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EE 370L

Controls Laboratory

Laboratory Exercise #7

Root Locus

Department of Electrical and Computer Engineering

University of Nevada, at Las Vegas

1. Learning Objectives

To demonstrate the concept of error and feedback

To demonstrate how to draw and analyze the root locus of a given system.

2. Equipment Usage

In this laboratory exercise, you will be exposed to the FEEDBACK control system. You will

use MATLAB and SIMULINK software to design a feedback control system.

3. Error and Feedback

Open Loop Systems: An open-loop controller, also called a non-feedback controller, is a type

of controller which computes its input into a system using only the current state and its model of

the system. A characteristic of the open-loop controller is that it does not use feedback to

determine if its input has achieved the desired goal. This means that the system does not observe

the output of the processes that it is controlling. Consequently, a true open-loop system cannot

engage in machine learning and also cannot correct any errors that it could make. It also may not

compensate for disturbances in the system.

Closed Loop Systems: A closed-loop or feedback control system is one in which an input forcing

function is determined in part by the system response. The measured response of a physical system

is compared with a desired response. The difference between these two responses initiates actions

that will result in the actual response of the system to approach the desired response. This in turn

drives the difference signal toward zero.

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Unity Gain Feedback System Response:

Transfer function of the system above can be defined as:

Assuming G = N/D, transfer function reduces to

If terms in numerator and denominator are not comprise, no cancellation of terms exists and

poles of transfer function can be restated as the roots of .

System stability:

For a feedback system to be stable all the poles of transfer function shall be located at the left

half plane. A simple technique to locate stable poles is root locus systems. A root locus

corresponding to G traces all the poles of H, as K varies from 0 to +. Thus, root locus is a family

of curves of the roots of .

A common technique in control system design is to plot the root locus corresponding to a plant

in order to discover where the closed-loop poles can be placed using a gain compensator. A typical

root locus is depicted below.

X O

p1 = 0 z1 = -0.5

X

p2 = -1

s = --1.45

X

p3 = -2

s = --1.25

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Second Order Systems Response:

In this Experiment, you are given G(s) having the form

( )

( )

( )

System is critically damped when

, is over-damped when

and under-damped

when

For under-damped case, systems response is as follows, where

Tr = rise time

Tp = peak time

Ts = settling time

Mp = Peak value

Figure 1. Step Response of an Under-damped Systems

Ideally, we would like the motor output to follow the input. Therefore, the design goal is to

have Tr and Ts as small value and Mp as close to unity.

Explicit formulas can be found for these parameters. First, note that the unit step response for t >

0 is given by

( ) { ( )

} [

( ) ]

{[ (

)

]

(

)}

From the above relationship, we can derive the following relationships:

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(

√ )

√ ( )

√ ( )

(

[

( ) ]

)

√ ( )

Steady state error to a step response is as follows:

| ( ) |

More often, we are also interested in steady state error for ramp response. A typical ramp

response is shown below:

Figure 2. Ramp Response of a closed loop Systems

ess1

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Steady state error to a ramp response is as follows:

| ( ) |

Using final value theorem, we get the following formula:

|

[ ( ) ]|

In addition to time response, we are also interested in frequency response parameters.

Frequency response of a closed loop systems is as follows:

| ( )|

√[ ( ) ] ( )

Figure below depicts |H(jω)| (with ω plotted on a logarithmic axis) for an under-damped

systems:

Figure 3. Frequency Response

Frequency Response

where,

m

b = bandwidth

Mm = Peak at resonance

From the above relationship of H(jω), we can derive the following relationships:

√ (

)

√ (

)

√(

)

( )

√ ( )

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Given the target specification of Tr, Tp, Ts, Mp, ess1, ωm, ωb, and Mm, the above formulas can

be used to determine whether a choice of K exists such that the specification goals are satisfied

and, if so, what range of values of K achieves the specifications.

4. Root Locus

Steps for sketching Root Locus:

1. Root locus begins at pole and ends at zero. The number of separate loci is equal to the number

of poles. m loci ends at zeros and n-m loci ends on infinities.

2. Root loci are symmetrical with respect to the horizontal real axis.

3. On the real axis, a point is on the root locus if the sum of poles and zeros on its right ride is an

odd number.

4. Asymptotes centroid is on the real axis as

∑ ∑

The angle of the asymptotes with the respect to the real axis is

( )

5. Points that locus crosses the imaginary axis can be derived by:

a. Routh-Hurwitz criterion.

b. Characteristic equation: ( ) ( ) (For low order equations only)

6. Breakaway points on the real axis:

( )

or

( ) ( ) ( ) ( )

7. The tangents to the loci at the breakaway point are equally spaced over .

8. The angle of locus departure from a pole:

∑( )

∑( )

( )

Matlab Commands

Plotting root locus:

First set the gain range.

Example. K = -50:.2:50; (this creates a vector varying from -50 to 50 with increment of 0.2)

Type r = rlocus(num, den,k), where num and den are the plant numerator and denominator.

To plot the root locus, type

plot(r)

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To find the gain and frequency corresponding to a point on the locus, type

k=rlocfind(num,den)

In this experiment we will calculate closed-loop response from open-loop data. Before

calculating either step response or Bode plots, the closed-loop transfer function must be found by

typing the following commands

nc=k0*num;

dc=den+k0*[0 0 num];

where “k0” is a specific gain. Note that two extra zeros are padded to ensure that “den” and “[0 0

num]” have the same length for addition.

The closed-loop unit step response can be generated by typing

step(nc,dc)

If you need more control over the time interval in which the computation is carried out, define

a vector “t” and type

step(nc,dc,t)

The magnitude of the closed-loop transfer function |H(jω)| versus ω can be plotted with the

commands

[mc,pc,wc]=bode(nc,dc);

semilogx(wc,mc)

Root Locus Example:

Consider the control system

with

s

sKsK P 2)(

,

1

1)(

ssG ,

4

4)(

ssH

Then

41

2

41

24)()()(

sss

sk

sss

sKsHsGsK P where the parameter PKk 4

Now draw the Root Locus:

R(s) C(s)

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1. There are 3 branches.

2. The real axis between –1<s<0 and –4<s<-2 is on the root locus because these areas are to the

left of an odd number of poles and zeros.

3. The Root Loci start (K 0 ) at the poles located at –4, -1 and 0 then end at the zero located at -2

as well as (2) zeros at s .

4. For large values of s, the Root Loci are asymptotic to asymptotes with angles,

OOO

a

k270,90

13

18012

5. The intersection of the asymptotes lies on the real axis at

5.1

2

3

2

25

1-3

2410

as

6. A Breakaway point occurs along the root locus 01 s at a relative maximum value of k.

Calculate 2

41

)(

)(

s

sss

sN

sDk and tabulate versus “s”.

s 2

41

)(

)(

s

sss

sN

sDk

-0.25

402.0

75.1

75.375.025.0

225.0

425.0125.025.0

k

-0.50

583.050.1

50.350.050.0

2

41

)(

)(

s

sss

sN

sDk

-0.75

488.025.1

25.325.075.0

2

41

)(

)(

s

sss

sN

sDk

-0.625

575.0375.1

375.3375.0625.0

2

41

)(

)(

s

sss

sN

sDk

-0.6

583.04.1

4.34.06.0

2

41

)(

)(

s

sss

sN

sDk

-0.55

589.045.1

45.345.055.0

2

41

)(

)(

s

sss

sN

sDk max

From the table it can be seen the maximum k is acquired around -0.55 which is the breakaway

point.

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s

-1 -4

-1.5

-2

707.0

js 65.065.0

k 1.5447 - 0.0531j 1.5

Breakaway @

s = -0.55

Root Locus can be calibrated to find gain at important points. Here

38.04

5.1

4

kKP

The Matlab commands to generate the root locus are:

>> Num=[1 2];Den=conv([1 0],conv([1 1],[1 4]));

>> GH=tf(Num,Den)

Transfer function:

s + 2

-----------------

s^3 + 5 s^2 + 4 s

>> rlocus(GH)

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More Root Locus Examples:

MatLab Program:

num = [1]

poles = [0;-1+j;-1-j;-3]

den = poly(poles)

rlocus(num,den)

Notes:

n=4 poles, m=0 zeros

asymptotes at ±45,±135

asymptote cg @ (-3-1-1+0)/4 = -1.25

angle of departure, d

( d + arctan(0.5) + 90 + 135) = 180

d = - 71.6 deg

Root Locus

Real Axis

Imagin

ary

Axis

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0-5

-4

-3

-2

-1

0

1

2

3

4

5

System: GH

Gain: 1.48

Pole: -0.625 + 0.63i

Damping: 0.704

Overshoot (%): 4.43

Frequency (rad/sec): 0.888

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MatLab Program:

zeros = [-1+3*j;-1-3*j]

num = poly(zeros)

poles = [0;-1+2*j;-1-2*j;-2]

den = poly(poles)

rlocus(num,den)

Notes:

n = 4 poles, m = 2 zeros

asymptotes at ±90 degrees

asymptote cg @ [(-2-1-1+0)-(-1-1)]/2 = -1.0

angle of departure, d

[( d + arctan(2)+(180-arctan(2))+ 90) -

(-90+90) ] = 180

d = -90 deg

Angle of Arrival, a

[(90+arctan(3)+(180-arctan(3))+90)-(-( a

+90)]=180

a = 90 deg

MatLab Program:

num = [1]

poles=[0;-1+4*j;-1-4*j;-2]

den = poly(poles)

rlocus(num,den)

Notes:

n = 4 poles, m = 0 zeros

asymptotes at ±45,±135

asymptote cg @ (-2-1-1+0)/4 = -1.0

angle of departure, d

( d + arctan(4) + (180-arctan(4)) + 90) = 180

d = -90 deg

5. Prelab

1. The loop transfer function of a single-loop negative feedback system is

( )

( )

a. Based on the second-order model and equations given above, find the value of K for

critical damping.

b. Using formulas given above, find the range of K for which the closed-loop system satisfies

the following specifications:

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Tr ≤ 100ms

Tp ≤ 120ms

Mp ≤ 1.1

Ts ≤ 150ms

ess1 ≤ .045

ωm ≥ 10r/s

ωb ≥ 50r/s

Mm ≤ 1.1.

2. The loop transfer function of a single-loop negative feedback system is

( ) ( )

( )( )

a. Analyze the system, do the hand calculation and plot root loci of the system step by step.

b. Using MATLAB, plot the root loci again and compare will your hand sketch.

c. Determine the range of feedback gain K such that the closed-loop system is stable. For

each value of K where instability begins, find the corresponding frequency of oscillation.

6. Experiment and Postlab

1. The loop transfer function of a single-loop negative feedback system is

( )

( )

a. Construct a Simulink model. You should plot the step input and the response of the system

on a single scope using the MUX block.

b. Verify the system is oscillatory for any K value in the corresponding interval by printing

the output of the scope block for this value of K.

c. Verify the system is stable for any K value in the corresponding interval by printing the

output of the scope block for this value of K.

d. Suppose we want the final value of our response to be within 0.01% of the final value of

our input (step) signal. That is, the steady state error is 0.01%. Further, let us assume the final

value is achieved for t ≥ 10 seconds. Using your Simulink block, determine for what values of K is

the steady state error 0.01% (for t ≥ 10 seconds)?

Notice that in designing a control system we first analyzed the stability of our open loop plant.

If our open loop plant is unstable, we use feedback to stabilize the system. Then we pick the

values for the parameters in our control law so our control objectives are achieved. Satisfying all

the design requirements is the goal of control theory.

2. The loop transfer function of a single-loop negative feedback system is

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( ) ( )

( )

a. Analyze the system, do the hand calculation and plot root loci of the system step by step.

b. Using MATLAB, plot the root loci again and compare will your hand sketch.

c. Find each value of K where system is critically damped.

d. Find each value of K where instability begins, find the corresponding frequency of

oscillation.