CONTROLLER DESIGN Objective ~ Type of compensator ~ Design of compensator in time response ~ Design of compensator in frequency response

Type of compensator Basic block diagram + - Actuator Sensor Input Output Compensator Actuator Plant + - Controller Sensor Actuator Takes low-energy signal to transform and amplify before going to the plant Sensor Takes high-energy signal from plant to transform to low-energy signal for further action

Forward path compensator Commonly used due to easy implementation + + + -

Feedback path compensator To avoid time delay and lag + + + -

Feed-forward path compensator To absorb for disturbance + + + + + -

Inner feedback loop Used to eliminate any minor disturbance + + + + - Also known as cascade control Used to eliminate any minor disturbance + + + + - -

Velocity feedback Also known as rate feedback + + + + - - To overcome the problem of feedback instantaneous change + + + + - -

Selection of compensator (1) Flow Very noisy, thus need derivative-action Overall gain less than, integral-action will ensure no steady-state error. (2) Level Normally is of type 1, thus just required proportional action (3) Temperature Thermal delay, thus normally required PID-action (4) Pressure Characteristic can be fast or slow depending on application, thus required PI or P.

PID compensator Proportional compensator (P) + - Use to improve steady state error type + R(s) E(s) A(s) Y(s) - B(s) Consider P-compensator transfer function as and where A(s) is the actuator signal. gives better steady state but poor transient response High Too high can cause instability.

Example: + - For a unit step input Final value

Integral compensator (I) Use to eliminate error for type 0 Consider the I-compensator and actuating signal of E(s) A(s) + R(s) Y(s) - B(s) Slow response, can be used with P-compensator to remedy this problem

+ - Example: Closed-loop transfer function For a unit step input, the response is

Derivative compensator (D) Consider the D-compenator as and actuating signal as R(s) + E(s) A(s) Y(s) - B(s) Quick response No effect at steady state because no error signal Useful for controlling type 2 together with a P-controller Response only to rate of change and no effect to steady state

, , e Example: where and Compensator Determine , if damping ratio 0.6 and Error detector Amplifier Servo motor e

, , , , New open loop transfer function Give closed loop For and . C.f. nominal second order dan Hence

Proportional-integral compensator (PI) Use for combination of fast convergence zero steady state error or

+ - Example: Consider a spped control system employing a PI-controller Closed-loop transfer function For unit step input

Proportional-integral-derivative compensator (PID)

PID Tuning There are many techniques for tuning PID

Zeigler-Nichols Step Response Suitable for process control where time lag is significant k 0.63k T a L The above response can be approximated by where k is static gain, T time constant, L time delay and

Based rule-of-thumb, for the controller gains can be appoximated by PID PI P 1/a 0.9/a 1.2/a 3L 2L L/2 Controller

Zeigler-Nichols Closed-Loop Method Using Routh-Hurwitz criteria, we can determined the margainally stable’s gain, and its period of oscilation, for a controller with a gain of K With rule-of-thumb of thumb Controller PID PI P 0.833T 0.5T 0.125T

Example Design a P, PI and PID controller for a plant with an open loop transfer function as follows With a gain of K connected in cascade to the system, its characteristic equaton becomes Forming the Routh array 1 11 6 6(1+K) 1+K

and its frequency of osccilation Thus and its frequency of osccilation which gives T= 1.895 s This gives Controller P PI PID 5 4.5 6 0.947 0.237 1.57 Their step response are

Effect of adding zero and pole Example Consider an open-loop transfer function By introducing a zero and normalizing the response The step response for uncompensated, a=4 and a=8.

Reduce rise time, peak time Increases overshoot The effect of introducing zero Reduce rise time, peak time Increases overshoot As the zero approach the orgin its contribution is more significant Bandwidth increase Improve gain margin For a compensator with a pole and normalizing the output The step response for uncompensated, a=4 and a=8.

The effect of introducing pole Reduce oscillation, and as its more dominant the response becomes sluggish

Design in time response Lag compensator Use for steady state improvement without affecting the transient response Consider a lag compensator transfer function where .

Dc gain without compensator Consider a plant Dc gain without compensator where and zeros and poles position from test point. With compensator As , thus with that the steady state error is reduced.

Lead compensator For transient response where Uncompensated . Uncompensated Compensated Thus

+ - Example: Use a lead compensator so that the overall system has undamped frequency of 4 rad.s-1 and rate of decay 0.5 s-1, if + - Open loop transfer function

Third pole is determined from Dominant poles and that give a damping ratio of and Third pole is determined from

s-plane c j3.97  j c -11-10 -1 -0.5  -j c -j3.97

Hence, the dc gain Which give the closed loop transfer function

Location of zero and pole of the system together with the compensator. Satah s s  j   O  -12 -10 -1 -0.5  -j

Using the angle condition on the closed loop poles of where the angle of from i.e. the angle contribution required by Thus the compensator transfer function

Design in frequency response Basic concepts in the design criteria Increase of phase margin, the overshoot will be reduced Increase of bandwith, the response will be faster Increase of low frequency magnitude, the steady state will be reduced

Gain adjustment Use to reduce the overshoot during transeint period by incresing the phase margin dB GM  (rad.s-1)  Plot LM PM o -180 PM’ Plot  Design Procedure (i) With a chosen gain, obtain the Bode plot and the gain crossover frequency, (ii) Determine the required phase margin (iii) Obtain the gain that is required for the new cossover frequency, This gain is the extension of the gain in (i).

+ - Example: Consider a type-1 system. Determine the uncompensated phase margin. If a P-compensator is cascaded to the system, determine the required gain so that the phase margin is 30. o + -

Solution: By replacing , the frequency response in corner frequency form is and and and

(rad.s-1) 1 10 500 (dB/dekad) -20 (dB/dec) Total -40 -60 Pole at origin Real pole Real pole slope and

 (rad.s-1) 1 50 100 5000 (deg/dec) -45 Total slope 45 -90 Real pole Real pole and

(rad.s-1) 1 10 500 10000 Total slope (dB/dec) -20 -40 -60 LM (dB) 12 -8 -76 -154

 (rad.s-1) 1 50 100 5000 10000 Total slope (deg/dec) -45 -90 (deg) -166 -193 -269 >> w1=[1 10 500 10000];LM=[12 -8 -76 -154]; >> w2=[1 50 100 5000 10000];ph=[-90 -166 -193 -269 -269]; >> subplot(2,1,1);semilogx(w1,LM); subplot(2,1,2);semilogx(w2,ph); From the Bode plot, and for uncompensated system

That gives the compensator gain as , For the new phase margin, d an increase of gain That gives the compensator gain as

Lag compensator The compensator can also be presented as where . The pole is chosen to be closed with the origin, example As the zero is far away from the pole, no obvious transient response will be affected Increased in the open-loop gain will improve the steady state error Asymtote approximation of the Bode plot

Example of actual plot for and 5, dB  (rad.s1)   (rad.s-1) Example of actual plot for and 5, , >> alpha1=2;T=0.1;alpha2=5 >> sys1=tf([1 1/T],[1 1/(alpha1*T)]); >> sys2=tf([1 1/T],[1 1/(alpha2*T)]); >> bode(sys1,sys2)

Gambarajah 6.47: Plot Bode bagi pemampas mengekor dengan penghampiran asimtot

gain so we can incresed the gain for steady state improvement Observation from the plot: At low frequency, the compensator will not have influenced on the open-loop gain so we can incresed the gain for steady state improvement Design Procedure Prosedur rekabentuk adalah seperti berikut: Determine the gain K for the required steady state error and draw the Bode plot (ii) Obtained the required phase margin and add, compensator phase angle, =5 - 12 o o (iii) Compensator’s zero will be a decade below of the new gain crossover frequency of (ii). Compensator’s pole is obtained by drawing a slope of -20dB/dec from the compensator’s zero until the line touched the 0dB.

Example If the new gain crossover frequency is, , and the LM at that frequency is dB. Compensator’s zero While, compensator’s pole is obtained by finding the frequency where the slope of -20dB/dec which begin from the compensator’s zero cross the 0dB-axis. dB  (rad.s-1) 20dB/dekad -Y (iv) Adjust the gain K for the specified error coefficient

Example: Design a lag compensator so that the ramp error coefficient is 10 ses-1, gain margin of at least 20 dB and phase margin of at least 45o. Assume the compensator’s phase angle is 5o. + - Open loop transfer function Consider and K=50, which give an open loop transfer function of

and >> w1=[0.1 1 5 20 100 1000];LM=[20 0 -28 -64 -120 -200]; >> w1a=[0.1 1000];LMa=[0 0]; >> subplot(2,1,1);semilogx(w1,LM,w1a,LMa,'-'); >> w2=[0.1 0.5 2 10 50 200 1000];ph=[-90 -121 -176 -270 -333 -360 -360]; >> w2a=[0.1 1000];pha=[-180 -180]; >> subplot(2,1,2);semilogx(w2,ph,w2a,pha,'-'); From the Bode plot

Given Open loop transfer function without compensator with initial value of Thus shift 20dB upward from original plot. New phase margim with =5o, PM’=45o+5o=50o. thus ’=0.65 rad.s-1. Hence the compensator must provide –20.

>> w1=[0.1 1 5 20 100 1000];LM=[40 20 -8 -44 -100 - 180]; >> w1a=[0.1 1000];LMa=[0 0]; >> subplot(2,1,1);semilogx(w1,LM,w1a,LMa,'-');title('PLOT BODE');ylabel('Magnitud (dB)') >> w2=[0.1 0.5 2 10 50 200 1000];ph=[-90 -121 -176 -270 -333 -360 -360]; >> w2a=[0.1 1000];pha=[-180 -180]; >> w2b=[0.1 1000];phb=[-130 -130]; >>subplot(2,1,2);semilogx(w2,ph,w2a,pha,w2b,phb,'-');xlabel('Frekuensi (rad/s)');ylabel('Sudut fasa (Darjah)') At ’, At ’,

The compensator Hence pole of the compensator dB 0.0065 0.065  (rad.s-1) -20dB/dekad -20 Gain of the compensator The compensator >> syms s; >> den=expand((s+0.0065)*s*(s+5)*(s+20)*(s+1)) den = s^5+52013/2000*s^4+125169/1000*s^3+1613/16*s^2+13/20*s >> num=expand(0.1*(s+0.065)*2) num = 1/5*s+13/1000 >> num=[1/5 13/1000];den=[1 52013/2000 125169/1000 1613/16 13/20 0]; >> bode(num,den)

>> syms s; >> den=expand((s+0.065) *s*(s+5)*(s+20)*(s+1)) den = s^5+5213/200*s^4+12669/100*s^3+865/8*s^2+13/2*s >> num=expand(0.1*(s+0.65)*2) num = 1/5*s+13/100 >> num=[1/5 13/100];den=[1 5213/200 12669/100 865/8 13/2 0]; >> bode(num,den) >> num=[1/5 13/100];den=[1 5213/200 12669/100 865/8 13/2 0]; bode(num,den) sys=tf(num,den); [GM,PM,Wg,Wp] = margin(sys) GM_dB = 20*log10(Gm) GM = 693.1782 PM = 73.8732 Wg = 1.0751 Wp = 0.0192 GM_dB = 56.8169