Basic Root Locus

8/2/2019 Basic Root Locus

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Topic #2

16.31 Feedback Control Systems

Basic Root Locus

Basic aircraft control concepts

Basic control approaches

Cite as: Jonathan How, course materials for 16.31 Feedback Control Systems, Fall 2007. MIT OpenCourseWare(http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].


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Fall 2007 16.31 21

Aircraft Longitudinal Control

Consider the short period approximate model of an 747 aircraft.

xsp = Aspxsp + Bspe

where e is the elevator input, and

xsp =

w

q

, Asp =

Zw/m U0

I1yy (Mw + MwZw/m) I1yy (Mq + MwU0)

Bsp = Ze/m

I1yy (Me + MwZe/m)

Add that = q, so s = q

Take the output as , input is e, then form the transfer function

(s)

e(s)=

1

s

q(s)

e(s)=

1

s

0 1

(sI Asp)

1Bsp

For the 747 (40Kft, M = 0.8) this reduces to:

(s)

e(s)=

1.1569s + 0.3435

s(s2 + 0.7410s + 0.9272) Ge(s)

so that the dominant roots have a frequency of approximately 1

rad/sec and damping of about 0.4

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.11

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

10.090.20.30.420.540.68

0.84

0.95

0.090.20.30.420.540.68

0.84

0.95

0.2

0.4

0.6

0.8

0.2

0.4

0.6

0.8

PoleZero Map

Real Axis

ImaginaryAxis

Figure 1: Pole-zero map for Gqe

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Fall 2007 16.31 22

Basic problem is that there are vast quantities of empirical data to

show that pilots do not like the flying qualities of an aircraft with

this combination of frequency and damping

What is preferred?

Figure 2: Thumb Print criterion

This criterion was developed in 1950s, and more recent data is

provided in MILSPEC8785C

Based on this plot, a good target: frequency 3 rad/sec and

damping of about 0.6

Problem is that the short period dynamics are no where near these

numbers, so we must modify them.

Could do it by redesigning the aircraft, but it is a bit late for

that...

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6

7

4

5

2

3

0

0.1 0.2 0.4 0.6 0.8 1 2 4

1

POOR

ACCEPTABLE

UNACCEPTABLE

SATISFACTORY

Damping ratio s

Undampednaturalfrequencysrad/sec

Figure by MIT OpenCourseWare.



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Fall 2007 16.31 23

First Short Period Autopilot

First attempt to control the vehicle response: measure and feed

it back to the elevator command e.

Unfortunately the actuator is slow, so there is an apparent lag

in the response that we must model

ce

4

s + 4

ae Ge(s)

k

c

Dynamics: ae is the actual elevator deflection, ce is the actuator

command created by our controller

= Ge(s)ae ;

ae = H(s)

ce; H(s) =

4s + 4

The control is just basic proportional feedback

ce = k( c)

which gives that

= Ge(s)H(s)k( c)

or that(s)

c(s)=

Ge(s)H(s)k1 + Ge(s)H(s)k

Looks good, but how do we analyze what is going on?

Need to be able to predict where the poles are going as a func-

tion ofk Root Locus

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Root Locus Basics

r

e Gc(s)

u Gp(s)

y

Assume that the plant transfer function is of the form

Gp = KpNpDp = K

pi(s zpi)

i(s ppi)

and the controller transfer function is

Gc(s) = KcNcDc

= Kc

i(s zci)i(s pci)

Signals are:

u control commands

y output/measurements

r reference input

e response error

Unity feedback form. We could add the controller Gc in the feed-

back path without changing the pole locations.

Will discuss performance and add disturbances later, but for now

just focus on the pole locations

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Basic questions:

Analysis: Given Nc and Dc, where do the closed loop poles

go as a function of Kc?

Synthesis: Given Kp, Np and Dp, how should we chose Kc, Nc, Dcto put the closed loop poles in the desired locations?

Block diagram analysis: Since y = GpGce and e = r y, theneasy to show that

y

r=

GcGp1 + GcGp

Gcl(s)

where

Gcl(s) =KcKpNcNp

DcDp + KcKpNcNpis the closed loop transfer function

Denominator called characteristic equation c(s) and the

roots ofc(s) = 0 are called the closed-loop poles (CLP).

The CLP are clearly functions ofKc for a given Kp, Np, Dp, Nc, Dc a locus of roots [Evans, 1948]

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Root Locus Analysis

General root locus is hard to determine by hand and requires Mat-

lab tools such as rlocus(num,den) to obtain full result, but wecan get some important insights by developing a short set of plot-

ting rules.

Full rules in FPE, page 260.

Basic questions:

1. What points are on the root locus?

2. Where does the root locus start?3. Where does the root locus end?4. When/where is the locus on the real line?5. Given that s0 is found to be on the locus, what gain is need for

that to become the closed-loop pole location?

Question #1: is point s0 on the root locus? Assume that Ncand Dc are known, let

Ld =NcDc

NpDp

and K = KcKp

c(s) = 1 + KLd(s) = 0

So values of s for which Ld(s) = 1/K, with K real are on theRL.

For K positive, s0 is on the root locus if

Ld(s0) = 180 l 360, l = 0, 1, . . .

If K negative, s0 is on the root locus if [0 locus]

Ld(s0) = 0

l 360

, l = 0, 1, . . .These are known as the phase conditions.

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Question #2: Where does the root locus start?

c = 1 + KNcNpDcDp

= 0

DcDp + KNcNp = 0

So if K 0, then locus starts at solutions of DcDp = 0 which

are the poles of the plant and compensator.

Question #3: Where does the root locus end?

Already shown that for s0 to be on the locus, must have

Ld(s0) = 1

K

So ifK , the poles must satisfy:

Ld =NcNpDcDp

= 0

There are several possibilities:

1. Poles are located at values ofs for which NcNp = 0, which are

the zeros of the plant and the compensator

2. If Loop Ld(s) has more poles than zeros

As |s| , |Ld(s)| 0, but we must ensure that thephase condition is still satisfied.

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More details as K :

Assume there are n zeros and p poles ofLd(s)

Then for large |s|,

Ld(s) 1

(s )pn

So the root locus degenerates to:

1 +1

(s )pn= 0

So n poles head to the zeros of Ld(s) Remaining p n poles head to |s| = along asymptotes

defined by the radial lines

l =180 + 360 (l 1)

p nl = 1, 2, . . .

so that the number of asymptotes is governed by the number

of poles compared to the number of zeros (relative degree). Ifzi are the zeros ifLd and pj are the poles, then the centroid

of the asymptotes is given by:

=

ppj

nzi

p n

Example: L(s) = s4

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Question #4: When/where is the locus on the real line?

Locus points on the real line are to the left of an odd number

of real axis poles and zeros [K positive].

Follows from the phase condition and the fact that the phase

contribution of the complex poles/zeros cancels out

Question #5: Given that s0 is found to be on the locus, whatgain is needed for that to become the closed-loop pole location?

Need

K 1

|Ld(s0)|=

Dp(s0)Dc(s0)Np(s0)Nc(s0)

Since K = KpKc, sign ofKc depends on sign of Kp

e.g., assume that Ld(s0) = 180, then need Kc and Kp to

be same sign so that K > 0

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Root Locus Examples

Figure 3: Basic Figure 4: Two poles

Figure 5: Add zero Figure 6: Three poles

Figure 7: Add zero

Examples similar to control design process: add compensator dynamics to

modify root locus and then chose gain to place CLP at desired location on

the locus.

September 2, 2007

Im

Re

Im

Re

Im

Re0

Im

Re

Im

Re0


Figure by MIT OpenCourseWare.Figure by MIT OpenCourseWare.

Figure by MIT OpenCourseWare.Figure by MIT OpenCourseWare.



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Fall 2007 16.31 211

Figure 8: Complex case

Figure 9: Very complex case

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Im

Re0 0

Asymptotes

Real line

Keys to a good sketch

0





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Performance Issues

Interested in knowing how well our closed loop system can track

various inputs Steps, ramps, parabolas

Both transient and steady state

For perfect steady state tracking want error to approach zero

limt

e(t) = 0

Can determine this using the closed-loop transfer function and

the final value theorem

limt

e(t) = lims0

se(s)

So for a step input r(t) = 1(t) r(s) = 1/s

y(s)

r(s) =G

c(s)G

p(s)

1 + Gc(s)Gp(s)

y(s)

e(s) = Gc(s)Gp(s)

e(s)

r(s)=

1

1 + Gc(s)Gp(s)

so in the case of a step input, we have

e(s) =r(s)

1 + Gc(s)Gp(s)

1/s

1 + Gc(s)Gp(s)

lims0

se(s) = l i ms0

s1/s

1 + Gc(s)Gp(s)=

1

1 + Gc(0)Gp(0) e()

So the steady state error to a step is given by

ess =1

1 + Gc(0)Gp(0)

to make the error small, we need to make one (or both) ofGc(0), Gp(0) very large

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Clearly ifGp(s) has a free integrator (or two) so that it resembles1

sn(s+)m with n 1, then

lims0 Gp(s) ess 0

Can continue this discussion by looking at various input types (step,

ramp, parabola) with systems that have a different number of free

integrators (type), but the summary is this:

step ramp parabola

type 0 11 + Kp

type 1 01

Kv

type 2 0 01

Ka

where

Kp = lims0

Gc(s)Gp(s) Position Error Constant

Kv = lims0

sGc(s)Gp(s) Velocity Error Constant

Ka = lims0

s2Gc(s)Gp(s) Acceleration Error Constant

which are a good simple way to keep track of how well your system

us doing in terms of steady state tracking performance.

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Dynamic Compensation

For a given plant, can draw a root locus versus K. But if desired

pole locations are not on that locus, then need to modify it usingdynamic compensation.

Basic root locus plots give us an indication of the effect of

adding compensator dynamics. But need to know what to add

to place the poles where we want them.

New questions:

What type of compensation is required?

How do we determine where to put the additional dynamics?

There are three classic types of controllers u = Gc(s)e

1. Proportional feedback: Gc Kg a gain, so that Nc =

Dc = 1

Same case we have been looking at.

2. Integral feedback:

u(t) = Ki

t0

e()d Gc(s) =Kis

Used to reduce/eliminate steady-state error

If e() is approximately constant, then u(t) will grow to bevery large and thus hopefully correct the error.

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Consider error response of Gp(s) = 1/(s + a)(s + b)

(a > 0, b > 0) to a step,

r(t) = 1(t) r(s) = 1/swhere

e

r=

1

1 + GcGp e(s) =

r(s)

(1 + GcGp)

To analyze error, use FVT limt e(t) = lims0 se(s)

so that with proportional control,

limt

ess = lims0

ss

11 + KgGp(s)

=1

1 +Kgab

so can make ess small, but only with a very large Kg

With integral control, lims0 Gc(s) = , so ess 0

Integral control improves the steady state, but this is at

the expense of the transient response

Typically gets worse because the system is less well damped

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Example #1: G(s) =1

(s + a)(s + b), add integral feedback

to improve the steady state response.

Figure 10: RL after adding integral FB

Increasing Ki to increase speed of the response pushes the

poles towards the imaginary axis more oscillatory re-

sponse.

Combine proportional and integral (PI) feedback:

Gc = K1 +K2s

=K1s + K2

s

which introduces a pole at the origin and zero at s = K2/K1

PI solves many of the problems with just integral control (I).

Figure 11: RL with proportional and integral FB

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Im

Re

Im

Re00




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3. Derivative Feedback: u = Kde so that Gc(s) = Kds

Does not help with the steady state

Provides feedback on the rate of change of e(t) so that thecontrol can anticipate future errors.

Example # 2: G(s) =1

(s a)(s b), (a > 0, b > 0)

with Gc(s) = Kds

Figure 12: RL with derivative FB

Derivative feedback is very useful for pulling the root locus

into the LHP - increases damping and more stable response.

Typically used in combination with proportional feedback to

form proportional-derivative feedback PD

Gc(s) = K1 + K2s

which moves the zero from the origin.

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0

Im

Re



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Controller Synthesis

First determine where the poles should be located

Will proportional feedback do the job?

What types of dynamics need to be added? Use main building

block

GB(s) = Kc(s + z)

(s + p)

Can look like various controllers, depending how Kc, p, and zpicked

If pick z > p, with p small, then

GB(s) Kc(s + z)

s

which is essentially a PI compensator, called a lag.

If pick p z, then at low frequency, the impact of p/(s + p)

is small, so

GB(s) Kc(s + z)

which is essentially PD compensator, called a lead.

Various algorithms exist to design the components of the lead andlag compensators

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Im

Re0

Im

Re0





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Classic Root Locus Approach

Consider a simple system Gp = s2 for which we want the closed

loop poles to be at 1 2i Will proportional control be sufficient? no

So use compensator with 1 pole.

Gc = K(s + z)

(s + p)

So there are 3 CLP.

To determine how to pick the p, z, and k, we must use the

phase and magnitude conditions of the RL

To proceed, pick evaluate the phase of the loop

Ld(s) =s + z

(s + p)s2

at s0 = 1 + 2i. Since we want s0 to be on the new locus, we

know that Ld(s0) = 180 360l

Figure 13: Phase condition

As shown in the figure, there are four terms in Ld(s0) the

three poles at the origin contribute 117 each

Given the assumed location of the compensator pole/zero, can

work out their contribution as well

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(-1,2i)

0

- p - z

Im

Re0




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Geometry for the real zero: tan = 2z1 and for the real pole:

tan = 2p1

Since we expect the zero to be closer to the origin, put it firston the negative real line, and then assume that p = z, where

typically 5 10 is a good ratio.

So the phase condition gives:

2(117) + = 180

arctan2

z 1 arctan

2

10z 1 = 53

but recall that

tan(A B) =tan(A) tan(B)

1 + tan(A)tan(B)

so( 2z1) (

210z1)

1 + ( 2z1)(2

10z1)= 1.33

which give z = 2.2253, p = 22.2531, kc = 45.5062

% RL design using angles

clear all

target = -1+2*j;

phi_origin= 180-atan(imag(target)/-real(target))*180/pi;

syms z M; ratio=10;

phi_z=(imag(target)/(z+real(target)));

phi_p=(imag(target)/(ratio*z+real(target)));

M=(phi_z-phi_p)/(1+phi_z*phi_p);

test=solve(M-tan(pi/180*(2*phi_origin-180)));

Z=eval(test(1));

P=ratio*Z;

K=1/abs((target+Z)/(target^2*(target+P)));

[Z P K]

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Pole Placement

Another option for simple systems is called pole placement.

Know that the desired characteristic equation is

d(s) = (s2 + 2s + 5)(s + ) = 0

Actual closed loop poles solve:

c(s) = 1 + GpGc = 0

s2(s + p) + K(s + z) = 0

s3 + s2p + Ks + Kz = 0

Clearly need to pull the poles at the origin into the LHP, so need

a lead compensator Rule of thumb: take p = 5 10z. Compare the characteristic equations:

c(s) = s2 + 10zs2 + Ks + Kz = 0

d(s) = (s2 + 2s + 5)(s + )

= s3 + s2( + 2) + s(2 + 5) + 5 = 0

givess2 + 2=10z

s 2 + 5=K

s0 5=zK

solve for , z, K

K =25

5 2z

; =5z

5 2z z = 2.23, = 20.25, K = 45.5

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25 20 15 10 5 020

15

10

5

0

5

10

15

20

Root Locus

Real Axis

ImaginaryAxis

Figure 14: CLP with pole placement

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Observations

In a root locus design it is easy to see the pole locations, and thus

we can relatively easily identify the dominant time response Caveat is that near pole/zero cancelation complicates the pro-

cess of determining which set of poles will dominate

Some of the performance specifications are given in the frequency

response, and it is difficult to determine those (and the correspond-

ing system error gains) in the RL plot

Easy for low-order systems, very difficult / time consuming forhigher order ones

As we will see, extremely difficult to identify the robustness margins

using a RL plot

A good approach for a fast/rough initial design

Matlab tool called sisotool provides a great interface for design-

ing and analyzing controllers

Basic Root Locus

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