Root Locus

Unit 7: Part 1: Sketching the Root Locus

Engineering 5821:Control Systems I

Faculty of Engineering & Applied ScienceMemorial University of Newfoundland

March 14, 2010

ENGI 5821 Unit 7: Root Locus Techniques

Root Locus

1 Root Locus

Vector Representation of Complex Numbers

Defining the Root Locus

Properties of the Root Locus

Sketching the Root Locus

ENGI 5821 Unit 7: Root Locus Techniques

Root Locus

The Root Locus is a graphical method for depicting location of theclosed-loop poles as a system parameter is varied. It is applicablefor first and second-order systems, but also to higher ordersystems.

Changes in a parameter, such as the gain K , affects the location ofthe equivalent closed-loop system poles. Root locus allows us todetermine the movement of these poles as K is varied.

Vector Representation of Complex Numbers

We will need to represent complex numbers and complex functionsF (s) as vectors. Typically the complex functions we are concernedwith have the following form:

F (s) =(s + z1)(s + z2) · · ·(s + p1)(s + p2) · · ·

Consider the complex number s + q, where q will stand in foreither a zero or a pole. We can represent s + q as a vector in thecomplex plane. The magnitude and angle of this vector are givenby the complex exponential representation,

s + q = re jθ

Where r is the vector length and θ is the angle from the real-axis.

Normally we would draw s + q as a vector out of the origin.

s q

s + q

s

-q

s + q

Normal: Complex number drawn fromits own zero:

However, we can recognize s = −q as a zero of s + q and draw thesame vector with its tail at −q (see above right).

F (s) =(s + z1)(s + z2) · · ·(s + p1)(s + p2) · · ·

We can replace each term s + zi or s + pi with their correspondingcomplex exponential forms:

F (s) =rz1eθz1 rz2eθz2 · · ·rp1eθp1 rp2eθp2 · · ·

=rz1rz2 · · ·rp1rp2 · · ·

eθz1+θz2+···−θp1−θp2−···

The magnitude and phase of F (s) are as follows:

|F (s)| =rz1rz2 · · ·rp1rp2 · · ·

=|s + z1||s + z2| · · ·|s + p1||s + p2| · · ·

∠F (s) = θz1 + θz2 + · · · − θp1 − θp2 − · · ·= ∠(s + z1) + ∠(s + z2) + · · · − ∠(s + p1)− ∠(s + p2)− · · ·

e.g. Evaluate the following complex function when s = −3 + j4,

F (s) =(s + 1)

s(s + 2)

The magnitude and phase of the zero is:√

20∠116.6o .

The pole at the origin evalutes to 5∠126.9o .

The pole at -2 evaluates to√

17∠104.0o .

|F (s)|∠F (s) =

√20

5√

17∠ [116.9o − 129.9o − 104.0o ]

Defining the Root Locus

Consider the following system, designed to track a visual target.

For a unity feedback system such as this we will refer to K1K2s(s+10) as

the open-loop transfer function (if the unity feedback signal werecut the system would be open-loop). Thus, the open-loop polesare at s = 0 and s = −10.

However, depending upon K we will obtain different closed-looppoles, which are the roots of s2 + 10s + K . We can utilize thequadratic formula to obtain these roots for values of K ≥ 0. Wecan then plot the positions of these poles...

The path of the closed-loop poles as K varies is the root locus.

Observations:

for K < 25 the system is overdamped

for K = 25 the system is critically damped

for K > 25 the system is underdamped

More Observations:

In the underdamped portion the real-part of the pole, σd isconstant. Therefore so is Ts = 4/σd .

The root locus never crosses the jω axis. Therefore thesystem is stable.

Properties of the Root Locus

The transfer function for a general closed-loop system is,

T (s) =KG (s)

1 + KG (s)H(s)

We are concerned with the poles of T (s). We will have a polewhenever KG (s)H(s) = −1. The root locus is the locus of pointsin the s-plane for which this is true. We can express this equalityas follows,

|KG (s)H(s)|∠KG (s)H(s) = 1∠(2k+1)180o k = 0,±1,±2,±3, . . .

A particular s is on the root locus if its magnitude is unity and itsangle is an odd multiple of 180◦. Satisfying both of theserequirements means that s is a closed-loop pole of the system.

e.g. Consider the following system,

We will test a couple of points to see if they are closed-loop poles.We evaluate KG (s)H(s) graphically for s = −2 + j3,

θ1 + θ2 − θ3 − θ4 = 56.31◦ + 71.57◦ − 90◦ − 108.43◦ = −70.55◦

Therefore s = −s + j3 is not on the root locus.

If we test a point and find it is on the RL (e.g. s = −2 + j√

2/2)we may then want to determine the corresponding value of K .Since |KG (s)H(s)| = 1 on the RL,

K =1

|G (s)H(s)|=

∏pole lengths∏zero lengths

Sketching the Root Locus

One possibility is to sweep through a sampling of points in thes-plane and test each one for inclusion in the RL.

It is much more preferable to utilize some insights about the RL toidentify its major characteristics and therefore obtain a roughsketch. The following rules will help achieve this...

1. Number of branches:

Consider a branch to be the path that one pole traverses.

There will be one branch for every closed-loop pole.

e.g. Two branches for the previous quadratic example.

2. Symmetry:

The poles for real physical systems either lie on the real-axis orcome in conjugate pairs. Hence...

The root locus is symmetrical about the real axis.

3. Real-axis segments:

Consider evaluating the anglular contribution of the open-looppoles and zeros at points P1, P2, P3, and P4 below:

The angular contribution of a pair of complex open-loop poles(or zeros) is zero

The contribution of poles or zeros to the left of the point iszero

Using only the real-axis poles or zeros to the right of the point,we find that the angle sum alternates between 0o and 180o .

On the real axis, for K > 0 the root locus exists to theleft of an odd number of real-axis, finite open-loop polesand/or finite open-loop zeros.

e.g. G (s) = K(s+3)(s+4)(s+1)(s+2) , H(s) = 1

4. Starting and ending points:

Where does the RL begin and end? It begins at K = 0 and ends atK =∞. Consider the closed-loop transfer function:

T (s) =KG (s)H(s)

1 + KG (s)H(s)

=K NG (s)

DG (s)NH(s)DH(s)

1 + K NG (s)DG (s)

NH(s)DH(s)

=KNG (s)NH(s)

DG (s)DH(s) + KNG (s)NH(s)

If we let K → 0 the poles of T (s) approach the combined poles ofG (s) and H(s). If K →∞ the poles of T (s) approach thecombined zeros of G (s) and H(s).

Thus, the RL begins at the open-loop poles of G (s)H(s)and ends at the zeros of G (s)H(s).

e.g. G (s) = K(s+3)(s+4)(s+1)(s+2) , H(s) = 1

Note that we don’t know yet what the exact trajectory of the rootlocus will be.

What if the number of open-loop poles and zeros is mismatched?e.g.

F (s) =K

s(s + 1)(s + 2)

A function can have both poles and zeros at infinity. For example,as s →∞.

F (s) ≈ K

s · s · sWe therefore consider F (s) to have three zeros at infinity. If weinclude both finite and infinite poles and zeros every function hasan equal number of poles and zeros.

The root locus for F (s) (i.e. KG (s)H(s) = F (s)) would start atthe three finite poles and go towards the zeros at infinity. Yet howdo we get to these zeros at infinity?

5. Behaviour at infinity:

The root locus approaches straight lines as asymptotesfor zeros at infinity. These asymptotes are defined aslines with real-axis intercept σa and angle θa.

σa =

∑finite poles−

∑finite zeros

#finite poles−#finite zeros

θa =(2k + 1)π

#finite poles−#finite zeros

where k is an integer and θa is the angle (in radians) tothe positive real-axis. We get as many asymptotes asthere are branches corresponding to zeros at infinity.

The derivation for these formulae can be found atwww.wiley.com/college/nise under Appendix L.1.

e.g. Sketch the RL for the following system:

We first apply rules 1-4:

We now compute the real-axis intercept and the angles of allasymptotes:

σa =

∑finite poles−

∑finite zeros

#finite poles−#finite zeros

=(0− 1− 2− 4)− (−3)

4− 1= −4/3

θa =(2k + 1)π

#finite poles−#finite zeros

= π/3 for k = 0

= π for k = 1

= 5π/3 for k = 2

Notice that we have one real-axis intercept but multiple angles.There are three asymptotes—one for each infinite zero. We obtainthree unique values for θa before the angles start to repeat.

The following is our complete RL sketch.