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Transcript of Modeling in the Frequency Domain - test bank and solution ... · PDF file 2-2 Chapter 2:...

  • T W O

    Modeling in the

    Frequency Domain

    SOLUTIONS TO CASE STUDIES CHALLENGES

    Antenna Control: Transfer Functions

    Finding each transfer function:

    Pot: θ

    i

    i

    V (s)

    (s) =

    π

    10 ;

    Pre-Amp: p

    i

    V (s)

    V (s) = K;

    Power Amp:

    a

    p

    E (s)

    V (s) =

    150

    s 150

    Motor: Jm = 0.05 + 5( 50 250

    ) 2

    = 0.25

    Dm =0.01 + 3( 50 250

    ) 2

    = 0.13

    t

    a

    K

    R =

    1

    5

    t b

    a

    K K

    R =

    1

    5

    Therefore: θm

    a

    (s)

    E (s) =

     

    t

    a m

    t b m

    m a

    K

    R J

    K K1 s(s (D ))

    J R

    = 0.8

    s(s 1.32)

    And:

    θO

    a

    (s)

    E (s) =

    θm

    a

    (s)1

    5 E (s) =

    0.16

    s(s 1.32)

    Transfer Function of a Nonlinear Electrical Network

    Writing the differential equation, 

     

       20 0 d(i i)

    2(i i) 5 v(t) dt

    . Linearizing i 2 about i0,

    δ δ δ δ δ 

           0

    2 2 2 2

    0 0 0 0 0 0 i i

    (i i) i 2i | i 2i i. Thus, (i i) i 2i i.

    Full file at http://TestBankSolutionManual.eu/Solution-Manual-for-Control-Systems-Engineering-7th-Edition-by-Nise

  • 2-2 Chapter 2: Modeling in the Frequency Domain

    Substituting into the differential equation yields, δd i

    dt + 2i0

    2 + 4i0i - 5 = v(t). But, the

    resistor voltage equals the battery voltage at equilibrium when the supply voltage is zero since

    the voltage across the inductor is zero at dc. Hence, 2i0 2 = 5, or i0 = 1.58. Substituting into the linearized

    differential equation, δd i

    dt + 6.32i = v(t). Converting to a transfer function,

    δi(s)

    V(s) =

    1

    s 6.32 . Using

    the linearized i about i0, and the fact that vr(t) is 5 volts at equilibrium, the linearized vr(t) is vr(t) = 2i 2 =

    2(i0+i) 2 = 2(i0

    2+2i0i) = 5+6.32i. For excursions away from equilibrium, vr(t) - 5 = 6.32i = vr(t).

    Therefore, multiplying the transfer function by 6.32, yields, rV (s)

    V(s)

     =

    6.32

    s 6.32 as the transfer function

    about v(t) = 0.

    ANSWERS TO REVIEW QUESTIONS

    1. Transfer function

    2. Linear time-invariant

    3. Laplace

    4. G(s) = C(s)/R(s), where c(t) is the output and r(t) is the input.

    5. Initial conditions are zero

    6. Equations of motion

    7. Free body diagram

    8. There are direct analogies between the electrical variables and components and the mechanical variables

    and components.

    9. Mechanical advantage for rotating systems

    10. Armature inertia, armature damping, load inertia, load damping

    11. Multiply the transfer function by the gear ratio relating armature position to load position.

    12. (1) Recognize the nonlinear component, (2) Write the nonlinear differential equation, (3) Select the

    equilibrium solution, (4) Linearize the nonlinear differential equation, (5) Take the Laplace transform of

    the linearized differential equation, (6) Find the transfer function.

    SOLUTIONS TO PROBLEMS

    1.

    a.

    00

    1 1 ( ) st stF s e dt e

    s s

    

        

    b. 2 20

    00

    ( 1) ( ) ( 1)

    st st

    st

    e st F s te dt st

    s s e

            

    Full file at http://TestBankSolutionManual.eu/Solution-Manual-for-Control-Systems-Engineering-7th-Edition-by-Nise

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  • Solutions to Problems 2-3

    Using L'Hopital's Rule

    3 2

    1 ( ) 0. Therefore, ( ) .

    stt t

    s F s F s

    s e s 

       

    c. 2 2 2 2

    0 0

    ( ) sin ( sin cos ) st

    st eF s t e dt s t t s s

        

     

          

     

    d. 2 2 2 2

    0 0

    ( ) cos ( cos sin ) st

    st e sF s t e dt s t t s s

         

          

     

    2.

    a. Using the frequency shift theorem and the Laplace transform of sin t, F(s) = ω

    ω2 2(s+a) + .

    b. Using the frequency shift theorem and the Laplace transform of cos t, F(s) = ω2 2

    (s+a)

    (s+a) + .

    c. Using the integration theorem, and successively integrating u(t) three times, dt = t; tdt = 2t

    2 ;

    2t 2

    dt = 3t

    6 , the Laplace transform of t3u(t), F(s) =

    4

    6

    s .

    3.

    a. Taking the sum of the voltages around the loop and assuming zero initial conditions yields:

    0

    ( ) 1 ( ) ( ) ( )

    t di t

    Ri t L i d v t dt C

       

    b. Applying Laplace transform and solving for I(s)/V(s) gives:

    ( ) 1 1

    1 1( ) ( )

    I s

    RV s Ls R L s

    Cs L LCs

     

       

    Substituting the values of R, L, and LC, we have:

    2

    ( ) 2 2

    16( ) 2 16 ( 2 )

    I s s

    V s s s s

    s

       

     

    Full file at http://TestBankSolutionManual.eu/Solution-Manual-for-Control-Systems-Engineering-7th-Edition-by-Nise

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  • 2-4 Chapter 2: Modeling in the Frequency Domain

    Solving for I(s) and noting that V(s) = 1/s, we get:

    2

    2 ( )

    2 16 I s

    s s 

     

    Observing that the denominator has complex roots, we re-write the above equation as:

    2 2

    2 ( )

    ( 1) ( 15) I s

    s 

     

    Applying the frequency shift theorem to the Laplace transform of sin t u(t), we find that the

    transform for ( ) sin( )atf t e t is 2 2

    ( ) ( )

    F s s a

     

      .

    Comparing F(s) to I(s), we conclude that in the latter: a = 1 and  15 . Thus, the current, i(t),

    may be given by:

    2 ( ) 15 sin( 15 )

    15

    ti t e t

    c.

    Full file at http://TestBankSolutionManual.eu/Solution-Manual-for-Control-Systems-Engineering-7th-Edition-by-Nise

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  • Solutions to Problems 2-5

    4.

    a. The Laplace transform of the differential equation, assuming zero initial conditions,

    is, (s+7)X(s) = 2 2 5s

    s 2 . Solving for X(s) and expanding by partial fractions,

       

      2 2 5 35 1 5 7 4

    53 7 53( 7)( 4) 4

    s s

    ss s s

    Or,

       

      2 2 5 35 1 5 7 2 4

    53 7 53( 7)( 4) 4

    s s

    ss s s

    Taking the inverse Laplace transform, x(t) = - 35

    53 e-7t + (

    35

    53 cos 2t +

    10

    53 sin 2t).

    b. The Laplace transform of the differential equation, assuming zero initial conditions, is,

    (s2+6s+8)X(s) = 2

    15

    s 9 .

    Solving for X(s)

       2 2

    15 X(s)

    (s 9)(s 6s 8)

    and expanding by partial fractions,

         2

    1 6s 9

    3 3 1 15 19 X(s)

    65 10 s 4 26 s 2s 9

    Taking the inverse Laplace transform,

    4t 2t18 1 3 15x(t) cos(3t) sin(3t) e e 65 65 10 26

         

    c. The Laplace transform of the differential equation is, assuming zero initial conditions,

    (s2+8s+25)x(s) = 10

    s . Solving for X(s)

    2

    10 X(s)

    s(s 8 s 25) 

     

    and expanding by partial fractions,

    2

    4 1(s 4) 9

    2 1 2 9 X(s) -

    5 5 s 4 9s

     

      

    Taking the inverse Laplace transform,

    42 8 2( ) sin(3 ) cos(3 ) 5 15 5

    tx t e t t  

         

    Full file at http://TestBankSolutionManual.eu/Solution-Manual-for-Control-Systems-Engineering-7th-Edition-by-Nise

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  • 2-6 Chapter 2: Modeling in the Frequency Domain

    5.

    a. Taking the Laplace transform with initial conditions, s2X(s)-4s+4+2sX(s)-8+2X(s) = 2 2

    2

    s 2 .

    Solving for X(s),

    X(s) =

    3 2

    2 2

    4 4 16 18

    ( 4)( 2 2)

    s s s

    s s s

      

       .

    Expanding by partial fractions

    2 2 2

    1 s 2

    1 1 21(s 1) 22X(s) 5 s 2 5 (s 1) 1

       