1

Fourier Transform

Z. AliyaziciogluElectrical & Computer Engineering Dept.

Cal Poly Pomona

ECE 307

Fourier TransformThe 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 tX x t e dtωω∞

−

−∞

= ∫

1( ) ( )2

j tx t X e dωω ωπ

∞

−∞

= ∫

2

Fourier TransformAlso, The Fourier transform can be defined in terms of frequency of Hertz as

and corresponding inverse Fourier transform is

2( ) ( ) j ftX f x t e dtπ∞

−

−∞

= ∫

2( ) ( ) j ftx t X f e dfπ∞

−∞

= ∫

Fourier TransformDetermine the Fourier transform of a rectangular pulse shown in the following figure

Example:

-a/2 a/2

h

t

x(t)

/ 22 2

/ 2

( )

sin( )2 2sin( )2

2

sinc2

a a aj jj t

a

hX he dt e ej

ah a ha a

aha

ω ωωω

ω

ωω

ωω

ωπ

−−

−

= = − −

= =

=

∫

3

Fourier TransformExample: To find in frequency domain,

( )

/ 2 2 22 2 2

/ 2

( )2sin( )sin( )

sinc

a fa faj jj ft

a

hX f he dt e ej f

h fafa haf fa

ha fa

π ππ

π

πππ 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

( ) 2sinc2

ha

X ωωπ

==

=

Fourier Transform

1,2

2( ) 2sinc2

ha

X ωωπ

==

=

>> 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

Fourier Transform

Determine the Fourier transform of the Delta function δ(t)

Example

0( ) ( ) 1j t jX t e dt eω ωω δ∞

− −

−∞

= = =∫

1

X(ω)

ω

Fourier TransformProperties 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 Xω ω+ ⇔ +Then,

If and

Proof:

[ ]1 1 2 2 1 1 2 2

1 1 2 2

( ) ( ) ( ) ( )

( ) ( )

j t j t j ta 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

Fourier TransformProperties of the Fourier Transform

2. Time Shifting

Then,

If

Proof:

( ) ( )x t X ω⇔

00( ) ( ) j tx t t X e ωω −− ⇔

0t tτ = − 0t tτ= + dt dτ=

0

0

0

( )0( ) ( )

( )

( )

j tj t

j t j

j t

x t t e dt x e d

e x e d

e X

ω τω

ω ωτ

ω

τ τ

τ τ

ω

∞ ∞− +−

−∞ −∞

∞− −

−∞

−

− =

=

=

∫ ∫

∫

Let then and

Fourier Transform

Let0( ) ( )y t x t t= −

0 0

0

( )

( ( ) )

( ) ( ) ( )

( )

j t j tj X

j X t

Y X e X e e

X e

ω ωω

ω ω

ω ω ω

ω

− −∠

∠ −

= =

=

0( ( ) )( )( ) ( ) j X tj YY e X e ω ωωω 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

Fourier TransformExample: Determine the Fourier transform of the following time

shifted rectangular pulse.

0 a

h

t

x(t)

2( ) sinc2

ajaX ha eωωω

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 Transform3. Time Scaling

then

Proof:

If , a>0 then

( ) ( )x t X ω⇔If

1( ) ( )x at Xa a

ω⇔

atτ = /t aτ= (1/ )dt a dτ=Let then and

1( ) ( )

1 ( )

jj t ax at e dt x e da

Xa a

ωτω τ τ

ω

∞ ∞−−

−∞ −∞

=

=

∫ ∫

If , a<0 then

1( ) ( )

1 1( ) ( )

jj t a

ja

x at e dt x e da

x e d Xa a a

ωτω

ωτ

τ τ

ωτ τ

∞ ∞−−

−∞ −∞

∞−

−∞

=

= =

∫ ∫

∫

7

Fourier TransformExample. if , then find the Fourier transform of the

following signals1( 2 ) ( )2 2

x t X ω−− ⇔

( / 5) 5 (5 )x t X ω⇔

21( 5( 2)) ( )5 5

jx t X e ωω −−− − ⇔

( ) ( )x t X ω⇔

a.

b.

c.

Example: Find the Fourier transform of the following signal.

1 1( ) ( ) ( ) sinc2

x t t X ωωπ

= ∏ ⇔ =

2 2 11 1( ) (5 ) ( ) ( ) sinc5 5 5 10

x t t X X ω ωωπ

= ∏ ⇔ = =

3 3 1( ) ( / 5) ( ) 5 (5 ) 5sinc0.4

x t t X X ωω ωπ

= ∏ ⇔ = =

a.

b.

c.

Fourier Transform4. Duality (Symmetry)

If then( ) ( )x t X ω⇔

( ) 2 ( )X t xπ ω⇔ − ( ) ( )X t x f⇔ −or

Proof: Since t and ω are arbitrary variables in the inverse Fourier transform

1( ) ( )2

j tx t X e dωω ωπ

∞

−∞

= ∫

we can replace ω with t and t with - ω to get

1( ) ( )2

j tx X t e dtωωπ

∞−

−∞

− = ∫{ }( ) ( ) 2 ( )j tX t X t e dt xF ω π ω

∞−

−∞

= = −∫

Therefore,

8

Fourier TransformSimilarly, if we can replace f with t and t with -f in the inverse

Fourier transform

2( ) ( ) j ftx t X f e dfπ∞

−∞

= ∫

2( ) ( ) j ftx f X t e dfπ∞

−

−∞

− = ∫

{ }( ) ( )X t x fF = −

to get

Therefore,

thenLet

Fourier TransformExample: Applying symmetry property,( ) ( ) ( ) 1x t t Xδ ω= ⇔ =

( ) 1 ( ) 2 ( ) 2 ( )x t X ω πδ ω πδ ω= ⇔ = − = ( is even function)( )δ ω

( ) 1 ( ) ( ) ( )x t X f f fδ δ= ⇔ = − =or

Example:

( ) ( ) sinc2

t ax t rect X aa

ωωπ

= ⇔ =

( ) sinc ( ) 2 22tax t a X rect rect

a aω ωω π π

π− = ⇔ = =

2acπ

= 2a cπ=

( ) 1( ) sinc ( ) 22 2

x t a ct X rect rectc c c

ω ωω ππ π

= ⇔ = =

9

Fourier TransformTime Reversal

If then( ) ( )x t X ω⇔

( ) ( )x t X ω− ⇔ −

Proof: Let . Then and t τ− = t τ= − dt dτ= −

( )( ) ( ) ( )j t jx t e dt x e d Xω ω ττ τ ω∞ ∞

− − −

−∞ −∞

− = − = −∫ ∫

Fourier TransformFrequency Shifting

If then( ) ( )x t X ω⇔

( ) ( )cj tcx t e Xω ω ω− ⇔ −

( )( ) ( ) ( )c cj t j tj tcx t e e dt x t e dt Xω ω ωω ω ω

∞ ∞− −−

−∞ −∞

= = −∫ ∫

Proof:

10

Fourier Transform

Determine the Fourier transform of andExample: cos ctω sin ctω

[ ]1 1( ) cos ( ) ( ) ( )2 2

c cj t j tc c cx t t e e Xω ωω ω π δ ω ω δ ω ω−= = + ⇔ = − + +

[ ]1 1 1( ) cos ( ) ( ) ( )2 2 2

c cj t j tc c cx t t e e X f f f f fω ωω δ δ−= = + ⇔ = − + +

or

f

1/2

fc-fc

The phase spectrum is zero everywhere.

X(f)

Fourier Transform

[ ]1 1( ) sin ( ) ( ) ( )2 2

c cj t j tc c cx t t e e X j

j jω ωω ω π δ ω ω δ ω ω−= = − ⇔ = − − − +

[ ]1 1( ) sin ( ) ( ) ( )2 2 2

c cj t j tc c c

jx t t e e X f f f f fj j

ω ωω δ δ− −= = − ⇔ = − − +

f

π/2

-fc

fc

-π/2

f

1/2

fc-fc

|X(f)|

θ(f)

11

Fourier Transform7. Modulation

If then

Proof:

( ) ( )x t X ω⇔

[ ]1( )cos( ) ( ) ( )2c c cx t t X Xω ω ω ω ω⇔ − + +

[ ]

( ) ( )

1( )cos( ) ( )2

1 ( ) ( )2

1 ( ) ( )2

c c

c c

j t j tj t j tc

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 ωω ω

ω ω ω ω

ω

ω ω ω ω

∞ ∞− −

−∞ −∞

∞ ∞− − − +

−∞ −∞

= +

= +

= − + +

∫ ∫

∫ ∫

Fourier Transform8. Time Differentiation:

If then

Proof:

( ) ( )x t X ω⇔

( ) ( )dx t j Xdt

ω ω⇔( ) ( ) ( )

nn

nd x t j X

dtω ω⇔

General case

Taking the derivative of the inverse Fourier transform

1( ) ( )2

j tx t X e dωω ωπ

∞

−∞

= ∫

( ) 1 ( )2

j tdx t j X e ddt

ωω ω ωπ

∞

−∞

= ∫

( ) ( )dx t j Xdt

ω ω⇔

we obtain

Therefore

12

Fourier Transform9. Time Differentiation:

If then

Proof:

( ) ( )x t X ω⇔ General case

Taking derivative of Fourier Transform

( )( ) dX jtx t jd

ωω

⇔( )( )

nn n

nd Xt x t j

dω

ω⇔

( ) ( ) j tX x t e dtωω∞

−

−∞

= ∫

( ) ( ) ( ) j tdX jt x t e dtd

ωωω

∞−

−∞

= −∫

with respect to ω, we obtain

( )( ) dX jtx t jd

ωω

⇔Therefore

Fourier Transform10 Conjugate

If then

Proof:

( ) ( )x t X ω⇔

If x(t) is real so that

* *( ) ( )x t X ω⇔ −

*

* ( ) *( ) ( ) ( )j t j tx t e dt x t e dt Xω ω ω∞ ∞

− − −

−∞ −∞

= = −

∫ ∫*( ) ( )x t x t= *( ) ( )X Xω ω= −

13

Fourier Transform11. Convolution

Proof:

Interchanging the 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 dτ τ τ∞

−∞

= = −∫

( ) ( ) ( )Y H Xω ω ω=

( ) ( ) ( ) j tY h x t d e dtωω τ τ τ∞ ∞

−

−∞ −∞

= −

∫ ∫

( ) ( ) ( ) j tY h x t e dt dωω τ τ τ∞ ∞

−

−∞ −∞

= −

∫ ∫ ( ) ( ) ( ) ( ) ( )

( ) ( )

j jY h X e d X h e d

X H

ωτ ωτω τ ω τ ω τ τ

ω ω

∞ ∞− −

−∞ −∞

= =

=

∫ ∫

Fourier Transform12. Multiplication

or

If , and

1 2 1 2 1 21 1( ) ( ) ( ) * ( ) ( ) ( )

2 2x t x t X X X v X v dvω 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

Fourier Transform13. Parseval’s Theorem

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 21( ) ( ) ( )2

E x t dt X d X f dfω ωπ

∞ ∞ ∞

−∞ −∞ −∞

= = =∫ ∫ ∫

2 * *1( ) ( ) ( ) ( ) ( )2

j tx t dt x t x t dt x t X e d dtωω ωπ

∞ ∞ ∞ ∞−

−∞ −∞ −∞ −∞

= =

∫ ∫ ∫ ∫

Interchanging the order of integration, we obtain

Fourier Transform

2 *

*

2

1( ) ( ) ( )2

1 ( ) ( )2

1 ( )2

j tx t dt X x t e dt d

X X d

X d

ωω ωπ

ω ω ωπ

ω ωπ

∞ ∞ ∞−

−∞ −∞ −∞

∞

−∞

∞

−∞

=

=

=

∫ ∫ ∫

∫

∫

Proof (cont)