The requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot --- in time domain

Transcription

1 Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot --- in time domain (d) phase margin or (e) gain margin ---- in frequency domain. If the given plant does not satisfy 2 or 3 conditions specified from the above, a compensator circuit may be necessary to modify the overall plant characteristics (i.e of G c(s) G f(s)) to satisfy the given specifications. The controllers used are typically (other than P, I, D combination controllers) Lead compensator (to speed up transient response, margin of stability and improve error constant in a limited way) Lag compensator (to improve error constant or steady-state behavior while retaining transient response) Lead Lag compensator (A combination of the above two i.e. to improve steady state as well as transient). Lead Compensator:- is so called since it adds phase angle so that the phase curve of the combined system is lifted above the line so much as to provide the necessary phase margin. A compensator 1 1 gives an angle Ф c. If 1, then Ф c is positive. (or lead). In pole - zero form, it is given as G c = = = with = Z c/ P c also gives a lead angle. 1

2 Lead compensator can be realized with electrical or mechanical components G c(s) Letting = R 1 C; and = R 2 /(R 1 +R 2) < 1 G c(s) = / / = = --- (1) with τ = B 1 / K 1 α = K 1 /( K 1 + K 2) Y(s)/X(s) = + 1) / 1) The phase difference for G c(s) given by eqn. (1). is Ø c or tan Ф = 1 /1 dф /d = 0 gives the frequency for maximum phase difference as ω - which is the geometric mean of corner frequencies. The max. phase lead is given as tan Ф or Ф Ф The maximum gain is 20 log 1/ at high frequency, and the gain at ω G c(ω = = or log G c(ω =10 log, db is the gain contribution by compensator at the mean frequency. or sin Ф 2

3 Design of Compensator for a system Design can be carried on in frequency domain or in time domain using Root-locus Both methods are explained for lead & lag compensator cases. Let us consider Frequency domain approach first in each case. Lead Compensator : Design procedure steps For the given plant determine K v or K a based on specification Draw Bode Plot and check whether adequate Ф is provided If is specified, find the required phase margin from approx relation 0.01 Ф margin Or Ф margin = becomes the specified phase margin Determine the phase lead to be introduced by controller from Ф e = Ф s Ф uc + Where Ф s is the specified phase margin, Ф uc - phase margin of uncompensated system, and to account for shift of cross-over frequency. Let Ф m = Ф l, then determine from Ф Ф (If Ø m is more than 60 0, it is desirable to provide two identical networks each providing ½ the phase) Draw a line 10 log1/ below 0, db on the uncompensated Bode plot and mark the corresponding frequency 1 = 1/ = m ; 2 = 1/ = m / ; G c(s) = (= / / ) Compensated system is G c(s) G f(s) G c G f Example (from Nagrath & Gopal) Lead Compensator in frequency domain To design a compensator for a system G f(s) = so that Acceleration Error Const. Ka = 10, and phase margin = 350. K a = s 2 G(s) s = K = 10. Hence choose K= 10 Draw Bode Plot for G f(s) =.. Phase margin is -33 0, system quite unstable. Phase lead to be introduced by network = 35 0 (-33 0 ) +. 3

4 Chose 12 to 14 0 as gain slope in high. Thus Ф e = = Being high, two networks in tandem are required With Ф m = (1-sin 41 0 )/(1+sin 41 0 ) = log 1/ for lead network occurs at the frequency where Ф is maximum. Hence check the frequency c2 where the gain of UC system is -10 log 1/ (actually 2 x -10log 1/ in this case as 2 networks are used). c2 = 6.3 rad/sec (at db) = m c2 = 6.3 rad/sec should be designed as the mean frequency ( m ) Thus 1 = m = 2.8 r/sec ; 2 = m = 14.0 r/sec ; Thus G e(s) =.. =.. and the compensated plant is (

5 Lag Compensator can be realized as shown Letting = R 2 C ; = (R 1 + R 2) /R 2 > 1 The transfer function becomes G c(s) = ( (also 1 ) Ø c which is ve as >1 Hence the contribution to phase angle of G f(c) is negative where ρ = (K 1+K 2)/K 1 >1 Note: Relations for Ф max (at the geometric mean of = & = ) etc. are similar to lead compensator. However, lag (or decreasing the phase difference further is undesirable). The characteristic taken advantage is high frequency gain attenuation (reduction by 20log decibels) with very little decrease in phase. 5

6 Lag Compensator Design procedure This can be used only when there is a range of Ф more than in G f(s), since phase lag increases due to lag compensator. The Compensator relies on reducing the gain of the compensated plant in the critical region to make it adequately stable. Find the open-loop gain necessary to satisfy the required condition Find the frequency ω c2 when uncompensated system makes Ø 2 =Ø s + where is 5 0 to 15 0 to account for phase lag due to network Measure the gain of UC system at ω c2, and equate it to 20 log and find. Choose upper corner frequency ω 2 (= 1/) of network one octave or decade below ω c2, (ω 2 = ω c2 /2 ie. Octave or ω c2 /10 for decade) Knowing &, ω 2 (= 1/); w 1 (= 1/); G c = ( Following example will clarify: Example: Design a lag compensator so that the plant G f(s) = the specifications: ξ = 0.4, t s = 1.0 sec; K v= 5; meets Solution: K v = s G(s) s o = k/4 => K=20 ; Using approximation Ø s = ξ/0.01 = 0.4/0.01 = 40 0 (Bode plot for the system) G f(s) = is as shown. When amplitude ratio, in log terms G f(jω) is zero, the Ф curve is 4 0 below , Ф 2 < Hence the system is unstable. To the required Ф s = 40 0, add = 10 0 to compensate for negative Ф s. Ф = =

7 Frequency where Ф = is 0.6 rad/sec (ω c2 ) choose ω 2 at 2 octaves less than ω c2 ω 2 = 0.6/2 2 = 0.15 rad/sec Also the gain at A has to be brought down to 0 when Ф = Hence 20 log = 19, or = = 1/ = 0.15/8.91 = Hence the compensator to be implemented is G c(s) = ( =... The compensated plant is G c(s) G f(s)type equation here. = Compensator Design Root Locus Approach (Time-domain) Lead Compensator design procedure:: Translate the desired specifications into a dominant pair of complex poles in the Root Locus. (complex plane). If the given system G f(s) is such that G f(s) at S d then a compensator needs to be designed such that G f(s) G c(s) at Sd or G c(s) d G f(s) d = Φ (A) ie. the pole-zero pair of compensator have to contribute Φ. From the above figure, we have from properties of s, z c / n = Φ Φ So that zc /pc = and pc/n = Φ Φ Φ and d/d = z c /p = sin 7

8 (to reduce the effect of gain due to compensator, it is desirable to have large ). which leads to = ½ ( Φ (B) Thus after locating Sd in the complex plane, knowing = Cos -1 ξ, and Ø from (A), locate z c from (B), as well as p c Knowing z c and p c, the compensator is (s+z c)/(s+p c) Example Lead Compensator design - Time domain Design a compensator is give ξ = 0.5 and n = 2 for the system G f(s) = Solution: Given the damping ratio ξ = 0.5(=Cos -1 ) and n of the dominant complex pole pair, the point S d to pass through the root locus of the compensated plant is Sd= - ξ n j n1 ξ 2 ) = -1 j 1.73 = Cos = 60 0 Φ = -180 G f(s d) = ( ) = 60 0 ν= ½ ( Φ) = ( 60 60/2= 30 0 However, locating oz so that ν =30 0 makes Z c coincide with the pole of (-1,0) of the given system. Hence choose Z c = -1.2 close to -1 Now draw OP c so that angle Z c O P c = Φ = 60 0 P c cuts the real axis at so that G c =. =. Also the gain at Sd(= 1+ j 1.73) for the compensated system is K = s. s+1. s+4. s / s+1.2 ==30, in which s, s+1 etc are the lengths of vectors which can be measured and substituted. Hence compensated system = G f G c = 30(s +1.2) / s(s+1) (s+4) (s+4.95) 8

9 Lag Compensator Time Domain Method: Lag compensator is designed when the steady state response needs improvement, while transient performance is satisfactory. As the transient performance is satisfactory, the technique lies in ensuring that the root locus passes through the same Sd as for the uncompensated system while the error constant is improved. Therefore pole P is chosen close to origin and zero to its left so that β =Z c /P c >1 where Φ = PSZ is 2 to 5 0 and OSZ < 10 0, We can establish value of β using the following reasoning: Gain of UC system at Sd = K uc.... (Sd) =. Similarly K c (Sd) =..... X Since a b, we have K uc (S d) = K c (S d) Now, Error Constant K e c = K c (Sd) x = K uc (Sd) x Ke uc. Hence β = = Ke c /Ke uc Thus β can be chosen as the ratio of the expected Error Constant to the Error Constant of the given system 9

10 Example: Design a lag compensator for the system G f (s) = to meet the following specs: Damping ratio = ζ = 0.5; settling time t s =10 secs & K v 5 sec -1 Solution: From t s = 10 = 4/ζ n n = 4/ 10 x 0.5 = 0.8 Thus Sd = - ζ n + j 1 ζ 2 n = j 0.7 Sd lies on the uncompensated system as well. At Sd, gain of the UC system is K uc = OS x SA x SB = 0.8 x 0.9 x 3.7 = 2.66 Velocity Error Constant of UC system K uc v = s G f (s) = 2.66/(1 x 4) = so Thus β = K v desired / K v existing = 5/0.666 = 7.5 Say β = 10 to counter the effect of negative phase difference due to lag compensator. Now to locate the pole and zero, Draw a line from S to Z c so that OSZ c = 6 0. (small value) Determine OZ e = 0.1 (measure) Since β = Z c/p c, P c = Z/10 = 0.1/10 = 0.01 Also as the gain compensated system may be treated as the same as of UC system in which case.. G f G c =, which is the Compensated plant transfer function.... Or get K = = 2.2 instead of 2.66, by measuring the line lengths from the figure. ** ** ** 10