Control SystemsTime domainAnalysisTime domainAnalysis of Control Systems of Control Systems0 2 4 6 8 1 0 1 200 .20 .40 .60 .811 .21 .41 .61 .8 wn* ty(t)ze ta=0 .1 , 0.2 , 0.4 , 0 .7 , 1 .0, 2 .02Control SystemsOutline Outline- Types of Responses- First Order Systems- Second Order Systems- Response Specifications- Effect of Zeros- Higher Order Systems- Effects of Additional Poles3Types of Responses Types of ResponsesNatural Response: Response due to the effects of the initial ConditionsForced Response:Response due to the effects of theinputsTransient Response: The initial part of the response prior to settling downSteady State Response:The final part of the response after settling down4Types of Responses Types of ResponsesNatural ResponseForced ResponseTransient ResponseSteady State Responsey(t)tT(s)r(t) y(t)r(0)5ojSystem Response & Pole Locations System Response & Pole Locations6First Order Systems First Order Systemsa(s + a)r(t) y(t)vA first order system without zeros can be described by the transform function shown.vIf the input is a step function , i.e. R(s) =1/s Then the Step Response is Y(s) = T(s) R(s) = Taking the inverse transform, the step response yields y(t) = 1 - e-at) ( a s sa+a =10 0.5 1 1.5 2 2.5 300.050.10.150.20.250.30.350.40.451ty(t)transient steady-statex-a7First Order Systems First Order SystemsvLet us examine the significance of a being the only parameter needed to describe the transient response. When t =1/a37 . 01/ 1= =

=

e ea tat63 . 0 37 . 0 1 1 ) (/ 1/ 1====

=a tata te t yv orv We now use the above three equations to define three transient response performance specifications8First Order Systems First Order SystemsSpecifications Specifications1. Time ConstantWe call 1/a the time constant of the response, where the time constant can be described as the timefor e-atto decay to 37% of its initial value. Alternately the time constant is the time it takes for the step response to riseto 63%of its final or steady state value.2. Rise TimeRise time is defined as the time for the waveform to go from0.1 to 0.9 of its final value. Thus, rise time isthe difference in time aty(t)=0.9 andy(t)=0.1. Hence ,Tr=2.2/a3. Settling TimeSettling time is defined as the time for the response to reach and stay within 2% of its final value.Letting y(t)=0.98and solving for time t in the step response, we find the settling time to be Ts=4/a9Second Order Systems Second Order Systems A general second-order system is given as follows The two important quantities responsible for the shape of the system response are the natural frequency and the damping ratio defined as follows Natural Frequency: It is the frequency of oscillation of the systemwithout damping.Damping Factor:It is the rate at which the system response decays to the steady state value S2+ a s + bbs T=S2+2 ns + n2n2=whereb = n2 and a = 2 n10System Poles Locations Responsey(t)Second Order Systems Second Order SystemsIntroductory Example9 992+ + s sy(t)=1 + 0.171e-7.854t+ 0.171 e-1.146tS1 = -7.854 , S2 = -1.1460 2 0 4 0 6 0 8 0 1 0 0 1 2 000 . 10 . 20 . 30 . 40 . 50 . 60 . 70 . 80 . 91x xS1=-1 + j2.82,S2=-1 - j2.82y(t)=1 - 1.06e-tcos (8 t -19.47)9 292+ + s s0 2 0 4 0 6 0 8 0 1 0 0 1 2 000 . 20 . 40 . 60 . 811 . 21 . 4xxPole Location:s1,2= - n+- n 2 lY(s)T(s)R(s) = 1/sS2+2 ns + n2n211System Pole Locations Responseb a s sb+ +2T(s)Y(s) R(s) = 1/sSecond Order Systems Second Order SystemsIntroductory Example .. Continued992+ sS1 = j3 , S2 =-j3 y(t )= 1 - cos 3t0 2 0 4 0 6 0 8 0 1 0 0 1 2 0 1 4 0 1 6 0 1 8 000 . 20 . 40 . 60 . 811 . 21 . 41 . 61 . 82xxy(t)=1 - 3 t e-3t- e-3t9 692+ + s sS1 = -3 ,S2 = -30 2 0 4 0 6 0 8 0 1 0 0 1 2 000 .10 .20 .30 .40 .50 .60 .70 .80 .91xxPole Location:s1,2= - n+- n 2 l12Second Order Systems Second Order Systems0 2 0 4 0 6 0 8 0 1 0 0 1 2 000 .10 .20 .30 .40 .50 .60 .70 .80 .91Overdamped0 2 0 4 0 6 0 8 0 1 00 1 2000 .20 .40 .60 .811 .21 .4Underdamped0 2 0 4 0 6 0 8 0 1 00 1 20 1 40 1 60 1 8000 .20 .40 .60 .811 .21 .41 .61 .82Undamped0 2 0 4 0 6 0 8 0 1 00 1 2000 .10 .20 .30 .40 .50 .60 .70 .80 .91Critically dampedPoles:real andequal at- nPoles:ComplexConjugate at- n+ j n1- 2and- n+ j n1- 2Poles:Imaginaryatj n and- j nPoles:real at- n+ n 2 land- n+n 2 l = 00l= l> lResponse as a function ofdamping factor13% Compute step response for a second-order system t=[0:0.1:12]; num=[1];zeta1=0.1; den1=[1 2*zeta11]; zeta2=0.2; den2=[1 2*zeta21];zeta3=0.4; den3=[1 2*zeta31]; zeta4=0.7; den4=[1 2*zeta41];zeta5=1.0; den5=[1 2*zeta51];zeta6=2.0;den6=[1 2*zeta61]; [y1,x,t]=step(num,den1,t); [y2,x,t]=step(num,den2,t);[y3,x,t]=step(num,den3,t); [y4,x,t]=step(num,den4,t);[y5,x,t]=step(num,den5,t); [y6,x,t]=step(num,den6,t);%plot(t,y1,t,y2,t,y3,t,y4,t,y5,t,y6)xlabel('wn*t'), ylabel('y(t)')title('zeta = 0.1,0.2,0.4,0.7,1.0,2.0'), gridSecond Order Systems0 1 2 3 4 5 6 7 8 9 1 0- 0 .8- 0 .6- 0 .4- 0 .200 .20 .40 .60 .81 w n* ty(t)/wnze ta=0 .1 , 0 .2 5 , 0 .5 ,1 .00.250.10.51.0Impulse ResponseThe effect of 14%Compute step response for a second-order system t=[0:0.1:12]; num=[1];zeta1=0.1; den1=[1 2*zeta11]; zeta2=0.2; den2=[1 2*zeta21];zeta3=0.4; den3=[1 2*zeta31]; zeta4=0.7; den4=[1 2*zeta41];zeta5=1.0; den5=[1 2*zeta51]; zeta6=2.0; den6=[1 2*zeta61][y1,x,t]=step(num,den1,t); [y2,x,t]=step(num,den2,t);[y3,x,t]=step(num,den3,t); [y4,x,t]=step(num,den4,t);[y5,x,t]=step(num,den5,t); [y6,x,t]=step(num,den6,t);plot(t,y1,t,y2,t,y3,t,y4,t,y5,t,y6)xlabel(' wn*t'), ylabel('y(t)')title('zeta = 0.1, 0.2, 0.4, 0.7, 1.0, 2.0'), gridSecond Order SystemsThe effect of 0 2 4 6 8 1 0 1 200 . 20 . 40 . 60 . 811 . 21 . 41 . 61 . 8 w n* ty(t)ze ta =0 .1 ,0 . 2 ,0 . 4 ,0 .7 , 1 .0 ,2 . 00.20.10.40.7Step Response1.02.015Response Specifications Response SpecificationsSecond Order Systems1. Peak Time Tp: It is the time required to reach the maximum peak value of the response. That isTp =taty(t) = ymaxPO =2. Percent Overshoot PO: The increase in response magnitude above the steady state value as a percentage of the steady state value. That is(ymax - yss ) yss16Response Specifications Response SpecificationsSecond Order Systems3. Rise Time Tr: It is the time required for the response to go from 10% to90% of its steady state value. That isTr =t90%- t10%4. Delay Time Td: It is the time required for the response to reach 50% of its steady state value. That isTd =t50%5. Settling Time Ts: It is the time required for the transient response to reachand stay within +-2% of its steady state valueTs =t2%0f ss170y(t)t11+H1+H.9. 1TpPeak TimeTsSettlingTimeSteady State ErroressTrRise TimeOvershoot.5MpPeakResponseTdDelay Time1.2Response Specifications Response SpecificationsSecond Order Systems18Response SpecificationsResponse Specifications - - Underdamped Case Underdamped CaseSecond Order SystemsTp =-----------Tn 1 - 2PO =e-( T/

1 - 2x l00Ts =-----4 nxxnU njn 1 - 2- jn 1 - 2= cos U19Response SpecificationsResponse Specifications - - Underdamped Case Underdamped CaseSecond Order SystemsTr =-----------------------1- 0.4167 + 2.917 2nTd =----------------------1.1- 0.125 + 0.469 2nxxnU njn 1 - 2- jn 1 - 2= cos U0 < < 120Response SpecificationsResponse Specifications - - Underdamped Case Underdamped CaseSecond Order Systemsn3n2n1n1 < n2< n3l= 0 214U2U3U4l