Chapter+6 Mechanics

8/10/2019 Chapter+6 Mechanics

http://slidepdf.com/reader/full/chapter6-mechanics 1/23

Feedback Control Systems. 6–

ROOT-LOCUS ANALYSIS

• Recall step response: we have seen that pole locations in the systemtransfer function determine performance characteristics; such as risetime, overshoot, settling time.

• We have also seen that feedback can change pole locations in thesystem transfer function and therefore performance is changed.

• Suppose that we have one variable parameter in our control system.Then, we make a parametric plot of pole locations as that parameterchanges. The poles are the roots of the denominator of the transferfunction ( a.k.a. the “characteristic equation.”) This plot is a plot of thelocus of the roots or the “ROOT LOCUS PLOT” (First suggested byEvans.)

VERY IMPORTANT NOTE : Root locus is a parametric plot (vs. K ) of theroots of an equation

1 +K b(s )a (s ) =0.

• A common control conguration is

r (t ) y( t )K G (s )

– Unity feedback, proportional gain.– We will generalize control conguration later.



ROOT-LOCUS ANALYSIS 6–

– Closed-loop transfer function

T (s ) =K G (s )

1 +K G (s ).

– Poles = roots of 1 +K G (s ) = 0.

• Assume plant transfer function G (s ) is rational polynomial:

G (s ) =b(s )a (s )

, such that

b(s ) = (s − z1)( s − z2) · · ·(s − zm) (b(s ) is monic .)

a (s ) = (s − p1)( s − p2) · · ·(s − pn ) n ≥m (a (s ) is monic .)

[a (s ) may be assumed monic without loss of generality. If b(s ) is notmonic, then its gain is just absorbed as part of K in 1 +K G (s ) =0]

• zi are zeros of G (s ) , the OPEN-LOOP transfer function.

• pi are poles of G (s ) , the OPEN-LOOP transfer function.

• CLOSED-LOOP poles are roots of equation

1 +K G (s ) = 0a (s ) +K b (s ) = 0,

which clearly move as a function of K .

• Zeros are unaffected by feedback.

EXAMPLE : G (s )

=1

s (s +2). Find the root locus.

• a (s ) = s (s +2) ; b(s ) =1.

• Locus of roots: (aside, stable for all K > 0 .)s (s +2) +K =0

s 2 +2s +K =0.




• For this simple system we can easily nd the roots.

s1,2 =−2 ±√ 4 −4 K 2

= −1

±√ 1

−K .

• Roots are real and negative for 0 < K < 1.

• Roots are complex conjugates for K > 1.

s1, K =0

s2, K =0K

=1

(s )

(s )

• Suppose we want damping ratio=0.707. We can recall that

s 2

+2ζ ω ns +ω2n = 0

s 2 +2s +K = 0K =2

or we can locate the point on the root locus where

| {poles }|=| {poles }|.| −1| = |√ 1 −K |

2

= K .

(or, K =0, not an appropriate solution).

How to Plot a Root Locus

• A value s =s1 is on the ( 180 ◦) root locus iff 1 +K G (s1) =0 for somereal value of K , 0 ≤ K ≤ ∞.




• [A value of s =s1 is on the ( 0◦ ) complementary root locus iff1 +K G (s1) =0 for some REAL value of K , −∞< K < 0.]

• So, K = −1G (s )

• Note, G (s ) is complex, so this is really two equations!

|K | =1

G (s )(6.1

G (s ) = −1K

. (6.2

• Since K is real and positive, K =0.

Therefore, G (s ) = 180 ◦ ± l360 ◦, l =0, 1, 2, . . .

• So once we know a point on the root locus, we can use themagnitude equation 6.1 to nd the gain K that produced it.

• We will use the angle equation 6.2 to plot the locus. i.e. , the locus ofthe roots = all points on s -plane where G (s ) =180 ◦ ± l360 ◦.

NOTES :1. We will derive techniques so that we don’t need to test every point on

the s-plane!

2. The angle criteria explains why the root locus is sometimes called the180 ◦ root locus.

3. Similarly, the complementary root locus is also called the 0◦ root

locus.

KEY TOOL : For any point on the s -plane

G (s ) = (due to zeros ) − (due to poles ).

EXAMPLE : G (s ) =(s − z1)

(s − p1)( s − p2).




θ p1θ p2θ z1

Test points1

(s )

(s )

G (s1) = θ z1 −θ p1 −θ p2 .

Locus on the Real Axis

• Consider a test point s1 on the real axis.

• If the point is right of all poles and zeros of G (s ) , then G (s ) =0. NOTON THE LOCUS .

z1 p1

Test points1

p2

¯ p2

(s )

(s )G (s ) = z1 − p1 − p2 − p

= 0 −0 − p2 − ¯ p2

= 0.

OBSERVATION : If the test point is on the real axis, complex-conjugateroots have equal and oposite angles which cancel and may beignored.

• If the test point is to the left of ONE pole or zero, the angle will be

−180 ◦ or +180 ◦ (= −180 ◦) so that point is on the locus.




• If the test point s1 is to the left of

– 1 pole and 1 zero: G (s1) =180 ◦ −(−180 ◦) =360 ◦ =0◦.– 2 poles: G (s1) = −180 ◦ −180 ◦ = −360 ◦ =0◦.–

2 zeros: G (s1) =180 ◦ +180 ◦ =360 ◦ =0◦.NOT ON THE LOCUS.

GENERAL RULE #1All points on the real axis to the left of an odd number of poles and zerosare part of the root locus.

EXAMPLE : G (s ) =1

s (s +4 +4 j)( s +4 −4 j).

(s )

(s )

EXAMPLE : G (s ) =s +8s +1

.

(s )

(s )




Locus Not on the Real Axis

OBSERVATION : Because we assume that system models arerational-polynomial with real coefcients, all poles must be either realor complex conjugate pairs. Therefore, THE ROOT LOCUS IS SYMMETRICAL WITH RESPECT TO THE REAL AXIS .

• What happens at K =0, ∞?

1 +K G (s ) =1 +K (s − z1)( s − z2) · · ·(s − zm )(s − p1)( s − p2) · · ·(s − pn ) = 0

(s − p1)( s − p2) · · ·(s − pn ) +K (s − z1)( s − z2) · · ·(s − zm ) = 0.

• At K =0 the closed loop poles equal the open-loop poles.

• As K approaches ∞(s − p1)( s − p2) · · ·(s − pn )

K +(s − z1)( s − z2) · · ·(s − zm) =0;the n poles approach the zeros of the open-loop transfer function,INCLUDING THE n −m ZEROS AT C ∞.

• Plug s = ∞into G (s ) and notice that it equals zero if m < n.

• The idea of ∞ in the complex plane is a number with innitemagnitude and some angle.

• To nd where the n −m remaining poles go as K → ∞, consider thatthe m nite zeros have canceled m of the poles. Looking back at theremaining n

−m poles (standing at C ∞), we have approximately

1 +K 1

(s −α) n−m,

or n −m poles clustered/centered at α .

• We need to determine α , the center of the locus, and the directionsthat the poles take.




• Assume s1 = Re jφ is on the locus, R large and xed, φ variable. Weuse geometry to see what φ must be for s1 to be on the locus.

• Since all of the open-loop poles are at approximately the same placeα

, the angle of 1 +K G (s

1) is 180◦ if the

n

−m

angles from α

to s

1 sumto 180.(n −m)φ l =180 ◦ +l360 ◦, l =0, 1, . . . , n −m −1

or

φ l =180 ◦ +360 ◦(l −1)

n −m, l =1, 2, . . . , n −m .

• So, if

n −m =1;φ = +180 ◦.

There is one pole going to C ∞along the negative real axis .

n −m =2;φ = ±90◦.

There are two poles going toC ∞ vertically .

n −m =3;φ = ±60◦, 180◦.

One goes left and the other twogo at plus and minus 60◦.

(etc)

• To nd the center, α , reconsider our approximation b(s )a (s ) ≈

1(s

−α) n−m

as |s | → ∞.

– Divide (by long division) b(s )a (s )

and 1

(s −α) n−m and match the top

two coefcients of s , the dominant ones.– An easier way...




– Note that roots of denominator of G (s ) satisfy:s n

+a 1s n−1 +a 2s n−2 +· · ·+a n =(s − p1)( s − p2) · · ·(s − pn )

a 1 = − pi .

– Note that the roots of the denominator of T (s ) are

s n

+a 1s n−1

+a 2s n−2

+· · ·+a n +K (s m

+b1s m−1

+· · ·+bm) =0.

If n −m > 1 then the (n −1)st coefcient of the closed loop systemis such that a 1 = − r i where r i are the closed-loop poles.

– We know that m poles go to the zeros of G (s ) , and assume the

other n

−m are clustered at

1

(s −α)n

−m

. Therefore, the asymptotic

sum of roots is −(n −m)α − zi .

– Putting this all together,

r i =(n −m)α + zi = pi

or

α

= pi − zi

(n −m).

GENERAL RULE #2

• All poles go from their open-loop locations at K =0 to:

– The zeros of G (s ) , or– To C ∞.

• Those going to C ∞ go along asymptotes

φl =180 ◦ +360 ◦(l −1)

n −m

centered at

α = pi − zi

n −m.



ROOT-LOCUS ANALYSIS 6–1

EXAMPLES :

Additional Techniques

• The two general rules given, plus some experience are enough tosketch root loci. Some additional rules help when there is ambiguity.(As in examples , )




Departure Angles, Arrival Angles

• We know asymptotically where poles go, but need to know how theystart, and how they end up there.

• Importance: One of the following systems is stable for all K > 0, theother is not. Which one?

(s )

(s )

(s )

(s )

• We will soon be able to answer this. Consider an example:

EXAMPLE : G (s ) =1

s (s +4 +4 j)( s +4 −4 j).

φ̄1

φ2

p1

¯ p1

p2

(s )

(s )

• Take a test point s0 very close to p1. Compute G (s0) .

• If on locus

−90◦ −φ1 −135 ◦ =180 ◦ +360 ◦l l =0, 1, . . .




whereφ̄1 = Angle from ¯ p1 to s0 ≈90◦.

φ1 = Angle from p1 to s0.

φ2

= Angle from p2 to s0

≈135 ◦.

φ1 = −45◦.

• Can now draw “departure of poles” on locus.

• Single-pole departure rule:

φdep = (zeros ) − (remaining poles ) −180 ◦ ±360 ◦l .

• Multiple-pole departure rule: (multiplicity q

≥1)

q φdep = (zeros ) − (remaining poles ) −180 ◦ ±360 ◦l .

• Multiple-pole arrival rule: (multiplicity q ≥1)

q ψ arr = (poles ) − (remaining zeros ) +180 ◦ ±360 ◦l .

• Note: The idea of adding 360◦l is to add enough angle to get theresult within

±180 ◦. Also, if there is multiplicity, then l counts off the

different angles.

Imaginary Axis Crossings

• Routh stability test can be run to nd value for K = K that causesmarginal stability.

• Substitute K and nd roots of a (s ) +K b(s ) =0.

• Alternatively, substitute K , let s = jω0, solve fora ( jω0) +K b( jω0) = 0. (Real and Imaginary parts)




Points with Multiple Roots

• Sometimes, branches of the locus intersect. (see , on pg.6–6–10).

• Computing points of intersection can clarify ambiguous loci.

• Consider two poles approaching each other on the real axis:

(s )

(s )

• As two poles approach each other gain is increasing.

• When they meet, they break away from the real axis and so K

increases only for complex parts of the plane.

• Therefore gain K along a branch of the locus is maximum atbreakaway point.

• Gain K is minimum along a branch of the locus for arrival points.

• Both are “saddle points” in the s-plane.

• So,d K ds = 0,

where 1

+K G (s )

=0

K = −1G (s )

or ,d ds

−1G (s ) s=s0

= 0 multiple root at s0.

[must verify that s0 is on the root locus. May be an extraneous result.]




• Some similar loci for which nding saddle points helps clarifyambiguity ...

(s )

(s )

(s )

(s )

(s )

(s )

(s )

(s )

Finding K for a Specic Locus Point

• Recall that the root locus is dened by the equation1 +K G (s ) =0,

which can be broken up into a magnitude-equation and aphase-equation (Eqs: 6.1 and 6.2).

|K

| =−1

G (s )

G (s ) = −1K

.

• The phase equation is used to plot the locus.




• The magnitude equation may be used to nd the value of K to get aspecic set of closed-loop poles.

• That is, K = −1G (s0)

is the gain to put a pole at s0, if s0 is on the locus.

Summary: Root-Locus Drawing Rules— 180 ◦ Locus

• The steps in drawing a 180 ◦ root locus follow from the basic phase denition. This is the locus of

1 +K b(s )a (s ) =0, K ≥0 phase of

b(s )a (s ) = −180 ◦ .

• They are

– STEP 1: On the s -plane, mark poles (roots of a (s )) by an × and zeros (roots of a (s )) with an ◦.There will be a branch of the locus departing from every pole and a branch arriving at every zero.

– STEP 2: Draw the locus on the real axis to the left of an odd number of real poles plus zeros.– STEP 3: Draw the asymptotes, centered at α and leaving at angles φ , where

n −m = number of asymptotes .

n = order of a (s )

m = order of b(s )

α = pi − zi

n −m = −a 1 +b1

n −m

φl =180 ◦

+(l

−1)360 ◦

n −m , l =1, 2, . . . n −m .

For n −m > 0, there will be a branch of the locus approaching each asymptote and departing toinnity. For n −m < 0, there will be a branch of the locus arriving from innity along eachasymptote.

– STEP 4: Compare locus departure angles from the poles and arrival angles at the zeros where

q φdep = ψ i − φ i −180 ◦ ± l360 ◦

q ψ arr = φi − ψ i +180 ◦ ± l360 ◦,

where q is the order of the pole or zero and l takes on q integer values so that the angles arebetween ±180 ◦ . ψ i is the angle of the line going from the i th zero to the pole or zero whoseangle of departure or arrival is being computed. Similarly, φ i is the angle of the line from the i thpole.

– STEP 5: If further renement is required at the stability boundary, assume s0 = jω0 andcompute the point(s) where the locus crosses the imaginary axis for positive K .

– STEP 6: For the case of multiple roots, two loci come together at 180◦ and break away at ±90◦Three loci segments approach each other at angles of 120 ◦ and depart at angles rotated by 60◦



ROOT-LOCUS ANALYSIS 6–1– STEP 7 Complete the locus, using the facts developed in the previous steps and making

reference to the illustrative loci for guidance. The loci branches start at poles and end at zeros orinnity.

– STEP 8 Select the desired point on the locus that meets the specications ( s0), then use themagnitude condition to nd that the value of K associated with that point is

K =1

|b (s0) / a (s0)|.

EXAMPLE :

r ( t ) y( t )K (s +1)( s +2)

s (s +4)

(s )

(s )

EXAMPLE : This example is NOT a unity-feedback case, so we need to

be careful.

r (t ) y( t )K s +2

s −10

s +3s +1

• Recall that the root-locus rules plot the locus of roots of the equation

1 +K b(s )a (s ) =0.




• Compute T (s ) to determine the characteristic equation as a functionof K .

T (s ) =K s+2

s−10

1

+K (s+2)( s+3)

(s

+1)( s

−10 )

=K (s +2)( s +1)

(s +1)( s −10 ) +K (s +2)( s +3)

• Poles of T (s ) at(s +1)( s −10 ) +K (s +2)( s +3) =0

or

1 +K (s

+2)( s

+3)

(s +1)( s −10 ) =0.

(s )

(s )

1) “Open loop” poles and zeros:

2) Real axis

3) Asymptotes

4) Departure angles

5) Stability boundary: Note, characteristic equation

= (s +1)( s −10 ) +K (s +2)( s +3)

= s 2 −9s −10 +K s 2 +5 K s +6K

= ( K +1)s 2

+(5K −9)s +6K −10




Routh Arrays 2 K +1 6K −10 K > −1

s 1 5K −9 K >95 = 1.8 ← stability criterion

s 0 6K −10 K >106 =1.66

When K =9/ 5, the s1 row of the Routh array is zero—top rowbecomes a factor of the characteristic equation. Imag.-axis crossingswhere

9

5 +1 s 2

+6

9

5 −10

=0

14s 2

+4 = 0 . . . s = ± j 27

.

(s )

(s )

6) Breakaway points

K (s ) = −(s +1)( s −10 )(s +2)( s +3)

=−(s 2 −9s −10 )

s2

+5s +6d ds

K (s ) = −(s +1)( s −10 )( 2s +5) −(−2s +9)( s +2)( s +3)(den )2 =0

= −2s 3 +18s 2 +20s −5s 2 +45s +50 −−2s 3 −10s 2 −12s +9s 2 +45s +54 =0




= (−2 +2)s 3 +(18 −5 +10 −9)s 2 +(20 +45 +12 −45 )s +(50 −54 ) = 0

= 14s 2 +32s −4 =0

s =−32 ± 322

−4(14 )( −4)28 =−32 ±√ 1248

28roots at {0.118 , −2.40}. Now we can complete the locus:

(s )

(s )

EXAMPLE :

r (t ) y( t )K 1

s +as +1

s 2

• Two design parameters: K , pole location −a .

G (s ) =s +1

s 2(s +a ).

• Test a =2, a =50 , a =9.

1) Poles and zeros of G (s ) .

(s )

(s )

2) Real axis.




3) Asymptotes:n −m =2

α = pi − zi

2 =0 +0 +(−a ) −(−1)

2 =1 −a

2

α = {−0.5, −24 .5, −4}φ l =

180 ◦ +360 ◦(l −1)2 = ±90◦.

4) Departure angles for two poles at s =0.

2φdep = (zeros ) − (remaining poles ) −180 ◦ −360 ◦l

= 0 −0 −180 ◦ −360 ◦l

φdep = −90◦ − 360 ◦2

l = ±90◦.

5) Will always be stable for K > 0.

6) Breakaway points:

K (s ) = −s 2(s +a )s

+1

d ds

K (s ) = −(s +1)( 3s 2 +2as ) −s 2(s +a )( 1)

(s +1)2 =0

= s 2s 2

+(a +3)s +2a = 0

breakaway at 0, −a +3 ± (a +3)2 −4(2)( 2a )2(2)

sa={2,50 ,9} = −1.25 ± j 7/ 16

not on locus, (−2.04 and −24 .46 ), −3 and 0




(s )

(s )

a =2 (s )

(s )

a =9

(s )

(s )

a =50

Matlab and Root Loci

• If G(s ) =b(s )a (s ) =

b0s m +b1s m−1 +· · ·+bm

a 0s n

+a 1s n−1

+· · ·+a n

.

b = [b0 b1 b2 . . . bm];

a = [a0 a1 a2 . . . an];

rlocus(b,a); % plots the root locus

• Also, k=rlocfind(b,a); returns the value of K for a specic point on theroot locus (graphical, with a mouse).

• Matlab is also able to draw space aliens!z=[-5+5j -5-5j -4.8+2j -4.8-2j 0.35+0.2j 0.35-0.2j ...

2j -2j 2j -2j 0.7 0.7 1.0+1.5j 1.0-1.5j 1.0+1.5j 1.0-1.5j 2.2];

p=[-4.5+3.2j -4.5-3.2j -0.15+0.8j -0.15-0.8j 0.3j -0.3j 0.4+0.3j ...

0.4-0.3j 0.45+0.6j 0.45-0.6j 1+1.3j 1-1.3j 1+1.3j 1-1.3j 2+0.15j ...

2-0.15j 2];

num=poly(z); den=poly(p); rlocus(num,den);




−6 −5 −4 −3 −2 −1 0 1 2 3−6

−4

−2

0

2

4

6

Real Axis

I m a g

A x

i s




• A system with a time delay has the formG (s ) = e−τ d s G (s )

where τ d is the delay and G (s ) is the non-delayed system.

• This is not in rational-polynomial form. We cannot use root locustechniques directly.

METHOD 1:

• Approximate e−τ d s by

b0s +b1

a 0s +1, a polynomial.

• Padé approximation.

e−τ d s ≈1 −(τ d s / 2)1 +(τ d s / 2)

First-order approximation

≈1 −τ d s / 2 +(τ d s )2/ 12

1 +τ d s / 2 +(τ d s )2/ 12Second-order approximation.

≈ 11 +τ d s

Very crude.

• Extremely important for digital control!!!

METHOD 2:

• Directly plot locus using phase condition.

• i.e. , G (s ) if s =σ + jω .

• i.e., look for places where

G (s ) = (−τ d ω) +G (s ) =180 ◦ +τ d ω +360 ◦l .

• Fix ω, search horizontally for locus.

Extensions to Root Locus - Time Delay

Chapter+6 Mechanics

Documents

Transcript of Chapter+6 Mechanics