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Advanced Control Systems (ACS) Lecture-8 S-plane Design Dr. Imtiaz Hussain Assistant Professor email: imtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/ 1 Lecture Outline 2 Lead Compensation • Generally Lead compensators are represented by following transfer function = +1 +1 , (0 < < 1) • or = 1 + 1 + , (0 < < 1) 3 = +1 3 +10 ( = 0.1) , Pole-Zero Map Bode Diagram 1 Magnitude (dB) 0 0.5 0 -0.5 -1 -10 -5 -10 -15 -20 60 Phase (deg) Imaginary Axis Lead Compensation -8 -6 -4 Real Axis -2 0 30 0 -2 10 -1 10 0 10 1 10 2 3 10 10 Frequency (rad/sec) 4 Example-1 • Consider the position control system shown in following figure. • It is desired to design an Electronic lead compensator Gc(s) so that the dominant closed poles have the damping ratio 0.5 and undamped natural frequency 3 rad/sec. 5 Step-1 (Example-1) • Determine the characteristics of given system using root loci. C ( s) 10 2 R( s) s s 10 • The damping ratio of the closed-loop poles is 0.158. • The undamped natural frequency of the closed-loop poles is 3.1623 rad/sec. • Because the damping ratio is small, this system will have a large overshoot in the step response and is not desirable. 6 Step-2 (Example-1) • From the performance specifications, determine the desired location for the dominant closed-loop poles. • Desired performance Specifications are: It is desired to have damping ratio 0.5 and undamped natural frequency 3 rad/sec. n2 C ( s) 9 2 2 2 R(s) s 2 n s n s 3s 9 s 1.5 j 2.5981 7 Step-3 (Exampl-1) • To calculate the angle of deficiency apply Angle Condition at desired closed loop pole. Desired Closed Loop Pole s 1.5 j 2.5981 d 180 120 100.8 -2 d 40.89 -1 100.8o -2 120o -1 8 Step-5 (Exampl-1) Solution-1 • Solution-1 – If we choose the zero of the lead compensator at s = -1 so that it will cancel the plant pole at s =-1, then the compensator pole must be located at s =-3. 40.89 9 Step-5 (Example-1) Solution-1 • The pole and zero of compensator are determined as 1 = + 1 + = • The Value of determined as +1 +3 can be 40.89 yields 1 =1 =1 yields 1 =3 = 0.333 10 Step-6 (Example-1) Solution-1 • The Value of Kc can be determined using magnitude condition. ( + 1) 10 + 3 ( + 1) 10 ( + 3) =1 =−1.5+2.5981 =1 40.89 =−1.5+2.5981 ( + 3) = 10 = 0.9 =−1.5+2.5981 +1 = 0.9 +3 11 Final Design Check Root Locus Solution-1 Root Locus 5 5 0.158 3.16 3 Imaginary Axis 0.5 0 0 0.5 3 0.158 3.16 -5 -4 -3 -2 -1 Real Axis 10 () = ( + 1) 0 1 -5 -4 -3 -2 -1 0 1 Real Axis 9 () = ( + 3) 12 Step-5 (Exampl-1) Solution-2 • Solution-2 -2 40.89 -1 90o 49.2o -3 -2 -1 13 Solution-2 Step-5 (Exampl-1) • Solution-2 -2 40.89 -1 90o 49.2o -3 -2 -1 + 1.5 = 1.03 + 3.6 14 Exampl-2 • Design a lead compensator for following system. 4 s ( s 2) • The damping ratio of closed loop poles is 0.5 and natural undamped frequency 2 rad/sec. It is desired to modify the closed loop poles so that natural undamped frequency becomes 4 rad/sec without changing the damping ratio. 15 Example-3 • Consider the model of space vehicle control system depicted in following figure. • Design lead compensator such that the damping ratio and natural undamped frequency of dominant closed loop poles are 0.5 and 2 rad/sec. 16 Lag Compensation • Lag compensation is used to improve the steady state error of the system. • Generally Lag compensators are represented by following transfer function = +1 +1 • Or = 1 1 + + , , (β > 1) (β > 1) • Where is gain of lag compensator. 17 Lag Compensation = +10 3 +1 ( = 10) , Bode Diagram Pole-Zero Map 30 Magnitude (dB) 1 0.5 20 15 10 5 0 Phase (deg) 0 -0.5 -1 -10 25 -8 -6 -4 Real Axis -2 0 -30 -60 -2 10 0 10 Frequency (rad/sec) 2 10 18 Lag Compensation • Consider a lag compensator Gc(s), where 1 = + 1 + , (β > 1) • If we place the zero and pole of the lag compensator very close to each other, then at s=s1 (where s1 is one of the dominant closed loop poles) then the magnitudes 1 1 + and 1 1 + are almost equal, or 1 + ≅ (1 ) = 1 + 19 Lag Compensation • To make the angle contribution of the lag portion of the compensator small, we require 1 + −5° < 1 + < 0° • This implies that if gain of the lag compensator is set equal to 1, the alteration in the transient-response characteristics will be very small, despite the fact that the overall gain of the open-loop transfer function is increased by a factor of , where >1. 20 Lag Compensation • If the open loop transfer function of the uncompensated system is G(s), then the static velocity error constant Kv of the uncompensated system is = lim () →0 • Then for the compensated system with the open-loop transfer function Gc(s)G(s) the static velocity error constant becomes = lim () = lim →0 →0 1 + = = lim 1 →0 + 21 Design Procedure • The procedure for designing lag compensators by the rootlocus method may be stated as follows. • We will assume that the uncompensated system meets the transient-response specifications by simple gain adjustment. • If this is not the case then we need to design a lag-lead compensator which we will discuss in next few classes. 22 Example-4 • Consider the system shown in following figure. • The damping ratio of the dominant closed-loop poles is 0.491. The undamped natural frequency of the dominant closed-loop poles is 0.673 rad/sec. The static velocity error constant is 0.53 sec–1. • It is desired to increase the static velocity error constant Kv to about 5 sec–1 without appreciably changing the location of the dominant closed-loop poles. 23 Example-4 (Step-1) • The dominant closed-loop poles of given system are s = -0.3307 ± j0.5864 24 Example-4 (Step-3) • The static velocity error constant of the plant ( ) is 1.06 = lim() = lim →0 →0 +1 +2 = 0.53 −1 • The desired static velocity error constant ( ) of the compensated system is 5 −1 . = lim () = lim →0 →0 1 + = = lim 1 →0 + 25 Example-4 (Step-3) 1 = = lim 1 →0 + + = 5 = 0.53 = 10 26 Example-4 (Step-4) • Place the pole and zero of the lag compensator = • Since = 10, therefore = 1 + 1 + 1 0.1 + + 27 Example-4 (Step-4) Solution-1 • Place the zero and pole of the lag compensator at s=–0.05 and s=–0.005, respectively. • The transfer function of the lag compensator becomes +0.05 = +0.005 • Open loop transfer function is given as +0.05 1.06 ()= +0.005 (+1)(+2) (+0.05) ()= (+0.005)(+1)(+2) ℎ = 1.06 28 Example-4 (Step-5) Solution-1 • Root locus of uncompensated and compensated systems. • New Closed Loop poles are = −0.31 ± 0.55 29 Example-4 (Step-6) Solution-1 • The open-loop gain K is determined from the magnitude condition. ( + 0.05) ( + 0.005)( + 1)( + 2) =1 =−0.31+0.55 = 1.0235 • Then the compensator gain is determined as = 1.06 = = 0.9656 1.06 30 Example-1 (Step-6) Solution-1 • Then the compensator transfer function is given as + 0.05 = 0.9656 + 0.005 31 Example-4 (Final Design Check) Solution-1 • The compensated system has following open loop transfer function. 1.0235(+0.05) ()= (+0.005)(+1)(+2) • Static velocity error constant is calculated as = lim () →0 1.0235 + 0.05 = lim [ ] →0 + 0.005 + 1 + 2 1.0235 0.05 = = 5.12 −1 0.005 1 2 32 Example-4 (Step-4) Solution-2 • Place the zero and pole of the lag compensator at s=–0.01 and s=–0.001, respectively. • The transfer function of the lag compensator becomes +0.01 = +0.001 • Open loop transfer function is given as +0.01 1.06 ()= +0.001 (+1)(+2) (+0.01) ()= (+0.001)(+1)(+2) ℎ = 1.06 33 Example-4 (Step-5) Solution-2 • Root locus of uncompensated and compensated systems. Root Locus 6 • New Closed Loop poles are Compensated System 4 Imaginary Axis = −0.33 ± 0.55 Actual System 2 0 -2 -4 -6 -8 -6 -4 -2 Real Axis 0 2 4 34 Example-5 • Design a lag compensator for following unity feedback system such that the static velocity error constant is 50 sec-1 without appreciably changing the closed loop poles, which are at = −2 ± 6. 35 Lag-Lead Compensation • Lag-Lead compensators are represented by following transfer function = 1 1 + 1 + 1 2 1 + 2 + , (γ > 1 β > 1) • Where Kc belongs to lead portion of the compensator. Lag-Lead Compensation = +1 +2 +0.4 +0.1 Example-6 (Case-1) • Consider the control system shown in following figure • The damping ratio is 0.125, the undamped natural frequency is 2 rad/sec, and the static velocity error constant is 8 sec–1. • It is desired to make the damping ratio of the dominant closed-loop poles equal to 0.5 and to increase the undamped natural frequency to 5 rad/sec and the static velocity error constant to 80 sec–1. • Design an appropriate compensator to meet all the performance specifications. Example-6 (Case-1) • From the performance specifications, the dominant closed-loop poles must be at = −2.50 ± 4.33 • Since 4 235 s ( s 0.5) s 2.50 j 4.33 • Therefore the phase-lead portion of the lag–lead compensator must contribute 55° so that the root locus passes through the desired location of the dominant closed-loop poles. Example-6 (Case-1) • The phase-lead portion of the lag–lead compensator becomes 1 + 1 + 1 +0.5 = +5.02 • Thus 1 = 2 and = 10.04. • Next we determine the value of Kc from the magnitude condition: ( + 0.5) 4 + 5.02 ( + 0.5) ( + 5.02) = 4 =1 =−2.5+4.33 = 5.26 =−2.5+4.33 Example-6 (Case-1) • The phase-lag portion of the compensator can be designed as follows. • First the value of is determined to satisfy the requirement on the static velocity error constant = lim () →0 80 = lim →0 1 25.04 + 2 + 5.02 1 + 2 80 = 4.988 = 16.04 Example-6 (Case-1) • Finally, we choose the value of 2 such that the following two conditions are satisfied: Example-6 (Case-1) • Now the transfer function of the designed lag–lead compensator is given by = 6.26 +0.5 +5.02 +0.2 +0.0127 PID • PID Stands for – P Proportional – I Integral – D Derivative 44 Four Modes of Controllers • Each mode of control has specific advantages and limitations. • On-Off (Bang Bang) Control • Proportional (P) • Proportional plus Integral (PI) • Proportional plus Derivative (PD) • Proportional plus Integral plus Derivative (PID) 45 On-Off Control • This is the simplest form of control. Set point Error Output 46 Proportional Control (P) • In proportional mode, there is a continuous linear relation between value of the controlled variable and position of the final control element. () () () - () = () () • Output of proportional controller is () = () • The transfer function can be written as () = () 47 Proportional Controllers (P) • As the gain is increased the system responds faster to changes in set-point but becomes progressively underdamped and eventually unstable. 48 Proportional Plus Integral Controllers (PI) • Integral control describes a controller in which the output rate of change is dependent on the magnitude of the input. • Specifically, a smaller amplitude input causes a slower rate of change of the output. 49 Proportional Plus Integral Control (PI) ∫ () () () () ()+ + () = + 50 Proportional Plus Integral Control (PI) = + • The transfer function can be written as () 1 = + () 51 Proportional Plus derivative Control (PD) () () () () ()+ + () () = + 52 Proportional Plus derivative Control (PD) () = + • The transfer function can be written as () = + () 53 Proportional Plus derivative Control (PD) • The stability and overshoot problems that arise when a proportional controller is used at high gain can be mitigated by adding a term proportional to the time-derivative of the error signal. The value of the damping can be adjusted to achieve a critically damped response. 54 Proportional Plus derivative Control (PD) • The higher the error signal rate of change, the sooner the final control element is positioned to the desired value. • The added derivative action reduces initial overshoot of the measured variable, and therefore aids in stabilizing the process sooner. • This control mode is called proportional plus derivative (PD) control because the derivative section responds to the rate of change of the error signal 55 Proportional Plus Integral Plus Derivative Control (PID) () () () () () + - + () + ∫ () = + () () + 56 Proportional Plus Integral Plus Derivative Control (PID) = + () () + () 1 = + + () 57 Proportional Plus Integral Plus Derivative Control (PID) • Although PD control deals neatly with the overshoot and ringing problems associated with proportional control it does not cure the problem with the steady-state error. Fortunately it is possible to eliminate this while using relatively low gain by adding an integral term to the control function which becomes 58 The Characteristics of P, I, and D controllers CL RESPONSE RISE TIME OVERSHOOT SETTLING TIME S-S ERROR Kp Decrease Increase Small Change Decrease Ki Decrease Increase Increase Eliminate Kd Small Change Decrease Decrease Small Change 59 Empirical Tuning (Zeigler-Nichol’s First Method) • In the first method, we obtain experimentally the response of the plant to a unit-step input. • If the plant involves neither integrator(s) nor dominant complexconjugate poles, then such a unit-step response curve may look S-shaped 60 Zeigler-Nichol’s First Method • This method applies if the response to a step input exhibits an S-shaped curve. • Such step-response curves may be generated experimentally or from a dynamic simulation of the plant. Table-1 61 Zeigler-Nichol’s Second Method • In the second method, we first set = ∞ and = 0. • Using the proportional control action only (as shown in figure), increase Kp from 0 to a critical value Kcr at which the output first exhibits sustained oscillations. • If the output does not exhibit sustained oscillations for whatever value Kp may take, then this method does not apply. 62 Zeigler-Nichol’s Second Method • Thus, the critical gain Kcr and the corresponding period Pcr are determined. Table-2 63 Example-7 C (s) K sL e R( s ) Ts 1 1 L t 64 Example-7 Step Response 10 8 Amplitude C( s ) 10 2s e R( s ) 3s 1 6 4 2 0 0 5 10 15 Time (sec) 65 Example-8 • Consider the control system shown in following figure. • Apply a Ziegler–Nichols tuning rule for the determination of the values of parameters , and . 66 Example-8 • Transfer function of the plant is 1 = ( + 1)( + 5) • Since plant has an integrator therefore Ziegler-Nichol’s first method is not applicable. • According to second method proportional gain is varied till sustained oscillations are produced. • That value of Kc is referred as Kcr. 67 Example-8 • Here, since the transfer function of the plant is known we can find using – Root Locus – Routh-Herwitz Stability Criterion • By setting = ∞ and = 0 closed loop transfer function is obtained as follows. () = () + 1 + 5 + 68 Example-8 • The value of that makes the system marginally unstable so that sustained oscillation occurs can be obtained as 3 + 6 2 + 5 + = 0 • The Routh array is obtained as • Examining the coefficients of first column of the Routh array we find that sustained oscillations will occur if = 30. • Thus the critical gain is = 30 69 Example-8 • With gain set equal to 30, the characteristic equation becomes 3 + 6 2 + 5 + 30 = 0 • To find the frequency of sustained oscillations, we substitute = into the characteristic equation. ()3 +6()2 +5 + 30 = 0 • Further simplification leads to 6(5 − 2 ) + (5 − 2 ) = 0 6(5 − 2 ) = 0 = 5 / 70 Example-8 = 5 / • Hence the period of sustained oscillations is 2 = = • Referring to Table-2 2 5 = 2.8099 = 0.6 = 18 = 0.5 = 1.405 = 0.125 = 0.35124 71 Example-8 = 18 = 1.405 = 0.35124 • Transfer function of PID controller is thus obtained as 1 () = (1 + + ) 1 () = 18(1 + + 0.35124) 1.405 72 Example-8 73 To download this lecture visit http://imtiazhussainkalwar.weebly.com/ END OF LECTURE-8 74