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KECE321 Communication Systems I(Haykin Sec. 2.1 - Sec. 2.2 and Ziemer Sec. 2.5)

Lecture #6, March 21, 2012Prof. Young-Chai Ko

12년 3월 21일 수요일

From Fourier Series to Fourier Transform

Fourier Series

x(t) =1X

n=�1Xne

j2⇡nf0t

=1X

n=�1

"1

T0

Z T0/2

�T0/2x(t)e�j2⇡nf0t

dt

#e

j2⇡nf0t

where x(t) is periodic signal with period T0

Now T0 ! 1 then nf0 ! f , 1/T0 ! df , andP1

n=�1 !R1�1


x(t) =1X

n=�1

"1

T0

Z T0/2

�T0/2x(t)e�j2⇡nf0t

dt

#e

j2⇡nf0t

x(t) =

Z 1

�1

Z 1

�1x(t)e�j2⇡ft

dt

�e

j2⇡ftdf

which means f0 ! 0Now T0 ! 1,

1

�1

f f

df

= X(f)

x(t) =

Z 1

�1X(f)ej2⇡ftdf


Fourier Transform

Fourier Transformx(t)

X(f) Inverse Fourier Transform x(t)

X(f)

X(f) =

Z 1

�1x(t)e�j2⇡ft

dt

x(t) =

Z 1

�1X(f)ej2⇡ftdf


• Notations

X(f) = F [x(t)]

x(t) = F�1[X(f)]

x(t) () X(f)


Properties of Fourier Transform

• Linearity

• Dilation

• Conjugation rule

• Duality property

• Time shifting property

• Frequency shifting property

• Area property

• Differentiation in the time domain

• Modulation theorem

• Convolution theorem

• Correlation theorem

• Rayleigh’s Energy theorem (or Parserval’s theorem)


Linearity

Let

then for all constants and

g1(t) () G1(f)

g2(t) () G2(f)

c1 c2

c1g1(t) + c2g2(t) () c1G1(f) + c2G2(f)


Example

Find the Fourier transform of .

g(t)

1 2�1�2

1

2

t

g(t)

g(t)

1 2�1�2

1

2

t

1�1

g1(t)

t

1/2

2�2

g2(t)

t1/4

g(t) = 2g1(t) + 4g2(t)


1�1

g1(t)

t

1/2 F1

2rect

✓t

2

◆�= sinc(2f)

2�2

g2(t)

t1/4 F

1

4rect

✓t

4

◆�= sinc(4f)

−10 −8 −6 −4 −2 0 2 4 6 8 10−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

f [Hz]

−10 −8 −6 −4 −2 0 2 4 6 8 10−0.2

0

0.2

0.4

0.6

0.8

1

1.2

f [Hz]


g(t) = 2g1(t) + 4g2(t) () G(f) = 2G1(f) + 4G2(f)

−10 −8 −6 −4 −2 0 2 4 6 8 10−1

0

1

2

3

4

f [Hz]

2G(f)4G(f)

−10 −8 −6 −4 −2 0 2 4 6 8 10−1

0

1

2

3

4

5

6

f [Hz]

G(f)


−10 −8 −6 −4 −2 0 2 4 6 8 100

2

4

6

−10 −8 −6 −4 −2 0 2 4 6 8 100

1

2

3

4

Amplitude spectrum

Phase spectrum

f [Hz]

f [Hz]


e�ateat

�eat

e�at


[Ref: Haykin & Moher, Textbook]


Dilation Property

g(at) () 1

|a|G✓f

a

◆

Reflection Property

g(�t) ! G(�f)


Conjugation Rule

Duality Property

g⇤(t) ! G⇤(�f)

g(t) ! G(f)

G(t) ! g(�f)then

If


• We have the following pair of the Fourier transform:

• Then, if the time function, given as

Example of Duality Property: Sinc Pulse

g(t) = A sinc(2Wt) ! G(f) =A

2Wrect

✓f

2W

◆

A

2Wrect

✓t

2W

◆ ! Asinc(�2Wf) = Asinc(2Wf)


Time Shifting Property

Frequency Shifting Property

g(t) ! G(f)

then

If

If

then

exp(j2⇡fct)g(t) ! G(f � fc)

g(t) ! G(f)

g(t� t0) ! G(f) exp(�j2⇡ft0)


Example of Frequency Shifting Property

s(t) = rect

✓t

T

◆S(f) = T sinc(Tf)F

F�1

g(t) = s(t) · cos(2⇡fct) G(f) =?

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1.5

−1

−0.5

0

0.5

1

1.5

t [sec]

Example for T=2 and fc=10 Hz1

fc= 0.1

g(t)


cos(2⇡fct) =1

2

⇥ej2⇡fct + e�j2⇡fct

⇤Euler’s formula

exp(j2⇡fct)g(t) ! G(f � fc)

Frequency shift property

g(t) = s(t) · cos(2⇡fct) =rect

�tT

�

2

ej2⇡fct +rect

�tT

�

2

e�j2⇡fct

G(f) =1

2S(f � fc) +

1

2S(f + fc)

=T

2sinc[T (f � fc)] +

T

2sinc[T (f + fc)]


−20 −15 −10 −5 0 5 10 15 20

0

0.5

1

T = 1 [sec], fc = 10 [Hz]

−100 −80 −60 −40 −20 0 20 40 60 80 100−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

T = 1 [sec], fc = 40 [Hz]


[Ref: Haykin & Moher, Textbook]


• In the special case of , that is, the frequency is large compared to the reciprocal of the pulse duration - we may use the approximate result

fcT >> 1 fcT

G(f) =

8<

:

T2 sinc[T (f � fc)], f > 00, f = 0,T2 sinc[T (f + fc)], f < 0


Example 2 for frequency shift property of FT

Given the amplitude and phase spectrum for the signal as follows:

|M(f)|

W�W 0

W

�W

ang[M(f)]

m(t)

Draw the amplitude and phase spectrum of for x(t) = m(t) cos(2⇡fct) fc >> W

1

x(t) =m(t)

2

⇥e

j2⇡fct + e

�j2⇡fct⇤

X(f) =M(f � fc)

2+

M(f + fc)

2

Solution:


ang[M(f)]

|X(f)|

fc�fc

1

2

fc +Wfc �W�fc �W �fc +W

fc�fc fc +Wfc �W�fc �W �fc +W


Example 3 for frequency shift property of FT

Given and signal is given as x(t) = m(t) cos(2⇡fct) y(t)

y(t) = x(t) · cos(2⇡fct)

Draw the amplitude spectrum of y(t)

Solution:y(t) = m(t) cos(2⇡fct) · cos(2⇡fct)

= m(t) cos2(2⇡fct)

= m(t)

1 + cos(4⇡fct)

2

�

=

m(t)

2

+

m(t) cos(4⇡fct)

2

cos(4⇡fct) =ej4⇡fct + e�j4⇡fct

2

=m(t)

2+

m(t)ej4⇡fct

4+

m(t)e�j4⇡fct

4


Y (f) =M(f)

2+

M(f � fc)

4+

M(f + fc)

4

Y (f) = F [y(t)]

W�W 0

1/2

1/4

|Y (f)|


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