Fourier Series and Transform
KEEE343 Communication Theory
Lecture #8, March 29, 2011
Prof. Young-Chai Ko
koyc@korea.ac.kr
Review
•
Properties of Fourier Transform•
Time shift, frequency shift, modulation properties•
Energy theorem•
Inverse relation between time and frequency•
Bandwidth•
Dirac delta function•
Gaussian pulseSummary
·• 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)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
11
(t) dt = 1
g(t) (t t
0) Z
11
g(t) (t t
0) dt = g(t
0) Z
11
g(⇥ ) (t ⇥ ) dt = g(⇥ )
= ) g(t) ⇤ (t) = g(t)
•
Fourier transform of the delta function•
Delta function as a limiting form of the Gaussian pulseF [ (t)] =
Z
11
(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
•
Fourier transform of Gaussian pulseG(f ) = exp( ⇥
2f
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?
•
DC signalUsing 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
11
exp( j2⇥f t) dt = (f )
Z
11
cos(2⇥f t) dt = (f )
•
Complex exponential function•
Sinusoidal functionsSimilarly,
exp(j2⇥f
ct) ! (f f
c)
cos(2 f
ct) = 1
2 [exp(j2 f
ct) + exp( j2 f
ct)]
cos(2⇥f
ct) ! 1
2 [ (f f
c) + (f + f
c)]
sin(2⇥f
ct) ! 1
2j [ (f f
c) (f + f
c)]
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 havesgn(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)
•
Fourier transform ofFourier 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]
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 )
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= 1c
nexp(j2 nf
0t)
cn = 1 T0
Z T0/2 T0/2
gT0(t) exp( j2 nf0t) dt where
f0 = 1 T0
•
Define as a function equals over one period•
Also, we have•
Accordingly we may rewrite the Fourier coefficients aswhich gives
g(t) gT0(t)
g(t) =
⇢ gT0(t), T20 t T20 0, elsewhere
g
T0(t) =
X
1 m= 1g(t mT
0)
cn = f0
Z 1
1
g(t) exp( j2 nf0t) dt = f0G(nf0)
g
T0(t) = f
0X
1 n= 1G(nf
0) exp(j2 nf
0t)
•
Sincewe can write
•
Using , the Fourier transform of the periodic signal can be written asg
T0(t) =
X
1 m= 1g(t mT
0)
X
1 m= 1g(t mT
0) = f
0X
1 n= 1G(nf
0) exp(j2 nf
0t)
F[exp(j2⇥nf0t)] = (f nf0)
X
1m= 1
g(t mT
0) ! f
0X
1 n= 1G(nf
0) (f nf
0)
•
Ideal sampling functionIdeal Sampling Theorem (Example)
T0
(t) =
X
1 m= 1(t mT
0)
T0 2T0
3T0 0 T0 2T0 3T0
•
Fourier transform of ideal sampling functionWe first note that so for all .G(f ) = 1 G(nf0) = 1
n
X
1 m= 1(t mT
0) ! f
0X
1 n= 1(f nf
0)
•
Pulse train•
We knowExample 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
•
Example of and Then,T = 1 T0 = 4
G(nf0) = G⇣ n 4
⌘ = sinc(nf0T ) = sinc⇣ n 4
⌘
−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]
Transmission of Signals Through Linear Systems
•
Time response for LTI systems•
Frequency responsex(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 )
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•
Amplitude modulation family•
Amplitude modulation•
Double sideband-suppressed carrier (DSB-SC)•
Single sideband (SSB)•
Vestigial sideband (VSB)•
Theory•
Consider a sinusoidal carrier wave•
Denote the message signal (information bearing signal) as•
Then an amplitude-modulated (AM) wave isAmplitude Modulation
c(t) = A
ccos(2 f
ct)
m(t)
s(t) = A
c[1 + k
am(t)] cos(2 f
ct)
[Ref: Haykin & Moher, Textbook]