class_12_properties_inversefourier.pdf

Chengbin Ma UM-SJTU Joint Institute

Class#12

(Relations between time domain and frequency domain) - Multiplication property (3.14)

- Parseval relationships (3.16)

- Time-bandwidth product (3.17)

- Duality (3.18)

- Finding inverse Fourier transforms by using partial-fraction expansions (3.13)

* check of Midterm#1 papers after this lecture.

* definition of sinc function: sinc(x)=sin(x)/x in mathematics, sinc(x)=sin(px)/px

in signal processing [refer to Equ. (3.24) in P223]

* cos(wt)/(jw) is odd function, therefore its integration from inf to inf is zero.

[refer to slide#27, class#10]

Slide 1

MichaelHighlight


Review of Previous Lecture

Properties:

Linearity, Symmetry, Time- and frequency-shift,

Scaling

Convolution property

Differentiation and integration properties

And thus transfer function

Slide 2


Class#12

- Multiplication property (3.14)



- Duality (3.18)


Slide 3


Multiplication Property

1 1FT ( ) ( ) ( )* ( ) ( ) ( ( ))

2 2

FS ( ) ( ) [ ]* [ ] [ ] [ ]l

x t y t X j Y j X j Y j d

x t y t X k Y k X l Y k l

p p

Enable to study the effects of truncating a time-domain

signal on its frequency-domain representation, i.e.

windowing.

][][][)(*)()( :FS

)()()()(*)()( :FT

kXkTHkYtxthty

jXjHjYtxthty

MichaelRectangle


Proof (FT)

Proof of multiplication property: FT

( )

( ) ( ) ( ) ( )

1( ) ( )

2

1( ) ( )

2

1 1( ) ( ( )) ( )* ( )

2 2

j t

j t j t

j t

x t y t x t y t e dt

X j e d y t e dt

X j y t e dt d

X j Y j d X j Y j

p

p

p p

p

dejXtx

dtetxjX

tj

tj

)(2

1)(

)()(


Proof (FS)

Proof of multiplication property: FS

0

0 0 0

0

0

( )

( )

1( ) ( ) ( ) ( )

1[ ] [ ]

1[ ] [ ]

1[ ] [ ]

[ ] [ ] [ ] [ ]

jk t

T

jl t jm t jk t

Tl m

j m l k t

Tl m

j m l k t

Tl m

l m

x t y t x t y t e dtT

X l e Y m e e dtT

X l Y m e dtT

X l Y m e dtT

X l Y m m l k X l

[ ] [ ]* [ ]l

Y k l X k Y k

k

tjkekXtx 0][)(


Example: Windowing

Slide 7

0 200 400 600 800 1000 1200-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Truncated Cosine

250 300 350 4000

10

20

30

40

50

60

70Overlay of both Spectra

truncated cosine spectrum

original cosine spectrum

Class12_windowing

Smooth and

oscillations


Class#12




- Duality (3.18)


Slide 8


Parseval Relationships (Parsevals Theorem)

2 2

2 2

1FT ( ) ( )

2

1FS ( ) [ ]

Tk

x t dt X j d

x t dt X kT

p

n

n

T

TT

nxE

dttxdttxE

2

22

][

)()(lim

HINT: Total energy of the signal:

Physical Meanings

In-class problem

energy -> nonperiodicpower -> periodic


Conservation of Energy

The Parseval relationships state that the energy (or

power) in the time-domain representation of a signal

is equal to the energy (or power) in the frequency-

domain representation.

The quantity of power/energy does relate to the

domain in which it is discussed.

2 2

2

1( ) ( )

2

1where ( ) is defined as the energy spectrum of ( )

2

xW x t dt X j d

X j x t

p

p

MichaelRectangle


Proof (FT)

Proof of Parsevals relation: FT

2 *

*

*

* *

2

( ) ( ) ( )

1 1( ) ( ) ( ) ( )

2 2

1 1( ) ( ) ( ) ( )

2 2

1( )

2

x

j t j t

j t

W x t dt x t x t dt

x t X j e d dt x t X j e d dt

X j x t e dtd X j X j d

X j d

p p

p p

p

p dejXtx tj)(

2

1)(


Proof (FS)

Proof of Parsevals relation: FS

0 0

0

2 *

*

*

* *

2

2

1 1( ) ( ) ( )

1 1( ) [ ] ( ) [ ]

1[ ] ( ) [ ] [ ]

[ ]

where [ ] is the average power in the th harmoi

xT T

jk t jk t

T Tk k

jk t

Tk k

k

P x t dt x t x t dtT T

x t X k e dt x t X k e dtT T

X k x t e dt X k X kT

X k

X k k

c component

of ( ).x t

k

tjkekXtx 0][)(


Class#12




- Duality (3.18)


Slide 13


Bandwidth

Bandwidth The bandwidth of a signal is the extent of the signals

significant frequency content.

It is difficult to define bandwidth, especially for signals having infinite frequency extent, because the meaning of the term significant is not mathematically precise.

Several definitions for bandwidth are in common use: Absolute bandwidth

Half-power bandwidth

Null-to-null bandwidth (Zero-crossing bandwidth)

Equivalent-noise bandwidth

102

103

104

105

106

-250

-200

-150

-100

Magnitu

de (

dB

)

Bode Diagram

Frequency (rad/sec)


Half-power bandwidth and Zero-crossing bandwidth

0

12

2T

Example


Effective Duration/Bandwidth

We formally define the effective duration and

bandwidth as (Assume a zero-centered low-pass signal such as a

power signal, otherwise the energy will be infinite!)

dttx

dttxt

Td2

22

)(

)(Effective duration (time-domain)

(Evaluate slower responses):

Bandwidth (frequency-domain)

(Evaluate wider passbands):

djX

djX

B2

22

|)(|

|)(|

Total energy

Distribution of energy:

Slower response has

larger energy when time

is large.

Similarly, the

distribution of

energy over

frequencies


Example: Effective Duration

Responses of two 2nd-order LTI systems: Td > Td

Slide 17


Time-Bandwidth Product

It can be shown that the time-bandwidth product for any

signal is lower bounded according to the relationship

(Uncertainty principle)

This bound indicates that we cannot simultaneously decrease the duration

(related with the speed of time response) and bandwidth of a signal, i.e.,

bandwidth and speed of the response are dependent, and they can reflect

each other in a different domain.

It is also known as the uncertainty principle.

The general nature of the relationship between time and frequency is

demonstrated by the scaling property.

1FT ( ) ( )

FS ( ), 0 [ ]

x at X ja a

x at a X k

variance.frequency is variance, timeis

4

1or

2

1

22

22

t

td BT

MichaelRectangle

MichaelLine

MichaelHighlight


Proof (1)

Proof of uncertainty principle.

2

2

2

2

2*

2

2

2

2

22

2

2

2

2

2

22

2

2

2

2

22

2

|)(||)(|

)()(

|)(|

)(|)(|

|)(|

|)(|

|)(|

|)(|

|)(|

|)(|

|)(|

|)(|

|)(|

|)(|

dttx

N

dttx

dttxdt

dttx

dttx

dttxdt

ddtttx

dttx

dttxdt

d

djX

djXj

djX

djX

dttx

dtttx

dttx

dttxt

t

t

2 2

2 2

1FT ( ) ( )

2

1FS ( ) [ ]

Tk

x t dt X j d

x t dt X kT

p

Energy conservation in

both domains

CauchySchwarz inequality

and |tx(t)| and |tx(t)d(x(t))/dt|

are all positive. Therefore

|x(t)y(t)dt|^2>= |x(t)y(t)|^2dt.

just think of (a+b)2 >= a2+b2

MichaelLine


Proof (2)

22* *

2 2*2

* *

*

( ) ( ) Re ( ) ( )

1Re ( ) ( ) ( )

4

1Re ( ) ( ) ( ) ( ) ( ) ( )

2

1(

2

d dN tx t x t dt tx t x t dt

dt dt

d dt x t x t dt t x t dt

dt dt

d d dx t x t x t x t x t x t

dt dt dt

dx

dt

2*1

) ( ) ( )2

dt x t x t

dt

it is indeed valid! try to expand it into series and examine each term

MichaelRectangle


Proof (3)

2 2 2 2

2

2 22 2 2

2

2 2

2 2

2 2

( ) ( ) ( ) ( )

( )

( ) lim ( ) 0

lim ( ) 0

1 1( ) ( )

4 4

1

4

tt

t

t w

dt x t dt td x t t x t x t dt

dt

x t dt

t x t dt t x t

t x t

N x t dt x t dt

MichaelRectangle

MichaelRectangle


Examples

Frequency and step responses of the 1st-order low pass filter

using Matlab.

For a rectangle pulse,

Slide 22

0

0

,0

,1)(

Tt

Tttx

3/0

2/12

0

0

0

0 Tdt

dttT

T

T

T

T

d

)2/(3)2/(1 0TTB d

Class12_bandwidthProduct

variance.frequency is variance, timeis

4

1or

2

1

22

22

t

td BT


Class#12




- Duality (3.18)


Slide 23


Duality

FT ( ) ( ), ( ) 2 ( )FT FT

x t X j X jt x p

Interchange time and frequency!


FT ( ) ( ), ( ) 2 ( )FT FT

x t X j X jt x p

Example

Slide 25

?)(1

1)(

jX

jttx

jjFtuetf t

1

1)()()(

)(21

1)( p

f

jtjtF

Therefore,

)(2)(2)( pp uefjX

)(*)(2

1)()( :tionMultiplica

)()()(*)( :nConvolutio

p

jYjXtytx

jXjHtxth


Proof

Proof of the duality property: FT

( ) ( )

1( ) ( ) 2 ( ) ( )

2

2 ( ) ( ) 2 ( ) ( )

2 ( ) ( ) ( )

( ) 2 ( )

j t j t

j jt

jt j t

FT

x t X j

x t X j e d x t X j e d

x X j e d x X jt e dt

x X jt e dt X jt e dt

X jt x

p p

p p

p

p

)()(

)()(

)(2

1)(

p

jXtx

dtetxjX

dejXtx

FT

tj

tj

MichaelRectangle


Class#12




- Duality (3.18)


Slide 27

Chengbin Ma UM-SJTU Joint Institute Slide 28

Inverse Fourier Transform

dtetxjX tj )()(

p

dejXtx tj)(2

1)(

Fourier Transform

Inverse

Fourier Transform


Ratios of Polynomials

Some times we dont have to use the following formula to

determine the IFT:

1( ) ( )

2

j tx t X j e dtp

The frequency response of a system described by a linear

constant-coefficient differential equation is given by a ratio of

two polynomials in j (transfer function).

In order to find inverse transforms for ratios of polynomials,

we use partial-fraction expansions the inverse of placing

a sum of fractions over a common denominator.

MichaelRectangle

MichaelHighlight

MichaelHighlight

MichaelHighlight


The frequency response of a LTI system is given by

ratio of two polynomials in j.

Slide 30

Transfer Function for a LTI System

( ) ( )* ( ) ( ) ( ) ( )

( )( )

( )

y t x t h t Y j X j H j

Y jH j

X j

01

1

1

01)(ajajaj

bjbjbjH

N

N

N

M

M


Partial-fraction expansion:

Slide 31

Partial-fraction Expansion

N

N

N

M

M

N

N

N

M

M

dj

C

dj

C

dj

C

djdjdj

jbjbb

ajajaj

bjbjbjH

2

2

1

1

21

10

01

1

1

01)(


Partial-fraction expansion (special cases):

1. deg P > deg Q

2. Complex roots

3. Multiple roots

Slide 32

Special Cases

r

r

N

N

ax

C

ax

C

ax

C

jQ

jP

xx

dcx

x

b

x

a

xxxx

x

jQ

jPjP

jQ

jP

dj

C

dj

C

dj

C

jQ

jPjH

2

21

22

2

11

2

2

1

1

.3

112112

1.2

.1

)(

MichaelRectangle

MichaelLine

MichaelLine

Chengbin Ma UM-SJTU Joint Institute Slide 33

FT of Exponential Functions

01

)(

dfordj

tueFT

dt

If d>=0, the FT does not converge.

dje

jd

dtedtetuejX

tjd

tjdtjdt

11

)()(

0

0

N

N

dj

C

dj

C

dj

CjH

2

2

1

1)(

MichaelRectangle


Therefore,

Slide 34

Solution

0,,,

)(

2121

2

2

1

1

01

1

1

01

21

N

td

N

tdtd

N

N

N

N

N

M

M

dddforeCeCeC

dj

C

dj

C

dj

C

ajajaj

bjbjbjH

N

Dynamics of a LTI system


Note: Homogenous Solution (2) 1. General form:

2. Let all the input x(t) related terms be zero,

3. The characteristic equation is

4. Suppose ri is the N roots of the characteristic equation,

5. ci are determined later to satisfy the initial conditions

Slide 35

)()()()()()( 1010 txdt

dbtx

dt

dbtxbty

dt

daty

dt

datya

M

M

MN

N

N

0)()()( 10 tydt

daty

dt

datya

N

N

N

0110 N

N raraa

tr

N

trtrh Necececty 21 21)( )(

)0(),0(),0( )1()1( Nyyy


Examples (1)

Use partial-fraction expansions to determine the time-domain

signals corresponding to the following FT:

2

5 12( )

( ) 5 6

jX j

j j

N

N

N

N

N

M

M

dj

C

dj

C

dj

C

ajajaj

bjbjbjH

2

2

1

1

01

1

1

01)(

01

)(

dfordj

tueFT

dt


Solution

Solution (Residue method):

2

1 2

1 2

2

2 3

3

2 3 2 3

1 2

5 12 5 12( )

( ) 5 6 ( 2)( 3)

2 3

5 12( 2) ( ) 2

3

5 12( 3) ( ) 3

2

( ) ( ) ( ) 2 ( ) 3 ( )

j

j

j

j

t t t t

j jX j

j j j j

A A

j j

jA j X j

j

jA j X j

j

x t A e u t A e u t e u t e u t

pay attention to this method of evaluating A1 and A2

MichaelRectangle


Examples (2)

For the case of multiple roots

3 2

2

3 5( )

( ) 4( ) 5 2

3 5

( 1) ( 2)

jX j

j j j

j

j j


Solution 1

Solution (Residue method):

2

31 2

2

2

1 11

2

2 210 0

3 22

2

3 5( )

( 1) ( 2)

( 1) 1 2

3 5( 1) ( ) 2

2

3 2 3( 1) (3 2)( 1) ( ) 1

( 1) 1 ( 1)

3 5( 2) ( ) 1

( 1)

(

jj

jv v

j

j

jX j

j j

AA A

j j j

jA j X j

j

d d v v vA j X j

d j dv v v

jA j X j

j

x

2 2

1 2 3) ( ) ( ) ( ) 2 ( ) ( ) ( )t t t t t tt A te u t A e u t A e u t te u t e u t e u t

it's still a constant here, not a polynomial like pjw+q

note that it's te-tu(t) here!

MichaelOval

MichaelLine

MichaelOval

MichaelRectangle

MichaelLine


Alternative Solution:

2

31 2

2

2

1 11

3 22

2

3 31 2 2 1

1 2 3

3 5( )

( 1) ( 2)

( 1) 1 2

3 5( 1) ( ) 2

2

3 5( 2) ( ) 1

( 1)

5 5Let 0, we have 1

2 2 2 2

( ) ( ) ( )

jj

j

j

t t

jX j

j j

AA A

j j j

jA j X j

j

jA j X j

j

A Aj A A A A

x t Ate u t A e u t A e

2 2( ) 2 ( ) ( ) ( )t t t tu t te u t e u t e u t

Solution 2


Homework

3.40(b)

3.43(a)

3.73(a) to (f)

3.76(a)(b)

3.81(a)(b)

Due 2:00PM, Thursday of next week

Slide 41

what is a slower response?

class_12_properties_inversefourier.pdf

Documents

mx t y t x t y t e dttx

e e dttx

y kktjkekxtx

jk ttjl t jm t jk ttl

time domain

timedomain signal

timedomain representation

function slide