LN5 Frequency Domain Representation Part2 FourierTransform v2

7/25/2019 LN5 Frequency Domain Representation Part2 FourierTransform v2

1/33

ilolu 1

FREQUENCY DOMAIN REPRESENTATION OF LTI SYSTEMS

DICRETE TIME FOURIER TRANSFORM (DTFT)

EXISTENCE

MEAN SQUARE CONVERGENCE

FOURIER TRANSFORM OF A CONSTANT SEQUENCEFOURIER TRANSFORM OF A COMPLEX EXPONENTIAL SEQUENCE

FOURIER TRANSFORM OF A SINUSOIDAL SEQUENCE

FOURIER TRANSFORM OF UNIT STEP SEQUENCE

SYMMETRY PROPERTIES OF FOURIER TRANSFORM

REAL SEQUENCES

FOURIER TRANSFORM THEOREMS

FOURIER TRANSFORM PAIRS

IDEAL LOWPASS FILTER (IMPULSE RESPONSE)MOVING AVERAGE FILTER

EXAMPLES

LCCDES AND FREQUENCY RESPONSE


2/33

ilolu 2

DICRETE TIME FOURIER TRANSFORM (DTFT)

The Fourier transform of a sequence x n is defined as

j j nn

X e x n e

If the FT exists (summation converges) the sequence can be obtained from its FT

as

1

2

j j n

x n X e e d

Fourier Transform is periodic with 2.


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ilolu 3

LTI SYSTEMS

The frequency response function, ()is the FT of the impulse response

[]


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ilolu 4

EXISTENCE

FT of a sequence []exists, i.e., []=

converges to a continuous function of ,

if []is absolutely summable.(sufficient condition)

Proof: Exercise

[]= []cos sin

=

[] cos

=

[] sin

=

Both sums have to converge


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ilolu 5

All stable LTI systems have frequency response functions.


6/33

ilolu 6

Ex: n

x n a u n

0 01

if 1 or 11

nj n j n j j

jn n

X e a e ae ae aae

cos21

1

sincos1

1

1

1

2

2

2

2

aa

ja

aeeX

j

j

cos1

sintan

sincos10

11

1

a

a

jaa

aeeX jj

-4 -3 -2 -1 0 1 2 3 40.5

1

1.5

2

2.5

3

3.5

4

4.5

5

jX e 8.0a

-4 -3 -2 -1 0 1 2 3 4-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

4

4

jeX


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ilolu 7

MEAN SQUARE CONVERGENCE

Some sequences, which are not absolutely summable but square summable

2

n

x n

can still be represented by Fourier Transform, but


8/33

ilolu 8

Ex: Ideal lowpass filter.

1

0

cj

c

H e

Lets find h n !

[] 1

2

12 ( ) sin

Note that,

[] sin is not absolutely summable!

Then, one may question the Fourier transform of [], sin

= ?

c

c

jH e

1


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ilolu 9

Define =

Even if you take M oscillations do not die to zero.

However 2

lim 0j j

MM

H e H e d

. This is called mean square convergence.

The oscillatory behavior aroundc

is called the Gibbs phenomenon.

MATLAB codeclear all; close all;

precision =0.0001

w = [-pi:precision:pi];

ideaL = zeros(1,length(w));

wc = pi/2;

orta = round(length(w)/2);

ideaL((orta-round(wc/precision)):(orta+round(wc/precision)))=1;

M = 10000;

H = 0;

forn = -M:-1

H = H+(sin(wc*n)/(pi*n))*exp(-i*w*n);

end

forn = 1:M

H = H+(sin(wc*n)/(pi*n))*exp(-i*w*n);

end

H = H+(wc/pi);

plot(w,H); hold on;

plot(w,idea,'r')

grid

detail

-4 -3 -2 -1 0 1 2 3 4-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1M 2

c

-4 -3 -2 -1 0 1 2 3 4-0.2

0

0.2

0.4

0.6

0.8

1

1.2

2c

4M

-4 -3 -2 -1 0 1 2 3 4-0.2

0

0.2

0.4

0.6

0.8

1

1.2

10M 2

c

-4 -3 -2 -1 0 1 2 3 4-0.2

0

0.2

0.4

0.6

0.8

1

1.2

100M 2

c

-4 -3 -2 -1 0 1 2 3 4-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1000M 2

c

-1.571 -1.5708-1.5706-1.5704 -1.5702 -1.57 -1.5698-1.5696 -1.5694

0.5

0.6

0.7

0.8

0.9

1

1.1

10000M


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ilolu 10

FOURIER TRANSFORM OF A CONSTANT SEQUENCE

1x n 2 2jr

X e r

not absolutely

summable

or we can write as

202 jeX

keeping in mind that FT is periodic with 2.

-4 -2 2 4

(2)


11/33

ilolu 11

FOURIER TRANSFORM OF A COMPLEX EXPONENTIAL SEQUENCE

0j nx n e 02 2jr

X e r

not absolutely

summable

or we can write as

202 0

j

eX

keeping in mind that FT is periodic with 2

0+4 0 -2 0 0+2 0+4

(2)


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ilolu 12

FOURIER TRANSFORM OF A SINUSOIDAL SEQUENCE

0 001

cos( )2

j n j nx n n e e

r

jrreX

22

00

not absolutely summable

or 2000

jeX since FT is periodic with 2


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ilolu 13

FOURIER TRANSFORM OF UNIT STEP SEQUENCE

x n u n 1

21

j

jr

X e re

not absolutely summable


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ilolu 14

SYMMETRY PROPERTIES OF FOURIER TRANSFORM

Definitions:

Conjugate symmetric (CS) sequence. x n x n

Conjugate antisymmetric (CaS) sequence. x n x n

Using the above definitions, any sequence can be written as

e o

x n x n x n

where

1

2e

x n x n x n

is the CS part

and

12

ox n x n x n

is the CaS part.


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ilolu 15

SYMMETRY PROPERTIES

Fundamental relations

Let j

x n X e

be a FT pair. Then, the following hold:

jx n X e since j j n

n

X e x n e

[] (

) since

(

) []

=

Above yields

jx n X e


16/33

ilolu 16

The two relations above also yield:

1)

Re2 2

j j

j

e

X e X ex n x nx n X e

( CS part of jX e )

2)

22Im

**

jjj

o

eXeXeX

nxnxnxj

( CaS part of jX e )

3)

Re2 2

j j

j

e

X e X ex n x nx n X e

( real part of jX e )

Therefore FT of an even seq. is real!

4) Im2 2

j jj

o

X e X ex n x nx n j X e

( imag. part of jX e )

Therefore FT of an odd seq. is purely imaginary!


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ilolu 17

Ex:Let []and []be two real sequences with their DTFTs()and(), respectively.Let

[] [] []Then,

() ( )()Note that

()is NOT the real part of

().

However,

() () ()2 Since

() ()() ()() Similarly,

(

) () ()

2


18/33

ilolu 18

Ex: (contd)

[]: [1 1]

[]: [1 1]

[]: [1 1 ]

() 1

(

) 1

() 1 () 1

() ()2 1 () ()2


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ilolu 19

REAL SEQUENCES

Based on the above relations, for real sequences ( x n x n ):

jj

eXeX

, conjugate symmetrywhich implies

Magnitude is even.... jj eXeX

Phase is odd... jj eXeX Real part is even.. jRjR eXeX Imaginary part is odd......................... jIjI eXeX


20/33

ilolu 20

Verification by an example

Ex: nx n a u n 1

11

j

jX e a

ae

a)FT is conjugate symmetric:

1

11

j j

jX e X e a

ae

b)Real part of FT is an even function:

21 cos

Re1 2 cos

j j j

R R

aX e X e X e

a a

c) jR eX is the FT of nxe :

2

1 cos

1 2 cos

j

R

aX e

a a

1

2

n n

ex n a u n a u n

d)Imaginary part of Ft is an odd function.

2sin

Im1 2 cos

j j j

I I

aX e X e X e

a a

n

e

x n

1 2

1

n

x n

1 2

01

10 a


21/33

ilolu 21

e) jI eX is the FT of nxo :

2

sin

1 2 cos

j

I

aX e

a a

1

2

n n

ox n a u n a u n

f) Magnitude of FT is an even function:

2

1

1 2 cos

j jX e X e

a a

g)Phase of FT is an odd function

jjeX

a

aeX

cos1

sintan

1

n

o

x n

1 2

1

4

-4 -3 -2 -1 0 1 2 3 40.5

1

1.5

2

2.5

3

3.5

4

4.5

5

jX e

8.0a

-4 -3 -2 -1 0 1 2 3 4-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

4

4

jeX


22/33

ilolu 22

FOURIER TRANSFORM THEOREMS

1,F

j j jx n X e x n F X e X e F x n

1) F

j jax n by n aX e bY e Linearity

2) 00F

j n jx n n e X e

Time-shift

3)

00

Fjj n

e x n X e

Freq. shift

4) F

jx n X e

Time reversal

5) jF dX e

n x n jd

Differentiation in frequency

domain

6) F

j jx n y n X e Y e

Convolution

7) 1

2

Fj j jy n x n w n Y e X e W e d

Modulation, windowing

Parsevalstheorem (prove as an exercise)

8) 22 1

2

j

n

x n X e d

energy of the signal

2

jX e

is called the energy density spectrum.

9) 1

2

j j

n

x n y n X e Y e d


23/33

ilolu 23

Proof of (6):

Let

[] [][ ]= [] []

() ?() [][ ]=

=

[] [ ]

=

=

[] []=

=

()()


24/33

ilolu 24

Proof of (8) using (6):

Let

[] []

[ ]

= []

[]

() ( )() ( )

[0] []

[]

= |[]|

=

[0] 12 ()

12 ()


25/33

ilolu 25

Proof of (9):

[]

[]

=

12 ()

[]=

1

2 (

)

[]

= ()

12 ()()


26/33

ilolu 26

FOURIER TRANSFORM PAIRS

n 1

0n n 0j ne

1 n 2 2jr

X e r

nx n a u n 1a 1

1 jae

x n u n 1

21

jk

ke

nn a u n 1a

2

1

j

j

ae

ae

1 nn a u n 1a

n n

a u n a u n

2

1

1 jae

2 12

n

n n a u n

3

1

1 jae

1 !1

1 ! !

nn k

a u nk n

11 2 1

1 !

nn k n k n a u n

k

1

1k

jae

1

sin 1sin

n

p

p

r n u n

1r

2 2

1

1 2 cos j j

pr e r e

show using

nx n a u n

sincn

n

1,

0,

cj

c

X e

1, 0

0, otherwise

n Mx n

/ 2

sin 1 / 2

sin / 2

j MM

e

0j n

e 02 2

r

r

0cos n

r

j

r

jrere 22

00

n0

sin

r

j

r

jrejrej 22

00


27/33

ilolu 27

Ex: IDEAL LOWPASS FILTER (IMPULSE RESPONSE)

Plots of sin sequence

They are infinitely long sequences in

< <

Plots are arbitrarily in 2 0 < < 2 0

-20 -15 -10 -5 0 5 10 15 20-0.05

0

0.05

0.1

0.15

0.2

-20 -15 -10 -5 0 5 10 15 20-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

-20 -15 -10 -5 0 5 10 15 20-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

sin cn

n

4c

c

2c

n

n

n

6c


28/33

ilolu 28

Ex: MOVING AVERAGE FILTER

Magnitude

Phase

DTFT functions are plotted in or in 20

-4 -3 -2 -1 0 1 2 3 40

1

2

3

4

5

6

-4 -3 -2 -1 0 1 2 3 40

2

4

6

8

10

12

-4 -3 -2 -1 0 1 2 3 4-3

-2

-1

0

1

2

3

5M

-4 -3 -2 -1 0 1 2 3 4-3

-2

-1

0

1

2

3

10M

/ 2sin 1 / 2

sin / 2

j MM

e

5M

1M

2

1M

10M


29/33

ilolu 29

M = 1

1

12

h n n n 21

1 cos2 2

jj j

H e e e

1

12

y n x n x n

M = 4

1

1 2 3 45

y n x n x n x n x n x n

1

1 2 3 45

h n n n n n n

2 3 4

2

11

5

11 2 cos 2 cos 2

5

j j j j j

j

H e e e e e

e

or

2 3 41

15

j j j j jH e e e e e

5 5 5

2 2 2

5

2 2 2

1 1 1

5 1 5

j j j

j

jj j j

e e ee

ee e e

2

5sin

1 2

5sin

2

je

-4 -3 -2 -1 0 1 2 3 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

frequency

|H|

-4 -3 -2 -1 0 1 2 3 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

frequency

|H|

2

5

4

5

2

5

4

5


30/33

ilolu 30

Ex: Express 1j

X e , 2 jX e , in terms of jX e , the DTFT of x n .

1 7 7x n x n x n x n x n 7

1

j j j jX e X e e X e

2 4x n x n x n 4

2

j j j j

X e X e e X e

-2 0 2 4 6 8 100

1

2

3

4

1x n

-2 0 2 4 6 8 100

1

2

3

4

2x n

-2 0 2 4 60

0.5

1

1.5

2

2.5

3

3.5

4

x n


31/33

ilolu 31

One can also write as

() ( ) 7() () 7() since []is real()() 7()

()7 7() 7()

2 Re {()+7}()7() ( ) ( 1 4)

()( ) ()2cos2 2 c o s2 ()()


32/33

ilolu 32

Ex: What is the inverse DTFT, [], of () 38?From the table 1 nn a u n

2

1

1 jae

for 1a

2 1

18

n

n u n 22

11

8

je

2

3

12 3

8

n

n u n

3

21

8

2

1

j

j

e

e

Why does the high frequency gain of MA filter decreaseas M increases?

Comment on the above illustrations.

-5 0 5 10 15 20-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-5 0 5 10 15 20-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1


33/33

LCCDEs AND FREQUENCY RESPONSE

[] [] [] () ( )() () ()

[ ]= [ ]

= ()=

()

=

() = ()

=

() () = =

()

Date post:	26-Feb-2018
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LN5 Frequency Domain Representation Part2 FourierTransform v2

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