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Created by Dr Grant Covic for paper: Systems and Control 1998: Document 2 of 3

Proportional & Integral Controllers

Proportional + Integral (PI) controllers were developed because of

the desirable property that systems with open loop transfer functions

of type 1 or above have zero steady state error with respect to a step

input.

The PI regulator is:s

KK

sE

sU IP +=

)(

)(

But can be realised easily in the following form:

Ex: assume we wish to apply PI regulator to a type 0 plant:

23

1)(

2 ++=

sssGp

Thus:

s

sG

TsKOLTF

p

I

P

)()

1( +=

IPP

IP

TKsKsss

TsK

sR

sCCLTF

++++

+

== )23(

)1(

)(

)(

2

+KP

IsT

1

E(s) U(s)

)(te)(tu

R1

R2C1

+

_

C(s)E(s)R(s)+

U(s)

23

12 ++ ss

)1

1(I

PsT

K +

1

2

1

1

)(

)(

R

RsC

SE

sU+

=

+=

112

1 1

RsCR

R

Gc(s) Gp(s)

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Tuning PI Controllers

General approach to tuning:

1. Initially have no integral gain (TI large)2. Increase KP until get satisfactory response3. Start to add in integral (decreasing TI) until the steady state error

is removed in satisfactory time

(may need to reduce KP if the combination becomes oscillatory)

Anti-windup in I & PI controllers

Under some operating conditions non-linearities in the plant or

controller can stop an Integral controller from removing the steadystate error. If the Integrator output is not limited, then during this

time the total of the integrated (summed) error {KIe(t)dt} willcontinue to build. Once the restrictions are finally removed,

problems can arise because this built up energy must be removed

before the integral control can act normally this can take a long

time. To avoid this, anti-windup circuits are added that place

limits on the integral total. These limits are usually placed on thesummed output of the P&I controller as well.

5V

t

u(t)

C(s)E(s)R(s)+

U(s)Gp(s)

R1

R2C1

+

_-1

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PID (three term) Controllers

Proportional action: responds quickly to changes in error deviation.

Integral action: is slower but removes offsets between the plants

output and the reference.

Derivative action: Speeds up the system response by adding in

control action proportional to the rate of change of the feedback

error. Consequently this is susceptible to noise in the error signal,

which limits the derivative gain. When present this allows larger

values of KP and KI (smaller TI) to be used than possible in pure PI

regulators, but large values of derivative gain (KD) will causeinstability.

The PID regulator is given by:

++=

dt

tdeTdtte

T

teKtu DI

P

)()(

1)()(

Here:I

P

IT

KK =

DPD TKK =

)(

)(

)(

)(2

sD

sN

s

KsKsKsK

s

KK

sE

sU IPDD

IP =

++=++=

NB: this transfer function is non-proper and is therefore difficult to

realise in practice.

Proper T.F.: Order N(s) Order D(s)Strictly proper T.F.: Order N(s) < Order D(s)

These are generally easier to realise, and also reduce the

susceptibility of the derivative action to noise.

Error

{e(t)}

t

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Practical PID controllers

+++= )1(1

)1(

1

)(

)(

DD

IP TssTsTKsE

sU

1. The 1/(1+sTD) term acts as an effective low-pass filter on theP+D regulator to attenuate noise in the derivative block.

2. If = 0 the original PID form is obtained3. Typically = 0.1 to place the filter as far away from the

derivative action as possible. It is generally impractical to movethe filter further away as the control action (effort) becomes too

much and can-not be realised (supply rail or limiters saturate the

control effort)

4. PD controllers could be realised on their own if the plant doesnot require an integrator.

Circuit realisation:

If the filter term is disregarded for KP then:

+

++=

+++

+

D

DP

I

PP

Ts

TsK

sT

KK

RsC

RsC

RsCC

C

R

R

sE

sU

1)1(

1

)(

)(

22

11

311

2

3

1

+

++

+=

2

2

1

1

3

1

1

1

11

)(

)(

sCR

sCR

R

sCR

SE

sU

+

+++=

)1(

)1(1

22

11

1

2

313

1

RsC

RsC

C

C

RsCR

R

)(te)(tu

R3

+

_

R2 C2 R1 C1

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A complete derivation results in:

+

+

+

+

+

+

+=

)1(

1

1

1

)(

)(

22

1

2

3

1

3

212

1

2

3

131

1

2

3

1

RsC

C

C

R

RR

RRC

s

C

C

R

RRsC

C

C

R

R

sE

sU

The Ziegler-Nichols tuning Technique for P, PI & PID controllers

(Golten & Verwer 7.7)

Proportional: Pc KsG =)(

PI:

+=

I

pcsT

KsG1

1)(

PID:

+

++=

D

D

I

PcTs

sT

sTKsG 1.01

11)(

1. Set TD = 0 & TI = 2. Increase KP until the system just starts to oscillate (KP = KPO).

The frequency of oscillation here is c and the period isTO=2/c.

Set controller gains as:

Type KP TI TD

P

PI

PID

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Notes:

1. Even though first and second order systems cannot in principlefully oscillate, practical systems always contain transport delays

and non-linearities that make them oscillate if the loop gain ishigh.

2. A disadvantage with the Z-N technique is that it really needs tobe performed in real-time in the actual plant. In some cases it

may be undesirable to have the plant/process oscillate even for

tuning purposes.

Example of a practical PD Controller

Rate Feedback with the DC Servo Drive

If we place a tachometer on the servo drive, this will output a voltage

proportional to the mechanical speed m. This can be feed back to

the regulator.

As m = d /dt, feeding back some measure of the m term is likefeeding back the derivative (hence termed rate feedback)

K C(s) = E(s)R(s) = ref

+

U(s)

)1(

+sKm

s

1m

K

Tacho Feedback

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What effect does this have on our performance specs?

Eg: damping?

Consider the inner loop as a non-unity feedback control system then:

resulting in the following system:

mm

m

ref KKKKKss

KK

sR

sC

+++==

)1()(

)(2

if K = 0 then the original transfer function arises.

Here: mn KK= andm

m

KK

KKK

2

1+=

As opposed to simple proportional control, K can be increased to

give improved response times (larger n), while maintaining gooddamping by increasing K.

C(s) = E(s)R(s) = ref

+

)1( KKKs

KK

m

m

++ s1m

KE(s)R(s)

+U(s)

)1(

+sKm m

K

Tacho Feedback)1(

1

)1(

++

+=

s

KKK

s

KK

m

m

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ROOT LOCUS

Root locus: is a plot of the paths of the characteristic equation for a

closed loop system (CLTF poles) in the complex plane as a functionof various loop gains.

Plotting Techniques

1. Determine the OLTF of the system.

clearly the OLTF; G(s)H(s) affects the system poles.

In general form G(s)H(s) = KN s

D s

( )

( )

N(s) is a polynomial in s of order "W"

D(s) is a polynomial in s of order "N"

N is also the order of the characteristic equation

K is called the loop sensitivity if the transfer function is

written so that the coefficients of the highest powers in sin both the numerator and denominator are unity.

2. Factorise G(s)H(s) into the form (s+a) where a maybe complex.

K

K

))((

)()()(

21

1

psps

zsKsHsG

+++

=

H(s)

C(s)E(s)R(s)

GP(s)+

Gc(s)

)()(1

)(

sHsG

sGCLTF

+=

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3. The characteristic equation is: 0)()(1 =+ sHsGso that the roots of this equation must satisfy the expression:

1))((

)()()(

21

1

=+++

= KK

psps

zsKsHsG

Because G(s)H(s) is a complex quantity, the above condition

{G(s)H(s) = 1} is in fact a composite of two conditions whichboth must be valid.

i. A magnitude condition

1)()( =sHsG (i)ii. An angle condition

)21()()( hsHsG += (ii)where h=0, 1, 2,

These conditions can be investigated graphically by plotting all ofthe poles and zeros of the OLTF in the complex plane and noting

that each of the terms (s+z1),(s+z2),...,(s+p1),(s+p2)... are vectorsterminating at an arbitary points in the complex plane.

Fors to be a point on the root loci it must satisfy both conditions:

Ex:

))((

)()()(

32

1

pspss

zsKsHsG

+++

=

similar to Ex fig 7.6

(DAzzo & Houpis Linear Control System Analysis & Design)

j

-p3 -p2-z1

s

-p1

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using (i) we have:

132

1 =++

+

pspss

zsK

1

32

zs

pspssK

+

++=

using (ii) we have:

3211180 =o

Further examples:

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Qualitative analysis

How do zeros and higher order poles in the OLTF (G(s)H(s)) affect

the root locus plot?

In order to answer this we'll first try to obtain a conceptual feel for

the effects of additional zeros and poles on simple and more complex

systems by looking at the following root locus plots.

Observations

1. The addition of a zero to the OLTF - pulls the root locus to the

left. This tends to enable poles to be selected that result in stable

faster response times (ie, the settling time is smaller).

2. The addition of a pole to the OLTF - pulls the root locus to the

right so that slower response times are possible, but the new root

locus paths may cross over into the RHP so that K must be

bounded to avoid instability.

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Ex fig 7.4


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Ex fig 7.5


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Root Locus Plotting Rules

The following eight rules enable fast sketches of root locus to be

accomplished that:- are a useful check for computer generated plots.

- enable a qualitative idea of the system performance.

1. Number of branches of the Root Locus

The number of branches (root loci paths) equals the number of

poles of the OLTF (N).

2. Real Axis Locus Values

Applying the angle condition to any search path along the real axis:

If the total number of real poles and zeros (in the OLTF) to the

right of a search point s on the real axis is odd, this point lies on the

root locus.

Note:

(i) The angular contribution of complex conjugate pairs (poles and zeros) tosuch a point is 360, and can therefore be ignored.

(ii) A point satisfying this condition may be part of more than one branch.

3. Root Locus Start and End Points

=

=

= W

h

h

N

c

c

zs

ps

K

1

1

when: s = pc, K = 0, and when: s = zh, K = .

also since for proper systems: NW, then: (NW) poles must existsuch that their root locus branches do not finish on a zero, but rather

go to infinity (s) as K, (this is equivalent to the effect of azero).

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The starting points (K=0) are the poles of the OLTF locus path.

The ending points (K=) are either zeros of the OLTF or points ats= that are considered equivalent zeros of multiplicity (N-W)).

4. Asymptotes of Locus as s approaches Infinity

N-W branches of the root locus have straight line asymptotes (ass )emulating from a single point whose angles (w.r.t the real

axis) are given by:

)(

)21(

WN

h

+=

h=0,1, 2

Proof:

1lim)()(lim == WNss sK

sHsG

must satisfy the magnitude and angle condition, hence:

)21()()( hsKsHsG WN +==

)21()( hsWN +=where =s

As N and W are constants for a given OLTF then as s(irrespective of magnitude) the angle constant. Thus thebranches are asymptotic to straight lines.

Examples:

(i) (N-W) = 1

j

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(ii) (N-W) = 2

(iii) (N-W) = 3

(iv) (N-W) = 4

5. Real Axis Intercept of the Asymptotes

The asymptotes intersect at a point o on the real axis

)()()( 110 WNzp

W

hh

N

cc

= ==

j

j

j

j

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6. Breakaway Point on the Real Axis

In many instances root loci must breakaway from the real axis (ie,

become complex) in order to reach the end points (K= ). If theseloci return to the real axis at some later stage there will also be abreak-in point.

Plotting the loop sensitivity (K) against the real axis demonstrateswhen these conditions arise.

Examples: If a locus exists on the real axis between:

(i) two poles: the locus starts at K=0 and as K increases mustbreakaway in order to finish on either a zero or a point of

infinity (both with K=).

Ex fig 7.11


(ii) two zeros or a zero & a point of infinity: 2 branches must

break-in from the complex region between them so as to

terminate at these end points.

K

K

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(iii)a pole and a zero: either pairs of breakaway and break-in points

occur, or none at all.

Breakaway and break-in points can be identified as follows.For any point on the root locus:

G(s)H(s) = KN(s)/D(s) = -1

hence

K = -D(s)/N(s)

so that

d(K)/ds = 0

gives the minimum & maximum value of K with varying s.

Only answers with pure real values are of interest (s=) and thesemust satisfy rule 2.

Ex:

K

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7. Complex Pole (or Zero) Angle of Departure

The angle condition can be used to determine the direction that a

locus branch leaves a complex pole, or enters a complex zero.

Recall that:

)21(3212111

hpzN

h

h

W

c

c+=++=

==

KK

A test point is chosen very close to, but not exactly on, the complex

root of interest (r) so that all angles between the remaining roots

and this test point are equivalent to the angles measured between

these roots and r.

Ex: Given))()((

)()()(

2

1

jbasjbaspss

zssHsG

+++++

=

Thus to find the angle of departure for complex pole P3 (-a+jb):

( ) o18043211 =+++

8. The Imaginary Axis Crossing PointPlace s = jc in the characteristic equation and solve for c.

j

-z1 -p2 -p1

-p3

-p4

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Control in the Frequency Domain

Bode and Polar Plots of Type 0-2 Transfer Functions

Type 0: Type 1:Bode:

)21)(21()()(

21

TjTj

KjHjG

++=

)21(

)21()()(

1

Tjj

TjKjHjG a

++

=

K=10, T1=2/10, T2=2/100 K=100, T a=2/100, T1=2/10

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Polar/Nyquist:

Type 2:

Bode: (Example)

Ex Fig 8.10


Polar: (Additional examples of typical type 2 polar plots)

(stable) (unstable)

Ex Fig 8.16, 8.18


90

0

90

180 1 0

90

180 1

90

)1()()()(

2

2

aTjj

KjHjG

+=

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Nyquist Stability in the Frequency domain

The Nyquist criterion applied to a CLTF states that:

For stability; none of the roots of the characteristic equation mustlie on or to the right of the imaginary axis in the s-plane.

This rule can be translated into the frequency domain using

Cauchies Theorem with a closed loop contour c incorporating the

whole RHP ( =c

residuesjdzzf 2)( ).

In practice this rule must applied to the solution of the CLTFs

characteristic equation. Thus if: 0)()(1)( =+= sHsGsB

Any zero of B(s) inside the RHP will create a +2 rotation, whileany pole of B(s) will create a 2 rotation. The total number ofrotations of the poles and zeros are respectively: PRb, ZRb.

There will be a net number of rotations related to the number of

poles and zeros of B(s) inside the RHP. NR= ZRb+ PRb.

Since B(s) cannot have any zeros (poles of CLTF) in the RHP for

system stability, the requirement for stability is:

0== RRbRb NPZ

However, the poles of B(s) are simply the poles of the OLTF, while

the zeros of B(s) are the poles of the CLTF.

Proof by example:

)1()()(

2sTs

KsHsG

+=

KsTs

KCLTF

++=

)1(2

0)1(

)1()()(1)(

2

2

=+

++=+=

sTs

KsTssHsGsB

So that our new stability rule can be rewritten as:0)()( == ROLTFRCLTFR NPP

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In the frequency domain this test for stability only requires a plot of

B(j) as " < < +" and we need to determine the net number ofrotations about the origin (B(j)=0).

Alternatively since B(j)1 = G(j)H(j), we can plot G(j)H(j)

and look at the net rotations about the 1 point:

Type 0 system stability.

Given: G j H jk

j T j T j T ( ) ( )

( )( )( )

=

+ + +1 1 11 2 3... (1)

plotting this results in:

N is determined by noting the direction of increasing at intersections of anarbitrary radial line from 1 with the plot of G(j)H(j).

A Condition on the Nyquist Stability TheoremA simple condition, necessary for valid application of the Nyquist

Theorem requires that:

The contour integral c must never pass through any poles or

zeros ofB(s).

In systems Type 1, stable poles lie at the origin. Thus the path ofthe contour in the s-plane must be modified so that it does not pass

through the origin. This is accomplished by taking an infinitely

small detour around the origin in the s-plane from j0 +j0.

NR =

from (1): PR(OLTF)= 0,thus: PR(CLTF)= 0

1180

90

90

0

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The effect of this in the frequency domain on a type n system is to

add n infinitely large cw semi-circles to the Nyquist plot connecting

the j0 and +j0 points.

Proof:

The modification to the contour s=ej (-90

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Type 2 examples

G j H jk j T

j j T j T

a ( ) ( )( )

( ) ( )( )

=+

+ +1

1 12

1 2

whereT T T T

a > + +1 2 3

Ex Fig 8.33(DAzzo & Houpis Linear Control System Analysis & Design)

Conditionally stable systems can also arise where both increases and

decreases in k will cause instability (as shown below)

Ex Fig 8.34(DAzzo & Houpis Linear Control System Analysis & Design)

Final notes:

Only the positive part of G(j)H(j) for 0

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