Chapter 16:

Discrete-Time Fourier Transform

Chapter 18:

System Function, the Frequency

Response

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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

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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 𝜃 𝜔 = ∠𝑋(𝜔)

13-4

DTFT is Periodic DTFT 𝑋(𝜔) is periodic with period 2π because

𝑋(𝜔) is specified by

𝑋 𝜔 − 𝜋 ≤ 𝜔 < 𝜋

(When the DTFT is expressed in terms of f

(𝜔 = 2𝜋𝑓), the period of f is 1 Hz and

𝑋 𝑓 = 𝑋(𝜔)𝜔=2𝜋𝑓

may be specified for −0.5 < 𝑓 < 0.5

13-5

IDTFT, The Inverse of DTFT

The inverse of the DTFT may be found from

Copyright © 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: 𝑋 𝜔 = 1

Example 16.1b. Find the DTFT and plot its

magnitude given

𝑥 𝑛 = 1

2,

0,

𝑛 = −1,1𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒

Solution:

𝑋 𝜔 =1

2𝑒𝑗𝜔 +

1

2𝑒−𝑗𝜔 = cos𝜔

13-7

𝑋 𝜔 = cos𝜔

𝑥(𝑛)

−𝜋 0 𝜋 𝜔

Example 16.1b.

12-8 Copyright © 2014 McGraw-Hill Education. Permission required for reproduction or display

Quiz 1 Find the DTFT given

𝑥 𝑛 = 0.5𝑑 𝑛 + 1 + 𝑑 𝑛 + 0.5𝑑 𝑛 − 1

𝑋 𝜔 = 1 + cos𝜔

−𝜋 0 𝜋

𝜔

Examples (Cont.)

Example 16.3a. Find the DTFT and its

magnitude and phase of 𝒙 𝒏 = 𝒂𝒏𝒖(𝒏)

Solution:

𝑋 𝜔 = 𝑎𝑛𝑒−𝑗𝜔𝑛

∞

𝑛=0

= 𝑎𝑒−𝑗𝜔𝑛

∞

𝑛=0

=1

1 − 𝑎𝑒−𝑗𝜔=

1

1 − 𝑎 cos𝜔 + 𝑗𝑎 sin𝜔

𝑋(𝜔) =1

(1 − 𝑎 cos𝜔)2+(𝑎 sin𝜔)2

𝜃 𝜔 = −𝑡𝑎𝑛−1𝑎 sin𝜔

1 − 𝑎 cos𝜔

Magnitude:

Phase:

13-10

j

1

1 0.8e

ω

ω

Example 𝛼 = 0.8

13-11

16.3 The DTFT and the z-Transform

Copyright © 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 𝑥 𝑛 = 0.8 𝑛𝑢(𝑛)

DTFT

(magnitude)

13-13

Phase response

1

1X(z)

1 0.8z

z-transform of 𝑥 𝑛 = 0.8 𝑛𝑢(𝑛)

DTFT (phase)

16.4 Inverse DTFT (IDTFT) Example

Example 16.8. Find the inverse DTFT of

Solution.

13-15 Figure 16.3. X(ω) (upper) and x(n) (lower).

−𝜋 0 𝜋

𝑋(𝜔)

𝑥(𝑛)

13-16

16.6 Properties of DTFT

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Linearity:

Time shift:

𝑎𝑥 𝑛 + 𝑏𝑦 𝑛

𝑥 𝑛 − 𝑘

𝑎𝑋 𝜔 + 𝑏𝑌 𝜔

𝑒−𝑗𝑘𝜔𝑋 𝜔

Convolution: 𝑥 𝑛 ∗ 𝑦 𝑛 𝑋 𝜔 ∙ 𝑌 𝜔

𝑥(𝑛)

∞

𝑛=−∞

= 𝑋(0) Zero frequency

Copyright © 2014 McGraw-Hill Education. Permission required for reproduction or display

Quiz 2

Find the DTFT of given signal

𝑥 𝑛 = 𝑑 𝑛 − 5 + 𝑑(𝑛 + 5)

13-18

1. Find the DTFT of

𝑥 𝑛 = 1,0,

𝑛 = −1,0,1𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒

and plot its amplitude for −𝜋 ≤ 𝜔 < 𝜋

16.14 Problems

6. a.

Find the DTFT of

𝑥 𝑛 = 1,0,

𝑛 = 0,1𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒

12. a. b.

Find the DTFT of 𝑥 𝑛 = 𝛼𝑛𝑢(𝑛) and plot its magnitude

for

a. 𝛼 = 0.95 , b. 𝛼 = 0.5

13-19

Chapter 18:

System Function, the Frequency Response

ℎ(𝑛) 𝑥(𝑛) 𝑦(𝑛)

■ 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

ℎ 𝑛 = 𝑎𝑛𝑢(𝑛). Find its system function

Solution

System Function

𝐻 𝑧 =1

1−𝑎𝑧−1

ℎ 𝑛 = 𝑎𝑛𝑢(𝑛)

13-22

Example 18.3 Find the system function of the LTI system with the

given input-output pair.

𝑥 𝑛 = 1, 1, 1 , y 𝑛 = 1, 3, 4, 3, 1

Solution : In the z-domain the input-output pair is

𝑋 𝑧 = 1 + 𝑧−1 + 𝑧−2 ,

𝑌 𝑧 = 1 + 3𝑧−1 + 4𝑧−2 + 3𝑧−3 + 𝑧−4

By long division we obtain the system function

𝐻 𝑧 =𝑌(𝑧)

𝑋(𝑧)=

1 + 3𝑧−1 + 4𝑧−2 + 3𝑧−3 + 𝑧−4

1 + 𝑧−1 + 𝑧−2

= 1 + 2𝑧−1 + 𝑧−2

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

𝑦 𝑛 = −0.3𝑦 𝑛 − 1 − 0.4𝑦 𝑛 − 2 + 𝑥(𝑛) Find the system function

Solution

By transforming both sides of the difference equation above

𝑌 𝑧 = −0.3𝑧−1𝑌(𝑧) − 0.4𝑧−2𝑌(𝑧) + 𝑋(𝑧) 1 + 0.3𝑧−1 + 0.4𝑧−2 𝑌 𝑧 = 𝑋 𝑧

Therefore, the system function is given by

𝐻 𝑧 =1

1 + 0.3𝑧−1 + 0.4𝑧−2

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

𝐻 𝑧 =𝑌(𝑧)

𝑋(𝑧)=

1 − 0.3𝑧−1

1 − 0.7𝑧−1 + 0.2𝑧−2

Solution

1 − 0.7𝑧−1 + 0.2𝑧−2 𝑌 𝑧 = (1 − 0.3𝑧−1)𝑋 𝑧

𝑌 𝑧 = 0.7𝑧−1𝑌 𝑧 − 0.2𝑧−2𝑌 𝑧 + 𝑋 𝑧 − 0.3𝑧−1𝑋(𝑧) Thus, we have 𝑦 𝑛 = 0.7𝑦 𝑛 − 1 − 0.2𝑦 𝑛 − 2 + 𝑥 𝑛 − 0.3𝑥(𝑛 − 1)

13-25 Copyright © 2014 McGraw-Hill Education. Permission required for reproduction or display

Quiz 3

An LTI system is described by the input-output

difference equation

𝑦 𝑛 = −𝑦 𝑛 − 1 + 𝑥 𝑛 − 𝑥(𝑛 − 1) Find the system function

Solution

By transforming both sides of the difference equation above

𝑌 𝑧 = −𝑧−1𝑌 𝑧 + 𝑋 𝑧 − 𝑧−1𝑋(𝑧) 1 + 𝑧−1 𝑌 𝑧 = (1 − 𝑧−1)𝑋 𝑧

Therefore, the system function is given by

𝐻 𝑧 =1 − 𝑧−1

1 + 𝑧−1

13-26

18.2 Poles and Zeros

• In this section the system function

represented by a function of 𝒛−𝟏 is

converted into the equivalent system

with H(z)=B(z)/A(z), where A(z) and B(z)

are polynomials in z.

• The roots of the numerator are called

zeros of the system.

• The roots of the denominator are called

poles of the system.

Copyright © 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. 𝐻 𝑧 = 1 + 𝑧−1 + 𝑧−2

b. 𝐻 𝑧 =1−0.3𝑧−1

1−0.7𝑧−1+0.2𝑧−2

Solution

a. 𝐻 𝑧 =𝑧2+𝑧+1

𝑧2

Zeros of H(z): 𝑧2 + 𝑧 + 1=0 𝑧 =−1± 12−4×1

2= −

1

2± 𝑗

3

2

Poles of H(z): 𝑧2 = 0 z=0 (double roots)

b. 𝐻 𝑧 =𝑧2−0.3𝑧

𝑧2−0.7𝑧+0.2

Zeros of H(z): 𝑧2 − 0.3𝑧=0 z=0, 0.3

Poles of H(z): 𝑧2 − 0.7𝑧 + 0.2=0 𝑧 =0.7± (−0.7)2−4×0.2

2

= 0.35 ± 𝑗0.28

13-28

Quiz 4

Find the poles and zeros of the following system function

𝐻 𝑧 =1 + 0.2𝑧−1

1 − 𝑧−1 + 0.5𝑧−2

13-29

18.3 The Frequency Response 𝑯(𝝎)

𝑯 𝝎 : frequency response of the system.

■ DTFT of the unit-sample response h(n) of

the system,

■It can be found from the system function H(z).

Definition

𝑥 𝑛 = 𝑒𝑗𝜔𝑛 𝑦 𝑛 = 𝐻(𝜔)𝑒𝑗𝜔𝑛

LTI system

𝐻(𝜔)

13-30

Response for sinusoidal

Copyright © 2014 McGraw-Hill Education. Permission required for reproduction or display

LTI system

𝐻(𝜔)

cos𝜔𝑛 𝐻(𝜔) cos(𝜔𝑛 + 𝜃)

13-31

Example The frequency response of an system with the system

function 𝐻 𝑧 =1

2+ 𝑧−1 +

1

2𝑧−2 (ℎ 𝑛 = {

1

2, 1,

1

2} )

is given by 𝐻 𝜔 = 𝑒−𝑗𝜔(1 + cos𝜔), where the

magnitude response: 𝐻(𝜔) = 1 + cos𝜔, θ 𝜔 = −𝜔.

Find the response of the system to the input

a. 𝑥 𝑛 = 1

b. 𝑥 𝑛 = cos(𝜋

2𝑛)

c. 𝑥 𝑛 = cos(𝜋𝑛)

13-32

Solution For the input 𝑥 𝑛 = cos(𝜔𝑛), the output of the system with magnitude

response of 𝐻(𝜔) and the phase response of 𝜃(𝜔) is given by

𝑦 𝑛 = 𝐻(𝜔) cos(𝜔𝑛 + 𝜃(𝜔))

a. Here 𝑥 𝑛 = 1 = cos 0 ∙ 𝑛, and 𝐻(0) = 1 + cos 0 = 2, θ 0 = 0,

𝑦 𝑛 = 𝐻(0) cos 0 ∙ 𝑛 + θ 0 = 2

b. From 𝐻(𝜋

2) = 1 + cos

𝜋

2= 1, 𝜃

𝜋

2= −

𝜋

2,

𝑦 𝑛 = 𝐻(𝜋

2) cos

𝜋

2𝑛 + θ

𝜋

2= cos(

𝜋

2𝑛 −

𝜋

2) = sin(

𝜋

2𝑛)

c. From 𝐻 𝜋 = 𝑒−𝑗𝜋 1 + cos 𝜋 = 0, 𝑦 𝑛 = 𝐻(𝜋) cos 𝜋𝑛 + θ 𝜋 = 0

13-33

20. (Modified in part)

a. Find the system function given the following difference

equation:

𝑦 𝑛 = 0.2𝑦 𝑛 − 1 + 𝑥(𝑛) b. Find the magnitude and phase of the frequency response

for 𝜔 = 𝜋.

c. Find the output to the input 𝑥 𝑛 = cos(𝜋𝑛).

21. a.

The unit-sample response of a discrete-time system is

ℎ 𝑛 = 1, 2, 1 . Find and plot 𝐻 𝜔 over −𝜋 ≤ 𝜔 < 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)