3 The Discrete-Time Fourier Analysis and Transform Part 1 · t f T x t A t a ( ) cos( ) : T:...

3

The Discrete-Time

Fourier Analysis and Transform

Part 1 Assoc.Prof.Dr. Peerapol Yuvapoositanon

Mahanakorn Institute of Innovation (MII)

Mahanakorn Institute of Innovation

Mahanakorn U. of Technology DSP3part1-1 EEET0485 Digital Signal Processing Assoc. Prof. Dr.

Peerapol Yuvapoositanon

Objectives

• Students understand the concepts of: The

Discrete-Time Fourier Transform; DTFT),


Mahanakorn U. of Technology DSP3part1-2 EEET0485 Digital Signal Processing

Assoc. Prof. Dr. Peerapol Yuvapoositanon

Continuous to Discrete-time Signals

( )ax t ( )x n




Continuous-time Sinusoidal Signal

• Let us define the function of the Continuous-

time sinusoidal signal as

• or

( ) cos(2 )ax t A ft time(sec)

Frequency(Hz)

Phase(radians)

t

f

( ) cos( )ax t A t

Angular Frequency (radians/second) Mahanakorn Institute of Innovation



Converting Continuous-time to

Discrete-time Sinusoidal Signals

Digital Frequency

(radians/sample)




Discrete-time Sinusoidal Signal

• The discrete-time signal is

( ) cos( )x n A n

sample

Digital Frequency (radians/sample)

Phase(radians)

n

T




Discrete-time Sinusoidal Signals

• Digital Angular frequency of Digital frequency

is related to the analogue frequency fa as

• Where fs is the sampling frequency

2

2

a

a

s

T

f T

f

f




Normalised Frequency

• Define , then

• Where f is Normalised frequency: i.e.

Normalised fa with respect to the sampling

frequency fs.

a

s

ff

f




1 period




Digital Frequency

• It is obvious that, for this signal, there are six

samples per cycle.

• One cycle has 2pi radians.

• Then each sample is separated by

• or

3

2

6

radians/sample

radians/sample




2

6

nn

n




N is the number of samples in 1 cycle

• From

• then

• Let N be the number if samples in 1 cycle, the

digital frequency is defined as

2

6n n

2

N

radians/sample

radians/sample 2

6




When f = 1/2, 1/4, 1/8 cycles per sample

fa = 1 kHz fa = 1 kHz fa = 1 kHz

f = 1/2 f = 1/4 f = 1/8




Periodic Signals

• For a signal with period N, its value at the

sample n+N is equal to its value at sample n.

• A signal is periodic iff frequency (f)* period

(N) must be an integer k

• Otherwise it is called aperiodic.

fN k




1) For a periodic signal,

k and N must be integer

• The interpretation is that the ratio

must be a rational number, i.e. k and N must

be integer numbers.

• Example

kf

N

1

5f Rational

1

2f Irrational

ตรรกยะ

อตรรกยะ




Proof

• Let x(n) and x(N+n) respectively be




For f *N is an integer

When f*N is an integer, this term is 1

Q.E.D.




Example 1

Samples of each cycle have the same values: Periodic

2cos( )

20n

N=20 N=20




• For

• So this signal is periodic.

kf

N

2 12 2

20 20f

1

20

f

N

2cos( )

20n

Rational




• Example 2

Samples of each cycle are different in values: Aperiodic

N=11.54

2cos( )

20 3n




• For

• So this signal is aperiodic.

kf

N

2 32 2

2020 3

f

N

3

20

f

N

2cos( )

20 3n

Irrational




3

16

N=32

2 3 6

2 16 32

N=32




4

16

N=8

2

8

N=8




5

16

N=32

2 5 10

2 16 32

N=32




N=16

6

16

6 3 2

16 16




N=32

7

16




N=4

8

16




Example 3.1

• Determine whether the following item is

periodic or not and if it is what is its period N?

4) ( ) cos( )

7

13) ( ) cos( )

12

5) ( ) cos( )

9

5) ( ) cos( )

10

na x n

nb x n

nc x n

nd x n

periodic.

periodic.

periodic.

aperiodic. Mahanakorn Institute of Innovation



4) ( ) cos( )

7

na x n

N=14




13) ( ) cos( )

12

nb x n

N=24




5) ( ) cos( )

9

nc x n

N=18




5) ( ) cos( )

10

nd x n




2 The maximum frequency is Pi radians

• For

• We can also write

• Where

( ) cos( )x n n




Frequency

• We say that the frequency is meaningful

when

• Beyond that, the value of the cosine is

repetitive

• Or we can say the range of frequency is




( ) cos( )4

x n n

cos( 0) 14

2cos( 1)

4 2

cos( 2) 04

2cos( 3)

4 2

cos( 4) 14

2cos( 5)

4 2

cos( 6) 0

4

2cos( 7)

4 2




Normalised Frequency f

• We write the frequency in terms of the

normalised frequency f as

• We can say the range of frequency f is




3 Sine signals with 2Pi radians apart are

the same signals

• From

• let

• We have




Discrete-Time Fourier Series

• Discrete-time periodic signals can be written

in the forms of Fourier Series




Discrete-Time Fourier Series

• The Synthesis Equation

• The Analysis Equation




Discrete-Time Fourier Transform

• For Aperiodic signals, we use modify the

signals as if its period is infinity then use the

Discrete-time Fourier Transform (DTFT).




Frequency Domain

Time Domain




• Consider the system with input

( )x n ( ) ( ) ( )

( )j n

y n x n h n

e h n

( ) cos sin j nx n n j n e




The result of h(n)*e^jwn




The Discrete-Time Fourier Transform

• The DTFT of h(n) is then

• Where is the digital frequency.




The delta function as the impulse

response

• Let us determine the output of a system

whose impulse response is described by the

delta function




The output of the system:

y(n) = x(n)

( 2)( ) ( ) ( )jy n H e x n

( 2) ( 2)( [ ( )] )j k j n

k

k e e

( 2)0 ( 2)(0) j j ne e

( 2)1 j ne

( 2)j ne

( )x nMahanakorn Institute of Innovation



The delta function as the impulse

response : DTFT

• By the notation of delta function, we have

• So the DTFT of delta function is one.

• This means the frequency response of the

system is a constant function.

• A highly preferable property of audio

amplifiers.




LabVIEW

1




The delta function and a unit delayed as

the impulse response




The Discrete-Time Fourier Transform

( ) ( )

[ ( ) ( 1)]

j j k

k

j k

k

H e h n e

k k e




The output of the system

( 2)( ) ( ) ( )jy n H e x n

( 2) ( 2)( [ ( ) ( 1)] )j k j n

k

k k e e

( 2)0 ( 2) ( 2)(0) (1 1)j j j ne e e

( 2) ( 2)1 j j ne e

( 2) ( 2) ( 2)j n j j ne e e ( 2)( ) ( )jx n e x n




The delta function and a unit delayed as

the impulse response: DTFT

• In general, for any particular frequency




LabVIEW




Output at a particular frequency for

• For a delta function as impulse response

• For a unit delayed delta function as impulse

response

( )n

( 1)n

( )x n

( )x n

( ) ( )y n x n

( ) ( )jy n e x n

( ) ( )h n n ( ) ( 1)h n n




Output at a particular frequency for

• For any k delayed delta function

( )n k ( )x n ( ) ( )jky n e x n

( ) ( )h n n k




Geometric Series and DTFT

• Infinite geometric series:

• Finite geometric series:

0

1, 1

1

n

n

a aa

1

0

1, 1

1

, 1

L

Ln

n

aa

a a

L a




Example 3.2: Determine the DTFT of

• Solution:

( ) ( )j j n

n

X e x n e

( ) (0.5) ( )j n j n

n

X e u n e

0

(0.5)n j n

n

e

0

(0.5 )j n

n

e

1

1 0.5 je

0.5

j

j

e

e

( ) (0.5) ( )nx n u n

From

Then

0

1, 1

1

n

n

a aa





Solution:

0.5, 0 1( )

0,

n Lx n

otherwise

1

0

, 1

1, 1

1

Ln L

n

L a

a aa

a

1

0

( ) (0.5)L

j j n

n

X e e

1

0

0.5 ( )L

j n

n

e

10.5

1

j L

j

e

e




Example 3.4: DTFT in general

• Determine the DTFT of

• Solution:

( ) {1,2,3,4,5}x n

Note the position of up-arrow symbol




For Sequences with the Unit Step

function

• From 1, 0( )

0, 0

nu n

n





• Solution: From the Finite geometric series:

5

0

6

( )

1

1

j j n

n

j

j

X e e

e

e

1

0

, 1

1, 1

1

Ln L

n

L a

a aa

a

( ) ( ) ( 6)x n u n u n





• Solution: 0

( )j j n

n

X e e

0

( )j n

n

e

( ) ( )x n u n

0

j m

m

e

1

1 je

0

1, 1

1

n

n

a aa

Change the

direction

of index

m=-n





• Solution: 0

0

0

( ) 2

2

2

j n j n

n

m j m

m

j m

mm

X e e

e

e

0

( )2

1

1 2

2

2

jn

n

j

j

e

e

e

( ) 2 ( )nx n u n




DTFT (x*h)= DTFT(x)DTFT(h)

• We usually compute the convolution in x(n)

and h(n) in the time domain.

• But if x(n) and h(n) each has DTFT

• then

{ ( ) ( )} { ( )} { ( )}

( ) ( )j j

F x n h n F x n F h n

X e H e

( ), ( )j jX e H e




Example 3.8: Find the result of

x(n)*h(n) using DTFT

• Solution: 1

1

( ) ( ) ( )j j j n

n

X e H e x n e

( ) ( ) {1,1,1}x n h n

( 1) (0) (1)1 1 1j j je e e

1j je e




• So we say that the output of the system

• Or we can write

2

2 2

( ) ( ) ( )

( 1 )

2 3 2

j j j

j j

j j j j

Y e X e H e

e e

e e e e

2 2( ) 1 2 3 2 1j j j j j jY e e e e e e




y(n) are the coefficients of the DTFT

result

• If we compare

• With

• This means that we can use the coefficients for

y(n) as

2 2( ) 1 2 3 2 1j j j j j jY e e e e e e

( ) {1,2,3,2,1}y n

2

2

( ) ( )j j n

n

Y e y n e




Homework 3.1

Consider the system whose impulse response is

And with input , determine the

output of the system by means of

( ) [1,2,3]x n

( ) [1,1,1]h n

{ ( ) ( )} { ( )} { ( )}

( ) ( )j j

F h n x n F h n F x n

H e X e




Frequency Responses (DTFT) of

Systems at a particular frequency

( )h n0( )j n

x n e

0( )j n

h n e

0Mahanakorn Institute of Innovation



( )jH e 0( )j n

x n e

( ) ( )j j n

n

H e h n e

( ) ( ) ( )j j jY e H e X e

0( ) ( )j njy n H e e




Example 3.9:

Determine from the

following system

( )h n( ) ( )x n n ( )y n

( ) {1,0.8,0.2}h n

( ), ( ), ( )j j jX e Y e H e




Solution 3.9

0 2( ) ( ) 1 0.8 0.2j j n j j j

n

H e h n e e e e

0( ) ( ) ( ) 1 1j j n j n j

n n

X e x n e n e e

2( ) ( ) ( ) 1 0.8 0.2j j j j jY e H e X e e e




Poles and Zeros

of the Frequency Response

• For a first-order system with impulse response

h(n), the DTFT (or Frequency Response) is

( ) ( )j j n

n

H e h n e

1

1

j

j

e z

e p




Poles and Zeros


• For a system with

• The DTFT (or Frequency Response) is

( ) {1, (1)}h n h

( ) ( ) 1 (1)j j n j

n

H e h n e h e

( ) ( (1))j j jH e e e h (1)

0

j

j

e h

e

1

1

j

j

e z

e p




1

je z

1

je z

1zje

1

j




1

je p

p1

1

je p

je

1

j




Example 3.10: Poles and Zeros



• The DTFT (or Frequency Response) is

( ) {1, 0.8}h n

( ) ( ) 1 0.8j j n j

n

H e h n e e

( ) ( 0.8)j j jH e e e 0.8

0

j

j

e

e

1

1

j

j

e z

e p




Z1=0.8

0.8je

1 0.8z

je

1

j




p1=0

1 0p

je

1

j







• The DTFT of this system is

( ) 0.8 ( 1) ( )y n y n x n

( ) 0.8 ( ) ( )j j j jY e e Y e X e

( ) 1( )

( ) 1 0.8 0.8

j jj

j j j

Y e eH e

X e e e

1

1

( )

( )

j

j

e z

e p




z1=0

( 0)je

1 0z

je

1

j




p1=0.8

( 0.8)je

0.8je

1 0.8p

je

1

j







• The DTFT of this system is

( ) 0.8 ( 1) ( ) 0.5 ( 1)y n y n x n x n

( ) 0.8 ( ) ( ) 0.5 ( )j j j j j jY e e Y e X e e X e

( ) 1 0.5 ( 0.5)( )

( ) 1 0.8 0.8

j j jj

j j j

Y e e eH e

X e e e

1

1

( )

( )

j

j

e z

e p




z1=-0.5

( 0.5)je

1 0.5z

je

1

j

0.5je




p1=0.8

( 0.8)je

0.8je

1 0.8p

je

1

j




( 0.5) / ( 0.8)j je e

0.8je

1 0.8p

je

1

j

1 0.5z

0.5je




Poles and Zeros in General Form

0

1

( ) ( )

1

Mj k

kj j n k

Nj kn

kk

b e

H e h n e

a e

1 20

1 2

( )( ) ( )

( )( ) ( )

j j j

M

j j j

N

e z e z e zbe p e p e p




The Magnitude of Frequency Response

• At a particular frequency, the magnitude is

1 20

1 2

( )( ) ( )( )

( )( ) ( )

j j jj M

j j j

N

e z e z e zH e b

e p e p e p

1 2

0

1 2

( ) ( ) ( )

( ) ( ) ( )

j j j

M

j j j

N

e z e z e zb

e p e p e p




The Phase of Frequency Response

• At a particular frequency, the phase is

( ) numerator of ( )

denominator of ( )

j j

j

H e H e

H e

1 Im{ ( )}( ) tan

Re{ ( )}

jj

j

H eH e

H e




Example 3.13: System with a zero at 0.8

• From

• DTFT

• The magnitude is

( ) {1, 0.8}h n

0.8( )

0

jj

j

eH e

e

0.8( )

0

j

j

j

eH e

e




At Frequencies

0 and pi/4 radians/sample

( 4)( 4)

( 4)

0.8( ) 0.713 0.916

jj

j

eH e

e

(0)(0)

(0)

0.8( ) 1 0.8 0.2 0

jj

j

eH e

e




At Frequencies

pi/2, 3pi/4 and pi radians/sample

( 2)

( 2)

( 2)

0.8( ) 1.28 0.674

jj

j

eH e

e

(3 4)(3 4)

(3 4)

0.8( ) 1.66 0.346

jj

j

eH e

e

( )( )

( )

0.8( ) 1.8 0

jj

j

eH e

e




0.8( )

0

j

j

j

eH e

e

0

4

2

3 4

0.2 0

0.713 0.916

1.28 0.674

1.66 0.346

1.8 0Mahanakorn Institute of Innovation



Frequency Response: Magnitude and

Phase

0.20

0.71

1.28

1.66 1.80

0 4 2 3 4

Magnitude Phase

0




Frequency Response from Poles and Zeros

Zero pole

0.8( )

0

j

j

j

eH e

e




Example for Frequency Response

( 4)( )

0.713 0.916

jH e

( 2)( )

1.28 0.674

jH e

(3 4)( )

1.66 0.346

jH e

4 2 3 4




Example 3.14: System with a pole at 0.8

• From

• DTFT

• The magnitude is

0( )

0.8

jj

j

eH e

e

0( )

0.8

j

j

j

eH e

e

( ) 0.8 ( 1) ( )y n y n x n




At Frequencies

0 and pi/4 radians/sample

( 4)( 4)

( 4)( ) 1.40 0.916

0.8

jj

j

eH e

e

(0)(0)

(0)

1( ) 5 0

0.8 1 0.8

jj

j

eH e

e




At Frequencies

pi/2, 3pi/4 and pi radians/sample

( 2)

( 2)

( 2)( ) 0.78 0.674

0.8

jj

j

eH e

e

(3 4)

(3 4)

(3 4)( ) 0.6 0.346

0.8

jj

j

eH e

e

( )( )

( )( ) 0.55 0

0.8

jj

j

eH e

e




0.8( )

0

j

j

j

eH e

e

0

4

2

3 4

5 0

1.40 0.916

0.78 0.674

0.6 0.346

0.55 0Mahanakorn Institute of Innovation



Frequency Response: Magnitude and

Phase

5

1.40

0.78 0.6 0.55

0 4 2 3 4

Magnitude Phase

0




Frequency Response from Poles and Zeros

Zero pole

( )0.8

j

j

j

eH e

e




Example for Frequency Response

4 2 3 4

1.40 0.916 0.78 0.674 0.6 0.346




Finding h(n) from DTFT of the single

pole system

• From

• Remember the infinite geometric formula

• And the DTFT of h(n)

0( )

0.8

jj

j

eH e

e

0

1, 1

1

n

n

a aa

1

1 0.8 je

( ) ( )j j n

n

H e h n e




1( )

1 0.8

j

jH e

e

0

(0.8 )j n

n

e

0

0.8n j n

n

e

0.8 ( )n j n

n

u n e

( )h n

( ) 0.8 ( )nh n u n

Invoke the unit step

property




Example 3.15:

• Determine the output y(n) of a system whose

impulse response is

• With input

1( ) ( )

2

n

h n u n

2( )n

j

x n Ae




Solution

• Notice that we are interested the response of

the system at the particular frequency

• So

• DTFT

( ) ( ) 0.5 ( )j j n n j n

n n

H e h n e u n e

0

2

1

1 0.5 je

0

1From , 1

1

n

n

a aa

2

2

1( )

1 0.5

j

j

H e

e

26.62

5

je1

1 0.5j




The Output y(n) at a Particular

Frequency

• From

• Then 26.6 2

26.62

2( )

5

2

5

nj

j

nj

y n e Ae

Ae

( ) ( ) ( )jy n H e x n




1

1 11

2 2

j j

jj j

e e

ee e

( ) ( )j j n

n

H e h n e

1

11

2

je




Frequency Response from Difference

Equations

• A difference equation is in the form of

• With , and

1 0

( ) ( ) ( )N M

l ml m

y n a y n l b x n m

( ) j nx n e ( ) ( )j j ny n H e e

( ) ( )

1 0

( ) ( )N M

j j n j j n l j n m

l ml m

H e e a H e e b e




The Frequency Response (DTFT)

• Then the DTFT is 0

1

( )

1

Mj m

mj m

Nj l

ll

b e

H e

a e

( ) ( )

1 0

( ) ( )N M

j j n j j n l j n m

l ml m

H e e a H e e b e

( ) ( )

1 0

1 ( ) ) ( )N M

j j l j j n j m j n

l ml m

a H e e H e e b e e




Example 3.16: Determine the frequency

Response of

with

• Solution:

( ) ( ) 0.8 ( 1) 0.2 ( 2)y n x n x n x n

( ) j nx n e

( 1) ( 2)( ) 0.8 0.2j j n j n j n j nH e e e e e

2( ) 0.8 0.2j j n j n j j n j j nH e e e e e e e

2( ) 1 0.8 0.2j j jH e e e




Homework 3.2:

• Determine the frequency Response of

with

( ) 0.2 ( 1) ( ) 0.8 ( 1) 0.2 ( 2)y n y n x n x n x n

( ) j nx n e




Example 3.17: Determine the output

y(n)

for

• Solution:

( ) 0.8 ( 1) ( )y n y n x n

( ) cos(0.05 ) ( )x n n u n

( 1)( ) 0.8 ( )j j n j j n j nH e e H e e e

( ) 0.8 ( )j j n j j n j j nH e e H e e e e 1

( )1 0.8

j

jH e

e




( ) cos(0.05 ) ( )x n n u n

0 0.05

0.05

0.05

0.5377

1 1( )

1 0.8 1 0.8 1 0.05

4.0928 1 0.5377

4.0928

j

j

j

H ee j

e

For

This means therefore

This means the system modifies the magnitude

and phase of input by 4.0928 and -0.5377 radians

respectively.




The output y(n)

• Now the output can be written as

• or the phase can be written in terms of

number of samples as

( ) 4.0928cos(0.05 0.5377)y n n

( ) 4.0928cos 0.05 ( 3.42)y n n

Frequency is the

same as the input




Testing

( ) cos(0.05 )x n n

1( ) ( )

1 0.8 jy n x n

e

Phase=3.42 samples

Magnitude=4.092

( )jH e ( )x n ( )y n




Example 3.18: Determine the output

y(n)

for

• Solution:

( ) 0.8 ( 1) ( ) 0.5 ( 1)y n y n x n x n

( ) cos(0.05 ) ( )x n n u n

( 1) ( 1)

( ) 0.8 ( 1) ( ) 0.5 ( 1)

( ) 0.8 ( ) 0.5

( ) 0.8 ( ) 0.5

j j n j j n j n j n

j j n j j n j j n j n j

y n y n x n x n

H e e H e e e e

H e e H e e e e e e




1 0.5( )

1 0.8

jj

j

eH e

e

( ) 0.8 ( ) 1 0.5j j j jH e H e e e




0.050.05

0.05

0.59

1 0.5 2.57 .03( )

1 0.8 0.244 .537

6.122

jj

j

j

eH e

e

e

( ) 6.122cos(0.05 0.59)

6.122cos 0.05 ( 3.756)

y n n

n

( ) cos(0.05 ) ( )x n n u n

0 0.05

For

This means therefore




Homework 3.3

• Consider the system with difference equation

• Determine the output of the system when

( ) cos(0.05 ) ( )x n n u n

( ) 0.8 ( 1) 0.2 ( 2) ( ) 0.1 ( 1)y n y n y n x n x n




3 The Discrete-Time Fourier Analysis and Transform Part 1 · t f T x t A t a ( ) cos( ) : T:...

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Transcript of 3 The Discrete-Time Fourier Analysis and Transform Part 1 · t f T x t A t a ( ) cos( ) : T:...