new_s_spect_07_09

8/3/2019 New_S_Spect_07_09

1/89

1

Spectral Analysis

Lecture #7-9

0 2000 4000 6000 8000 10000 1 2000 1 4000-4

-3

-2

-1

0

1

2

3

4

Time m sec

M a g n

i t u

d e

i n

V o

l t s

8/3/2019 New_S_Spect_07_09

2/89

2

Table 1 Summary of FT PropertiesIntegration in time domain

)f (G)t(g

Diff. in freq. domain

Diff. in time domain ) 2 ( )( j f d

g t f d

Gt

2 ( ) ( )d

G f d

j g tf

t

8/3/2019 New_S_Spect_07_09

3/89

3

Proof

( )( )g t G f

Diff. in time domain) 2 ( )( j f

d

g t f d Gt

Duality ( )d G f df

2 ( ) ( ) j t g t


( )2 ( ) j t g t d G f df

8/3/2019 New_S_Spect_07_09

4/89

4

Examples

Find the Fourier transform (FT) of -atg(t) = t e u(t)

Solution

-ate u(t) 1a + j 2 f


( )2 ( ) j t g t d G f df

2 ( )at j t e u t ( ) 12d a j f df

+

***

8/3/2019 New_S_Spect_07_09

5/89

5

( )att e u t ( )212a j f +

***

8/3/2019 New_S_Spect_07_09

6/89

6

Example

Find the Fourier transform (FT) of 2 -atg(t) = t e u(t)

( )att e u t ( )2

1

1 2 j f +Diff. in freq. domain

( )2 ( ) j t g t d G f df

Solution

8/3/2019 New_S_Spect_07_09

7/89

7

2 ( )att e u t

( )3

2

1 2 j f +

***

8/3/2019 New_S_Spect_07_09

8/89

8

Table 1 Summary of FT Properties

)f (G)t(g

Area under G(f)

Area under g(t) ( ) (0)g t d Gt

=

( ) (0)G f d gf

=

8/3/2019 New_S_Spect_07_09

9/89

9

Area under g(t)

2( )) ( j f tg tG f e dt

=

( )G f df

Substituting f=0

(0)G = ( )g t dt

Area under G(f)2( )( ) j f tG f dt e tg

= Substituting t=0

(0)g =

8/3/2019 New_S_Spect_07_09

10/89

10

Example

Find

tA tri

( )2sinA c f Solution

( )2sin 4c f df

( )2

sin 4c f 14 4

ttri

( )( 0)G f d f g

=

( )2sin 4c f df

= 14

=

***

8/3/2019 New_S_Spect_07_09

11/89

11

Example

Find

( )2222

a t aa

ef

+

Solution2

11

dtt

+

a=1

( )22

1 2 t+f e

***

8/3/2019 New_S_Spect_07_09

12/89

12

( )22

1 2 t +f

e

( ) 1 f Ga a

g at

4- Time scaling

12

a

=

( )22

1 t

+

( )21

1 t

+

( ) (0)g t d Gt

= 211 dtt

=+ (0)G =

***

8/3/2019 New_S_Spect_07_09

13/89

13


)f (G)t(g

Conjugate function ( ) ** ( )g G f t =

Proof 2( )) ( j f tg tG f e dt

= Taking conjugate

** 2( )) ( j f tG f g t e dt

= 2* *( ) ( ) j f tG g t dtf e

= ( ) ** ( )g G f t =

8/3/2019 New_S_Spect_07_09

14/89

14


)f (G)t(g

Real function ( ) * ( )G f G f =

Real & magnitude parts of G(f) have evensymmetry

Imajinary & phase parts of G(f) have oddsymmetry

( ) ** ( )g G f t =

8/3/2019 New_S_Spect_07_09

15/89

15


)( ()g t G f

Even function ( ) ( )G f G f =

Proof

Scaling in time domain

( ) 1 f Ga a

g at

a = -1

( ) 11 1

f Gg t

( ) ( )g Gt f

)( ()g t G f

8/3/2019 New_S_Spect_07_09

16/89

16


is real & eveng(t) is real & even ( )G f

)( ()g t G f

( ) ( )G f G f = ( )( )g t g t= Even functionReal function ( ) * ( )G f G f = ( )* ( )G f G f =

Real & even function

( )*

( )G f G f =

G(f) is Real & even

Proof

8/3/2019 New_S_Spect_07_09

17/89

17


is pure imaginary &

odd

g(t) is real & odd ( )G f

)( ()g t G f

( ) ( )G f G f = ( )( )g t g t= odd functionReal function ( ) * ( )G f G f =

Real & odd function ( ) * ( )G f G f =

G(f) is pure imaginary & odd

Proof

( )* ( )G f G f =

8/3/2019 New_S_Spect_07_09

18/89

18


Refer to Table 2 (FT pairs) and identify :

a) Real & even signals: Verify that G(f) isreal & even.

b) Real & odd signals: Verify that G(f) ispure imaginary & odd.

8/3/2019 New_S_Spect_07_09

19/89

19

Table 2 Summary of FT Pairs

G (f)g (t)

( )sinA c f t

A rect t

A tri

( )2sinA c f

( )ate u t12a j f +

( )ate u t 12a j f

8/3/2019 New_S_Spect_07_09

20/89

20


1

1

G (f)g (t)

a t

e

( )22

2

2

a

a f +( )t

( )f

( )0cos 2 f t ( ) ( )0 012

f f f f + +

( )0sin 2 f t ( ) ( )0 012

f f f f j

+

8/3/2019 New_S_Spect_07_09

21/89

21


G (f)g (t)1

j f ( )sgn t

( )u t ( )1 12 2

f j f

+

2te

1t ( )sgn f

2f e

8/3/2019 New_S_Spect_07_09

22/89

22


Ex) Find the FT of the Gaussian pulse

( )2tg t e

=

Solution

( )2tg t e

= ( )G f Property of Diff. in time domain

( ) 2 ( )d j f dt

f t Gg

22 ( )t

dd

f e jt

f G

8/3/2019 New_S_Spect_07_09

23/89

23

Sol.

Property of Diff. in freq. domain

( ) ( )2 j t g t d G f df

22 ( )td

df e j

tf G

( )2

2 ( )2t

e t j f G f

2)2 (t j t

dG f e

df

2

(2 2 )t j t e f G f

* -j

8/3/2019 New_S_Spect_07_09

24/89

24

Sol.

2)2 (t j t d G f e

df

2 (2 2 )t j t e f G f

( )

d

G f df =

2 ( )f G f

( )G f =2f e

2 2f te e

8/3/2019 New_S_Spect_07_09

25/89

25


11( () )g Gt f

Multiplication intime domain

( ) ( ) ( ) ( )1 2g t g t g Gt f =

22 ( () )g Gt f

1 2( )* ( )G G f d

=

( ) 1 2( ) ( )G f G f G f =

8/3/2019 New_S_Spect_07_09

26/89

26


11( () )g Gt f

Convolution intime domain

( ) ( ) ( ) ( )1 2g t g t Gg f t=

22 ( () )g Gt f

( ) ( )1 2 1 2( ) ( )g t g t g g t d

=

( ) 1 2( ) ( )G f G f G f =

8/3/2019 New_S_Spect_07_09

27/89

27

Linear Time Invariant (LTI) Systems

An LTI system is characterized by its impulseresponse (system output when the input

h(t) is the impulse response of the system .

LTI Systemx(t) y(t)

x ).(t)= (t)

LTI Systemx(t)= (t) y(t)=h(t)

8/3/2019 New_S_Spect_07_09

28/89

28

Linear Time Invariant (LTI) Systems

Given the impulse response of an LTI system, thesystem response (output y(t) ) to an arbitraryinput x(t) can be calculated by two methods :

1) Time domain method.2) Frequency domain method.

LTI Systemx(t) y(t)=?


8/3/2019 New_S_Spect_07_09

29/89

29

Time Domain method


(t) h (t)

(t- ) h (t- )TI

( )x (t- ) ( )x h (t- )LTI

( ) ( )-

x t = x (t- ) d

( )y t

( ) ( )-

y t = x h(t- ) d

( )=x t h(t)

8/3/2019 New_S_Spect_07_09

30/89

30

Time Domain method

LTI Systemx(t) y(t)

( ) ( )x t )y t = h(t

( ) ( ) ( )-

y t = =x x h(t-t )h d(t)

( )=h t x(t)

( ) ( ) ( )-

y t = =h h x(t-t )x d(t)

Convolution processis cumulative

8/3/2019 New_S_Spect_07_09

31/89

31

Time Domain method

LTI Systemx(t) y(t)

( ) ( ) ( )-

y t = =x x h(t-t )h d(t)

1) Plot x( )2) Plot h(- ) inversion of h( )

3) Plot h(t-) shift h(-

) by t

4) y(t) = area under the product ( )x h(t- )

8/3/2019 New_S_Spect_07_09

32/89

32

Frequency Domain method

LTI Systemx(t) y(t)

( ) ( ) ( )-y t = =x x h(t-t )h d(t)

H(f) Transfer function of the system.

( )Y(f)=X f H(f )

X (f)

H (f)

y(t) 4) Y(f)IFT

( ) ( )x3 t) )y t = h(t FT2) h (t) FT

1) x (t) FT

8/3/2019 New_S_Spect_07_09

33/89

33

Proof

( ) ( ) ( )-

y t =x t h(t)= x h(t- ) d

( ) ( )2 j f t

- y t=f e dY t

( ) 2

-

j f t

-

= ex h(t- ) d dt

=

( )-

x d =

2( ) f H f e

***

8/3/2019 New_S_Spect_07_09

34/89

34

Proof

( ) ( ) 2( )-

f x dY f = H ef

( )= H f

( )( ) = H f X f

( ) 2-

j f ex d

8/3/2019 New_S_Spect_07_09

35/89

35

Example

An LTI system has an impulse response

Is excited by an input . Find the system

output y(t) using:a) Multiplication in frequency domain

b) Convolution in time domain.

Solution

234

th(t) = rect 22

4t

x(t) = rect

8/3/2019 New_S_Spect_07_09

36/89

36

b) Frequency Domain method

( )Y(f) = X f 3) H(f)

X (f) =

t-4y(t)=24 tri 4

( ) 82 j f Y(f)= 96 sinc 4f e4)

4t-2

1) x (t) = 2 rect

FT

4t-2

2) h (t) = 3 rect

H (f) =

( ) 82 j f = 96 sinc 4f e

IFT

FT

***

8/3/2019 New_S_Spect_07_09

37/89

37

a) Frequency domain method

4t-2

x (t) = 2 rect

4t-2

h (t) = 3 rect

x(t)

4t

2

t

h(t)3

4

4t-4y (t) = 24 tri

8

y(t)

t

24

4

8/3/2019 New_S_Spect_07_09

38/89

38

b) Time Domain method

4t-2

x (t) = 2 rect

4t-2

h (t) = 3 rect

( ) ( ) ( )-

y t = = xx t h h(t- ) d(t)

x(t)

4t

2

t

h(t)3

4

8/3/2019 New_S_Spect_07_09

39/89

39


h( )3

4

( ) ( )-

y t = x h(t- ) d

x( )

4

2

h(- )3

-4

tt-4

h(t- )3

8/3/2019 New_S_Spect_07_09

40/89

40

Case 1 : No overlapping

( ) ( )-y t = x h(t- ) d

0t

=

x( )

4

2

tt-4

h(t- )3

***

8/3/2019 New_S_Spect_07_09

41/89

41

Case 2 : Partial overlapping

( ) ( )-

y t = x h(t- ) d

0t 4 0t

40 t

t

x( )h(t- )

6

=

x( )

4

2

h(t- )

tt-4

3

***

8/3/2019 New_S_Spect_07_09

42/89

42


( ) ( )-

y t = x h(t- ) d

4 4t 4t

=

84 t

x( )h(t- )

t-4 4

6

=

x( )

4

2

tt-4

3

***

8/3/2019 New_S_Spect_07_09

43/89

43


( ) ( )-

y t = x h(t- ) d

h(t- )

tt-4

3

4 4t

8t

=

x( )

4

2

***

8/3/2019 New_S_Spect_07_09

44/89

44

b) Time domain method

( )

0 0t

y t =

y(t)

4

t

24

6 0 4t t

48 6 4 8t t 0 8t

424 4

ttri

=

8

8/3/2019 New_S_Spect_07_09

45/89

45

Notes on convolution process

Start of y =start of x

+ start of h

=0+0 = 0

End of y = end of x

+ end of h

=4+4 = 8

x()

4

2

t

h(t)3

4y(t)

84t

24

8/3/2019 New_S_Spect_07_09

46/89

4646

impulse response

of an LTI system

LTI Systemx(t)= (t) y (t)=h(t)

h (t) impulse response

LTI Systemx(t) y(t)=x(t) h(t)

= (t) h(t)

(t) h(t)=h(t) h(t) for all

A h l i

8/3/2019 New_S_Spect_07_09

47/89

4747

Another solution

x(t)

4t

2

t

h(t)3

4

dx(t)=dt

t

dx(t)dt2

04

-2

dx(t) h(t)=

dt

t6

4

dx(t) h(t)

dt

-6

8

***

A th l ti

8/3/2019 New_S_Spect_07_09

48/89

4848

Another solution

y(t) x(t) h(t)=

t

-

d= x( ) h( ) ddt

t

6

4

dx(t) h(t)dt

-6

8

y(t)

4

t

24

8

***

l

8/3/2019 New_S_Spect_07_09

49/89

49

Example

An LTI system has an impulse response

Is excited by an input . Find the system

output y(t) using:a) Convolution in time domain.

b) Multiplication in frequency domain

Solution

234

th(t) = rect 32

6t

x(t) = rect

) Ti D i h d

8/3/2019 New_S_Spect_07_09

50/89

5050

x(t)

6t

2d x(t)=dt

2

06 t

dx(t)dt

-2

dx(t) h(t)=

dt

t

h(t)3

4

t6

6

dx(t) h(t)

dt

-6

104

a) Time Domain method ***

8/3/2019 New_S_Spect_07_09

51/89

) Ti D i th d ( th l ti )

8/3/2019 New_S_Spect_07_09

52/89

52

a) Time Domain method (another solution)

6t-3

x (t) = 2 rect

4

t-2h (t) = 3 rect

t

x(t)2

6

( ) ( ) ( )-

y t = = xx t h h(t- ) d(t)

4t

h(t)3


8/3/2019 New_S_Spect_07_09

53/89

5353


( ) ( )-

y t = x h(t- ) d

h( )

4

3 h(- )

-4

3

tt-4

h(t- )3

x( )2

6

8/3/2019 New_S_Spect_07_09

54/89

5454


( ) ( )-

y t = x h(t- ) d

0t

=

x( )2

6

tt-4

h(t- )3

***

8/3/2019 New_S_Spect_07_09

55/89

5555


( ) ( )-

y t = x h(t- ) d

0t 4 0t

40 t

x( )h(t- )

t

6

x( )2

6

tt-4

h(t- )3

***

=

8/3/2019 New_S_Spect_07_09

56/89

5656

Case 3 : complete overlapping

( ) ( )-y t = x h(t- ) d

4 0t 6t

64 t

x( )h(t- )

t

6

t-4

x( )2

6

tt-4

h(t- )3

***

=

=

8/3/2019 New_S_Spect_07_09

57/89

5757


( ) ( )-

y t = x h(t- ) d

4 6t 6t

106 t

6

x( )h(t- )

6

t-4

x( )2

6

tt-4

h(t- )3

***

=

=


8/3/2019 New_S_Spect_07_09

58/89

5858


( ) ( )-

y t = x h(t- ) d

4 6t

10t

x( )2

6

t

x( )h(t- )

6

t-4

***

=

a) Time domain method

8/3/2019 New_S_Spect_07_09

59/89

59


( )

0 0t

y t =

6 0 4t t

60 6 6 10t t 0 10t

y(t)

10t

24

24 4 6t

64

No Overlapping

No Overlapping

Partial Overlapping

Partial Overlapping

Complete Overlapping


8/3/2019 New_S_Spect_07_09

60/89

60

b) Show that

5( ) 30

5t

y t tri =

56

1t

tri


( ) ( )2 2 10( ) 150 sin 5 6 sin j f Y f c f c f e =

y(t)

10t

24

64

30

Example

8/3/2019 New_S_Spect_07_09

61/89

6161

Example

An LTI system has an impulse response h(t) shown Isexcited by an input x(t) shown. Find the system output y(t) .

Solution21

t

h(t)

1

1

t

x(t)1

-1

2

-1

Solution x(t)***

8/3/2019 New_S_Spect_07_09

62/89

6262

Solution

t

dx(t)

dt1

-1

2

1

d x(t)=dtd

x(t) h(t)=dt

1t

( )1

-12

-1

1

-2

21t

h(t)1

dx(t) h(t)

dt

t1

12

3

-2

4

Solutiond

***

8/3/2019 New_S_Spect_07_09

63/89

6363

y(t) x(t) h(t)=

dx(t) h(t)

dt

21

t1

3 4

-2

t

-

d= x( ) h( ) d

dt

y(t)

21t

1

3 4

-1

Example

8/3/2019 New_S_Spect_07_09

64/89

6464

p

An LTI system has an impulse response h(t) shown. ItIs excited by an input x(t) shown. Find an expression for

the system output in the interval .

Solution

2 t

h(t)

3

2 t

x(t)4

4

2 t 4

( )

8/3/2019 New_S_Spect_07_09

65/89

6565

( ) ( )-

y t = x h(t- ) d

2

x( )4

4

-2

h(- )

32

h( )

3

***

8/3/2019 New_S_Spect_07_09

66/89

6666

Complete overlapping

( ) ( )-

y t = x h(t- ) d

42 t

2

x( )4

4

2

2

2 * 3t

= d

t

h(t- )3

t-2( )

2

8 2 * 3t

+ d

=

***

8/3/2019 New_S_Spect_07_09

67/89

6767

26= t +36 t-36

( )y t =

=

=

Example***

8/3/2019 New_S_Spect_07_09

68/89

68

An LTI system has an impulse responseIs excited by an input . Find:a) The system transfer function (H(f))

b) The system output y(t).

Solution

( )- t

h(t) = e tu( )- tx(t) = e tu

) ( ) ( )F

tT

a h t e u t

= ( ) =H f

) ( ) ( )F

tT

b x t e u t

= ( ) =X f

Solution***

8/3/2019 New_S_Spect_07_09

69/89

69

( ) =H f ( ) =X f

( ) ( )( )Y f X f H f =

( ) ( )1

2 2 j f j f =

+ +

( ) ( )2 2C D

j f j f = +

+ +

( )=y t

Partial fraction

Solution1

***

8/3/2019 New_S_Spect_07_09

70/89

70

( ) ( )1

( ) 2 2Y f j f j f = + + ( ) ( )2 2C D

j f j f = ++ +

C =( )

1

=

D =

( )

1

=

( ) ( )1 t ty t e e u t

=

Example

8/3/2019 New_S_Spect_07_09

71/89

71

An LTI system has an impulse responseIs excited by an input . Find: the systemoutput y(t) using time domain analysis.

Solution

( )- t

h(t) = e tu( )- tx(t) = e tu

x( )

8/3/2019 New_S_Spect_07_09

72/89

72

( ) ( )-

y t = x h(t- ) d

x( )1 - e

h( )

1 - e

h(- )1

0

h(t- )1

t

( )- t e


8/3/2019 New_S_Spect_07_09

73/89

73

( ) ( )-y t = x h(t- ) d

x()1 - e


=

0t

h(t- )1

t

( )- t e


***

8/3/2019 New_S_Spect_07_09

74/89

74

( ) ( )-

y t = x dh(t- )

x()1 - e

pp g

t

h(t- )1( )- t e

0t

( )

0

tt

= ee d

=


***

8/3/2019 New_S_Spect_07_09

75/89

75

x()1 - e

pp g

t

h(t- )1( )- t e

0t

( )( )

0

tt

y t = e e d

( )y t =

( )1 t t= e e

8/3/2019 New_S_Spect_07_09

76/89

76

( )( )

0

10t t

0 t

y t =e e t

( ) ( )1 t ty t e e u t

=

System PropertiesMemory & memoryless systems

8/3/2019 New_S_Spect_07_09

77/89

77

Memory & memoryless systems

A system is memoryless if its output at agiven time depends only at the input at thistime.

Resistive circuits are memoryless circuits.

RC circuits are memory circuits.


8/3/2019 New_S_Spect_07_09

78/89

78

Examples of memoryless systems:

The simple identity system y(t)=x(t)The amplifier system y(t) = A x(t)

Examples of memory systems:

The delay system y(t)=x(t-t 0)

The accumulator :

( )( )t

y t x d

=

System Causality

8/3/2019 New_S_Spect_07_09

79/89

79

Causal system: Output at any time depends onlyon the input at present and last time (not futuretime).

Called non-predictive system

y(t) = x(t)+ x(t-1) is a causal system

y(t) = x(t)+ x(t+1) is a non- causal system

All memoryless system are causal systems.

System Causality***

8/3/2019 New_S_Spect_07_09

80/89

80

Find whether the following systems are causal or not

1) y(t) = x(-t)

2) y(t) = x(t) cos (t+1)

System Stability

8/3/2019 New_S_Spect_07_09

81/89

81

Stable system: Bounded input bounded output

Find whether the following system is stable or not :

( ) ( )t

y t x d

= The integrator system

System Stability***

8/3/2019 New_S_Spect_07_09

82/89

82

3) The integrator system

is a non-stable system

Let x(t)=u(t)

( ) ( )t

y t x d

= ( )x

1

t

( )y t( ) =y t

8/3/2019 New_S_Spect_07_09

83/89


8/3/2019 New_S_Spect_07_09

84/89

8484

A system is memoryless if its output at agiven time depends only at the input at this time.

An LTI system is memoryless if h(t)= k (t)

y(t) = k x(t)

LTI Systemx(t)= (t) y (t)=h(t)

( ) ( ) ( ) ( ) ( )y(t)= x t h t t h t h t = =

LTI Systemx(t) y (t)=x(t) ( ) ( )h t t=

LTI Systemx(t) y (t)=k x(t) ( ) ( )h t k t=

System Causality

8/3/2019 New_S_Spect_07_09

85/89

8585

A DT system is causal if its output at any timedepends only on the input at present and lasttime (not future time).

For an LTI system y(t) = x(t) h(t)

If h( ) exists at any ve value of , y(t) will dependon future input.

An LTI system is causal if, h(t) =0 t < 0

( )-

x(t= h - ) d

System Stability

8/3/2019 New_S_Spect_07_09

86/89

8686

Stability Bounded input Bounded output

It can be proved that, an LTI system is stable if

h(t) is absolutely summable .

Ex) The following are the impulse response of a CTLTI system. Determine whether each system is

causal and/or stable or not. Justify your answer.

( )h t d t

<

4) ( ) ( 2)ta h t e u t= 4) ( ) ( 1 )tb h t e u t

=

Solution4 t h(t)

***

8/3/2019 New_S_Spect_07_09

87/89

8787

) ( ) ( 2)= a h t e u tt

2

( )

= h t d t 42

te d t

4

24

=

te 8

4

=

e e8

4

= e =bounded

Causality

h(t)=0 for all t < 0Stability

Solutionh(t)4t

***

8/3/2019 New_S_Spect_07_09

88/89

8888

( )

= h t d t

Unstable

Causality

Stability

t-1

) ( ) ( 1 )b h t e u t=

8/3/2019 New_S_Spect_07_09

89/89

89

Conclusions

new_s_spect_07_09

Documents