Question

Q4. 1 2 3 G 10 pts. Use MATLAB and plot the step response of the following systems G3 2s+1 figure. Gy on the same 2s+1 2s+1 Explain the similarities (at least 1) and differences (at least 1) between these responses. E_ figure. G, G 3 10 pts. Use MATLAB and plot the impulse response of the following systems Explain the similarities and differences between these responses. on the same 25+1 10 pts. Find the time constant (Te), pole(s), DC gain (Kac) of each of these three systems. 10 pts. Use the final value theorem to find the response of each of these three systems to a unit step input, as well as unit impulse input. Hint: Laplace (impulse function) =1 10 pts. Find the transfer function of a second order system with poles at -1 +j.Calculate the associated damping ratio, natural frequency, percentage overshoot, and time constant of this system.

Answer 1

Ans 1 :- Step response using MATLAB :-

C:Program Files MATLAB MATLAB Production Server R2015a bin Command Window Figure 1 >G1=tf ([1], [2 11); >G2-tf ([2], [2 1]) >>G3-tf ([3] , [2 1]) >>step (G1, G2, G3) > legend ('G1, 'G2','G3') fx> Help File Edit View Insert Tools Desktop Window Step Response G1 G2 2.5 G3 2 0.5 00 16 2 4 8 10 12 14 18 Time (seconds) Amplitude

bitnen these 3 step espone plbts Jimalarty eyall instant of ()Time Constant r all 3 yespones are as same atahing stoady state value at the same Diffrences betneen these 3 step ponse plots steady state Valuese fr plots have diffent ò A 3 for Ga itis 3. 9 it is 1 for Gg itis 2
Ans 2 :- Impulse response using MATLAB :-

C: Program Files MATLAB MATLAB Production Server R2015a bin Command Window AFigure 1 X >G1-tf ([1], [2 11) >G2=tf([2] , [2 1]); >G3=tf ([3], [2 1]) >impulse (G1, G2,G3) > legend ('G1', 'G2','G3 ) File Edit View Insert Tools Desktop Window Help Impulse Response 1.5 G1 G2 G3 0.5 2 10 12 8 Time (seconds) Amplitude 88

Siwalari ty impeke responses between these 3 all functons G,92& Gs is exponatay )mpulbe espone f drjiey with tine impeelse Yesponses Differnce betheen these 3 impule ses 0 The ain at t-o is diffenut for all 3 le for G it is 0s, for 42 it is ,foy Gts s st nwrse 0.Se Ame3 G pbee 2(st)Tfm 2stl mpulke responje , kt) Kee0se cnstt

St 3 て dc 夏( Ansa4 frunit stp inpit, R-1 ft stp inpot YG= H)R) (sy )9) G=G for G sH) S Vsing final val Teoan Lin s SCs+) S-0 S-0

st T ( S (2st) Lim sYs) Using final Value theoren, y() S2st) Qst Us in final Value Theorem (25H) 3 . Cst) S-20 RoIfor Unit impuke imput for unt npule inpt D T=19 stl for G Usiy final vale Theorem, ()Lis YG sos+)

for 42G-G2 Y)2 st 2s+ Using tina value Theorem, yfo)1im sYs = Lin 2s 2s+ So So Y)= 3 s+ for 43 G3-G = st! Usin final value Theran, yoosY6 in 3s S0 st SO 1t, - Ans5 Transfer funcbon poles, Tranfer function 4) = - S- (stl-j+i) 6H- 2stl+ 2 Trenfer fureton Ge) = sst

Chancteridteg etin Denominaty of Go- s4s+2 Comparing it with stindard. "doder charactenistic eatn S+2ntn hatiral frereny where e Yatio Damping n w ewn 0 707 Wn Damping o, o 707 Natorel freenyn- J 44adsec perent overshet e tot 0-1 4.33 433% Percent overshoot

sec Time Constat wn Time coustant T= 1 sec

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