Documents
Resources
Learning Center
Upload
Plans & pricing Sign in
Sign Out
Get this document free

Effects of Addition of Poles or Zeros to the open loop T.F

VIEWS: 133 PAGES: 13

									   Effects of Addition of Poles or Zeros to the open loop T.F of a
                          feedback control system
Case#1 Add a pole to G(s)




                                Figure: Effect of Adding Poles
Adding a pole to a system has the effect of pulling the Root Locus branches to the right, thereby,
making the system less stable

ω n → Decrease, then Ts → increase, the system is slow responding system.

Case#2 add a zero to G(s)




                                Figure: Effect of Adding Zero
Adding a zero to a system has the effect of pulling the Root Locus branches to the left, thereby,
making the system more stable
ω n → Increase, then Ts → Decrease, the system is fast responding system.

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

                                                                                  k
Example          consider a unity feedback control system with G ( s ) =                 plot the RL
                                                                            s 2 ( s + 2)
This system is unstable system. In order to stabilize this system a zero must be added, so that
           k ( s + a)
G (s) =                 . Find the range of a that the system is absolute stable
           s 2 ( s + 2)
Solution
                 S 3 + 2 S 2 + ks + ka = 0
S3              1               k
S2              2               ak
S1              2k − ak / 2     0
S0              ak              0

ak f 0 → a f 0
2k − ak / 2 f 0 → a p 2
                                Control System Design
compensator or controller placed in the forward path of a control system will modify the shape
of the loci (i.e. reshaping the root loci) if it contains additional poles and zeros.




The purpose of reshaping the R.L. generally falls into one of the following categories:

   1. A given system is stable & its transient response is satisfactory but its steady state
      error is too large.
   2. A given system is stable & its steady state error is satisfactory but its transient
      response is unsatisfactory.
   3. A given system is stable but both its transient response & its steady state error is
      unsatisfactory.
   4. A given system is unstable for all values of gain k.
Common Dynamic Controllers
Several common dynamic controllers appear very often in practice. They are known as PD, PI,
PID, phase-lag, phase-lead, and phase-lag-lead controllers.
 Compensator                                      Characteristics
      PD                                 One additional zero
      PI                       One additional zero and a pole at origin
     PID                  Two additional zeros and a additional pole at origin
   Phase lag                       One additional zero and a pole
  Phase lead                       One additional zero and a pole
 Phase lead-lag                    Two additional zeros and poles

1) PD Controller
PD stands for a proportional and derivative controller. The output signal of this
controller is equal to the sum of two signals: the signal obtained by multiplying the
input signal by a constant gain Kp and the signal obtained by differentiating and
multiplying the input signal by KD, i.e. Its transfer function is given by




                     E1 ( s )                            K
          Gc ( s ) =          = K p + K D S = K D (S + p ) = K D (S + Z )
                     E ( s)                              KD
                                                K
          GT ( s ) = Gc ( s )G ( s ) = K D ( S + p )G ( s )
                                                KD
When transient response of a feedback control system is unsatisfactory but its ess is
low, therefore, the dominant poles must be removed to the left of the S-plane. A PD
controller has the effect of adding a zero to the system, therefore, pulling the root
locus branches to the left thereby improving response parameters.

2) PI Controller
Similarly to the PD controller, the PI controller produces as its output a weighted
sum of the input signal and it's integral. Its transfer function is




                                                            KI
                                                 K p (S +      )
             E1 ( s )       K   K S + KI                    Kp         K p (S + Z )
Gc ( s ) =            = Kp + I = p       =                         =
             E (s)           S     S                    S                   S

When the transient response of a feedback control system is considered satisfactory
but its steady state error is too large it's possible to eliminate the error by increasing
system type. This must be accomplished without changing the dominant roots of
characteristic equation.

A PI controller is used to improve the system response steady state errors since it
increases the control system type by one.

In this system the quantity e1 (t) continuous to increase as long as the error e (t) is
present. E1 (t) becomes large enough to produce an output c (t) equal to the input r
(t). The error e (t) is then equal to zero. The location of the poles and zero of this
compensator must be chosen so that the transient response parameters remains
unchanged, since the pole alone would move the root locus to the right & the zero
alone would move the root locus to the left, therefore pole & zero must be near the
origin.

3) PID Controller
 The PID controller is a combination of PD and PI controllers; hence its transfer
function is given by
                         KI KDS + K pS + KI
                                2

Gc ( s ) = K p + K D S +   =
                         S         S
                     K     K
          K D (S 2 + P S + I )
                     KD    KD    K ( S + Z 1 )( S + Z 2 )
       =                       = D
                     S                    S
The PID controller can be used to improve both the system transient response and
steady state errors. This controller is very popular for industrial applications

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                      Where τ = R1C and α =
                                                                                         R2
                                                                                               < 1.
           =              =
  Ei ( s ) R + R1 / Cs      R1 R2Cs + R1 + R2                                          R1 + R2
               R1 + 1/ Cs
             2


                     s + 1/ R1C            s + zc   s + 1/ τ
        =                                =        =                       zc
            s + [( R1 + R2 ) / R2 ] / R1C s + pc s + 1/ ατ     ;             =α
                                                                          pc

The pole-zero configurations is shown in figure above on the right side. The zero frequency gain
is cancelled by an amplifier of gain 1/ α .

B. Lag Compensator
                           R2 + 1/ Cs        1 + R2Cs
            Gc ( s ) =                  =
                         R1 + R2 + 1/ Cs 1 + ( R1 + R2 )Cs
                        s + 1/ R2C       R2          s + 1/ R2C
                   =                  =
                              R + R2
 where,             1/ R2C + 1       s R1 + R2 s + ⎛ R2 ⎞ / R2C
                                                   ⎜          ⎟                            τ = R2C and
                                R2                 ⎝ R1 + R2 ⎠
                 R1 + Rs
             β = = 1 2 + 1/ τ
                          >1
                    β
                   R2 s + 1/ βτ




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




C. Lag-lead Compensator


                                                       ⎛    1 ⎞⎛         1 ⎞
                                                       ⎜s+      ⎟⎜ s +      ⎟
                                                       ⎝   R1C1 ⎠⎝     R2C2 ⎠           ⎛ s + 1/ τ 1 ⎞⎛ s + 1/ τ 2 ⎞
                                    Gc ( s ) =                                         =⎜             ⎟⎜             ⎟;
                                                  ⎛ 1
                                               s +⎜      +
                                                            1
                                                                +
                                                                    1 ⎞          1      ⎝ s + 1/ βτ 1 ⎠⎝ s + 1/ ατ 2 ⎠
                                                                        ⎟s+
                                                2

                                                  ⎝ R1C1 R2C2 R2C1 ⎠         R1 R2C1C2

                                                              where, β > 1, α < 1.
Comparing left and right side we get,

         R1C1 = τ 1 ; R2C2 = τ 2 ; R1 R2C1C2 =αβτ 1τ 2 . Therefore, αβ = 1 .
           1         1     1       1      1     1       β
And,           +       +        =      +     =      + .
          R1C1 R2C2 R2C1 βτ 1 ατ 2 βτ 1 τ 2
The pole-zero configuration is shown in figure below.
                                                              jω



                   zc2               zc1

                                                                     σ
   pc2                                          pc1



Cascade Compensation in Time Domain

Here the design specifications are converted to ζ and ωn of a complex conjugate pairof closed-
loop 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 jω -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 ) = ±180o
        or, ∠Gc ( sd ) = φ = ±180o − ∠G f ( sd )
For a given φ 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, φ . This
       angle will be contributed by the lead compensator.
                          1
                      s+
    3. Gc ( s ) = K c     T ; α <1
                          1
                      s+
                         αT
    4. Locate zc and pc so that the lead compensator will contribute necessary φ .
    5. Determine K of the compensated system from magnitude condition.

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

                                      K
Example 01 Let, G f ( s) =                    . Compensate the system so as to meet the transient
                                 s ( s + 1.5)
                                   2


response specifications: settling time ≤ 4 second. Peak overshoot for step input ≤ 20% .

This specifications imply that, ζ ≥ 0.45 and ζωn ≥ 1 . ⎡ M P = e −πζ / 1 − ζ 2 and ts = 4 / ζωn ⎤
                                                       ⎣                                        ⎦
The desired dominant roots lie at     sd = −1 ± j 2.   ⎡ −ζω ± jω 1 − ζ 2 ⎤
                                                       ⎣     n       n
                                                                              ⎦
                                                          ζ-line




                                                                         sd

                                                                                  2
                                                              Φ


   -19.8
                                                                  -1.5
The angle contribution required for the lead compensator is,
       φ = ±180 − (−2 ×117 − 75) = 129o .
As φ is large, a double lead network is appropriate. Each section of double lead network will
then contribute an angle of 64.5o 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 φ . 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
Example 02            G f (s) =               . ζ = 0.5, ωn = 2.
                           s ( s + 1)( s + 4)
The desired dominant closed-loop poles are located at, sd = −1 ± j1.73 . The angle condition
required from the lead compensator pole-zero pair is,     φ = ±180 − (−120 − 90 − 30) = 60o.




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 −1 + j1.73 j1.73 j1.73 + 3 j1.73 + 3.95
                K=                             =
                              sd + 1.2                         j1.73 + 0.2
                     1.99 × 1.73 × 3.463 × 4.312
or,             K=                               = 29.527 ≈ 30
                                1.741

  zc = 1/ τ       ⇒ τ = 0.833 ⎫      z
                               ⎬ α = c = 0.2424
 pc = 1/ ατ       ⇒ ατ = 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
                                                     K ∏ ( s + zi )
                                 G f (s) =                    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.
                                                            The gain of the uncompensated
                                                            system at sd is given by
                              sd               jω
                                                                                    n
        Less than 10°                                                          sd ∏ sd + p j
                                                                                  r

                                                                                                                                j = r +1
                                                                                                     K uc ( sd ) =             m
                                                                                                                                                      .
                                                    a                                                                     ∏s  i =1
                                                                                                                                          d   + zi

                                                                                              For the compensated system, the
                                                                                              system gain at sd is
                         b                                                                                                        n

                                                                                                                               ∏           sd + p j
                                                                                                                          r
                                                                                                                     sd
                                                                                 σ                                             j = r +1               a
                                                                                                     K ( sd ) =
                                                                                                       c
                                                                                                                              m

                                                                                                                          ∏s
                                                                                                                                                      b
                                                                                                                                      d       + zi
                                                                                                                          i =1
.
As a ≈ b , K c ( sd ) ≈ K uc ( sd ) . The error constant of the compensated system is given by
                                   m                                          m

                                ∏z         i
                                                   zc                      ∏z         i
                                                                                              zc          z
                K = K ( sd )
                  c
                  e
                         c        i =1
                                  n
                                                      ≈ K uc ( sd )          i =1
                                                                             n
                                                                                                 ≈ K euc × c
                                ∏p                                         ∏p
                                                   pc                                         pc          pc
                                               j                                          j
                                j = r +1                                   j = r +1

                                  zc   K ec
Following the above equation, β =    =      . Thus, β of the lag compensator is nearly equal
                                  pc K euc
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 ζ-line from sd .
    6. The compensated pole may be located at pc = zc / β . The pole-zero pair should
        contribute an angle γ less than 5° at sd .


                                                          K
Example Consider the system with G f ( s) =                         . The system is to be compensated
                                                 s ( s + 1)( s + 4)
to meet the following specifications: Damping ratio ζ = 0.5, Settling time ts = 10 sec.,
                                      Velocity error constant K v ≥ 5 sec-1 .
                                  4
Using the above data, ωn =             = 0.8 rad/sec . Thus, the desired dominant closed-loop
                              10 × 0.5
poles are required to be located at, sd = −ζωn ± jωn 1 − ζ 2 = −0.4 ± j 0.7 .


                                                                                 ′
                                                                                sd

                       Uncompensated system




  -4                                              -1                                    σ

                        compensated system

                                                                       2.66
Now, K uc = 0.8 × 0.9 × 3.7 = 2.66 ; and K vuc = lim sG f ( s ) =           = 0.666 .
                                                       s →0            1× 4
             5
Now, β =         = 7.5 . We take β as 10.
           0.666
Locate zc on the root locus taking an angle of 6° from sd. From plot, zc = −0.1 .
Thus, pc = −0.1/10 = −0.01.
                                                     ′
Now, for the compensated system locate the point sd that lies on the ζ-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,
                                                       2.2( s + 0.1)
                                     G ( s) =                               .
                                                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 A K f Km       K
           G f (s) = 2          = 2          .
                     s ( s + 10) s ( s + 10)

                         KA                 Kf                   Km
                                          s + 10                 s2

   (a)

                                                                                         M P = 10%,
                                                                                         ts ≤ 4sec.
                                                        K
                                                   s ( s + 10)
                                                    2




   (b)                                                αs




                                                                 K
Reducing the minor feedback loop we get, G ( s) =                          .
                                                         s( s + 10s + Kα )
                                                         2


                                                              K
The characteristic equation of the system is 1 + 3                     = 0.
                                                      s + 10s 2 + Kα s
                              Kα ( s + 1/ α )
It may be rewritten as,    1+ 2                = 0.
                                s ( s + 10)
                                       K ′( s + 1/ α )
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/α.
The root locus plot of the uncompensated system is shown in figure below.
                                                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
φ = −180 − (−264) = 84o. The 84o 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
                        K v = lim 2                  = 1/ α
                              s →0 s + 10 s + K α

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.



                                                   Km K f
                           KA
                                                 s ( s + 10)
                                                   2




                                                       αs




Examples
                                                                                  K
A control system has the open-loop transfer function,         G f (s) =                      ; k = 1.
                                                                          s ( s + 2)( s + 5)
A PD compensator of the form    Gc ( s ) = K1 ( s + a ) is to be introduced in the forward path to
achieve the performance specification:

               Overshoot less than 5%, settling time less than 2seconds

Determine the values of   K1 and a   to meet the specification.
Original Controller




                                                              For ζ = 0.5,        P.O. = 16.3%
                                                              For ζ = 0.7,        P.O. = 4.6%

                                                              ζ = 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


                                                                                   K1 ( s + 1)
                                                            G ( s) H ( s) =
                                                                              s ( s + 2)( s + 5)


                                                              ζ = 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.




                                                                                  K1 ( s + 2)
                                                            G ( s) H ( s) =
                                                                              s ( s + 2)( s + 5)


                                                            ζ = 0.7, K1 = 12.8, ts = 1.7 sec,
                                                            P.O. = 4.1%
                                                                                 K1 ( s + 3)
                                                            G (s) H ( s) =
                                                                             s ( s + 2)( s + 5)
                                                            ζ = 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.

								
To top