7.1 Root Locus (RL) PrincipleWe introduce the RL through an example.Consider servo motor system shown bellow

The closed loop transfer function is

KssK

ssK

ssK

sT

2

)2(1

)2()( 2

motorcompensator

K)2(

1ss

R(s)A(s)

–

+

The CE iss2 + 2s + K = 0

The system is stable for K > 0. It is not evident how K affect the transient response.

To have more understanding how K affect the system characteristic we will plot the locus of the roots in the s plane as K is varied from 0 to infinity. The roots are

KKs

112

442

K=0. K=0.

s

K=1–1–2

K=2

K=2

450

j

–j

RL is a plot of the closed loop poles as some parameters of the system is varied.

7.1 Root Locus Principle

We generally consider a system as in the following figure

The CE of this system is

1+KG(s)H(s)=0 (1)A value s1 is a point in the RL if and only if satisfies (1) for real 0<K<.

For this system we call KG(s)H(s) the open loop function (OLF)

Equation (1) can be written as

K= –1/G(s)H(s) (2)

Since G(s) and H(s) are generally complex and K is real then there are 2 condition must be met

|G(s)H(s)|=1/|K| and (3)

G(s)H(s) = r, r = ±1, ±3, ±5 … (4)

We call (3) the magnitude criterion of the RL, and (4) the angle criterion.

Any value s on the RL must meet both criterions.

K–

+ G(s)

H(s)

s

G(s) includes plant and compensator TF. The closed loop TF is

)()(1)()(

sHsKGsKGsT

For 2nd order system the RL appears as a family of 2 paths or branches traced out by 2 roots of CE.

7.1 Root Locus PrincipleTo illustrate this criterion let us consider

))(()()()(

21

1

pspszsKsHsKG

s1

1

s1- z1

2

s1- p13

s1- p2s

z1 p2 p1

Suppose that s1on the RL. Thus the angle condition becomes

1 – 2 – 3= ±

The locus of all points meet this condition forms the complete RL of the system

From the preceding discussion, it is seen that the condition of a point to be on the RL is that

(all zero angles) – (all pole angles) = r

where zero angle is the angle of s – zk and pole angle is the angle of s – pk . zk and pk are zero and pole of the OLF

Calculating and plotting RL digitally is available and it is a convenient way to do it.

However a good knowledge of the rule for plotting the RL will offer insight effect of changing parameter and adding poles and/or zeros in the design process.

From mason gain formula, the CE is

(s) = 1 + F(s) = 0

where F(s) is the OLF.

Many design procedures are based on the OLF.

7.2 Root Locus RulesSince the TF is real function then its complex roots must exist in conjugate pairs, then RL is symmetrical with respect to the real axis

Rule 1.RL is symmetrical with respect to the real axis

With the zk and pk are zeros and poles of the OLF. The CE may be expressed as

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

21

11

pspszszsKbsHsKG m

0))(())(( 1121 zszsKbpsps m

Therefore for K = 0 the roots of CE are simply the poles of OLF RL start at the poles of G(s)H(s).

It is always that the number of zeros, NZ,, equals the number of poles, NP.

If the number of finite zeros < NP then the rest of zeros lies at infinity. If K but s remain finite then the branches of RL must approach the zeros of the OLF, else there must be zeros at infinity. This mean that RL finish at zeros of G(s)H(s).

Rule 2.RL originates on the poles of G(s)H(s) for K=0 and terminates on the zeros of G(s)H(s) as K, including zeros at infinity.

When there are zeros at infinity then the RL will extend to infinity as well. The asymptotes can be determined. If there is zeros at infinity then the OLF can be written as

11

11()()(

mn

m

mm

mm

sassbsbKsHsKG

7.2 Root Locus Rules

11

11()()(

mn

m

mm

mm

sassbsbKsHsKG

As s , the polynomial becomes

sKbsHsKG m

sslim)()(lim

Remember that the values of s must satisfy the CE 1-KG(s)H(s)= 0 regardless the value of s hence we have

0lim1)()(lim1 sKbsHsKG m

ss

hence we can write for large s

s + Kbm=0

The angle of the roots are principal values of angle (asymptote angles)

= r/ r = ±1, ±3,…

The asymptote intersect the real axis at

asymptote angle

0 No asymptote

1 1800

2 ±900

3 ±600,1800

4 ±450, ± 1350

zpa NN

ZP

Where P is the finite poles and Z is thefinite zeros

7.2 Root Locus RulesRule 3If OLF has zeros at infinity, the RL approach asymptote as K approach infinity. The asymptote intersect real axis at a with angle .

Example.Consider the OLF

)2()()(

ssKsHsKG

There are 2 zeros at infinity, hence there are 2 asymptotes with angle ±900.

The asymptote intersect the real axis at

102

0)]2(0[

zp

a NNZP

Consider point a to the right of p1. The angle condition states that if a is on the RL then

(all zero angles) – (all pole angles) = r.

Since this not the case then any points on the real axis to the right of p1 are not on the RL.By the same argumentations we conclude that the points between p1 and p2 are on the RL, the points between p2 and z1 are not on the RL,the points to the left of z1 are on the RL.Rule 4The RL includes all points to the left of an odd number of real critical frequencies (poles and zeros)

7.3 Additional Techniques

a

z1p2 p1

s

-

7.3 Additional Techniques

Multiple roots on the real axis.

If CE has multiple roots a on the real axis then we can write the characteristic polynomial as 1+KG(s)H(s) =(s-a)kQ(s) (1)Differentiating (1) with respect to s gives

For s =a it must be

)(')()()( 1 sQassQask kk

0)()()()( sHsGdsdsHsKG

dsd

(2)

Hence the multiple roots a can be found by solving (2).

Since G(s)H(s) is rational we can represent G(s)H(s)= N(s)/D(s)

And (2) can be written as

N(s)D’(s)- N’(s)D(s)=0 (3)

The multiple points are breakaway points, they are points at which branches of RL leave or enter the real axis.

Rule 5.The breakaway points on RL will appear among the roots of polynomial obtained either from (2) or (3).

Since on the RL

K = –1/ G(s)H(s) = – D(s)/N(s)

we conclude that

K’= N(s)D’(s)- N’(s)D(s)=0

Hence at breakaway points K is either maximum or minimum.

7.3 Additional Techniques

Angle of Departure from Complex Poles

Suppose the open loop poles are as shown in the figure below, and s1 is on the RL.

1

2

s1

3

p1

p2

p3

We are interested in finding the angle 1 when is very small. This is the angle of departure.From angle criterion we know that

– 1 – 2 – 3=1800

For very small , 2=900, hence

1= 1800 – 900 – 3

Rule 6Loci will depart from a pole pj (arrive at zj) at angle of d(a) where

d = zi pi + r

a = pi zi + r

7.3 Additional Techniques

Example 1

We consider a system with OLF

)3)(2)(1()()(

sssKsHsKG

-4/3-0.132j

1-1-2-3

-j

with 3 finite poles and 3 infinite zeros.

Rule 2. The RL start from pole s =1, s = – 2, and s = – 3.

Rule 3. The asymptote intersect the real axis

at (1-2-3)/(3-0) = – 4/3, with angle ±600, and 1800.

Rule 4. The RL occurs on the real axis for – 2< s <1

and s<–3.

As K is increased p1 and p3 move left, but p2 move right.

Rule 5. Determine the breakaway point by differentiating (1) and set the result to zero.

3s2+8s+1=0 s1 = -0.132 and s2 = -2.54 (ignore s2)

p1p2p3

(1)

Routh-Hurwitz method can be used to find values of K for stability, this is found to be 6 < K< 10.

K=10

K=6

7.3 Additional Techniques

Example 2

2

)1()()(ssKsHsKG

-1-2

The RL start from double pole s =0 and stop at zero s = 1 and s =

One asymptote at 1800.

The RL occurs on real axis for s < 1

Breakaway point at s = 0 and s = 2

Example 3A simplified ship steering system modeled as a system of order 2

ssKsHsKG

1.0)()( 2

The RL start from s =0 and s =0.1 and stop at s = .

Two asymptotes at 900 intersect the real axis at s =0.05

The RL occurs on real axis for 0.1< s <1

Breakaway point at s = 0.05

1 0.05 0.1

7.3 Additional TechniquesExample 4The order-3 model of ship steering system is

sssK

sssKsHsKG a

2.01.22.01.201.0)()( 223

The RL start from poles s =0, s =0.1, & s =2 and stop at 3 zeros at s =

3 asymptotes intersect real axis at s =0.7

The real axis RL, for 0.1<s<0 and s<2

Breakaway point at s = 0.0494

1 2 0.1s =0.7

j0.4472Ka=42

aKsssKGH 01.02.01.201 23

The CE is

The Routh array is s3 1 0.2 s2 2.1 0.01Ka

s1 (0.420. 01Ka)/2.1 Ka< 42

s0 0.01Ka Ka> 42

For stability 0<Ka<42.

The auxiliary polynomial for Ka=42 is

Qa = 2.1 s2 + 0.01(42) = 2.1 (s2 + 0.2)

Setting Qa to zero we find s = ±j0.4472.

At this point the system is oscillating with frequency = 0.4472 rad/s

7.3 Additional TechniquesExample 5: The order-4 model of ship steering system is

ssssK

ssssKsHsKG a

42.421.22)1.0)(2)(20(2.0)()( 234

Breakaway point at s = 0.0493, s = 15.19 For stability 0<Ka<38.03.

j0.4254Ka=38.03

2 0 20 15.19 s =5.525 450

1350

The RL start from poles s =0, s =0.1, s =2, & s =2 and stop at 4 zeros at s =

4 asymptotes intersect real axis at s =5.525. The real axis RL, for 0.1<s<0 and 20<s<2

For Ka=38.03 the system is oscillating with frequency 0.4254.

7.3 Additional TechniquesTable 1. Stabilities of the last 3 example

exampleKa for

marginal stability

Oscillation frequency

3 (order 2) Always stable …

4 (order 3) 42 0.4472

5 (order 4) 38.03 0.4254

The last 3 example illustrate RL construction and the reduction of order of system.From table 1 and 2 we see that for small Ka (=1) model of order 2 is adequate.For moderate Ka (= 20) model of order 3 is adequate. For large Ka (= 40) model of order 4 is required.Six rules for RL construction have been developed many additional rule are available for accurate RL graph, however we should use digital computer to draw accurate RL.

ex Ka =1 Ka =20 Ka =40

3 –0.05±j0.05 –0.05±j0.312 –0.05±j0.44

4–0.04±j0.05

– 2.003–0.025±j0.31

– 2.050–0.002±j0.4

– 2.0096

5 –0.04±j0.05

– 2.002–19.99997

–0.022±j0.31– 2.055

–19.9994

–0.023±j0.4– 2.106

–19.9990

Table 2. Poles of example 3, 4, and 5.

7.4 Additional RL propertiesSketching RL relies on experiences, followings are some low order RL

p1

1

1ps

p1

))((1

21 psps

p2 z1 p1

))(()(

21

1psps

zs

p2 z1 p1

))(()(

21

1psps

zs

p2

p1 p2 p3

))()((1

321 pspsps

p1

p2

p3

))()((1

321 pspsps

p1

p2

p3

))()((1

321 pspsps

p1

))()(()(

321

1pspsps

zs

z1 p1 p2

7.4 Additional RL properties

The CE can be written as

D(s) + KN(s) =0

For a given K1 we have

D(s) + K1 N(s) =0

Suppose that K is increased by K2 from K1

then the CE becomes

D(s) + (K1 + K2)N(s) =0

which can be expressed as

0)()(1

)()()(1

12

12

sDsNK

sNKsDsNK

With respect to K2 the locus appears to originate on roots as placed by letting K = K1

but still terminates on the original zeros. -1-2

Example suppose we have OLF

The CE can be written as s2 + K(s+1)=0For K =K1=2, we have s2+2s+2=(s+1)2+1=0 With roots s = 1±j.Thus the RL of OLF

is the same with RL of (1) but start from 1±j.

2

)1()()(ssKsHsKG

(1)

22)1()()( 2

sssKsHsKG (2)

7.4 Additional RL properties

The CE can be written as

1+ KG(s)H(s) =0 (1)

Consider replacing s with s1

1+ KG(s-s1)H(s-s1) =0 (2)

then for K0, (s-s1) = s0 satisfied (2).

This is the efect of shifting all poles and zeros of the OLF by a constant amount s1.

Example. The RL of OLF

-1-2

2

)1()()(ssKsHsKG

(3)

is as shown below

The RL of OLF

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

ssKsHsKG (3)

-2-3-4-3

is as shown below

The last property is that a RL has symmetry at breakaway points. Suppose 4 branches come together at a common breakaway point, the the angle between 4 branches will be 3600/4=900.

7.5 Other Configuration

So far we plot RL of CE

1+ KG(s)H(s) =0 (1)

by varying K from 0 to .

Other system configuration should be converted to this form first.

As an example let us vary from 0 to for the following CE

0)(

51

ss

(2)

or052 ss

We rewrite the equation to yield

05

1 2

s

s

0)5( 2 ssand

(3)

Now (3) is the same form as (1) with K= and G(s)H(s) = s/(s2+5)

In general the procedure is as follows1. Write the CE in a polynomial of s2. Grouped the terms that are multiplied by

and that are not, that is we express

Dc(s) + Nc(s)=0

Example Let us design PI controller using RL.

motorcompensator

)1.0(25.0

s–

+s

KK IP

7.5 Other Configuration

The system CE is

We have to rewrite this equation as

The sketch of th RL is as follow

Supposed that KI =1, and we want to vary KP and plot the RL. Eq. (1) becomes

025.025.01.02 sKss P

(2)025.01.0

25.01 2

sssKP

0-0.5

j0.5

-j0.5

Now the RL is available, it is easier to specify the roots.Suppose we want critical damping condition then s =-0.5 and from (2) KP is

6.3125.045.0

25.025.01.0

5.0

2

s

P sssK

01.0

25.01

ssKK I

P (1)

The required PI compensator is then Gc(s) = 3.6 + 1/s

and the design is complete