VIEWS: 133 PAGES: 13 CATEGORY: College POSTED ON: 6/28/2012
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.