4_z Transform_2013.pdf

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5/18/2018 4_zTransform_2013.pdf-slidepdf.com http://slidepdf.com/reader/full/4z-transform2013pdf 1/28 1 1 The Transform The z transform generalizes the Discrete-time Fourier Transform for the entire complex plane. For the complex variable is used the notation: { } 2 2 ; , arg  j  z x j y r e x y  z r x y  z Ω = + = = = + Ω = 2 notation: For any discrete-time LTI system: -the eigen function: -the eigen value: The Discrete LTI System Response to a Complex Exponential [ ] 0 0 0  j n n n  x n z r e  Ω = = System h[n] input output [ ] [ ] [ ] [ ]  ( ) 0 0 0 0 0 0 n n k n k n k  y n h n z h k z z h k z z H z =−∞ =−∞ = = = = [ ] [ ]  ( ) 0 0 0 n n  y n h n z z H z = = ( )  [ ]   H z h k =−∞ = ( ) 0  H z 0 n  z 

Transcript of 4_z Transform_2013.pdf

  • 11

    The z Transform

    The z transform generalizes the Discrete-time Fourier Transform for the entire complex plane.For the complex variable is used the notation:

    { }2 2

    ; ,

    arg

    jz x j y r e x y

    z r x y

    z

    = + = = = + =

    \

    2

    notation:

    For any discrete-time LTI system:

    -the eigen function:

    -the eigen value:

    The Discrete LTI System Response to a Complex Exponential

    [ ] 00 0 j nn nx n z r e = =System

    h[n]

    input output

    [ ] [ ] [ ] [ ] ( )0 0 0 0 0 0n n k n k nk k

    y n h n z h k z z h k z z H z = =

    = = = = [ ] [ ] ( )0 0 0n ny n h n z z H z= =

    ( ) [ ] kk

    H z h k z =

    =

    ( )0H z0nz

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    fast computation of the response of any discrete-time LTI system to linear combinations of complex exponentials of the form:

    transfer function, z transform of the impulse response h[n].

    [ ] ( )0 0ny n z H z=

    [ ] nk kk

    x n c z= System

    h[n]

    input output[ ] ( ) nk k k

    ky n c H z z=

    ( ) [ ] kk

    H z h k z =

    =

    4

    Bilateral z TransformThe bilateral z transform of the signal x[n] :

    generalization of the discrete-time Fourier transform

    Discrete-time Fourier transform = particular case of z transform on the unit circle |z|=1

    [ ]{ }( ) ( ) [ ] , , 0jnn

    Z x n z X z x n z z r e r =

    = = =

    [ ]{ }( ) [ ]( ) [ ]{ }( ); 0j nn nn

    Z x n z x n r e r x n r =

    = = F

    [ ]{ }( ) [ ]{ }( )1 jz r Z x n e x n= = = F|z|=1

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    Examples

    region of convergence (ROC) = set of values of z for which X(z) is convergent

    rational function, zeros = roots of numerator ; poles = roots of denominator.

    Pole/zero plot or constellation (PZC).

    [ ] [ ]1. , 1nx n a n a=

    { } [ ] ( ) [ ]0 0 00 0 n n m m n nn mn m m

    Z x n n x n n z x m z z x m z = +

    = = = = = =

    18

    3. Modulation in time

    Proof.

    More generally:

    Homework - Prove it.

    [ ]{ } [ ] [ ]( ) (0 0 0 0nj n j n j jnn n

    Z e x n e x n z x n e z X e z = =

    = = =

    [ ][ ]

    00 0

    00

    , n

    nu

    z zz x n X ROCz z

    zz x n Xz

    [ ] ( )0 0 , j n je x n X e z z ROC

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    19

    4. Time reversal

    Proof.

    [ ]{ } [ ] [ ] 1 1mm nnn m

    Z x n x n z x m Xz z

    == =

    = = =

    [ ] ( )1 1, x n X z ROCz

    20

    5. Differentiation in time

    Proof. direct application of the definitions and previous properties. Homework. Prove these properties.

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

    1

    1

    1 1 ;

    1 1 1u

    x n x n z X z z ROC

    x n x n z X z x

  • 11

    21

    6. Addition in time

    Proof.

    [ ] ( ) [ ]1

    1 ; 11

    unk

    k

    X z x kx k z

    z

    ==

    + >

    [ ] ( ) ( )1 ; 11n

    k

    X z zx k X z z ROC ROCzz

    =

    =

    [ ] [ ] [ ] ( ) ( ) ( ) [ ] [ ] [ ]111 1 1 ; 1uk

    x n y n y n X z z Y z y y x k=

    = = = [ ] [ ] [ ] [ ] [ ] ( ) ( ) ( )11 1n

    ky n x k x n y n y n X z z Y z

    == = =

    22

    7. Differentiation in z domain

    Proof.

    By direct application of definitions.

    Homework. Prove it.

    [ ] ( )

    [ ] ( );

    u

    dX znx n z z ROC

    dzdX z

    nx n zdz

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    23

    8. Complex conjugation in time domain

    Proof.

    By direct application of definitions.

    Homework. Prove it.

    [ ] ( )[ ] ( )

    ;

    u

    x n X z z ROC

    x n X z

    24

    9. Time convolution (convolution theorem)

    Proof.

    [ ] [ ]{ } [ ] [ ]( ) [ ] [ ][ ] [ ] ( )

    [ ] [ ] ( ) ( )

    n n

    n n k

    n kk

    n kn k m m k

    n k

    Z x n y n x n y n z x k y n k z

    x k z y n k z

    y m z x k z Y z X z

    = = = = =

    = = =

    = =

    =

    = =

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

    ,

    ; at leastx y

    u u

    x n y n

    x n y n X z Y z z ROC ROC

    x n y n X z Y z

    ^

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    25

    10. Product theorem or convolution theorem in the complex domain

    Proof.

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

    1 , 21 ,

    2 u u

    z dux n y n X u Y ROCj u u

    z dux n y n X u Y ROCj u u

    vv

    : ; : and

    :

    y y y yx x x xx y

    y yx x

    zROC R z R ROC R z R R u R R R

    u

    ROC R R z R R

    + + + +

    + +

    < < < < < < < = < =

    [ ] ( ) [ ] ( ) [ ]0.5 : 4 0.5 1 4 0.25n nz y n n n< < =

    Causal signal y[n]

    38

    3. Power series expansion of function Y (z)

    Expansion of function Y(z) into a power series

    Example #a

    ( )

    [ ]

    1

    1 1 2

    0

    , : 0

    1 1 1 111! 2! ! !1!

    n n

    n

    z

    z

    Y z e ROC z

    e z z z zn n

    y nn

    =

    = >= + + + + + =

    =

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    39

    Example #b

    ( )

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

    2

    00 1

    , : 0,

    1 1 1 111! 2! ! !

    1 1 1! !

    1 11! 1 !

    m m

    m

    n n

    n n

    n

    z

    z

    Y z e ROC z z

    e z z z zm m

    z zn n

    y n n n nn n

    =

    = =

    = = + + + + + =

    = = + = =

    40

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

    [ ] ( )[ ] ( ) [ ]

    1

    1 1

    1

    1

    ln 1 , :

    1 1; 1

    initial value theorem:

    0 lim ln 1 0

    1 1

    n nn nn

    n

    zn

    n

    Y z az ROC z a

    a aY z z y n n

    n n

    y az

    y n a nn

    + + =

    = + > = =

    = + ==

    causal signal

    Example #c

  • 21

    41

    Example #d

    ( ) [ ] [ ]11 , 1 nY z z a y n a naz= > =

    42

    Example #e, same transform as #d, different ROC

    ( )

    ( ) ( ) [ ] [ ]1

    11

    1

    1 , 1

    anticausal signal, z transform contains only powers of z

    1m n n n

    m n

    zY z z az aaz

    Y z a z a z y n a n

    = =

    = =

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    43

    Discrete LTI Systems Described by Linear Constant Coefficient Difference Equations

    [ ] [ ] 00 0

    ; 0N M

    k kk k

    a y n k b x n k a= =

    = ( ) ( ) ( ) ( )( )00 0

    0

    transfer function

    Mk

    N M kk k k

    k k Nkk k

    kk

    b z N za z Y z b z X z H z

    D za z

    =

    = ==

    = = =

    Transfer function of a discrete LTI system : rational function in z or z-1. Zeros : roots of numerator N(z); Poles : roots of denominator D(z)

    44

    Causal and stable system: poles inside unit circle |z|=1

    The initial value theorem can be applied:

    The degree of the numerator of the transfer function of a

    causal and stable system is smaller or equal with the

    degree of its denominator, M N

    1pkz Partial impulse response increases in timeInstability.

    ( )sin p pA n +

    1pr =

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    47

    #2. A pair of double complex conjugate poles

    Contribution of the two poles in the impulse response :

    ( ) ( ) ( )1 1 2 22 21 1 1 1 and

    1 1 1 1

    j jp p p p

    j j j jp p p p

    p p

    p p p p

    z r e z r e

    a a a aH z

    r e z r e z r e z r e z

    = == + + + + +

    ( ) ( ) [ ]1 2sin sinn np p p p p pA r n A nr n n + + + 1pr < Partial impulse response decreases in time

    No instability from these poles. 1pr > Partial impulse response increases in time

    Instability.

    48

    difference equation, non-zero initial conditions unilateral Z transform

    Causal input signal, x[-n]=0 for n >0:

    Response of a Discrete LTI System Described by a Difference Equation

    [ ] [ ] [ ]{ } [ ]{ }00 0 0 0

    , 0 N M N M

    k k k u k uk k k k

    a y n k b x n k a a Z y n k b Z x n k= = = =

    = =

    ( ) [ ] ( )0 1 0

    N k Mk n k

    k u k uk n k

    a z Y z y n z b z X z = = =

    + =

    ( ) [ ] ( ) [ ]1 10 0

    N k M kk n k n

    u uk kn nk k

    a z Y z y n z b z X z x n z = == =

    + = +

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    49

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

    [ ]

    [ ] [ ] ( ) [ ] .1 , 1 ; 1.

    11

    11

    11

    11

    11

    .1 ; 1

    1

    ,01 , , 1

    Example

    0

    0

    0

    000

    0

    0

    0

    0

    111

    111

    111

    11

    >=+==

    +

    ++

    zanea

    keea

    kayany

    azay

    azeaka

    zeeake

    azay

    azzekzY

    zze

    kzyzYazzY

    ynkenxnxnayny

    j

    nj

    j

    nn

    jjj

    j

    ju

    juu

    nj

    50

    First Order Systems0pole/zero plot 0; 0.5pz z= = [ ] [ ] [ ]

    ( )[ ] [ ]

    1

    1

    ; 1

    , 1 np

    y n ay n kx nk kzH z z ROCaz z a

    z a z a h n ka n

    = = = > = < =

    ( ) ( ) ( )1

    0 1, max. freq. response for 0, 1 0, max. for ,

    Positive low-pass filterNegative high-pass filter

    j jOAH e ePA PA

    aa

    aa

    = =

    < < = < < =

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    51

    Second Order Systems[ ] [ ] [ ] [ ] ( ) 21 2 1 2 2

    1 2 1 2

    1 2 1

    k kzy n a y n a y n kx n H za z a z z a z a

    + + = = =+ + + +

    1,2

    21 2

    21 2

    2 21 2 1

    2

    1 2

    4 complex conjugated poles

    4 real poles . The system is considered causal and stable system.

    41 magnitude

    2The conditions imposed on ,

    jp

    a az e

    a a

    a a aa

    a a

    <

    =

    + = =

    > >

    1.

    2.

    21 1 2

    0 1,2

    40 ,

    2pa a a

    z z = =

    52

    ( ) ( ).1 : system theof responsefrequency Thepole. real double single a of existence the toscorrespond parabola The

    212_____

    2_____

    1

    = jeAPAP

    kH

    Magnitude: even; Phase: odd

  • 27

    53

    Transfer Function of Equivalent System for Serial or Parallel Interconnections of two

    discrete LTI Systems

    [ ] [ ] [ ]( ) ( ) ( ).zHzHzH

    nhnhnh

    e

    e

    21

    21 ; +=

    +=[ ] [ ] [ ]( ) ( ) ( ).zHzHzH

    nhnhnh

    e

    e

    21

    21 ; =

    =

    54

    Digital Filters Implementation Forms Obtained Using the z

    Transform[ ] [ ] [ ] [ ] [ ]. ,1 ,

    100

    00====

    ==== Nk

    kM

    kk

    M

    kk

    N

    kk knyaknxbnyNMaknxbknya

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    55

    Direct form I (N+M adders). First form of implementation using z transform.

    Each adder has 2 inputs. One adder with 2N inputs (M=N, a0=1).

    56

    Direct form II (2N adders M=N). Second form of implementation using z transform.

    Each adder has 2 inputs. Two adders with N inputs each (M=N, a0=1).