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Chapter 16:
Discrete-Time Fourier Transform
Chapter 18:
System Function, the Frequency
Response
Copyright Β© 2014 McGraw-Hill Education. Permission required for reproduction or display .
Scope SMJE2053:
Ch.16, 16.1~4, 16.6, Problems16.14: Questions 1., 6.a., 12.a. b.
Ch.18, 18.1~4, 16.6, Problems18.13: Questions 20.(Mod.), 21.a.
Week 12
(SMJE2053;2014/2015_2)
Signal Transform Family
Copyright Β© 2014 McGraw-Hill Education. Permission required for reproduction or display
Signals
Continuous-time
Discrete-time
non-periodic
Fourier Transform
Discrete-time Fourier Transform
(DTFT)
finite,
periodic
Fourier Series
Discrete Fourier Transform
(DFT), FFT*
System/
Control
Laplace transform
X(s)
z-transform
X(z)
*: FFT is a fast algorithm of DFT
( ) j n
n
X x n e
DTFT
16.1 Definition
The DTFT is a complex function and can be
represented by its magnitude and phase
Magnitude Phase π π = β π(π)
13-4
DTFT is Periodic DTFT π(π) is periodic with period 2Ο because
π(π) is specified by
π π β π β€ π < π
(When the DTFT is expressed in terms of f
(π = 2ππ), the period of f is 1 Hz and
π π = π(π)π=2ππ
may be specified for β0.5 < π < 0.5
13-5
IDTFT, The Inverse of DTFT
The inverse of the DTFT may be found from
Copyright Β© 2014 McGraw-Hill Education. Permission required for reproduction or display
16.2 Examples of DTFT
Example 16.1a. Find the DTFT of the unit-
sample signal x(n)=d(n)
Solution: π π = 1
Example 16.1b. Find the DTFT and plot its
magnitude given
π₯ π = 1
2,
0,
π = β1,1πππ ππ€βπππ
Solution:
π π =1
2πππ +
1
2πβππ = cosπ
13-7
π π = cosπ
π₯(π)
βπ 0 π π
Example 16.1b.
12-8 Copyright Β© 2014 McGraw-Hill Education. Permission required for reproduction or display
Quiz 1 Find the DTFT given
π₯ π = 0.5π π + 1 + π π + 0.5π π β 1
π π = 1 + cosπ
βπ 0 π
π
Examples (Cont.)
Example 16.3a. Find the DTFT and its
magnitude and phase of π π = πππ(π)
Solution:
π π = πππβπππ
β
π=0
= ππβπππ
β
π=0
=1
1 β ππβππ=
1
1 β π cosπ + ππ sinπ
π(π) =1
(1 β π cosπ)2+(π sinπ)2
π π = βπ‘ππβ1π sinπ
1 β π cosπ
Magnitude:
Phase:
13-10
j
1
1 0.8e
Ο
Ο
Example πΌ = 0.8
13-11
16.3 The DTFT and the z-Transform
Copyright Β© 2014 McGraw-Hill Education. Permission required for reproduction or display
DTFT:
z-transform:
13-12
Magnitude response 1
1X(z)
1 0.8z
z-transform of π₯ π = 0.8 ππ’(π)
DTFT
(magnitude)
13-13
Phase response
1
1X(z)
1 0.8z
z-transform of π₯ π = 0.8 ππ’(π)
DTFT (phase)
16.4 Inverse DTFT (IDTFT) Example
Example 16.8. Find the inverse DTFT of
Solution.
13-15 Figure 16.3. X(Ο) (upper) and x(n) (lower).
βπ 0 π
π(π)
π₯(π)
13-16
16.6 Properties of DTFT
Copyright Β© 2014 McGraw-Hill Education. Permission required for reproduction or display
Linearity:
Time shift:
ππ₯ π + ππ¦ π
π₯ π β π
ππ π + ππ π
πβππππ π
Convolution: π₯ π β π¦ π π π β π π
π₯(π)
β
π=ββ
= π(0) Zero frequency
Copyright Β© 2014 McGraw-Hill Education. Permission required for reproduction or display
Quiz 2
Find the DTFT of given signal
π₯ π = π π β 5 + π(π + 5)
13-18
1. Find the DTFT of
π₯ π = 1,0,
π = β1,0,1πππ ππ€βπππ
and plot its amplitude for βπ β€ π < π
16.14 Problems
6. a.
Find the DTFT of
π₯ π = 1,0,
π = 0,1πππ ππ€βπππ
12. a. b.
Find the DTFT of π₯ π = πΌππ’(π) and plot its magnitude
for
a. πΌ = 0.95 , b. πΌ = 0.5
13-19
Chapter 18:
System Function, the Frequency Response
β(π) π₯(π) π¦(π)
β Besides of unit-sample response h(n)
System function and Frequency Response
are also specify the LTI system
LTI system
13-20
18.1 The System Function H(z)
Fact 1: H(z) is the z-transform of unit-sample
response h(n)
Fact 2: H(z) is the ratio Y(z)/X(z)
Fact 3: H(z) is found from the Difference
Equation and vice versa.
13-21 21
Fact 1: H(z) is the z-transform of unit-sample
response h(n)
Example 18.1A
The unit-sample response of an LTI system is
β π = πππ’(π). Find its system function
Solution
System Function
π» π§ =1
1βππ§β1
β π = πππ’(π)
13-22
Example 18.3 Find the system function of the LTI system with the
given input-output pair.
π₯ π = 1, 1, 1 , y π = 1, 3, 4, 3, 1
Solution : In the z-domain the input-output pair is
π π§ = 1 + π§β1 + π§β2 ,
π π§ = 1 + 3π§β1 + 4π§β2 + 3π§β3 + π§β4
By long division we obtain the system function
π» π§ =π(π§)
π(π§)=
1 + 3π§β1 + 4π§β2 + 3π§β3 + π§β4
1 + π§β1 + π§β2
= 1 + 2π§β1 + π§β2
Fact 2: H(z) is the ratio Y(z)/X(z)
13-23
Example 18.5A An LTI system is described by the input-output difference
equation
π¦ π = β0.3π¦ π β 1 β 0.4π¦ π β 2 + π₯(π) Find the system function
Solution
By transforming both sides of the difference equation above
π π§ = β0.3π§β1π(π§) β 0.4π§β2π(π§) + π(π§) 1 + 0.3π§β1 + 0.4π§β2 π π§ = π π§
Therefore, the system function is given by
π» π§ =1
1 + 0.3π§β1 + 0.4π§β2
Fact 3: H(z) is found from the Difference
Equation and vice versa.
13-24
Example 18.6 (modified)
Find the input-output difference equation of an LTI system
with the system function
π» π§ =π(π§)
π(π§)=
1 β 0.3π§β1
1 β 0.7π§β1 + 0.2π§β2
Solution
1 β 0.7π§β1 + 0.2π§β2 π π§ = (1 β 0.3π§β1)π π§
π π§ = 0.7π§β1π π§ β 0.2π§β2π π§ + π π§ β 0.3π§β1π(π§) Thus, we have π¦ π = 0.7π¦ π β 1 β 0.2π¦ π β 2 + π₯ π β 0.3π₯(π β 1)
13-25 Copyright Β© 2014 McGraw-Hill Education. Permission required for reproduction or display
Quiz 3
An LTI system is described by the input-output
difference equation
π¦ π = βπ¦ π β 1 + π₯ π β π₯(π β 1) Find the system function
Solution
By transforming both sides of the difference equation above
π π§ = βπ§β1π π§ + π π§ β π§β1π(π§) 1 + π§β1 π π§ = (1 β π§β1)π π§
Therefore, the system function is given by
π» π§ =1 β π§β1
1 + π§β1
13-26
18.2 Poles and Zeros
β’ In this section the system function
represented by a function of πβπ is
converted into the equivalent system
with H(z)=B(z)/A(z), where A(z) and B(z)
are polynomials in z.
β’ The roots of the numerator are called
zeros of the system.
β’ The roots of the denominator are called
poles of the system.
Copyright Β© 2014 McGraw-Hill Education. Permission required for reproduction or display
13-27
Example 18.10 (modified)
Find the poles and zeros of the following system function.
a. π» π§ = 1 + π§β1 + π§β2
b. π» π§ =1β0.3π§β1
1β0.7π§β1+0.2π§β2
Solution
a. π» π§ =π§2+π§+1
π§2
Zeros of H(z): π§2 + π§ + 1=0 π§ =β1Β± 12β4Γ1
2= β
1
2Β± π
3
2
Poles of H(z): π§2 = 0 z=0 (double roots)
b. π» π§ =π§2β0.3π§
π§2β0.7π§+0.2
Zeros of H(z): π§2 β 0.3π§=0 z=0, 0.3
Poles of H(z): π§2 β 0.7π§ + 0.2=0 π§ =0.7Β± (β0.7)2β4Γ0.2
2
= 0.35 Β± π0.28
13-28
Quiz 4
Find the poles and zeros of the following system function
π» π§ =1 + 0.2π§β1
1 β π§β1 + 0.5π§β2
13-29
18.3 The Frequency Response π―(π)
π― π : frequency response of the system.
β DTFT of the unit-sample response h(n) of
the system,
β It can be found from the system function H(z).
Definition
π₯ π = ππππ π¦ π = π»(π)ππππ
LTI system
π»(π)
13-30
Response for sinusoidal
Copyright Β© 2014 McGraw-Hill Education. Permission required for reproduction or display
LTI system
π»(π)
cosππ π»(π) cos(ππ + π)
13-31
Example The frequency response of an system with the system
function π» π§ =1
2+ π§β1 +
1
2π§β2 (β π = {
1
2, 1,
1
2} )
is given by π» π = πβππ(1 + cosπ), where the
magnitude response: π»(π) = 1 + cosπ, ΞΈ π = βπ.
Find the response of the system to the input
a. π₯ π = 1
b. π₯ π = cos(π
2π)
c. π₯ π = cos(ππ)
13-32
Solution For the input π₯ π = cos(ππ), the output of the system with magnitude
response of π»(π) and the phase response of π(π) is given by
π¦ π = π»(π) cos(ππ + π(π))
a. Here π₯ π = 1 = cos 0 β π, and π»(0) = 1 + cos 0 = 2, ΞΈ 0 = 0,
π¦ π = π»(0) cos 0 β π + ΞΈ 0 = 2
b. From π»(π
2) = 1 + cos
π
2= 1, π
π
2= β
π
2,
π¦ π = π»(π
2) cos
π
2π + ΞΈ
π
2= cos(
π
2π β
π
2) = sin(
π
2π)
c. From π» π = πβππ 1 + cos π = 0, π¦ π = π»(π) cos ππ + ΞΈ π = 0
13-33
20. (Modified in part)
a. Find the system function given the following difference
equation:
π¦ π = 0.2π¦ π β 1 + π₯(π) b. Find the magnitude and phase of the frequency response
for π = π.
c. Find the output to the input π₯ π = cos(ππ).
21. a.
The unit-sample response of a discrete-time system is
β π = 1, 2, 1 . Find and plot π» π over βπ β€ π < π
18.13 Problems
13-34
HOMEWORK (Due: 18/May/2015 13:00PM)
Hamada office 05.38.01 MJIIT Level 5
Chapter 16
Problem 6.a
Chapter 18
Problem 20. (Modified)
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