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    2. Z-transform and theorem

    Gc(z) ZOH GP(s)R(z)E(z) M(z)

    GHP(z)

    Computer system

    Y(z)

    Plant

    A/D D/A

    Gc(s) GP(s)

    R(s) E(s) M(s) Y(s)

    Controller Plant

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    2. Z-transform and theorem

    time

    f(t)

    Time kT

    f(kT)

    A/D

    Time kT

    f(kT)

    D/A

    Time kT

    f(kT)

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    2. Z-transform and theorem

    How can we represent the sampled data

    mathematically?

    For continuous time system, we have a

    mathematical tool Laplace transform. It

    helps us to define the transfer function of a

    control system, analyse system stability anddesign a controller. Can we have a similar

    mathematical tool for discrete time system?

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    2.1 Z-transform

    For a continuous signal f(t), its sampled data can bewritten as,

    Then we can define Z-transform of f(t) as

    where z-1 represents one sampling period delay intime.

    0 )()2()()0()()( ns nTfTfTffkTftf

    21

    0

    )2()()0(

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

    zTfzTff

    znTfnTfZtfZzFn

    n

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    2.1 Z-transform

    Solution:

    Example 1: Find the Z-transform of unit step

    function. f(t)

    t

    kT

    f(kT)

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    2.1 Z-transform

    Apply the definition of Z-transform, we have

    1

    21

    32

    21

    0

    1

    11)(

    1

    1

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

    zzzzF

    q

    aaqaqaqa

    zz

    zkTfkTfZtfZzF k

    k

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    2.1 Z-transform

    Another method

    1

    1

    3212111

    21

    0

    1

    1)(1)()(

    )1()(

    1

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

    zzFzFzzF

    zzzzzzzFz

    zz

    zkTfkTfZtfZzF k

    k

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    2.1 Z-transform

    Example 2: Find the Z-transform of a

    exponential decay.

    Solution:

    f(t)

    t

    1

    221

    0

    1

    1)(

    1

    )]([)(

    )(

    ze

    zF

    zeze

    zekTfZzF

    ekTf

    aT

    aTaT

    k

    kakT

    akT

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    2.1 Z-transform

    Exercise 1: Find the Z-transform of a

    exponential decay f(t)=e-at using other

    method. f(t)

    t

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    2.1 Z-transform

    Example 3: Find the Z-transform of a cosine

    function.

    Solution: As

    2sin;2cos

    sincos;sincos

    j

    eet

    eet

    tjtetjte

    tjtjtjtj

    tjtj

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    2.1 Z-transform

    21

    1

    21

    1

    211

    11

    11

    11

    11

    1

    cos21

    cos1

    )cos21(2cos22

    )(1)(2

    21

    )1)(1(

    11

    2

    1

    1

    1

    1

    1

    2

    1)(

    1

    1

    ][

    ])[][(2

    1

    2][cos)(

    zTz

    Tz

    zTzTz

    zzezezeze

    zeze

    zeze

    zezezF

    zeeZ

    eZeZee

    ZkTZzF

    TjTj

    TjTj

    TjTj

    TjTj

    TjTj

    aT

    at

    tjtjtjtj

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    2.1 Z-transform

    Exercise 2: Find the Z-transform for decayed

    cosine function tetf at cos)(

    221

    1

    cos21

    cos1)(

    zeTez

    TezzF

    aTaT

    aT

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    2.1 Z-transform

    Example 4: Find the Z-transform for

    Solution:

    atetf 1)(

    )1)(1(

    )1(

    1

    1

    1

    1)(

    1

    1

    ][;1

    1

    ][]1[

    ][]1[1)]([)(

    11

    1

    11

    11

    zez

    ze

    zezzF

    zeeZzstepZZ

    eZZeZkTfZzF

    aT

    aT

    aT

    aT

    at

    atat

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    2.1 Z-transform

    Exercise 3: Find the Z-transform for

    attetf )(

    21

    1

    )1()(

    aT

    aT

    ez

    eTzzF

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    2.2 Z-transform theorems

    Linearity: If f(t) and g(t) are Z-transformable

    and and are scalar, then the linear

    combination f(t)+g(t) has the Z-transform

    Z[f(t)+g(t)]= F(z)+ G(z)

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    2.2 Z-transform theorems

    Shifting Theorem:

    Given that the Z-transform of f(t) is F(z), find

    the Z-transform for f(t-nT).

    f(t)

    t

    f(t-nT)

    tnT

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    2.2 Z-transform theorems

    If f(t)=0 for t

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    2.2 Z-transform theorems

    Defining m=k-n, we have

    Since f(mT)=0 for m

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    2.2 Z-transform theorems

    Final value theorem:Suppose that f(t), where f(t)=0

    for t

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    0

    )()]([)(k

    kzkTftfZzF

    )()()]()([ 22112211 zFkzFktfktfkZ

    )()]([ aTat zeFtfeZ

    )()]([a

    zFtfaZ t

    )()]([ zFzkTtfZ k

    ])()([)]([1

    0

    n

    k

    kk zkTfzFzkTtfZ

    )()1()]1()([ 1 zFztftfZ

    10 1

    )()(

    z

    zFdfZ

    t

    )()1(lim)( 11

    zFzfz

    )(lim)0( zFfz

    Theorem Name

    Definition

    Linearity

    Multiply by e-at

    Multiply by at

    Time Shift 1

    Time Shift 2

    Differentiation

    Integration

    Final Value

    Initial Value

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    s

    111

    1z

    2

    1

    s 21

    1

    )1(

    z

    Tz

    as

    111

    1 ze aT

    )( ass

    a

    )1)(1(

    )1(11

    1

    zez

    zeaT

    aT

    ab

    ee btat

    ))((

    1

    bsas

    )1)(1(

    )(111

    1

    zeze

    zee

    ab bTaT

    bTaT

    22

    s21

    1

    cos21

    sin

    zTz

    Tz

    22s

    s

    21

    1

    cos21

    cos1

    zTz

    Tz

    22)(

    as 221

    1

    cos21

    sin

    zeTez

    TezaTaT

    aT

    22

    )(

    as

    as

    221

    1

    cos21

    cos1

    zeTez

    TezaTaT

    aT

    f(t) F (s) F(z)

    (t) 1 1

    u(t)

    t

    e-at

    1e-at

    sint

    cost

    e-atsint

    e-atcost

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    2.3 Z-transform examples

    Example 1: Assume that f(k)=0 for k

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    2.3 Z-transform examples

    )1()21(

    2

    13

    211

    )21(9]3[]2[)]2(9[)(

    )21(

    9

    1)-(z/2

    )2/(

    2

    9]2[

    2

    9]229[)]2(9[

    211

    12/2/]12[]2[

    )1()1(

    Tz Z[t];

    1z-1

    1 Z[1];)]([

    ]3[]2[)]2(9[]32)2(9[)(

    121

    2

    1121

    11

    21

    1

    2

    11

    1

    221

    1-

    1-

    11

    zz

    z

    zzzzZZkZzF

    z

    zzkZkZkZ

    zzzZZ

    z

    Tz

    zz

    z

    a

    zFtfaZ

    ZZkZkZzF

    kk

    kkk

    kk

    t

    kkkk

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    2.3 Z-transform examples

    Example 2: Obtain the Z-transform of the

    curve x(t) shown below.

    0 1 2 3 4 5 6 7 8

    1

    t

    f(t)

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    2.3 Z-transform examples

    Solution:From the figure, we have

    K 0 1 2 3 4 5 6

    f(k) 0 0 0 1/3 2/3 1 1Apply the definition of Z-transform, we have

    )1(313

    2z

    )1(3

    2z

    3

    2

    3000)()(

    1

    543

    1

    543-

    21543-

    6543

    0

    z

    zzz

    z

    zz

    zzzz

    zzzz

    zkfzF

    k

    k

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    2.3 Z-transform examples

    Example 3: Find the Z-transform of

    Solution: Apply partial fraction to make F(s) as a

    sum of simpler terms.

    )()(

    2

    2

    ass

    asF

    )1()1(

    ])1()1[(

    1

    1

    1

    1

    )1(

    ]1

    []1[][)]([)(

    11

    )()(

    121

    11

    1121

    1

    2

    2

    32

    2

    1

    2

    2

    zez

    zzaTeeeaT

    zezz

    aTz

    asZ

    sZ

    s

    aZsFZzF

    asss

    a

    as

    k

    s

    k

    s

    k

    ass

    asF

    aT

    aTaTaT

    aT

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    2.4 Inverse Z-transform

    The inverse Z-transform: When F(z), the Z-

    transform of f(kT) or f(t), is given, the operation

    that determines the corresponding time sequencef(kT) is called as the Inverse Z-transform. We

    label inverse Z-transform as Z-1.

    n

    no

    m

    mo

    n

    no

    m

    mo

    zazaza

    zbzbzbbZzFZkTf

    zazaza

    zbzbzbbtfZkTfZzF

    2

    1

    1

    2

    2

    1

    111

    2

    1

    1

    2

    2

    1

    1

    1)]([)(

    1

    )]([)]([)(

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    2.4 Inverse Z-transform

    Z-transform

    =Inverse Z-transform

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    2.4 Inverse Z-transform

    The inverse Z-transform can yield the

    corresponding time sequence f(kt) uniquely.

    However, it says nothing about f(t). There mightbe numerous f(t) for a given f(kT).

    f(t)

    t0 T 2T 3T 4T 5T 6T

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    2.4 Inverse Z-transform

    x(kT) f(t)Zero-orderHold

    Low-pass

    Filter

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    2.5 Methods for Inverse Z-transform

    How can we find the time sequence for a

    given Z-transform?

    1) Z-transform table

    Example 1: F(z)=1/(1-z-1), find f(kT).

    F(z)=1+z-1+z-2+z-3+

    f(kT)=Z-1[F(z)]=1, for k=0, 1, 2,

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    2.5 Inverse Z-transform examples

    Example 2: Given ,

    Find f(kT).

    Solution: Apply partial-fraction-expansion tosimplify F(z), then find the simpler terms from

    the Z-transform table.

    Then we need to determine k1and k2

    )1)(1(

    )1()(

    11

    1

    zez

    zezF

    aT

    aT

    12

    11

    11

    1

    11)1)(1(

    )1()(

    ze

    k

    z

    k

    zez

    zezF aTaT

    aT

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    2.5 Inverse Z-transform examples

    Multiply (1-z-1) to both side and let z-1=1, we have

    1

    2

    1

    1

    11

    1

    11)1)(1(

    )1()(

    ze

    k

    z

    k

    zez

    zezF

    aTaT

    aT

    11

    )1(

    1

    )1(

    1

    )1(

    11)1(

    )1)(1(

    )1()1(

    1

    1

    1

    11

    2

    1

    11

    1

    1

    2

    1

    11

    11

    11

    1

    z

    aT

    aT

    aTaT

    aT

    aTaT

    aT

    ze

    zek

    ze

    kzk

    ze

    ze

    ze

    k

    z

    kz

    zez

    zez

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    2.5 Inverse Z-transform examples

    Similar as the above, we let multiply (1-e-aTz-1) to

    both side and let z-1=eaT, we have

    Finally, we have

    11

    )1(

    1

    )1(

    1

    )1(

    11)1()1)(1(

    )1()1(

    1

    1

    1

    221

    1

    1

    1

    1

    12

    111

    11

    1

    1

    aTez

    aTaTaT

    aTaT

    aT

    aT

    aT

    z

    zekk

    z

    kze

    z

    ze

    ze

    k

    z

    kzezez

    zeze

    0,1,2,k,1)(

    1

    1

    1

    1

    )1)(1(

    )1()(

    1111

    1

    akT

    aTaT

    aT

    ekTf

    zezzez

    zezF

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    2.5 Inverse Z-transform examples

    Exercise 4: Given the Z-transform

    Determine the initial and final values of f(kT), the

    inverse Z-transform of F(z), in a closed form.

    Hint: Partial-fraction-expansion, then use Z-transform table, and finally applying initial &

    final value theorems of Z-transform.

    )4.03.11)(1(

    )(211

    1

    zzz

    zzF

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    2.5 Inverse Z-transform examples

    2) Direct division method

    Example 1:F(z)=1/(1+z-1), find f(kT).

    1

    1

    11

    1

    1

    1

    -z

    z

    z

    1

    2

    21

    1

    1

    1 1

    1

    11

    z

    z

    zz-z

    z

    z

    -

    21

    2

    21

    1

    1

    11

    1

    11zz

    z

    zz

    -z

    z

    z

    -

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    2.5 Inverse Z-transform examples

    Finally, we obtain: F(z)=1-z-1+z-2-z-3+

    K = 0 1 2 3

    F(kT)= 1 -1 1 -1

    Example 2: Given ,

    Find f(kT).Solution: Dividing the numerator by the

    denominator, we obtain

    21

    1

    )1(

    21)(

    z

    zzF

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    21

    54

    543

    43

    432

    32

    321

    21

    21

    121

    741

    1013

    102010

    710

    7147

    47

    484

    4

    21

    2121

    zz

    zz

    zzz

    zz

    zzz

    zz

    zzz

    zz

    zz

    zzz

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    2.5 Inverse Z-transform examples

    Finally, we obtain: F(z)=1+ 4z-1 + 7z-2 + 10z-3+

    K = 0 1 2 3

    F(kT)= 1 4 7 10

    Exercise 5: ,

    Find f(kT).Ans. :k 0 1 2 3 4 5

    f(kT) 0 0.3679 0.8463 1 1 1

    21

    4321

    3679.03679.11

    05659.002221.0343.03679.0)(

    zz

    zzzzzF

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    2.5 Inverse Z-transform examples

    3) Computational method using Matlab

    Example: Given find f(kT).

    Solution:

    num=[1 2 0]; den=[12 1]

    Say we want the value of f(kT) for k=0 to 30u=[1 zeros(1,30)]; F=filter(num, den, u)

    1 4 7 10 13 16 19 22 25 28 31

    21

    1

    )1(

    21)(

    z

    zzF

    12

    2

    )1(

    2

    )1(

    21)(

    2

    2

    2

    2

    21

    1

    zz

    zz

    z

    zz

    z

    zzF

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    2.5 Inverse Z-transform examples

    Exercise 6: Given the Z-transform

    Use 1) the partial-fraction-expansion method and 2)

    the Matlab to find the inverse Z-transform of

    F(z).

    Answer: x(k)=-8.3333(0.5)k+8.333(0.8)k-2k(0.8)k-1

    x(k)=0;0.5;0.05;0.615;1.2035;-1.6257;-1.8778

    211

    11

    )8.01)(5.01(

    )5.0()(

    zz

    zzzF

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    Reading

    Study book

    Module 2: The Z-transform and theorems

    Textbook

    Chapter 2 : The Z-transform (pp23-50)

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    TutorialExercise:The frequency spectrum of a continuous-

    time signal is shown below.

    1) What is the minimum sampling frequency for

    this signal to be sampled without aliasing.

    2) If the above process were to be sampled at 10

    Krad/s, sketch the resulting spectrum from 20

    Krad/s to 20 Krad/s.

    -8 -4 4 8

    Krad/s

    F()

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    TutorialSolution: 1) From the spectrum, we can see that the

    bandwidth of the continuous signal is 8 Krad/s.

    The Sampling Theorem says that the sampling

    frequency must be at least twice the highest

    frequency component of the signal. Therefore,

    the minimum sampling frequency for this signal

    is 2*8=16 Krad/s.

    -8 -4 4 8

    Krad/s

    F()

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    Tutorial

    2) Spectrum of the sampled signal is formed by

    shifting up and down the spectrum of the

    original signal along the frequency axis at i

    times of sampling frequency. As s=10 Krad/s,for i =0, we have the figure in bold line. For i=1,

    we have the figure in bold-dot line.

    4 8

    Krad/s

    F()

    122 6 14 181610-4-8

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    Tutorial

    For I=-1, 2,we have

    4 8

    Krad/s

    122 6 14 181610

    -18

    F()

    4 8

    Krad/s

    122 6 14 181610-2-4-6-8-14

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    Tutorial

    Exercise 1: Find the Z-transform of a

    exponential decay f(t)=e-aT using other

    method.f(t)

    t

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    Tutorial

    1

    1

    33221

    22111

    221

    0

    11)(1)()(

    )1()(1

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

    zezFzFzezF

    zezeze

    zezezezFzezeze

    kTfkTfZtfZzF

    aT

    aT

    aTaTaT

    aTaTaTaT

    aTaT

    k

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    Tutorial

    Exercise 2: Find the Z-transform for a

    decayed cosine function

    Solution 1:

    tetf at cos)(

    221

    1

    21

    1

    21

    1

    cos21

    cos1

    cos21cos1]cos[

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

    )(cos21

    cos1cos

    zeTze

    Tze

    zTzTzteZ

    zeFtfeFzFtfZ

    zFzTz

    TztZ

    aTaT

    aT

    zez

    at

    aTat

    aT

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    Tutorial

    Solution 2:

    221

    1

    11

    1

    cos21

    cos1

    11

    11

    21)(

    1

    1][

    ])[][(2

    1

    2]cos[)(

    zeTze

    Tze

    zezezF

    zeeZ

    eZeZ

    eeZteZzF

    aTaT

    aT

    TjaTTjaT

    aT

    at

    tjattjat

    tjattjatat

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    Tutorial

    Exercise 3: Find the Z-transform for

    Solution:

    attetf )(

    21

    1

    21

    1

    21

    1

    )1(

    )1(][

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

    )()1(

    ze

    zTe

    zTzteZ

    zeFtfeFzFtfZ

    zFz

    TztZ

    aT

    aT

    zez

    at

    aTat

    aT

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    Tutorial

    Exercise 4: Given the Z-transform

    Determine the initial and final values of f(kT), the

    inverse Z-transform of F(z), in a closed form.

    Solution: Apply the initial value theorem and thefinal value theorem respectively, we have

    )4.03.11)(1(

    )(211

    1

    zzz

    zzF

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    Tutorial

    Exercise 5: Given

    Find f(kT) using direct-division method.

    Solution:

    21

    4321

    3679.03679.11

    05659.002221.0343.03679.0)(

    zz

    zzzzzF

    1

    432

    321

    432121 3679.0

    0565.01576.00.8463

    1354.05033.03679.0

    0565.002221.0343.03679.03679.03679.11

    z

    zzz

    zzz

    zzzzzz

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    Tutorial

    Continuous

    4321

    4321

    54

    543

    43

    432

    432

    321

    432121

    8463.03679.0)(

    8463.03679.0

    3679.0

    3679.03679.1

    3679.0

    3114.01576.18463.0

    0565.01576.00.8463

    1354.05033.03679.0

    0565.002221.0343.03679.03679.03679.11

    zzzzkf

    zzzz

    zz

    zzz

    zz

    zzz

    zzz

    zzz

    zzzzzz

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    Tutorial

    Exercise 6: Given the Z-transform

    Use 1) the partial-fraction-expansion method and 2)

    the Matlab to find the inverse Z-transform of

    F(z).

    Solution1: To make the expanded terms more

    recognizable in the Z-transform table, we

    usually expand F(z)/z into partial fractions.

    211

    11

    )8.01)(5.01(

    )5.0()(

    zz

    zzzF

    )150()50(11

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    2)5.0(

    15.0

    5.0

    )8.0(

    5.0

    )8.0(

    5.0

    15.0

    8.0)8.0()5.0()8.0(

    )8.0)(5.0(15.0)8.0(

    333.8)8.0(

    15.00.5zlet

    8.0

    )5.0(

    )8.0(

    )5.0(

    )8.0(

    15.0

    8.0)8.0()5.0()5.0(

    )8.0)(5.0(

    15.0)5.0(

    8.0)8.0()5.0()8.0)(5.0(

    15.0)(

    )8.0)(5.0(

    )15.0(

    )8.01)(5.01(

    )5.0()(

    8.0

    23

    2

    2

    1

    3

    2

    212

    2

    2

    5.0

    21

    32212

    3

    2

    21

    2

    3

    2

    21

    2

    2211

    11

    z

    z

    z

    zk

    z

    zkk

    z

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    k

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    k

    zz

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    z

    zF

    zz

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    zzzF

    kkkz 150

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    kkk

    zz

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    zz

    z

    zzF

    zzzzz

    z

    z

    zF

    zzz

    zzk

    zz

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    z

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    z

    z

    derivativez

    zk

    kz

    zk

    z

    z

    z

    k

    z

    k

    z

    kz

    zz

    zz

    )8.0(333.8)8.0(2)5.0(333.8)(

    8.01

    333.8

    )8.01(

    2

    )5.01(

    333.8)(

    8.0

    333.8

    )8.0(

    2

    )5.0(

    333.8

    )8.0)(5.0(

    15.0)(

    333.83.0

    6.03.0*5.0)5.0(

    )15.0()5.0(5.05.015.0

    8.0let;)5.0(

    )8.0()5.0(0

    5.0

    )8.0(

    5.0

    15.0

    5.0

    )8.0(

    5.0

    )8.0(

    5.0

    15.0

    8.0)8.0()5.0()8.0(

    )8.0)(5.0(

    15.0)8.0(

    1

    121

    1

    1

    22

    2

    8.0

    2

    8.0

    '

    3

    33

    '2

    1

    '

    3

    2

    2

    1

    3

    2

    212

    2

    2

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    Tutorial

    Partial fraction for inverse Z-transform

    If F(z)/z involve s a multiple pole, eg. P1, thenipz

    ii

    n

    n

    n

    mm

    mm

    nn

    nn

    mm

    mm

    z

    zFpza

    pz

    a

    pz

    a

    pz

    apzpzpz

    bzbzbzb

    azazaz

    bzbzbzb

    z

    zF

    )()(;

    )())((

    )(

    2

    2

    1

    1

    21

    1

    1

    10

    1

    1

    0

    1

    1

    10

    11

    )()(;

    )()(;

    )()(

    )()()()()(

    2

    12

    2

    11

    3

    3

    1

    22

    1

    1

    3

    2

    1

    1

    1

    10

    pzpzpz

    ii

    n

    n

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    mm

    mm

    z

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    dz

    dc

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    pza

    pza

    pzc

    pzc

    pzpzpzbzbzbzb

    zzF

    i

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    Tutorial

    Solution 2: Expand F(z) into a polynomial form

    Num=[0 0.51 0];

    Den=[12.1 1.440.32];

    U==[1 zeros(1,40)];F=filter(Num, den,U)

    0 0.5 0.05 -0.615 -1.2035

    321

    21

    211

    11

    32.044.11.21

    5.0

    )8.01)(5.01(

    )5.0()(

    zzz

    zz

    zz

    zzzF