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Fourier Series and Transform

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Fourier Series and Transform

KEEE343 Communication Theory

Lecture #8, March 29, 2011

Prof. Young-Chai Ko

koyc@korea.ac.kr

(2)

Review

Properties of Fourier Transform

Time shift, frequency shift, modulation properties

Energy theorem

Inverse relation between time and frequency

Bandwidth

Dirac delta function

Gaussian pulse

(3)

Summary

·• Fourier transform

·•

Fourier transform of special function (Sec. 2.4)

·•

Delta function

·•

Signum function

·•

Unit step function

·•

Fourier transform of periodic signals (Sec. 2.5)

·•

Transmission of signals through linear systems: Convolution revisited (Sec. 2.6)

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Dirac Delta Function

Dirac delta function having zero amplitude everywhere except at t=0, where it infinitely large in such a way that it contains unit area under its curve.

Integral of the product

Convolution

(t) = 0, t 6= 0 Z

1

1

(t) dt = 1

g(t) (t t

0

) Z

1

1

g(t) (t t

0

) dt = g(t

0

) Z

1

1

g(⇥ ) (t ⇥ ) dt = g(⇥ )

= ) g(t) ⇤ (t) = g(t)

(5)

Fourier transform of the delta function

Delta function as a limiting form of the Gaussian pulse

F [ (t)] =

Z

1

1

(t) exp( j2⇥f t) dt = 1

g(t) = 1

⇥ exp

✓ t2

2

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

0 0.5 1 1.5 2 2.5 3 3.5 4

t

g(t)

=0.25

=0.5

=1

=2

(6)

Fourier transform of Gaussian pulse

G(f ) = exp( ⇥

2

f

2

)

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

=2

=1

=0.5

=0.25

f

G(f)

Can you prove this?

(7)

DC signal

Using the duality property we can show that

which gives the following relation

Recognizing that the delta function is real valued, we can simplify this relation as

Application to the Delta Function

1 ! (f)

Z

1

1

exp( j2⇥f t) dt = (f )

Z

1

1

cos(2⇥f t) dt = (f )

(8)

Complex exponential function

Sinusoidal functions

Similarly,

exp(j2⇥f

c

t) ! (f f

c

)

cos(2 f

c

t) = 1

2 [exp(j2 f

c

t) + exp( j2 f

c

t)]

cos(2⇥f

c

t) ! 1

2 [ (f f

c

) + (f + f

c

)]

sin(2⇥f

c

t) ! 1

2j [ (f f

c

) (f + f

c

)]

(9)
(10)

Signum Function

Definition

The signum function does not satisfy the Dirichlet conditions, and therefore, strictly speaking, it does not have a Fourier transform.

Consider the following odd-symmetric double exponential pulse :

Then we have

sgn(t) = 8<

:

+1, t > 0 0, t = 0 1, t < 0

g(t) = 8<

:

exp( at), t > 0 0, t = 0 exp(at), t < 0

alim!0g(t) = sgn(t)

(11)

Fourier transform of

Fourier transform of signum function can be obtained by taking a limit on G(f) for a to approach to zero.

That is,

G(f ) = j4 f a2 + (2 f )2

g(t)

F[sgn(t)] = lima

!0

4j f

a2 + (2 f )2 = 1 j f

sgn(t) ! 1 j f

[Ref: Haykin & Moher, Textbook]

(12)

Unit Step Function

Definition

Unit step function can be expressed using the signum function:

Therefore, the Fourier transform of the unit step function is u(t) =

8<

:

1, t > 0

1

2, t = 0 0, t < 0

u(t) = 1

2 [sgn(t) + 1]

u(t) ! 1

j2⇥f + 1

2 (f )

(13)
(14)

Fourier Transform of Periodic Signals

The periodic signals do not satisfy the Dirichlet conditions so, strictly speaking, they do not have Fourier transform. However, using the delta function, we can obtain the Fourier transform of the periodic signals.

Periodic signal with fundamental period :

Then we can express it in Fourier series form given as T0 gT0(t)

g

T0

(t) =

X

1 n= 1

c

n

exp(j2 nf

0

t)

cn = 1 T0

Z T0/2 T0/2

gT0(t) exp( j2 nf0t) dt where

f0 = 1 T0

(15)

Define as a function equals over one period

Also, we have

Accordingly we may rewrite the Fourier coefficients as

which gives

g(t) gT0(t)

g(t) =

⇢ gT0(t), T20  t  T20 0, elsewhere

g

T0

(t) =

X

1 m= 1

g(t mT

0

)

cn = f0

Z 1

1

g(t) exp( j2 nf0t) dt = f0G(nf0)

g

T0

(t) = f

0

X

1 n= 1

G(nf

0

) exp(j2 nf

0

t)

(16)

Since

we can write

Using , the Fourier transform of the periodic signal can be written as

g

T0

(t) =

X

1 m= 1

g(t mT

0

)

X

1 m= 1

g(t mT

0

) = f

0

X

1 n= 1

G(nf

0

) exp(j2 nf

0

t)

F[exp(j2⇥nf0t)] = (f nf0)

X

1

m= 1

g(t mT

0

) ! f

0

X

1 n= 1

G(nf

0

) (f nf

0

)

(17)

Ideal sampling function

Ideal Sampling Theorem (Example)

T0

(t) =

X

1 m= 1

(t mT

0

)

T0 2T0

3T0 0 T0 2T0 3T0

(18)

Fourier transform of ideal sampling function

We first note that so for all .G(f ) = 1 G(nf0) = 1

n

X

1 m= 1

(t mT

0

) ! f

0

X

1 n= 1

(f nf

0

)

(19)

Pulse train

We know

Example of Pulse Train

T0

T0 T /2 0 T /2

. . . . . .

−6 −4 −2 0 2 4 6

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

sinc(f T ) T rect

✓ t T

x axis is normalized by 1/T

(20)

Example of and Then,

T = 1 T0 = 4

G(nf0) = G⇣ n 4

⌘ = sinc(nf0T ) = sinc⇣ n 4

(21)

−6 −4 −2 0 2 4 6 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Normalized by 1/T [Hz]

(22)

Transmission of Signals Through Linear Systems

Time response for LTI systems

Frequency response

x(t)

LTI

y(t)

h(t)

y(t) = x(t) ⇤ h(t) =

Z 1

1

h( )x(t ) d

=

Z 1

1

x( )h(t ) d

F[h(t)] = H(f), F[x(t)] = X(f) = Y (f ) = X(f )H(f )

(23)

Introduction to Amplitude Modulation

Consideration of communication system design

-

Complexity

-

Two primary communication resources

Transmit power

Channel bandwidth

Carrier

Carrier is a signal to move the baseband signal to the passband signal

A commonly used carrier is a sinusoidal wave

(24)

Amplitude modulation family

Amplitude modulation

Double sideband-suppressed carrier (DSB-SC)

Single sideband (SSB)

Vestigial sideband (VSB)

(25)

Theory

Consider a sinusoidal carrier wave

Denote the message signal (information bearing signal) as

Then an amplitude-modulated (AM) wave is

Amplitude Modulation

c(t) = A

c

cos(2 f

c

t)

m(t)

s(t) = A

c

[1 + k

a

m(t)] cos(2 f

c

t)

(26)

[Ref: Haykin & Moher, Textbook]

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