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• Yazdan Rastegar

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Effects of Addition of Poles and Zeros

Figure: Effect of Adding Poles

Figure: Effect of Adding Zero

1. Addition of poles pulls the root locus to the right 2. Additional zero pulls the root-locus to the left

Controller Design A compensator or controller placed in the forward path of a control system will modify the shape of the loci if it contains additional poles and zeros. In compensator design, hand calculation is cumbersome, and a suitable computer pakage is generally used.

Compensator PD PI PID

Characteristics One additional zero One additional zero and a pole at origin Two additional zeros and a additional pole at origin

Example A control system has the open-loop transfer function, A PD compensator of the form Gc ( s)  achieve the performance specification:

G f ( s) 

K ; k  1. s( s  2)( s  5)

K1 (s  a) is to be introduced in the forward path to

Overshoot less than 5%, settling time less than 2seconds Determine the values of

K1 and a

to meet the specification.

Original Controller

  0.5, For   0.7, For

P.O.  16.3% P.O.  4.6%

  0.7, corresponds to the controller gain of K  7.13 . Settling time condition is not met.

PD compensator design When PD compensator is used, we actually add a zero in the open-loop transfer function. Potential locations of zero include: 1. At s  1,

2. At s  2 ,

3. At s  3

G(s) H (s) 

K1 ( s  1) s( s  2)( s  5)

  0.7, K1  15, ts  3.9 sec, Here, the closed-loop pole in the root locus branch between 0 and -1 dominate the time response with oscillatory 2nd-order response superimposed.

G(s) H (s) 

K1 ( s  2) s( s  2)( s  5)

  0.7, K1  12.8, ts  1.7 sec, P.O.  4.1%

G(s) H (s) 

K1 ( s  3) s( s  2)( s  5)

  0.7, K1  5.3, ts  3.1 sec, P.O.  5.3%

Summary Of the three compensators considered, only option 2 meets the performance specifications. The recommended compensator is therefore, Gc (s)  12.8(s  2) . The time-domain responses for the four conditions are shown in Figure below.

Realization of Compensator using Passive Components It is not possible to design isolated zero or pole at origin using passive components. In that case a pair of pole and zero is produced. Compensators may be of four types: Lead compensator, Lag compensator, Lag-lead compensator, and feedback compensator. [Cascade compensator]* A. Lead Compensator

Eo ( s) R2 R1Cs  1   Ei ( s ) R  R1 / Cs R1 R2Cs  R1  R2 2 R1  1/ Cs 

s  zc s  1/ R1C s  1/    s  [( R1  R2 ) / R2 ] / R1C s  pc s  1/ 

where   R1C and  

;

R2  1. R1  R2

zc  pc

The pole-zero configuration is shown in figure above on the right side. The zero frequency gain is cancelled by an amplifier of gain 1/  . B. Lag Compensator

Gc ( s ) 

R2  1/ Cs 1  R2Cs  R1  R2  1/ Cs 1  ( R1  R2 )Cs

s  1/ R2C R2 s  1/ R2C  R  R2 1/ R2C  1 s R1  R2 s   R2  / R2C R2  R1  R2  1 s  1/    s  1/ 



The pole-zero configuration is shown in figure (a) above.

where,   R2C and



R1  R2 1 R2

C. Lag-lead Compensator

 1  1  s  s   R1C1  R2C2   s  1/  1  s  1/  2   Gc ( s)    ; s  1/  1  s  1/  2   1 1 1  1 2  s    s R1R2C1C2  R1C1 R2C2 R2C1 

where,   1,   1. Comparing left and right side we get,

And,

R1C1  1; R2C2   2 ; R1R2C1C2 =1 2 . Therefore,   1. 1 1 1 1 1 1        . R1C1 R2C2 R2C1 1  2 1  2

The pole-zero configuration is shown in figure below. j

zc2

zc1  pc1

pc2

Cascade Compensation in Time Domain Here the design specifications are converted to  and n of a complex conjugate pairof closedloop poles based on the assumption that the system will be dominated by these two complex poles and therefore its dynamic behavior can be approximated by that of a second-order system. A compensator is designed so that closed-loop poles other than the dominant poles are located very close to the open-loop zeros or far away from the j -axis so that they make negligible contribution to the system dynamics. Lead Compensation Let us consider a unity feedback system with a forward path transfer function G f ( s) . Let, sd be the location of dominant complex closed-loop pole. If the angle condition at that point is not met for uncompensated system, a compensator has to be designed so that the compensated root locus passes through sd . Let the compensator has transfer function Gc ( s) . Applying angle condition we get Gc ( sd )G f ( sd )  Gc ( sd )  G f ( sd )  180

or, Gc ( sd )    180  G f ( sd ) For a given  there is no unique location for the pole-zero pair.

Procedure for designing lead compensator is as follows: 1. From specification determine sd . 2. Draw the root-locus plot of uncompensated system and see whether only gain adjustment can yield the desired closed-loop poles. If not, calculate the angle deficiency,  . This angle will be contributed by the lead compensator.

1 T ;  1 3. Gc ( s )  K c 1 s T 4. Locate zc and pc so that the lead compensator will contribute necessary  . 5. Determine K of the compensated system from magnitude condition. s

If large error constant is required, cascade a lag network.

K . Compensate the system so as to meet the transient s ( s  1.5) response specifications: settling time  4 second. Peak overshoot for step input  20% .

Example 01 Let, G f ( s) 

2

This specifications imply that,   0.45 and n  1 .  M P  e / 1   2 and ts  4 / n      j 1   2  The desired dominant roots lie at sd  1  j 2. n n   -line

sd 2



-19.8

-1.5

The angle contribution required for the lead compensator is,   180  (2 117  75)  129 . As  is large, a double lead network is appropriate. Each section of double lead network will then contribute an angle of 64.5 at sd . Let us now locate compensator zero at s  1.7. Join the compensator zero to sd and locate the compensator pole by making an angle of  . The pole is found to be at -19.8. The open-loop transfer function of the compensated system thus becomes, 8.30( s  1.7)2 . G(s)  2 s ( s  1.5)( s  19.8)2 Dominance of the closed-loop poles (-1 ± j2) is preserved.

K .   0.5, n  2. s( s  1)( s  4) The desired dominant closed-loop poles are located at, sd  1  j1.73 . The angle condition G f ( s) 

Example 02

  180  (120  90  30)  60 .

required from the lead compensator pole-zero pair is,

Place a compensator zero close to the pole -1 at s = -1.2. Join the zero to sd and make an angle of 60 to the left of the line. The compensator pole will be found at -4.95. The open-loop transfer function of the transfer function becomes, K ( s  1.2) . G(s)  s( s  1)( s  4)( s  4.95) The gain K can be evaluated using magnitude condition at sd .

sd sd  1 sd  4 sd  4.95

K or,

zc  1/  pc  1/ 

K

sd  1.2

=

1  j1.73 j1.73 j1.73  3 j1.73  3.95 j1.73  0.2

1.99 1.73  3.463  4.312  29.527  30 1.741

   0.833  zc  0.2424      0.202 pc

We can select R1 , R2 , C to any suitable value. Lag Compensation A lag compensator is used to improve the steady-state behavior of a system while preserving satisfactory transient response. This compensation scheme is useful where there is satisfactory transient response but unsatisfactory steady-state response. Let us consider a unity feedback system with a forward path transfer function of m

G f ( s) 

K  ( s  zi ) i 1 n

s

r

 (s  p )

j  r 1

.

j

The desired closed-loop pole location sd is indicated in figure below. It is required to improve the system error constant to a specified value K ec without impairing its transient response. To accomplish this a lag compensator with pole-zero pair close to each other is required such that it contribute a negligible angle at sd . Apart from being closed to each other, the pole-zero pair is also located close to origin.

sd

The gain of the uncompensated system at sd is given by

j

Less than 10

r

sd K ( sd ) 

n



sd  p j

j  r 1 m

uc

.

s

a

 zi

d

i 1

For the compensated system, the system gain at sd is

b

sd



K ( sd )  c

r

n



j  r 1 m

s i 1

d

sd  p j  zi

a b

. As a  b , K c (sd )  K uc (sd ) . The error constant of the compensated system is given by m

K  K ( sd ) c e

c

m

 zi

zc  K uc ( sd ) pc

i 1 n

p

j  r 1

Following the above equation,

j



z i 1 n

i

p

j  r 1

zc z  K euc  c pc pc j

zc Kc  uce . Thus,  of the lag compensator is nearly equal pc K e

to the ratio of the specified error constant to the error constant of the uncompensated system. Procedure for designing lag compensator is as follows: 1. Draw the root locus plot of uncompensated system. 2. Translate the transient response specifications into a pair of complex dominant roots. Locate these roots in the uncompensated root locus plot. 3. Calculate the gain of the uncompensated system at the dominant root sd , and also evaluate the error constant. 4. Determine the factor by which the error constant of the uncompensated system should be increased to meet the specified value. Select a larger value. 5. Select zero of the compensator sufficiently close to the origin. As a guide we may construct a line making an angle less than 10 with the -line from sd . 6. The compensated pole may be located at pc  zc /  . The pole-zero pair should contribute an angle  less than 5 at sd .

K . The system is to be compensated s( s  1)( s  4) to meet the following specifications: Damping ratio  = 0.5, Settling time ts = 10 sec., Velocity error constant Kv  5 sec-1 . 4  0.8 rad/sec . Thus, the desired dominant closed-loop Using the above data, n  10  0.5

Example Consider the system with G f ( s) 

poles are required to be located at, sd  n  jn 1   2  0.4  j 0.7 .

sd Uncompensated system

-4



-1 compensated system

Now, K uc  0.8  0.9  3.7  2.66 ; and Kvuc  lim sG f ( s)  s 0

2.66  0.666 . 1 4

5  7.5 . We take  as 10. 0.666 Locate zc on the root locus taking an angle of 6 from sd. From plot, zc  0.1 .

Now,  

Thus, pc  0.1/10  0.01. Now, for the compensated system locate the point sd that lies on the -line and the compensated root locus using angle criterion. For the compensated system at sd K c (sd )  2.2. Thus, the open-loop transfer function of the compensated system is,

G(s) 

2.2( s  0.1) . s( s  1( s  4)( s  0.01)

Feedback Compensation Though cascade compensation is quite satisfactory and economical in most cases, feedback compensation may be warranted due to the following factors: 1. In nonelectrical systems, suitable cascade devices may not be available. 2. Feedback compensation often provides greater stiffness against load disturbances. The net effect of applying feedback is to apply a zero to the open-loop transfer function, which is the principal of lead compensation.

Example Consider the system with open-loop transfer function, K K K K . G f (s)  2 A f m  2 s ( s  10) s ( s  10) KA

Km s2

Kf s  10

(a)

M P  10%, K 2 s ( s  10)

(b)

ts  4sec.

s

K . s( s  10s  K ) K The characteristic equation of the system is 1  3  0. s  10s 2  K s K ( s  1/  ) It may be rewritten as, 1 2  0. s ( s  10) K ( s  1/  ) From this equation we have, G( s)  2 . s ( s  10) The previous equation shows that the net effect of rate feedback is to add a zero at s = -1/. The root locus plot of the uncompensated system is shown in figure below. Reducing the minor feedback loop we get, G( s) 

2

84

-1.1

The desired dominant roots from the given specifications are calculated as, -1 ± j1.34. (sd-) The angle contribution of the open-loop poles at this point is -2 x 128 - 8 = -264. Therefore for the point sd on the root locus, the compensating zero should make an angle of

  180  (264)  84 . The 84 line cuts the real axis at s = -1.1. Thus the open-loop transfer function of the compensated system becomes, K ( s  1.1) G(s) H (s)  2 . s ( s  10) The value of K  at sd is found as, 17.4. The velocity error constant K v is given by, K Kv  lim 2  1/  s 0 s  10s  K If this K v is acceptable, then the design is complete; otherwise an amplifier is to be introduced in the forward path outside the minor feedback loop.

KA

Km K f s 2 ( s  10)

A

s