Chapter 3
Chapter 3
Chapter 3
CHAPTER OBJECTIVES
Upon a successful completion of this chapter, the student will be able to:
LECTURE # 5
3.3 RELATIONSHIP BETWEEN FREQUENCY RESPONSE AND DIFFERENCE
EQUATION FOR LTIS
EXAMPLE 3.3
EXAMPLE 3.4
Lecture # 4 2
3.1 DISCRETE –TIME FOURIER TRANSFORM (DTFT)
xn
n n
xn X e
1
X e .e dw
1 jw jw jwn
2
Notes:
• X(ejw) is a complex valued continuous function
• w = 2π f [rad/sec]
Lecture # 4 3
3.1.1 PROPERTIES OF DTFT
3.1.1.1 PERIODICITY
X e jw
Is periodic with period 2π
X e jw X e j w2 X e jwe j 2 X e jw .1 X e jw
Lecture # 4 4
3.1.1 PROPERTIES OF DTFT (Contd.)
3.1.1.2 LINEARITY
x1 n x2 n x1 n x2 n
xn k X e e
jw jwk
xne jwo n X e j wwo
Lecture # 4 5
3.1.1 PROPERTIES OF DTFT (Contd.)
3.1.1.5 SYMMETRY
For real-valued is conjugate symmetric
xn, X e jw
X e jw X e jw
ReX e ReX e
jw jw Even Symmetry
ImX e ImX e
jw jw Odd Symmetry
X e jw X e jw Even Symmetry
X e X e
jw jw
Odd Symmetry
x n X e jw
3.1.1.7 CONVOLUTION
This is one of the most useful properties that make the system convenient for analysis
Time Domain X e jw
H(ejw)
Y e jw X e jw H e jw
Frequency Domain
Lecture # 4 7
3.1.1 PROPERTIES OF DTFT (Contd.)
3.1.1.8 MULTIPLICATION
This is dual to the convolution property
x1 n.x2 n x1 n x2 n
3.1.1.9 ENERGY
The energy of the sequence x(n) can be written as :
X e
E xn
1 jw 2
2
dw
2
This is known as the Parseval’s Theorem
Lecture # 4 8
EXAMPLE 3.1
Consider the signal xn 0.5 U n
n
Solution:
a- Since the signal x(n) is absolutely summable, therefore, DTFT exists:
xne
jwn
0.5 e jwn
jw n
X e
n 0
0.5 e
jw n 1
0
1 0.5 e jw
cosw j sin w cosw j sin w
X e jw
e jw
jw
e 0.5 cosw j sin w 0.5 cosw 0.5 j sin w
Lecture # 4 9
X e jw
cosw j sin w cosw 0.5 j sin w
cosw 0.52 sinw2
X e
jw cosw 0.5cosw sinw jsinwcosw sinwcosw 0.5 sinw
2 2
X e
jw 1 0.5 cosw j 0.5 sin w
1.25 cosw
1 0.5 cosw
X e jw
1 0.5 cosw2 0.5 sinw2
Re X e jw
1.25 cosw
1.25 cosw2
0.5 sin w
X e jw tan 1
0.5 sin w
Im X e jw
1.25 cosw
1 0.5 cosw
Lecture # 4 10
Magnitude Response
Magnetude Part Real PartPart
Magnetude
2 2
1.8 1.8
1.6 1.6
Magnetude
1.4
Magnetude
1.4
1.2
1.2
1
1
0.8
0.8
0.6
0.6 0 0.5 1 1.5 2 2.5 3 3.5
0 0.5 1 1.5
Frequency 2
[rad\sec] 2.5 3 3.5 Frequency
Frequency in Pi units
[rad/sec]
frequency in pi unitsude Imaginary
Phase Response
Angle Part ImaginaryPart
Part
0 0
-0.1 -0.1
-0.2 -0.2
Magnetude
Magnetude
-0.3 -0.3
-0.4 -0.4
-0.5 -0.5
-0.6 -0.6
n hn
H e
LTI System
jw
1
hne
jwn
H e jw
n
For an arbitrary summable sequence x(n) as an input with impulse response h(n)
xn hn
yn xn hn
X e jw
He jw
Y e jw X e jw H e jw
Hint: yn Y e
1 jw
Lecture # 4 12
3.2 Frequency Domain Representation of LTI System (Contd.)
Special Cases :
Case 1:
If xn Ae jwo n complex exponential
yn A H e e jwo j wo n H ( e jwo )
Case 2:
yn A H e cosw n H e
jwo
0
jwo
hne
a- Using DTFT we find :
0.9 e jwn
jwn
jw n
He
1 n n 0
0.9e
jw n 1
n 1 0.9e jw
1 0.9 cosw j 0.9sin w
He
jw 1
1 0.9 cosw 0.9 sinw
Hence 2 2
Lecture # 4 14
H e jw 1
1
1 0.9 cosw2 0.9 sinw2
1 2 0.9 cosw 0.9 cosw sin w
2 2 2
H e jw
1
1.81 1.8 cosw
-0.1
7
6
Phase in Radians
-0.15
|H|
5
-0.2
4
-0.25
3
-0.3
2
-0.35 1
0
-0.4 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Frequency in pi units
Frequency in pi units
Lecture # 4 15
The steady state response [ at A = 0.1, θ = 0, wo = 0] is given by :
yss n A H e jw cos wo n H e jwo
yss n 0.1 H e cos 0 n 0 H e
0
jw0
yss n 0.1 H e0 0.110 1
Lecture # 4 16
3.3 RELATIONSHIP BETWEEN FREQUENCY RESPONSE
AND DIFFERENCE EQUATION FOR LTIS
When a LTIS is represented by the Difference Equation (DE) :
M N
yn bi xn i ak yn k
i 0 k 1
Take the DTFT for both sides :
b X e e e
M N
Ye jw
i
jw jwi
ak Y e jw jwk
i 0 k 1
N M
Y e jw 1 ak e jwk X e jw bi e jwi
k 1 M
i 0
a e jwi
jw i
Ye
He jw
i 0
X e jw N
1 bk e jwk
k 1
Lecture # 5 17
EXAMPLE 3.3
A LTI system is specified by the difference equation as:
a- Take the DTFT for both sides or apply the above equation:
Y e jw 0.8Y e jw e jw X e jw
H e
jw Y e jw
1
X e 1 0.8e jw
jw
Lecture # 5 18
b- When xn cos0.05n
wo 0.05 and 0
yn A H e jw cos wo n H e jwo
at wo = 0.05 π
H e
1 j 0.5377
jw
4.0928e
1 0.08e j 0.05
Therefore,
yn 4.0928 cos0.05n 0.5377
4.0928 cos0.05 n 3.42
This means that at the output, the sinusoidal is scaled by 4.0928 and
1.2 3
2
1
Phase in Radians
1
0.8
|H|
0
0.6
-1
0.4
-2
0.2 -3
0 -4
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Lecture # 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 20
0.9 1
Frequency in pi units Frequency in pi units
3.4 SAMPLING OF ANALOG SIGNALS
• The sequence x(nT) is obtained from a continuous time signal x(t) by sampling
• This is done multiplying a periodic impulse train s(t) [sampling function] by x(t)
• The period Ts is called the sampling period and Fs =1/Ts is the sampling frequency
ws 2Fs
xs t st .xt
xt
st t nT s
n Sampling Function
st t
st
Ts
xt xs t
xs nTs t
.
w
.0
.
w
.
wM
.0
.
w M
M M
S jw S jw
2 2
.
2w
. Ts
.0 . . . . . . . . . . . .w
Ts
4ws 3ws 2ws ws 0 ws 2ws 3ws 4ws
s ws
w
s
2w s w
X s jw 1
X s jw
1 Ts .
Ts
. .2w . . .w . w. .0 w. . w. . . 2.w . . .4.w . . . . . w. ...0 w... . . . . . .
s s M
M s s w s M
M
w
ws wM
w
Lecture # 5 ws wMs 23
ws wM ws wM
If ws 2wM The signal can be recovered exactly from xs t
by means of Low pass filter with gain Ts
pt t nT
n
s
Cut off frequency greater than wM
x f t
xt x p t
H jw
X p jw
X jw
1
.. .
wM0 wM w
X p jw
1 Ts ws 2wM
. 2.w. . w. . . .0 w. .w. . .2.w.
s s M s s w
ws
H jw
wM wc ws wM Ts w
Ts
He jw
2
wc
. .0 w.
X f jw
c w
0 otherwise
. .
wM 0 wM w Lecture # 5 24
3.5 SAMPLING THEOREM
Let xt be a band –limited signal with X jw 0 for w wM .
Then xt Is uniquely determined by its samples xnTs , n 0, 1, 2,......if