Lecture Wk07 No1

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    ENGN3223-Control System:

    Root-Locus Design

    Department of Engineering

    Australian National University

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    ENGN3223, 2009

    Controller Design using Root locus

    We will now look at a graphical approach, known as the rootlocus method, for designing control systems

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    ENGN3223, 2009 3/37

    This slide is also from Week 4

    What will happen in pole locationsas K increases?

    Lets try some cases, K=0, s=0, -2

    K=1, s=-1, -1

    K=2, s=-1j1

    K= , s=-1j

    So we can predict the timeresponses now.

    H(s) =K

    s2+ 2s+ K

    =

    K

    (s+1)2+ (K1)

    -2

    more overshoot

    faster rise time

    0

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    ENGN3223, 2009 4/37

    Selecting the proportional feedback gain

    1. The time-domainspecifications can be

    converted into s-plane ones:

    2. The locations of the poles are

    3. Thus,

    tr 1.8/n

    Mp 15 (%)

    ts 4.6/

    n1.8/ t

    r=1.8/1.2 =1.5 (rad/s)

    0.5

    4.6/ ts= 4.6/5 = 0.92

    n =1.5

    =0.92

    =0.5

    -1-2

    1 K1 3

    1

    1 i K1

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    ENGN3223, 2009 5/37

    Selecting the proportional feedback gain

    1. The time-domainspecifications can be

    converted into s-plane ones:

    2. Draw a root locus (thisis the goal of this week)

    3. Look at overlap of 1 and 2.That gives you an appropriatefeedback gain K.

    tr 1.8/n

    Mp 15 (%)

    ts 4.6/

    n1.8/ t

    r=1.8/1.2 =1.5 (rad/s)

    0.5

    4.6/ ts= 4.6/5 = 0.92

    n =1.5

    =0.92

    =0.5

    -1-2

    1

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    ENGN3223, 2009

    Rewrite the system

    +-

    G(s)

    KH(s)

    Closed-loop transfer function Y(s) =L(s)K

    1+ L(s)KR(s)

    +-

    G(s)H(s)

    = L(s)

    K

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    ENGN3223, 2009

    What is a root locus?

    T(s) =L(s)K

    1+ L(s)K

    +-

    G(s)H(s)

    = L(s)

    K

    The root locus is defined by the characteristic equation of T(s)

    1+KL(s) = 0

    Now we want to see the behavior of the closed-loop transferfunction T(s) as a function ofK

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    ENGN3223, 2009

    Example

    T(s) =L(s)K

    1+ L(s)K=

    K

    s2+ 2s+K

    +-

    G(s)H(s)

    = L(s)

    K

    G =1

    s+ 2

    H=1

    s

    1st order

    Integral

    Gain K

    Closed loop transfer function

    The root locus is defined by

    s2+ 2s+K= 0

    s = 1 1K

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    ENGN3223, 2009 9/37

    This slide is also from Week 4

    What will happen in pole locationsas K increases?

    Lets try some cases, K=0, s=0, -2

    K=1, s=-1, -1

    K=2, s=-1j1

    K= , s=-1j

    So we can predict the timeresponses now.

    H(s) =K

    s2+ 2s+ K

    =

    K

    (s+1)2+ (K1)

    2

    more overshoot

    faster rise time

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    ENGN3223, 2009

    Remark

    +-

    G(s)

    KH(s)

    Closed-loop transfer function Y(s) =G(s)

    1+ L(s)KR(s)

    The transfer function is different from the unit feedback.

    However, the denominator is the same.

    The system has the same poles and the same root locus.

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    ENGN3223, 2009

    Lets see the following properties

    of root locus

    (1)K=0(2) K

    (3) |s|Asymptotes

    (4) Multiple roots(5) Symmetry

    (6) Locus on the real axis

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    Properties of root locus

    T(s) =KL

    (s

    )1+KL(s)

    L(s) B(s)

    A(s)

    The characteristic equation (denominator=0)

    1+ KB(s)

    A(s)= 0

    The root locus is defined by this equation as a function of K.

    We want to see how to draw the root locus when K:0

    Closed loop transfer function

    Let us defineA(s) = s

    n+ a

    1sn1

    + = (s p1)(s p2)(s pn )

    B(s) = sm+ b

    1sm1

    + = (sz1)(sz

    2)(szm )

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    Property (1) (2)

    (K

    0) the root locus starts from the poles of L(s)=G(s)H(s).

    (K)the root locus goes to zeros of L(s)=G(s)H(s).

    1+KB(s)

    A(s)= 0

    A(s)+KB(s) = 0 K0 A(s) = 0

    1

    KA(s)+ B(s) = 0

    K B(s) = 0

    A(s) = sn+ a

    1sn1

    + = (s p1)(s p2)(s pn )

    B(s) = sm + b1sm1 +

    = (sz1)(sz2)

    (szm )

    pi : pole

    zi : zero

    For causal systems, mn

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    Property (1) (2)

    Check the real-line whether it is a part of the root-locus withthe starting and ending points of the CL poles

    K=0K=0

    K=K=

    pole

    zero

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    Properties of root locus

    A(s) = sn+ a

    1

    sn1

    + = (s p1

    )(s p2

    )(s pn

    )

    B(s) = sm+ b

    1sm1

    += (sz1)(sz

    2)(szm )

    pi:

    polezi : zero

    For causal systems, mn

    s 0 = A + KB sn + a1sn1

    + K(sm+ b

    1sm1

    )

    K=sn+ a1s

    n1

    s

    m

    + b1s

    m1

    = snm s

    m+ a1s

    m1

    sm+ b1s

    m1

    = snm

    1+a1

    s

    1+b1

    s

    snm 1+a1 b1s

    s+a1 b1n m

    nm

    1

    1+ 1

    (1+ )n

    1+ n

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    Asymptotes of root locus

    K= s+ a1 b1n m

    nm

    s = a1 b

    1

    n m+ (K)

    1

    nm

    =

    a1 b

    1

    n m +K

    1

    nm

    e

    i2l+1

    nm

    Equation (1) give the asymptotes of the root locus.

    In fact, Equation (1) represents (n-m) lines from

    l= 0,1,

    n m 1

    a1 b

    1

    n m

    (1)

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    Asymptotes of root locus

    Equation (1) give the asymptotes of the root locus.

    In fact, Equation (1) represents (n-m) lines from a1 b1n m

    =

    (poles) (zeros)n m

    A(s) = sn+ a

    1sn1

    + = (s p1)(s p

    2)(s p

    n

    )

    B(s) = sm+ b

    1sm1

    += (sz1)(sz

    2)(szm )

    pi: pole

    zi : zero

    From this expression, we have

    a1 = (p1 + p2 ++ pn ) = (poles)

    b1 = (z1 + z2 ++ zm ) = (zeros)

    Thus, we can also express as

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    Lets see the following properties

    of root locus

    (1)K=0(2) K

    (3) |s|Asymptotes

    (4) Multiple roots(5) Symmetry

    (6) Locus on the real axis

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    Multiple root

    (s p)2= 0 s = p

    d

    ds(s p)

    2= 0 s = p

    A simple exampleLet p be a multiple root

    A(s)+KB(s) = 0K= A(s)

    B(s)

    A'(s)+KB'(s) = 0

    A'(s)A(s)

    B(s)B'(s) = 0

    A'(s)

    A(s)B

    '(s)

    B(s)= 0

    1

    s pll=1

    n

    1

    szll=1

    m

    = 0

    The multiple root can be foundfrom these two equations

    We apply this idea to

    the characteristic

    equation

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    ENGN3223, 2009

    Lets see the following properties

    of root locus

    (1)K=0(2) K

    (3) |s|Asymptotes

    (4) Multiple roots(5) Symmetry

    (6) Locus on the real axis

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    Symmetry

    The root locus is always symmetric with respect to the realaxis.

    This is trivial because the roots of the characteristic

    equation are of the form

    s =Re iIm

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    ENGN3223, 2009

    Lets see the following properties

    of root locus

    (1)K=0(2) K

    (3) |s|Asymptotes

    (4) Multiple roots(5) Symmetry

    (6) Locus on the real axis

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    Root Locus on the real axis

    From the characteristic equation we have1+KB(s)

    A(s)= 0

    L(s) = K( ) =

    -

    -

    L(s) =(s0+ z)(s

    0+ p)

    pole

    zero

    L(s) = () ()

    =

    contribution from symmetric poles & zeros = 0

    |#{zeros}-#{poles}|=odd

    =L(s)

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    Root Locus on the real axis

    If there are more poles than zeros, (n-m) > 0 is called the poleexcess, then there are (n-m) branches of the root locus that

    diverge to infinity (zeros at infinity).

    K=0K=0

    K=K=

    n-m=1 (there is a zero at infinity)

    K=

    pole

    zero

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    Example

    L(s) = (s+ 3)s(s+1)(s+ 2)(s+ 4)+-

    pole

    zero

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    Example

    L(s) = (s+ 3)s(s+1)(s+ 2)(s+ 4)

    pole

    zero

    (1) K=0(2) K

    (3) |s|Asymptotes

    (4) Multiple roots

    (5) Symmetry

    (6) Locus on the real axis

    s =

    a1 b

    1

    n m+ K

    1

    nm

    e

    i2l+1

    nm

    = 7 3

    4 1+ K

    1

    41ei2l+1

    41

    = 4

    3+K

    1

    3ei2l+1

    3

    =

    sm+ b

    1sm1

    +

    sn+ a

    1sn1

    +

    l= 0,1,2where

    This represents three lines (red)

    -4/3

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    Example

    L(s) = (s+ 3)s(s+1)(s+ 2)(s+ 4)

    pole

    zero

    (1) K=0(2) K

    (3) |s|Asymptotes

    (4) Multiple roots

    (5) Symmetry

    (6) Locus on the real axis

    -4/3

    The root locus starts from the four poles.

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    Example

    L(s) = (s+ 3)s(s+1)(s+ 2)(s+ 4)

    pole

    zero

    (1) K=0(2) K

    (3) |s|Asymptotes

    (4) Multiple roots

    (5) Symmetry

    (6) Locus on the real axis

    -4/3

    The root locus ends at the four zeros.

    (three are at infinity)

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    ENGN3223, 2009

    Example

    (1) K=0(2) K

    (3) |s|Asymptotes

    (4) Multiple roots

    (5) Symmetry

    (6) Locus on the real axis

    |#{zeros}-#{poles}|=odd (to the right)

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    Example

    (1) K=0(2) K

    (3) |s|Asymptotes

    (4) Multiple roots

    (5) Symmetry

    (6) Locus on the real axis

    |#{zeros}-#{poles}|=odd (to the right)