Chapter 4 General Force Vibration

43
Dr.-Ing. Nantawatana Weerayuth Page 163 บทที4 การสั่นสะเทือนแบบบังคับของระบบที่มีลาดับขั ้นเสรี เท่ากับหนึ่ง ภายใต้แรงกระทาแบบทั่วไป จุดประสงค์การเรียนรู้ 1. สามารถหาผลตอบสนองของระบบ 1-DOF ภายใต้แรงกระทาแบบคาบเวลาที่มี รูปแบบทั่วไป โดยใช้อนุกรมฟูเรียร์ได้ 2. สามารถใช้อนุกรมฟูเรียร์เพื่อหาผลตอบสนองของระบบ 1-DOF ภายใต้แรงกระทา แบบเป็นคาบซ า ที่ไม่สามารถสร้างเป็นสมการทั่วไปได้ 3. สามารถใช้ระเบียบวิธี “Convolution” หรือ “Duhamel” ในการหาผลเฉลยของปัญหา การสั่นสะเทือนภายใต้แรงกระทาแบบทั่วไปได้ 4. สามารถใช้การแปลงลาปลาซในการแก้ปัญหาการสั่นสะเทือนเบื ้องต ้นได้

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Transcript of Chapter 4 General Force Vibration

  • Dr.-Ing. Nantawatana Weerayuth Page 163

    4

    1. 1-DOF

    2. 1-DOF

    3. Convolution Duhamel

    4.

  • Dr.-Ing. Nantawatana Weerayuth Page 164

    4.1 (Periodic Function)

    ( )F t Periodic

    ( ) ( )F t k F t (4.1)

    2

    1,2,...,k

    (4.1) ( )F t

    4.1 4.2

    ( ) ( 2 ) ( 3 ) .... ( )F t F t F t F t (4.2)

    4.1

    4.2

  • Dr.-Ing. Nantawatana Weerayuth Page 165

    ( )F t

    ( Fundamental Period ) ( Fundamental Frequency)

    2

    (4.3)

    Periodic: ( )F t

    0

    1 1

    ( ) cos( ) sin( )2

    n n

    n n

    aF t a n t b n t

    (4.4)

    0a na nb

    4.2

    (Periodic Force) 2 (4.4)

    0

    1 1

    ( ) cos( ) sin( )2

    n n

    n n

    aF t a n t b n t

    0a na nb

    0

    2( )cos( ) ; 0,1,2,...na F t n t dt n

    (4.5)

    0

    2( )sin( ) ; 1,2,3,...nb F t n t dt n

    (4.6)

    4.2.1

    ( )F t (Even Function)

    (Odd Function) 0a na nb

  • Dr.-Ing. Nantawatana Weerayuth Page 166

    Index: n

    1. (Even Function) :

    ( ) ( ),f x f x x (4.7)

    2( ) ,f x x 2( ) 1f x x ( ) cos( )f x nx [ y ]

    4.3

  • Dr.-Ing. Nantawatana Weerayuth Page 167

    2. (Odd Function) :

    ( ) ( ),f x f x x (4.8)

    ( )f x x , 3( )f x x , ( ) sinh( )f x x ( ) sin( )f x nx [

    180 ]

    4.4

  • Dr.-Ing. Nantawatana Weerayuth Page 168

    1. () () = () 2. () () = ()

    3. () () = () 4. () () = ()

    5. () () = ()

    6. ( )f x ( ) 0c

    c

    f x dx

    7. ( )h x 0

    0 0

    ( ) ( ) ( ) 2 ( )

    c c c

    c c

    h x dx h x dx h x dx h x dx

    8. ( )f x : [ 0na ]

    0

    2 1( )cos( ) ( )cos( ) 0 ; 0,1,2,...n

    t Odd function Odd function

    a f x n x dx f x n x dx n

    0

    2( )sin( ) 0 ; 0,1,2,...n

    t even function

    b f x n x dx n

    9. ( )f x : [ 0nb ]

    0

    2( )cos( ) 0; 0,1,2,...neven function

    a f x n x dx n

    0

    2 1( )sin( ) ( )sin( ) 0; 0,1,2,...n

    t odd function odd function

    b f x n x dx f x n x dx n

    (Even Function)

    (Odd Function)

  • Dr.-Ing. Nantawatana Weerayuth Page 169

    4.1 (Periodic Function)

    4.5

    4.5

    0

    0 02

    ( )

    02

    for t

    f t

    F for t

    ( )f t

    na nb (4.5) (4.6)

    0

    2( )cos( ) ; 0,1,2,...na F t n t dt n

    0

    2( )sin( ) ; 1,2,3,...nb F t n t dt n

    1. na : 0n 0

    02 2

    0

    002 2

    2 2( ) ( ) ( )

    F

    a F t dt F t dt F t dt

    2

    00 0 0

    0

    22( )( )

    2

    Fa F dt F

  • Dr.-Ing. Nantawatana Weerayuth Page 170

    1,2,...n

    0

    02 2

    0 002 2

    2 2 2( )cos( ) ( )cos( ) ( )cos( ) ( )cos( )n

    F

    a F t n t dt F t n t dt F t n t dt F t n t dt

    2

    0 20 0

    0

    22cos( ) sin( )n

    Fa F n t dt n t

    n

    1 2 3, , ,...a a a

    20 0

    0

    0 ( 1,2,3,..)2 2( ) sin( ) sin( )2

    n

    for all nF Fa n t n

    n n

    2. nb : 1,2,3,...n

    0

    02 2

    0 002 2

    2 2 2( )sin( ) ( )sin( ) ( )sin( ) ( )sin( )n

    F

    b F t n t dt F t n t dt F t n t dt F t n t dt

    2

    0 20 0

    0

    22sin( ) cos( )n

    Fa F n t dt n t

    n

    1 2 3, , ,...b b b

    2

    0 0

    0

    0

    0 (2,4,6,..)2 2

    ( ) cos( ) 1 cos( ) 22 (1,3,5,...)

    n

    if n is evenF F

    b n t n Fn n if n is odd

    n

    (4.4) ( )f t

    0 0

    1,3,5,...

    2( ) ( )sin( )

    2 n

    F Ff t n t

    n

  • Dr.-Ing. Nantawatana Weerayuth Page 171

    4.2 (Periodic Function)

    4.6

    4.6

    Periodic

    0

    02 4

    ( )4 4

    04 2

    for t

    f t F for t

    for t

    4.6 y

    ( ) ( )f t f t Even Function

    0nb 1,2,3,...n

    na ( 1,2,3,...n )

    0

    2( )cos( )na F t n t dt

  • Dr.-Ing. Nantawatana Weerayuth Page 172

    0

    4 4 2

    0 0 02 4 4

    2 2( )cos( ) ( )cos( ) ( )cos( ) ( )cos( )n

    F

    a F t n t dt F t n t dt F t n t dt F t n t dt

    4

    0 040

    4

    4

    2 22 2 2cos( ) sin( ) ( ) sin( ( )) sin( ( ))

    2 4 4n

    F Fa F n t dt n t n n

    n n

    0 0 022 2sin( ( )) sin( ( )) sin( ) sin( ) sin( )4 4 2 2

    n

    F F Fa n n n n n

    n n n

    0 ( 1)

    02

    0 ( 2,4,6,...)2

    sin( ) 22 ( 1) ( ) ( 1,3,5,...)

    nn

    if n is even nF

    a n Fn if n is odd n

    n

    ( )f t

    ( 1)

    0 02

    1,3,5,...

    2( ) ( 1) ( )cos( )

    2

    n

    n

    F Ff t n t

    n

    4.2.

    - 2

    2 3 2

    ( )mx cx kx f t (4.9)

  • Dr.-Ing. Nantawatana Weerayuth Page 173

    (4.4)

    0

    1 1

    ( ) cos( ) sin( )2

    n n

    n n

    af t a n t b n t

    (4.9)

    0

    1 1

    ( )

    cos( ) sin( )2

    n n

    n n

    f t

    amx cx kx a n t b n t

    (4.10)

    (4.10) 0( 2)a

    (1 1

    cos( ) sin( )n nn n

    a n t b n t

    )

    (Linear System)

    (Principle of Super Position)

    ( (4.10))

    0

    2

    amx cx kx (4.11)

    1

    cos( )nn

    mx cx kx a n t

    (4.12)

    1

    sin( )nn

    mx cx kx b n t

    (4.13)

    (4.11) (4.12) (4.13)

    3

    1. (Steady State Solution) (4.11)

    1( )px t C (4.11) 02a

    Ck

    01( )

    2p

    ax t

    k (4.14)

  • Dr.-Ing. Nantawatana Weerayuth Page 174

    2. (Steady State Solution) (4.12)

    3

    22 2 2

    ( ) cos( ); 0,1,2,...(1 ( ) ) (2 )

    np n

    a kx t n t n

    nr nr

    (4.15)

    3. (Steady State Solution) (4.13)

    2

    32 2 2

    ( ) sin( ); 1,2,3,...(1 ( ) ) (2 )

    np n

    b kx t n t n

    nr nr

    (4.16)

    n

    r

    n

    1

    2

    2tan

    1 ( )n

    nr

    nr

    (4.17)

    (4.10)

    0

    2 2 20

    2 2 21

    ( ) cos( )..2 (1 ( ) ) (2 )

    sin( )(1 ( ) ) (2 )

    np n

    n

    nn

    n

    a a kx t n t

    k nr nr

    b kn t

    nr nr

    (4.18)

    (4.18)

    ( 1,2,..,n )

    3 n 1nr

    Amplitude nr

    (4.18)

    2n 3

    Periodic

  • Dr.-Ing. Nantawatana Weerayuth Page 175

    4.3 (Steady State Solution) -

    1 DOF

    (Periodic Function) 4.7

    4.7

    ( )mx cx kx f t

    4.2 ( )f t

    ( 1)

    0 02

    1,3,5,...

    2( ) ( 1) ( )cos( )

    2

    n

    n

    F Ff t n t

    n

    1,3,5,...n

    (4.10)

    ( 1)

    0 02

    2 2 21,3,5,..

    2( ) ( 1) cos( )

    2 (1 ( ) ) (2 )

    n

    p n

    n

    F F knx t n t

    k nr nr

  • Dr.-Ing. Nantawatana Weerayuth Page 176

    4.4 (Steady State Solution) -

    4.8

    ( )f t

    4.8

    4.3

    ( )mx cx kx f t

    4.1 ( )f t

    0 0

    1,3,5,...

    2( ) ( )sin( )

    2 n

    F Ff t n t

    n

    1,3,5,...n

    (4.10)

    0 0

    2 2 21,3,5,..

    2( ) sin( )

    2 (1 ( ) ) (2 )p n

    n

    F F knx t n t

    k nr nr

  • Dr.-Ing. Nantawatana Weerayuth Page 177

    4.5 (Steady State Solution)

    Hydraulic Valve -

    4.9 (a) ( )p t 4.9

    (b) 2500k N/m c=10 N-s/m m=0.25 kg

    4.9

    ( )mx cx kx f t

    m ( ) ( )f t Ap t A

    ( )p t ( m )

    2 26 2(50) 625 10

    4 4

    dA m

    4.9 (b) ( )p t 2 ( )f t

    m

  • Dr.-Ing. Nantawatana Weerayuth Page 178

    (Fundamental Frequency) 2 2

    2

    rad/s ( )p t 4.9 (b)

    y ( )p t (Even Function) ( ) ( )f t Ap t

    0nb

    n

    ( )f t

    50,000 02

    ( )

    50,000 ( )2

    At for t

    f t

    A t for t

    na ( 0,1,2,3,...n )

    0

    2( )cos( )na F t n t dt

    0n

    2 2

    0

    0 0 0

    2 2

    2 2 2( ) ( ) ( ) ( ) ( )a F t dt F t dt F t dt F t dt F t dt

    2

    0

    0

    2

    250,000 50,000 ( ) 50,000a At dt A t dt A

    1,2,3,...n

    2

    0 0

    2

    2 2( )cos( ) ( )cos( ) ( )cos( )na F t n t dt F t n t dt F t n t dt

  • Dr.-Ing. Nantawatana Weerayuth Page 179

    2

    0

    2

    250,000 cos( ) 50,000 ( )cos( )na At n t dt A t n t dt

    By Part Integration : 1cos( ) sin( ) sin( )tt n t dt n t n t dtn n

    2

    0

    2

    2

    0

    250,000 sin( ) sin( ) 50,000 cos( )

    250,000 sin( ) sin( )

    n

    ta A n t n t dt A n dt

    n

    tA n t n t dt

    n

    2

    02

    2

    2 1 150,000 sin( ) cos( ) 50,000 sin( )

    2 150,000 sin( ) cos( )

    n

    ta A n t n t A n t

    n n n

    tA n t n t

    n n

    5

    2 22 2

    2 10(1,3,5,...)50,000

    2cos( ) cos( 2 ) 1)

    0 (2,4,6,..)

    n

    A if n is oddAa n n n

    nif n is even

    5

    0

    2 21,3,5,...

    2 10( ) cos( )

    2 n

    Ff t A n t

    n

    1,3,5,...n

    (4.10)

    5 2 2

    2 2 21,3,5,..

    25,000 2 10( ) cos( )

    (1 ( ) ) (2 )p n

    n

    A A knx t n t

    k nr nr

  • Dr.-Ing. Nantawatana Weerayuth Page 180

    3 0,1,n 3

    5 2 5 2

    1 32 2 2 2 2 2

    25,000 2 10 2 10 9( ) cos( ) cos(3 )

    (1 ( ) ) (2 ) (1 (3 ) ) (6 )p

    A A k A kx t t t

    k r r r r

    2500 1000.25

    n

    k

    m rad/s

    2

    rad/s

    0.031416100n

    r

    100.2

    2 2(0.25)(100)c n

    c c

    c m

    1

    1 2

    2tan 0.0125664

    1 ( )

    r

    r

    rad

    1

    2 2

    6tan 0.0380483

    1 (3 )

    r

    r

    rad

    ( )px t

    ( ) 0.019635 0.015930cos( 0.0125664) 0.0017828cos(3 0.0380483)px t t t m

  • Dr.-Ing. Nantawatana Weerayuth Page 181

    4.5

    4.10

    3

    4.10

    4.3

    ( )f t

    (Constant Sampling Time)

    1 2, ,..., Nt t t

  • Dr.-Ing. Nantawatana Weerayuth Page 182

    1 1 2 2( ), ( ),..., ( )N Nf f t f f t f f t N (

    2,4,6,...) 1 ( ) N tN

    N t 4.11

    4.11

    Trapezoidal Rule

    0

    1

    2 N

    i

    i

    a fN

    (4.19)

    1

    2 2cos( ); 1,2,...

    N

    n i i

    i

    a f n t nN

    (4.20)

    1

    2 2sin( ); 1,2,...

    N

    n i i

    i

    b f n t nN

    (4.21)

    (4.18) 0

    2 2 20

    2 2 21

    ( ) cos( )..2 (1 ( ) ) (2 )

    sin( )(1 ( ) ) (2 )

    np n

    n

    nn

    n

    a a kx t n t

    k nr nr

    b kn t

    nr nr

  • Dr.-Ing. Nantawatana Weerayuth Page 183

    (4.17)

    1

    2

    2tan

    1 ( )n

    nr

    nr

    2n

    r

    4.6 Hydraulic Valve

    4.1 0.01

    1

    0.12 ( )p t

    0.12 s 2 2 52.360.12

    rad/s

    0.01t 0.12 120.01

    Nt

  • Dr.-Ing. Nantawatana Weerayuth Page 184

    (4.19)-(4.21)

    12

    0

    1

    268166.7

    12i

    i

    a p

    12

    1

    2 2cos( )

    12 0.12n i i

    i

    a p n t

    12

    1

    2 2sin( )

    12 0.12n i i

    i

    b p n t

    4.2

    2

    ( ) 34083.3 26996.0cos(52.36 ) 8307.7sin(52.36 )...

    1416.7cos(104.72 ) 3608.3sin(104.72 )

    5833.3cos(157.08 ) 2333.3sin(157.08 ) ...

    p t t t

    t t

    t t N m

  • Dr.-Ing. Nantawatana Weerayuth Page 185

    4.4

    - Periodic

    (Impact Force)

    1. (Convolution Integral) Duhamel Integral

    2. (Laplace Transform)

    3. (Numerical Methods)

    (Convolution Integral)

    Periodic

    (Impulsive Force)

    0t

    (Impulse)

    2 1F t mx mx (4.22)

    (Impulse) F t F

    t t

    tFdt

    F (4.23)

    1 Unit Impulse

    0lim 1

    t t

    ttFdt Fdt

    f (4.24)

  • Dr.-Ing. Nantawatana Weerayuth Page 186

    (Unit Impulse) 1f 0t

    Dirac Delta Function

    ( ) ( )t t f f (4.25)

    F 0t

    ( )tF F (4.26)

    t ( )t

    1. ( ) 0t t

    2. 0

    ( ) 1t dt

    3. 0

    ( ) ( ) ( )F t t dt F

    4.4.1 (Unit Impulse Response) :

    - 4.12(a) 0t

    4.12 (b)

    4.12

  • Dr.-Ing. Nantawatana Weerayuth Page 187

    0mx cx kx (4.27)

    0 00( ) cos( ) sin( )

    nt nd d

    d

    x xx t e x t t

    (4.28)

    (4.28)

    2 n

    c

    m

    (4.29)

    n

    k

    m (4.30)

    2

    212

    d n

    k c

    m m

    (4.31)

    m 0x x 0t

    0t

    (0) (0 ) 1unit impulse mx mx f (4.32)

    0(@ 0) 0x t x (4.33)

    1(@ 0)x t

    m (4.34)

    (4.33) (4.34) (4.28)

    ( Unit Impulse Response) ( )g t Unit Impulse Response Function

    1( ) ( ) sin( )n

    t

    d

    d

    x t g t e tm

    (4.35)

    4.12(c)

  • Dr.-Ing. Nantawatana Weerayuth Page 188

    F

    (@ 0)x tm

    F

    (4.36)

    ( Impulse Response)

    ( ) sin( ) ( )nt dd

    x t e t g tm

    F

    F (4.37)

    F t 4.13(a)

    t (4.36) ( )x tm

    F

    4.13

  • Dr.-Ing. Nantawatana Weerayuth Page 189

    ( ( ) 0x t ) ( Impulse Response)

    ( )( ) sin( ( )) ( ) ;n

    t

    d

    d

    x t e t g t for tm

    FF (4.38)

    (4.38)

    t (4.37)

    (4.38) 0t

    4.13 (b)

    4.7 Load Cell

    (Impact Force) Impulse 4.14 (a)

    4.14

  • Dr.-Ing. Nantawatana Weerayuth Page 190

    5m kg, 2000k N/m 10c N-s/m

    Impulse 1 20F N-s 2 10F N-s 0t 0.2t

    Impulse 1 2( ) ( ) ( )F t t t F F

    4.14 (b)

    200020

    5n

    k

    m rad/s

    100.05

    2 2(5)(20)n

    c

    m

    2 21 20 1 (0.05) 19.975d n rad/s

    Impulse 0t ( 4.37)

    ( ) sin( ) ( )nt d

    d

    x t e t g tm

    F

    F

    0.05(20)

    1

    20( ) sin(19.975 )

    5(19.975)

    0.020025 sin(19.975 ) 0

    t

    t

    x t e t

    e t m for t

    E.1

    Impulse 0.2t s ( 4.38)

    ( )

    ( ) sin( ( )) ( ) ;nt

    d

    d

    x t e t g t for tm

    FF

    0.05(20)( 0.2)

    2

    ( 0.2)

    10( ) sin(19.975( 0.2))

    5(19.975)

    0.100125 sin(19.975( 0.2)) ; 0.2

    t

    t

    x t e t

    e t for t

    E.2

  • Dr.-Ing. Nantawatana Weerayuth Page 191

    Super Position

    ( 0.2)0.020025 sin(19.975 ) 0 0.2

    ( )0.020025 sin(19.975 ) 0.100125 sin(19.975( 0.2)) ; 0.2

    t

    t t

    e t for tx t

    e t e t for t

    E.3

    E.3 4.15

    4.15

  • Dr.-Ing. Nantawatana Weerayuth Page 192

    4.4.2 ( Response to a General Forcing

    Condition)

    ( )F t Periodic

    Function 4.16 ( Impulse )

    4.16

    t ( )F

    ( )F t t

    Impulse ( )F F

    ( ) ( ) ( )x t F g t (4.39)

    (0, )t ( )

    Impulse

    ( ) ( ) ( )x t F g t (4.40)

  • Dr.-Ing. Nantawatana Weerayuth Page 193

    (4.40) 0t

    0

    ( ) ( ) ( )

    t

    x t F g t d (4.41)

    ( )g t Unit Impulse Response Function (4.35)

    (4.41)

    ( )

    0

    1( ) ( ) sin( ( ))n

    t

    t

    d

    d

    x t F e t dm

    (4.42)

    (4.41) (4.42) Convolution Integral Duhamel

    Integral

    4.8 -

    (Step Function) 0F : ( 0( )F t F ) 4.17

    Duhamel Integral Convolution Integral

    4.17

  • Dr.-Ing. Nantawatana Weerayuth Page 194

    0F (4.42)

    ( )

    0

    1( ) ( ) sin( ( ))n

    t

    t

    d

    d

    x t F e t dm

    0( ) ; 0F t F t

    ( )0

    0

    ( ) sin( ( ))nt

    t

    d

    d

    Fx t e t d

    m

    ( )0

    2 2

    0

    0

    2

    sin ( ) cos ( )( )

    ( )

    11 . cos( ) ; 0

    1

    n

    n

    t

    t n d d d

    d n d

    t

    d

    F t tx t e

    m

    Fe t t

    k

    4.18

    1

    2tan

    1

    4.18

  • Dr.-Ing. Nantawatana Weerayuth Page 195

    4.9 -

    (Step Function) 0F 0t

    ( 0 0( ) ( )F t F t t ) 4.19

    4.19

    4.8

    0F 0t

    ( )0

    2 2

    0

    0

    2

    sin ( ) cos ( )( )

    ( )

    11 . cos( )

    1

    n

    n

    t

    t n d d d

    d n d

    t

    d

    F t tx t e

    m

    Fe t

    k

    0t

    t 0t t

    00 (

    0

    )

    02

    1( ) 1 . c ( )os( ) ;

    1

    n

    d

    t tt

    Ftx t e t t

    k

  • Dr.-Ing. Nantawatana Weerayuth Page 196

    4.10 -

    0F 00 t t 4.20 (a)

    4.20

  • Dr.-Ing. Nantawatana Weerayuth Page 197

    4.20(a) Step Function 1( )F t 0F

    0t Step Function 2 ( )F t 0F 0t t

    4.8 4.9

    0 0F F

    0

    0

    2

    0

    2

    ( )

    0

    1( ) 1 . cos( ) ...

    1

    11 . cos( ) ;(

    1)

    n

    n

    t

    d

    t

    d

    t

    Fx t e t

    k

    F

    kte t

    02

    0( ) cos( ( )) cos( )1

    nt

    d d t tF

    x t e tk

    1

    2tan

    1

  • Dr.-Ing. Nantawatana Weerayuth Page 198

    4.5

    (Laplace Transform)

    (Initial Conditions)

    ( )f t

    0

    ( ) ( ) ( )stf t e f t dt F s

    L (4.43)

    1 2 1 2( ) ( ) ( ) ( )f t f t F s F s L (4.44)

    (Linear Vibration System)

    (Linear Ordinary Differential Equation) 2

    1. Laplace Transform 1

    0

    00

    ( ) ( )( ( ))

    ( ) ( ) (0) ( )

    st

    st st

    df t df tf t e dt

    dt dt

    e f t se f t dt f sF s

    L L

    (4.45)

    ( )f t 0t (Initial Condition) (0)f

  • Dr.-Ing. Nantawatana Weerayuth Page 199

    2. Laplace Transform 2

    2

    2

    0

    0 0

    2

    0

    ( ( )/ )

    ( ) ( )( ( ))

    ( ) ( )

    ( )(0) (0) (0) ( )

    st

    st st

    st

    df t dt

    d f t d df tf t e dt

    dt dt dt

    df t df te se dt

    dt dt

    df tf s e dt f sf s F s

    dt

    L

    L L

    (4.46)

    ( )f t ( )f t 0t (Initial Condition) (0)f

    (0)f

    3. Laplace Transform n

    1 2 (1) ( 1)( ) ( ) ( ) (0) (0) ... (0)n

    n n n n n

    n

    Initial Conditions

    d f tf t s F s s f s f sf

    dt

    L L (4.47)

    ( )nf t n ( )f t

    4.11 1 DOF

    ( )mx cx kx f t

    (4.47)

    ( ( ))mx cx kx f t L L

    ) ( ) ( ( ( ))mx cx kx f t L L L L

    ) ( ) ( ( )m x c x k x F s L L L

  • Dr.-Ing. Nantawatana Weerayuth Page 200

    2 ( ) (0) (0) ( ) (0) ( ) ( )m s s sx x c s s x k s F s

    2 ( ) ( ) (0) (0)ms cs k s F s mx ms c x

    (0) (0) 0x x (Zero Initial Conditions)

    2 2 2( ) 1 1

    ( )( ) 2 n n

    sT s

    F s ms cs k m s s

    4.3

    4.3

  • Dr.-Ing. Nantawatana Weerayuth Page 201

    4.3

    (Laplace Transform : ( ) ( )f t F sL (S-Domain)

    (Inverse Laplace Transform : 1 ( ) ( )F s f t L )

    (Time Domain)

    11

    ( ) ( ) ( )2

    s i

    st

    s i

    F s e F s ds f ti

    L (4.48)

    (4.48)

    4.3

  • Dr.-Ing. Nantawatana Weerayuth Page 202

    4.12 ( )cx kx f t

    ( )f t bt 4.21

    4.21

    Ramp Function : ( )f t bt

    1 1( ) ( )

    a d

    kx x f t bt

    c c c

    x bx dt

    x bx dt L L

    2( ( ) (0)) ( )

    ds s x a s

    s

    (0) 0x

    2 2( )

    ( ) ( )

    d A B Cs

    s s a s a s s

  • Dr.-Ing. Nantawatana Weerayuth Page 203

    Partial Fraction Expansion 2( )s s a

    2 ( ) ( )As Bs s a C s a d

    2( ) ( )A B s Ba C Ca d

    2,

    dA

    a

    2

    dB

    a dC

    a

    2 2 2 2 2

    1 1 1 1 1 1( ) ( ) ( ) ( )

    ( ) ( )

    d d d d ds

    a s a a s a s a s a s a s

    4.3

    1 2( ) ( ) 1atd ds x t e t

    a a

    L

    Time Domain

    2( ) (1 )at

    dx t at e

    a

    2

    ( ) (1 )kt

    cc k

    x t t ek c

  • Dr.-Ing. Nantawatana Weerayuth Page 204

    4.13 - 4.22

    ( )f t 0(0)x x 0(0)x x

    4.22

    ( )mx kx F t ; 0t

    ( )mx kx F t L L

    ( )mx kx F t L L L

    2 ( ) (0) (0) ( ) ( )ms s msx mx k s F s

    2 2 2 2 2 2( ) (0) (0) 1 ( ) (0) (0)

    ( )F s msx mx F s sx x

    sk k kmms k ms k ms k

    s s sm m m

    2 2 2 2 2 21 ( ) (0) (0)

    ( )n n n

    F s sx xs

    m s s s

  • Dr.-Ing. Nantawatana Weerayuth Page 205

    4.3

    1 1 0 0

    2 2 2 2 2 2

    1 ( )( ) ( )

    n n n

    sx xF sx t s

    m s s s

    L L

    0 0

    1 1 10 0

    2 2 2 2 2 2

    cos( )sin( )

    1 ( )( )

    nn

    n

    n n n

    Convolution x t xt

    sx xF sx t

    m s s s

    L L L

    00

    0

    1( ) ( )sin( ( )) cos( ) sin( )

    t

    n n n

    n n

    xx t F t d x t t

    m