frequency resposne

8/10/2019 Frequency Resposne

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1. Inverse Z-transform - Partial Fraction


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Find the inverse Z-transform of G (z ) =

2z 2 + 2 zz 2 + 2 z 3


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2z 2 + 2 zz 2 + 2 z 3

G (z )z =

2z + 2(z + 3)( z 1)


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2z 2 + 2 zz 2 + 2 z 3

G (z )z =

2z + 2(z + 3)( z 1)

= Az + 3

+ Bz 1


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2z 2 + 2 zz 2 + 2 z 3

G (z )z =

2z + 2(z + 3)( z 1)

= Az + 3

+ Bz 1

Multiply throughout by z + 3 and let z = 3 to get


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2z 2 + 2 zz 2 + 2 z 3

G (z )z =

2z + 2(z + 3)( z 1)

= Az + 3

+ Bz 1


A = 2z + 2

z 1 z = 3


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2z 2 + 2 zz 2 + 2 z 3

G (z )z =

2z + 2(z + 3)( z 1)

= Az + 3

+ Bz 1


A = 2z + 2

z 1 z = 3=

4

4= 1

Digital Control 1 Kannan M. Moudgalya, Autumn 2007


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G (z )z

= Az + 3

+ Bz 1


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G (z )z

= Az + 3

+ Bz 1

Multiply throughout by z 1 and let z = 1


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G (z )z

= Az + 3

+ Bz 1

Multiply throughout by z 1 and let z = 1 to get

B = 44

= 1


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G (z )z

= Az + 3

+ Bz 1


B = 44

= 1

G (z )z =

1z + 3 +

1z 1 | z | > 3

G (z ) = zz + 3

+ zz 1

| z | > 3


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G (z )z

= Az + 3

+ Bz 1


B = 44

= 1

G (z )z =

1z + 3 +

1z 1 | z | > 3

G (z ) = zz + 3

+ zz 1

| z | > 3

( 3)n

1( n ) + 1( n )Digital Control 2 Kannan M. Moudgalya, Autumn 2007


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3. Partial Fraction - Repeated Poles


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G (z ) = N (z )(z ) pD 1(z )


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G (z ) = N (z )(z ) pD 1(z )

not a root of N (z ) and D 1(z )


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G (z ) = N (z )(z ) pD 1(z )


G (z ) = A1z

+ A2(z )2

+ + A p(z ) p

+ G 1(z )

G 1(z ) has poles corresponding to those of D 1(z ) .Multiply by (z ) p


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G (z ) = N (z )(z ) pD 1(z )


G (z ) = A1z

+ A2(z )2

+ + A p(z ) p

+ G 1(z )

G 1(z ) has poles corresponding to those of D 1(z ) .Multiply by (z ) p

(z ) pG (z ) = A1(z ) p 1 + A 2(z ) p 2 + + A p 1(z ) + A p + G 1(z )( z ) p



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Substituting z = ,


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Substituting z = ,

A p = ( z ) pG (z ) | z =


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Substituting z = ,

A p = ( z ) pG (z ) | z =

Differentiate and let z = :

l d l


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Substituting z = ,

A p = ( z ) pG (z ) | z =


A p 1 = ddz (z ) pG (z ) | z =


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4 P ti l F ti R t d P l


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Substituting z = ,

A p = ( z ) pG (z ) | z =


A p 1 = ddz (z ) pG (z ) | z =

Continuing, A1 = 1

( p 1)!d p 1

dz p 1(z ) pG (z ) | z =



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5 Repeated Poles an Example


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5. Repeated Poles - an Example

G (z ) = 11z 2 15z + 6(z 2)( z 1) 2 =



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G (z ) = 11z 2 15z + 6(z 2)( z 1) 2 =

A1

z 1 + A

2(z 1) 2 +

Bz 2



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G (z ) = 11z 2 15z + 6(z 2)( z 1) 2 =

A1

z 1 + A

2(z 1) 2 +

Bz 2

(z 2) ,



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G (z ) = 11z 2 15z + 6(z 2)( z 1) 2 =

A1

z 1 + A

2(z 1) 2 +

Bz 2

(z 2) , let z = 2 , to get B = 20 .

5 Repeated Poles - an Example


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G (z ) = 11z 2 15z + 6(z 2)( z 1) 2 =

A1z 1 +

A2(z 1) 2 +

Bz 2

(z 2) , let z = 2 , to get B = 20 . (z 1) 2,



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5. Repeated Poles an Example

G (z ) = 11z 2 15z + 6(z 2)( z 1) 2 =

A1z 1 +

A2(z 1) 2 +

Bz 2

(z 2) , let z = 2 , to get B = 20 . (z 1) 2,

11 z2

15z + 6z 2 = A1(z 1) + A 2 + B (z 1)2

z 2



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G (z ) = 11z 2 15z + 6(z 2)( z 1) 2 =

A1z 1 +

A2(z 1) 2 +

Bz 2

(z 2) , let z = 2 , to get B = 20 . (z 1) 2,

11 z2

15z + 6z 2 = A1(z 1) + A 2 + B (z 1)2

z 2With z = 1 , get A2 = 2.



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G (z ) = 11z 2 15z + 6(z 2)( z 1) 2 =

A1z 1 +

A2(z 1) 2 +

Bz 2

(z 2) , let z = 2 , to get B = 20 . (z 1) 2,

11 z2

15z + 6z 2 = A1(z 1) + A 2 + B (z 1)2

z 2With z = 1 , get A2 = 2.Differentiating with respect to z and with z = 1 ,



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G (z ) = 11z 2 15z + 6(z 2)( z 1) 2 =

A1z 1 +

A2(z 1) 2 +

Bz 2

(z 2) , let z = 2 , to get B = 20 . (z 1) 2,

11 z2

15z + 6z 2 = A1(z 1) + A 2 + B (z 1)2


A 1 =


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p p

G (z ) = 11z 2 15z + 6(z 2)( z 1) 2 =

A1z 1 +

A2(z 1) 2 +

Bz 2

(z 2) , let z = 2 , to get B = 20 . (z 1) 2,

11z 2

15z

+ 6z 2 = A1(z 1) + A 2 + B (z

1)2


A 1 = (z 2)(22 z 15) (11 z 2 15z + 6)

(z 2) 2 z =1= 9


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6. Important Result from Differentiation


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p



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Problem 4.9 in Text:


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Problem 4.9 in Text: Consider

1( n )a n zz a

=

n =0

a n z n , az 1 < 1

Differentiating with respect to a ,



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1( n )a n zz a

=

n =0

a n z n , az 1 < 1


z(z a )2


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1( n )a n zz a

=

n =0

a n z n , az 1 < 1


z(z a )2

=

n =0

na n 1z n , na n 11( n ) z

(z a )2


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1( n )a n zz a

=

n =0

a n z n , az 1 < 1


z(z a )2

=

n =0

na n 1z n , na n 11( n ) z

(z a )2

Substituting a = 1 , can obtain the Z-transform of n :

n 1( n ) z

(z 1) 2



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1( n )a n zz a

=

n =0

a n z n , az 1 < 1


z(z a )2

=

n =0

na n 1z n , na n 11( n ) z

(z a )2

Substituting a = 1 , can obtain the Z-transform of n :

n 1( n ) z

(z 1) 2

Through further differentiation,


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G (z ) = 9z 1

2(z 1) 2

+ 20z 2



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G (z ) = 9z 1

2(z 1) 2

+ 20z 2

zG (z ) = 9z

z 1

2z

(z 1) 2 +

20z

z 2



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G (z ) = 9z 1

2(z 1) 2

+ 20z 2

zG (z ) = 9z

z 1

2z

(z 1) 2 +

20z

z 2 ( 9 2n + 20 2n )1( n )



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G (z ) = 9z 1

2(z 1) 2

+ 20z 2

zG (z ) = 9z

z 1

2z

(z 1) 2 +

20z

z 2 ( 9 2n + 20 2n )1( n )

Use shifting theorem:G (z )



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G (z ) = 9z 1

2(z 1) 2

+ 20z 2

zG (z ) = 9z

z 1

2z

(z 1) 2 +

20z

z 2 ( 9 2n + 20 2n )1( n )

Use shifting theorem:G (z ) ( 9 2( n 1) + 20 2n 1)1( n 1)


8. Properties of Cont. Time Sinusoidal Signals


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u a ( t ) = A cos ( t + )


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u a ( t ) = A cos ( t + )u a ( t ) = A cos (2 F t + ) , < t <

A : amplitude


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A : amplitude: frequency in rad/s : phase in rad


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A : amplitude: frequency in rad/s : phase in radF : frequency in cycles/s or Hertz

= 2 F

T p = 1F




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u a ( t ) = A cos ( t + )


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With F xed, periodic with period T p

u a [t + T p]



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u a [t + T p] = A cos (2 F ( t + 1F

) + )



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u a [t + T p] = A cos (2 F ( t + 1F

) + )

= A cos (2 + 2 F t + )



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u a [t + T p] = A cos (2 F ( t + 1F

) + )

= A cos (2 + 2 F t + )= A cos (2 F t + )


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Cont signals with different frequencies are different


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Cont. signals with different frequencies are different


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u 1 = cos 2 t8, u 2 = cos 2 7t8


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u 1 = cos 2 t8, u 2 = cos 2 7t8

0 0.5 1 1.5 2 2.5 3 3.5 41

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

t

c o s w t

Different wave forms for different frequencies.


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11. Discrete Time Sinusoidal Signals


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u (n ) = A cos ( wn + ) , < n < w = 2 f


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u (n ) = A cos ( wn + ) , < n < w = 2 f

n integer variable, sample numberA amplitude of the sinusoid


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12. Discrete Time Sinusoids - Identical Signals


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Discrete time sinusoids whose frequencies are sepa-


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rated by integer multiple of 2 are identical, i.e.,cos (( w0 + 2 )n + ) = cos ( w0n + ) , n




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rated by integer multiple of 2 are identical, i.e.,cos (( w0 + 2 )n + ) = cos ( w0n + ) , n All sinusoidal sequences of the form,




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u k (n ) = A cos( wk n + ) ,




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u k (n ) = A cos( wk n + ) ,wk = w0 + 2 k ,




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u k (n ) = A cos( wk n + ) ,wk = w0 + 2 k , < w 0 < ,




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u k (n ) = A cos( wk n + ) ,wk = w0 + 2 k , < w 0 < ,are indistinguishable or identical.




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Only sinusoids in < w 0 < are different




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Only sinusoids in < w 0 < are different

< w 0 < or 12

< f 0 < 12


13. Sampling a Cont. Time Signal - Preliminaries


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Let analog signal u a [t ] have a frequency of F Hz


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u a [t ] = A cos(2 F t + )Uniform sampling rate (T s s) or frequency (F s Hz)

t = nT s = n

F s




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t = nT s = n

F su (n ) = u a [nT s ] < n <




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t = nT s = n

F su (n ) = u a [nT s ] < n <

= A cos (2 F T s n + )




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t = nT s = n

F su (n ) = u a [nT s ] < n <

= A cos (2 F T s n + )

= A cos 2 F F s

n +


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In our standard notation,


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u (n ) = A cos (2 fn + )It follows that

f = F

F sw =

F s= T s




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f = F

F sw =

F s= T s

Reason to callf normalized frequency, cycles/sample.




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f = F

F sw =

F s= T s

Reason to callf normalized frequency, cycles/sample.Apply the uniqueness condition for sampled signals


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( ) ( f )


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f = F

F sw =

F s= T s

Reason to callf normalized frequency, cycles/sample.Apply the uniqueness condition for sampled signals

12 < f ?

w1 = w0

15. Properties of Discrete Time Sinusoids - Alias

As w0 , freq. of oscillation, reaches maximum

at w0


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at w0 = What if w0 > ?

w1 = w0w2 = 2 w0



at w0 =


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w1 = w0w2 = 2 w0u 1(n ) = A cos w1n



at w0 =


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w1 = w0w2 = 2 w0u 1(n ) = A cos w1n = A cos w0n



at w0 =


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u 2(n ) = A cos w2n



at w0 =


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u 2(n ) = A cos w2n = A cos(2 w0)n


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u 1[t ] = cos (2 t8

) ,


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u 1[t ] = cos (2 t8

) , u 2[t ] = cos (2 7t8

) ,


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u 1[t ] = cos (2 t8

) , u 2[t ] = cos (2 7t8

) , T s = 1


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u 1[t ] = cos (2 t8

) , u 2[t ] = cos (2 7t8

) , T s = 1

u 2(n ) = cos (2 7n )


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u 2(n ) cos (2 8)


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u 1[t ] = cos (2 t8

) , u 2[t ] = cos (2 7t8

) , T s = 1

u 2(n ) = cos (2 7n8) = cos 2 (1 1

8)n


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2( ) ( 8) (

8)

= cos (2 28

)n = cos (2n

8) = u 1(n )


u 1[t ] = cos (2 t8

) , u 2[t ] = cos (2 7t8

) , T s = 1

u 2(n ) = cos (2 7n8) = cos 2 (1 1

8)n


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( ) (8

) (8

)

= cos (2 28

)n = cos (2n

8) = u 1(n )

0 0.5 1 1.5 2 2.5 3 3.5 41

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

t

c o s w t



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17. Fourier Transform of Aperiodic Signals

X (F ) =

x ( t )e j 2F t

dt


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( ) ( )x ( t ) =

X (F )e j 2F t dF

If we let radian frequency = 2 F

x ( t ) = 12

X []e j t d,


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18. Frequency Response


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Apply u (n ) = e jwn to g(n ) and obtain output y:


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y (n ) = g(n ) u (n ) =

k= g(k )u (n k )


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=

k=

g(k )e jw (n k )



y (n ) = g(n ) u (n ) =

k= g(k )u (n k )


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=

k=

g(k )e jw (n k ) = e jwn

k=

g(k )e jwk



y (n ) = g(n ) u (n ) =

k= g(k )u (n k )


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=

k=


k=

g(k )e jwk

Dene Discrete Time Fourier Transform



y (n ) = g(n ) u (n ) =

k= g(k )u (n k )


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=

k=


k=

g(k )e jwk


G (e jw ) =

k=

g(k )e jwk



y (n ) = g(n ) u (n ) =

k= g(k )u (n k )


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=

k=


k=

g(k )e jwk


G (e jw ) =

k=

g(k )e jwk =

k=

g(k )z k

z = e jw



y (n ) = g(n ) u (n ) =

k= g(k )u (n k )


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=

k=


k=

g(k )e jwk


G (e jw ) =

k=

g(k )e jwk =

k=

g(k )z k

z = e jw

= G (z ) | z = e jw



y (n ) = g(n ) u (n ) =

k= g(k )u (n k )


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=

k=


k=

g(k )e jwk


G (e jw ) =

k=

g(k )e jwk =

k=

g(k )z k

z = e jw

= G (z ) | z = e jw

Provided the sequence converges absolutelyDigital Control 18 Kannan M. Moudgalya, Autumn 2007



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y (n ) = e jwn

k= g(k )e jwk

= e jwn

G (e jw

)


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y (n ) = e jwn

k= g(k )e jwk

= e jwn

G (e jw

)


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Write in polar coordinates: G (e jw ) = |G (e jw ) | e j - isphase angle:

y (n ) = e jwn |G (e jw ) | e j


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y (n ) = e jwn

k= g(k )e jwk

= e jwn

G (e jw

)

j j j


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y (n ) = e jwn |G (e jw ) | e j = |G (e jwn ) | e j (wn + )

1. Input is sinusoid output also is a sinusoid with followingproperties:


y (n ) = e jwn

k= g(k )e jwk

= e jwn

G (e jw

)

j j j


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y (n ) = e jwn |G (e jw ) | e j = |G (e jwn ) | e j (wn + )

1. Input is sinusoid output also is a sinusoid with followingproperties:

Output amplitude gets multiplied by the magnitude of G (e jw )


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2. At where |G (e jw

) | is large, the sinusoid gets amplied


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) | is large, the sinusoid gets ampliedand at where it is small, the sinusoid gets attenuated.


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) | is large, the sinusoid gets ampliedand at where it is small, the sinusoid gets attenuated. The system with large gains at low frequencies and small

gains at high frequencies are called low pass lters


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) | is large, the sinusoid gets ampliedand at where it is small, the sinusoid gets attenuated. The system with large gains at low frequencies and small

gains at high frequencies are called low pass lters Similarly high pass lters


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20. Discrete Fourier Transform


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20. Discrete Fourier TransformDene Discrete Time Fourier Transform


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G (e jw

)

=

k= g(k )e

jwk=

k= g(k )z

k

z = e jw


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G (e jw

)

=

k= g(k )e

jwk=

k= g(k )z

k

z = e jw

= G (z ) | jw


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G (z ) | z = e jw



G (e jw

)

=

k= g(k )e

jwk

=

k= g(k )z

k

z = e jw

= G (z ) | jw


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G (z ) | z = e jw

Provided, absolute convergence

k=

| g(k )e jwk | <



G (e jw

)

=

k= g(k )e

jwk

=

k= g(k )z

k

z = e jw

= G (z ) | z = e jw


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G (z ) | z = e jw


k=

| g(k )e jwk | <

k=

| g(k ) | <


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G (e jw

)

=

k= g(k )e

jwk

=

k= g(k )z

k

z = e jw

= G (z ) | z = e jw


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( ) | z = e j


k=

| g(k )e jwk | <

k=

| g(k ) | <

For causal systems, BIBO stability.


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21. FT of Discrete Time Aperiodic Signals - De-nition


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U (e j ) =

n =

u (n )e jn


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U (e j ) =

n =

u (n )e jn

1


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u (m ) = 1

2

U (e j )e jm d


U (e j ) =

n =

u (n )e jn

1


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u (m ) = 1

2

U (e j )e jm d

= 1/ 2

1/ 2U ( f )e j 2f m df


22. FT of a Moving Average Filter - Example


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y (n ) = 1

3[u (n + 1) + u (n ) + u (n 1)]


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y (n ) = 1

3[u (n + 1) + u (n ) + u (n 1)]

y (n ) =

k=

g(k )u (n k )


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= g( 1) u (n + 1) + g(0) u (n )


y (n ) = 1

3[u (n + 1) + u (n ) + u (n 1)]

y (n ) =

k=

g(k )u (n k )


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= g( 1) u (n + 1) + g(0) u (n ) + g(1) u (n 1)


y (n ) = 1

3[u (n + 1) + u (n ) + u (n 1)]

y (n ) =

k=

g(k )u (n k )


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= g( 1) u (n + 1) + g(0) u (n ) + g(1) u (n 1)

g( 1) = g(0) = g(1) = 13


y (n ) = 1

3[u (n + 1) + u (n ) + u (n 1)]

y (n ) =

k=

g(k )u (n k )


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= g( 1) u (n + 1) + g(0) u (n ) + g(1) u (n 1)

g( 1) = g(0) = g(1) = 13

G e jw =

n =

g(n )z n | z = e jw


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y (n ) = 1

3[u (n + 1) + u (n ) + u (n 1)]

y (n ) =

k=

g(k )u (n k )


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= g( 1) u (n + 1) + g(0) u (n ) + g(1) u (n 1)

g( 1) = g(0) = g(1) = 13

G e jw =

n =

g(n )z n | z = e jw

= 13

e jw + 1 + e jw = 13

(1 + 2 cos w )




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G e j = 13

(1 + 2 cos w )


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G e j = 13

(1 + 2 cos w ) , |G e jw | = |13

(1 + 2 cos w ) |


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G e j = 13

(1 + 2 cos w ) , |G e jw | = |13

(1 + 2 cos w ) |

Arg (G ) =0 0 w < 2

3 23 w <

1 / / U p d a t e d ( 1 8 7 0 7 )


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2 / / 5 . 3

3

4 w = 0 : 0 .0 1 : %pi ;5 subplot (2 , 1 , 1 ) ;6 plot2d1 ( g l l ,w,abs (1+2 cos (w) ) / 3 , s t y l e = 2 ) ;7 l a b e l ( ,4 , , Magni tude , 4 ) ;8 subplot (2 , 1 , 2 ) ;9 plot2d1 ( g ln ,w,phasemag (1+2 cos (w) ) , s t y l e = 2 , r e c t =[0

10 l a b e l ( ,4 , w , Phase ,4)Digital Control 24 Kannan M. Moudgalya, Autumn 2007


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|G e jw | = |13

(1 + 2 cos w ) |


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|G e jw | = |13

(1 + 2 cos w ) | ,

Arg (G ) =0 0 w < 2

3 23 w <

101

100

e


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102

101

100

101

103

102

10

a g n u

102

101

100

101

0

50

100

150

200

w

P h a s e


25. Additional Properties of Fourier Transform


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Symmetry of real and imaginary parts for real valued sequences

G e jw =

n =

g(n )e jwn


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G e jw =

n =

g(n )e jwn

=

n =

g(n ) cos wn j

n =

g(n )sin wn


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G e jw =

n =

g(n )e jwn

=

n =

g(n ) cos wn j

n =

g(n )sin wn


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G e jw =

n =

g(n )e jwn

=

n =

g(n ) cos wn + j

n =

g(n ) sin wn

Re G e jw = Re G e jw



G e jw =

n =

g(n )e jwn

=

n =

g(n ) cos wn j

n =

g(n )sin wn


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G e jw =

n =

g(n )e jwn

=

n =

g(n ) cos wn + j

n =

g(n ) sin wn

Re G e jw = Re G e jw

Im G e jw = Im G e jwDigital Control 26 Kannan M. Moudgalya, Autumn 2007



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Recall

G e jw = G e jw


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Recall

G e jw = G e jw

Symmetry of magnitude for real valued sequences


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Recall

G e jw = G e jw


|G e jw | = G e jw G e jw1/ 2

= G e jw G e jw1/ 2


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G e G e


Recall

G e jw = G e jw



= G e jw G e jw1/ 2


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= |G (e jw ) |


Recall

G e jw = G e jw



= G e jw G e jw1/ 2


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= |G (e jw ) |

This shows that the magnitude is an even function.


Recall

G e jw = G e jw



= G e jw G e jw1/ 2


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= |G (e jw ) |

This shows that the magnitude is an even function. Similarly,


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Recall

G e jw = G e jw



= G e jw G e jw1/ 2


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= |G (e jw ) |

This shows that the magnitude is an even function. Similarly,

Arg G e jw

= Arg G e jw

Bode plots have to be drawn for w in [0, ] only.


27. Sampling and Reconstruction


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u (n ) = u a (nT s ) , < n <


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u (n ) = u a (nT s ) , < n < Fast sampling:


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u (n ) = u a (nT s ) , < n < Fast sampling:

U a (F )1

u a ( t )

t F

F s2

F 0 F s


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F s2

T sF t

B B

F s2

32

F s3

2F s

u a (nT ) = u (n ) U F F sF s

F 0 + F sF 0


28. Slow Sampling Results in Aliasing


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29. What to Do When Aliasing Cannot be Avoided?


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30. Sampling Theorem

Suppose highest frequency contained in an analogsignal u a ( t ) is F max = B .


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It is sampled at a rate F s > 2F max = 2 B .


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It is sampled at a rate F s > 2F max = 2 B . u a ( t ) can be recovered from its sample values:


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( t )

( T )sin

Ts( t nT )


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u a ( t ) =n =

u a (nT s ) T s sT s ( t nT s )




( t )

( T )sin

Ts( t nT

s)


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u a ( t ) =n =


If F s = 2 F max , F s is denoted by F N , theNyquist rate.




( t )

( T )sin

Ts( t nT

s)


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u a ( t ) =n =


If F s = 2 F max , F s is denoted by F N , theNyquist rate.

Not causal: check n > 0Digital Control 31 Kannan M. Moudgalya, Autumn 2007

31. Rules for Sampling Rate Selection


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Minimum sampling rate = twice band width


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Minimum sampling rate = twice band width Shannons reconstruction cannot be implemented,

have to use ZOH


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have to use ZOH Solution:


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have to use ZOH Solution: sample faster


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Number of samples in rise time = 4 to 10


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Number of samples in rise time = 4 to 10




Number of samples in rise time = 4 to 10S l 10 30 i b d id h


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Number of samples in rise time = 4 to 10 Sample 10 to 30 times bandwidth




Number of samples in rise time = 4 to 10S l 10 30 i b d id h


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Number of samples in rise time 4 to 10 Sample 10 to 30 times bandwidth Use 10 times Shannons sampling rate




Number of samples in rise time = 4 to 10S l 10 t 30 ti b d idth


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Sample 10 to 30 times bandwidth Use 10 times Shannons sampling rate

cT s = 0 .15 to 0.5, where,c = crossover frequency


frequency resposne

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