2 First and Second Order Linear Differential Equations -...

download 2 First and Second Order Linear Differential Equations - nptel.ac.in/courses/101108047/module8/Lecture 17.pdfhomogeneous linear differential equations. However, if 0 (Homogeneous)

of 34

  • date post

    14-Mar-2018
  • Category

    Documents

  • view

    221
  • download

    0

Embed Size (px)

Transcript of 2 First and Second Order Linear Differential Equations -...

  • Lecture 2First and Second Order

    Linear Differential Equations

    Dr. Radhakant PadhiAsst. Professor

    Dept. of Aerospace EngineeringIndian Institute of Science - Bangalore

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    2

    Linear Differential Equations

    1 2

    1 (Non homogeneous)1 cos and 1 sin both are solutions.

    x xx t x t+ == + = +

    Principle of superposition:If )(1 tx and )(2 tx are solutions then )()( 2211 txtx +is also a solution for any scalars 1 and 2

    Example:

    But2(1 cos ) or [(1 cos ) (1 sin )] are not solutions.t t t+ + + +

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    3

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    4

    Linear Differential Equations

    Lesson: Principle of superposition holds good only for

    homogeneous linear differential equations.

    However, if

    0 (Homogeneous) then both of the above are solutions.x x+ =

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    5

    Homogeneous First-order EquationsSystem dynamics:

    Solution:

    Initial condition:

    Hence,

    0 00, ( ) , 1/x kx x t x k + = = =

    ( )

    ( )

    /ln lnln /

    k t

    dx x k dtx k t cx c k t

    x e c

    =

    = +

    =

    =0 0

    0 0,kt ktx e c c e x= =

    0( )0( )

    k t tx t e x =

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    6

    Note:

    ( )

    2 2 3 3

    0

    /0 0

    1) 12! 3!

    2) If 0, then the solution is

    ( )

    at

    tk t

    a t a te at

    t

    x t e x e x

    = + + + +

    =

    = =

    Homogeneous First-order Equations

    (no overshoot)t

    0x

    00.63 x

    (0,0)

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    7

    Non-homogeneous First-order Equations: Response to Step Inputs

    1 ( : constant)

    Homogeneous solution:1 0

    Particular solution:

    1Substitute:

    h h

    t

    h

    p

    p

    x x A A

    x x

    x e cx B

    B A

    x B

    + =

    + =

    =

    =

    =

    = A=

    Consider:

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    8

    Non-homogeneous First-order Equations: Response to Step Inputs

    ( )

    0 0

    0

    ( )

    ( )

    ( )lim ( ) (Note: )

    h p

    t

    t

    ss sst

    x t x x

    x t e c Ax c A c x A

    x t e x A Ax x t A x A

    = +

    = += + =

    = +

    = =

    ( )0Complete solution: ( )t

    x t e x A A

    = +

    [ ] ( )Steady state error ( . input ): ( ) 1ss t

    wrt Ae A x t A A A

    = = =

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    9

    First-order Equations in Special Form: Unit Step Response

    ( )1 1 Special case in text books: 1

    Homogeneous:1 0

    Particular solution:

    1 1Substitute:

    h h

    t

    h

    p

    x x A A

    x x

    x e c

    x B

    B A

    + = =

    + =

    =

    =

    =

    ( )

    p

    B A

    x B A

    =

    = =

    Consider:

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    10

    First-order Equations in Special Form: Unit Step Response

    ( )

    0 0

    0

    ( )

    ( )

    ( )lim ( )

    h p

    t

    t

    ss t

    x t x x

    x t e c Ax c A c x A

    x t e x A Ax x t A

    = +

    = += + =

    = +

    = =

    ( )0Complete solution: ( )t

    x t e x A A

    = +

    Steady state error ( . input ): 0ss sswrt A e A x A A= = =

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    11

    Unit Step Response of a First-Order System

    Settling time: 4 4sT T = =

    T =1A =

    ( )x t

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    12

    Second Order System

    Homogeneous:2

    2

    2

    2

    1,2

    0, ( 2 , )( )

    ( ) 00

    = 0 : characteristic equation

    4 2

    n n

    t

    t

    t

    x ax bx a bx t e

    a b ee

    a b

    a a b

    + + = = =

    =

    + + =

    + +

    =

    Guess:

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    13

    Second Order SystemNote: , Arises three cases

    a b

    R

    Case 1: Distinct and real roots

    Case 2: Complex conjugate roots

    Case 3: Real and repeated roots

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    14

    Case-1: Distinct and real roots

    1 2

    1 2

    1 2

    1 1 2 2

    1 2

    ,( ) ( ) ( )

    ( )

    t t

    t t

    x e x ex t c x t c x t

    x t c e c e

    = == +

    = +

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    15

    Case-2: Complex conjugate roots

    1 2

    ( )1

    ( )2

    ( ) ( )1 2

    ,(c o s s in )(c o s s in )

    ( )

    j t t

    j t t

    j t j t

    j jx e e t j tx e e t j tx t c e c e

    +

    +

    = + =

    = = +

    = =

    = +

    However notice that:1 2

    1 2

    c o s2

    s i n2

    t

    t

    x x e t

    x x e tj

    +=

    =

    Linearly independentterms

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    16

    Case-2: Complex conjugate roots

    ( )

    1 2 1 2( )2 2

    ( ) cos sint

    x x x xx t A Bj

    x t e A t B t

    + = +

    = +

    So, we can also write the solution as :

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    17

    Complex conjugate roots: Typical response plot

    ( )x t

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    18

    Case-3: Real and repeated roots

    tetx

    =

    ==

    )(1

    21How to find )(2 tx

    Method of variation of parameters

    2 1

    2 1 1

    2 1 1 1 1

    Assume: ( ) ( ) ( )Then

    x t u t x t

    x u x u xx u x u x u x u x

    =

    = += + + +

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    19

    Case-3: Real and repeated roots

    ( ) ( )

    ( ) ( )

    ( )

    1 1 1 1 1

    1 1 1

    1 1 1

    1 1

    1 1

    2 0

    However, 0 since is a solution

    Moreover, since 2 2

    2 0

    This leads to 0. Moreover 0. Hence

    t

    t

    u x a x b u x a x u x

    x a x b x

    a ax e x x

    x a x

    u x x e u

    + + + + + =

    + + =

    = = = =

    + =

    = = 0=

    Substituting back and rearranging the terms, it leads to

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    20

    Case-3: Real and repeated rootsOne solution

    ttetxttxttu

    ===

    )()()(

    12

    )()()( 2211 txctxctx +=

    1 2( ) ( )tx t c c t e= +

    += ttcctx as)( 21Corollary: When 0= (double poles at origin)

    Double pole (in general, multiple poles) at origin is de-stabilizing..!

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    21

    Summary

    Real double Roots 3

    Complex Conjugate 2

    Distinct & Real 1

    General SolutionBasicsRoots of characteristic equationCase

    21 ,

    j+=

    tt ee 21 , tt ecec 21 21 +

    tete

    t

    t

    sincos )sincos( tBtAe t +

    tt tee ,tetcc )( 21 +

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    22

    Example

    044 =+ xxx 1)0(,3)0( == xx2 4 4 0 2, 2 + = = (Real, repeated)

    tetcctx 221 )()( +=tt ecetcctx 22

    221 )(2)( ++=

    53 21 == cctettx 2)53()( =

    Solving for the initial conditions

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    23

    Particular Solution: Method of Undetermined Coefficients

    Choice of xp(t)Terms in f(t)ptek ptecntk 11 1 0n nn nk t k t k t k+ + + +

    tk cos tBtA sincos +

    tk sin tBtA sincos +

    Substitute these expressions back in the differential equation and determine the unknown coefficients

    )(tfbxxax =++

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    24

    Example

    Find )(tx p for

    Choose tp cektktx3

    01 )()( ++=

    21,3,2 01 === ckk

    Substitute and equate the coefficients

    tp ettx

    3

    21)32()( ++=

    tetxxx 3423 +=+

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    25

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    26

    Second order system in Standard form

    ( )( )

    2 2

    2

    2 2

    2: Natural frequency: Damping ratio

    Transfer function (will be studied later):

    2

    n n n

    n

    n

    n n

    c c c u

    C sU s s s

    + + =

    =+ +

    Roots: poles

  • ADVANCED CONTROL SYSTEM DESIGN Dr. Radhakant Padhi, AE Dept., IISc-Bangalore

    27

    Unit Step Response of aSecond-order System

    =0

    0