω ω π ∞ −∞ = ∫ we can replace ω with t and t with - ω to get 1 ( ) ( ) 2 j t x X t e dt ω ω π ∞ − −∞ − = ∫ { } ( ) ( ) 2 ( ) j t X t X t e dt x F ω π ω ∞ − −∞ = = − ∫ Therefore,
- 8. 8 Fourier Transform Similarly, ifwe can replace f with t and t with -f in the inverse Fourier transform 2 ( ) ( ) j ft x t X f e df π ∞ −∞ = ∫ 2 ( ) ( ) j ft x f X t e df π ∞ − −∞ − = ∫ { } ( ) ( ) X t x f F = − to get Therefore, then Let Fourier Transform Example: Applying symmetry property, ( ) ( ) ( ) 1 x t t X δ ω = ⇔ = ( ) 1 ( ) 2 ( ) 2 ( ) x t X ω πδ ω πδ ω = ⇔ = − = ( is even function) ( ) δ ω ( ) 1 ( ) ( ) ( ) x t X f f f δ δ = ⇔ = − = or Example: ( ) ( ) sinc 2 t a x t rect X a a ω ω π = ⇔ = ( ) sinc ( ) 2 2 2 ta x t a X rect rect a a ω ω ω π π π − = ⇔ = = 2 a c π = 2 a c π = ( ) 1 ( ) sinc ( ) 2 2 2 x t a ct X rect rect c c c ω ω ω π π π = ⇔ = =
- 9. 9 Fourier Transform Time Reversal Ifthen ( ) ( ) x t X ω ⇔ ( ) ( ) x t X ω − ⇔ − Proof: Let . Then and t τ − = t τ = − dt dτ = − ( ) ( ) ( ) ( ) j t j x t e dt x e d X ω ω τ τ τ ω ∞ ∞ − − − −∞ −∞ − = − = − ∫ ∫ Fourier Transform Frequency Shifting If then ( ) ( ) x t X ω ⇔ ( ) ( ) c j t c x t e X ω ω ω − ⇔ − ( ) ( ) ( ) ( ) c c j t j t j t c x t e e dt x t e dt X ω ω ω ω ω ω ∞ ∞ − − − −∞ −∞ = = − ∫ ∫ Proof:
- 10. 10 Fourier Transform Determine theFourier transform of and Example: cos ct ω sin ct ω [ ] 1 1 ( ) cos ( ) ( ) ( ) 2 2 c c j t j t c c c x t t e e X ω ω ω ω π δ ω ω δ ω ω − = = + ⇔ = − + + [ ] 1 1 1 ( ) cos ( ) ( ) ( ) 2 2 2 c c j t j t c c c x t t e e X f f f f f ω ω ω δ δ − = = + ⇔ = − + + or f 1/2 fc -fc The phase spectrum is zero everywhere. X(f) Fourier Transform [ ] 1 1 ( ) sin ( ) ( ) ( ) 2 2 c c j t j t c c c x t t e e X j j j ω ω ω ω π δ ω ω δ ω ω − = = − ⇔ = − − − + [ ] 1 1 ( ) sin ( ) ( ) ( ) 2 2 2 c c j t j t c c c j x t t e e X f f f f f j j ω ω ω δ δ − − = = − ⇔ = − − + f π/2 -fc fc -π/2 f 1/2 fc -fc |X(f)| θ(f)
- 11. 11 Fourier Transform 7. Modulation Ifthen Proof: ( ) ( ) x t X ω ⇔ [ ] 1 ( )cos( ) ( ) ( ) 2 c c c x t t X X ω ω ω ω ω ⇔ − + + [ ] ( ) ( ) 1 ( )cos( ) ( ) 2 1 ( ) ( ) 2 1 ( ) ( ) 2 c c c c j t j t j t j t c j t j t c c x t t e dt x t e e e dt x t e dt x t e dt X X ω ω ω ω ω ω ω ω ω ω ω ω ω ∞ ∞ − − −∞ −∞ ∞ ∞ − − − + −∞ −∞ = + = + = − + + ∫ ∫ ∫ ∫ Fourier Transform 8. Time Differentiation: If then Proof: ( ) ( ) x t X ω ⇔ ( ) ( ) dx t j X dt ω ω ⇔ ( ) ( ) ( ) n n n d x t j X dt ω ω ⇔ General case Taking the derivative of the inverse Fourier transform 1 ( ) ( ) 2 j t x t X e d ω ω ω π ∞ −∞ = ∫ ( ) 1 ( ) 2 j t dx t j X e d dt ω ω ω ω π ∞ −∞ = ∫ ( ) ( ) dx t j X dt ω ω ⇔ we obtain Therefore
- 12. 12 Fourier Transform 9. TimeDifferentiation: If then Proof: ( ) ( ) x t X ω ⇔ General case Taking derivative of Fourier Transform ( ) ( ) dX j tx t j d ω ω ⇔ ( ) ( ) n n n n d X t x t j d ω ω ⇔ ( ) ( ) j t X x t e dt ω ω ∞ − −∞ = ∫ ( ) ( ) ( ) j t dX jt x t e dt d ω ω ω ∞ − −∞ = − ∫ with respect to ω, we obtain ( ) ( ) dX j tx t j d ω ω ⇔ Therefore Fourier Transform 10 Conjugate If then Proof: ( ) ( ) x t X ω ⇔ If x(t) is real so that * * ( ) ( ) x t X ω ⇔ − * * ( ) * ( ) ( ) ( ) j t j t x t e dt x t e dt X ω ω ω ∞ ∞ − − − −∞ −∞ = = − ∫ ∫ * ( ) ( ) x t x t = * ( ) ( ) X X ω ω = −
- 13. 13 Fourier Transform 11. Convolution Proof: Interchangingthe order of integration, we obtain If , , and ( ) ( ) x t X ω ⇔ ( ) ( ) h t H ω ⇔ ( ) ( ) y t Y ω ⇔ ( ) ( ) * ( ) ( ) ( ) y t h t x t h x t d τ τ τ ∞ −∞ = = − ∫ ( ) ( ) ( ) Y H X ω ω ω = ( ) ( ) ( ) j t Y h x t d e dt ω ω τ τ τ ∞ ∞ − −∞ −∞ = − ∫ ∫ ( ) ( ) ( ) j t Y h x t e dt d ω ω τ τ τ ∞ ∞ − −∞ −∞ = − ∫ ∫ ( ) ( ) ( ) ( ) ( ) ( ) ( ) j j Y h X e d X h e d X H ωτ ωτ ω τ ω τ ω τ τ ω ω ∞ ∞ − − −∞ −∞ = = = ∫ ∫ Fourier Transform 12. Multiplication or If , and 1 2 1 2 1 2 1 1 ( ) ( ) ( ) * ( ) ( ) ( ) 2 2 x t x t X X X v X v dv ω ω ω π π ∞ −∞ ⇔ = − ∫ 1 1 ( ) ( ) x t X ω ⇔ 2 2 ( ) ( ) x t X ω ⇔ 1 2 1 2 1 2 ( ) ( ) ( ) * ( ) ( ) ( ) x t x t X f X f X v X f v dv ∞ −∞ ⇔ = − ∫
- 14. 14 Fourier Transform 13. Parseval’sTheorem Proof If , then total normalized(based on one ohms resistor) energy E of and x(t) is given by 1 1 ( ) ( ) x t X ω ⇔ 2 2 2 1 ( ) ( ) ( ) 2 E x t dt X d X f df ω ω π ∞ ∞ ∞ −∞ −∞ −∞ = = = ∫ ∫ ∫ 2 * * 1 ( ) ( ) ( ) ( ) ( ) 2 j t x t dt x t x t dt x t X e d dt ω ω ω π ∞ ∞ ∞ ∞ − −∞ −∞ −∞ −∞ = = ∫ ∫ ∫ ∫ Interchanging the order of integration, we obtain Fourier Transform 2 * * 2 1 ( ) ( ) ( ) 2 1 ( ) ( ) 2 1 ( ) 2 j t x t dt X x t e dt d X X d X d ω ω ω π ω ω ω π ω ω π ∞ ∞ ∞ − −∞ −∞ −∞ ∞ −∞ ∞ −∞ = = = ∫ ∫ ∫ ∫ ∫ Proof (cont)