6 003 Signals and Systems Lecture 9 6 003 Signals and Systems October 8 2009 Last Time We analyzed a mass and spring system Second Order Systems x t F K x t y t M y t y t x t y t K M y t A A y t 1 K A2 Y MK X 1 M A2 October 8 2009 Last Time Second Order Systems We also analyzed a leaky tanks system r0 t Today Look more carefully at growth and decay of oscillatory responses by studying an analogous electrical circuit h1 t 1 r 1 t r0 t r1 t r1 t h2 t r0 t 1 1 r 1 t R vi 2 r 2 t r1 t r2 t L C vo r2 t A r1 t r 2 t 1 2 r2 t A R2 A 1 A 2 R0 1 A 1 1 A 2 But First Check Yourself The canonical forms for CT and DT differ X A Y s0 s0 t A s plane What if we had used the DT canonical form for CT Y z0 Y A H X 1 s0 A h t e X u t X Delay h n z plane A What is the impulse response of this system z0n u n 1 s0 es0 t u t R z1 s plane Y s0 Y 1 H X 1 z0 R 1 s 2 st0 u t z plane 3 1 s0 es0 t u t 4 t s0 es0 t u t 5 none of the above 1 6 003 Signals and Systems Lecture 9 October 8 2009 Second Order Systems Second Order Systems Today Look more carefully at growth and decay of oscillatory responses by studying an analogous electrical circuit Solve with state variable approach R vi L State variables represent the minimum knowledge of the past t t0 needed to propagate the output into the future t t0 vo C Check Yourself Second Order Systems State variable approach determine expressions for derivatives of state variables in terms of undifferentiated state variables iL iC vC iC C vL dvC dt vL di vL L L dt R vi L iL iC v C vo C Which of the following can be state variables 1 vC and vL 3 iC and iL 5 iC and vC and iL and vL 2 4 6 1 1 dvC iC iL dt C C iC and vL vC and iL none of above diL 1 1 vL vi RiL vC dt L L Second Order Systems R vi L iL C iC v C vo Alternatively determine system functional from block diagram dvC 1 iL dt C dvC 1 iL dt C diL 1 vi RiL vC dt L vi w1 1 diL vi RiL vC dt L w2 1 L w3 Use first equation to eliminate iL from the second equation C KVL Check Yourself Determine the system functional vL KCL A w4 1 C w5 y A R 1 d2 vC dv vi RC C vC L dt dt2 Integrate twice ignoring initial conditions why 1 CvC A2 vi RCAvC A2 vC L Which node corresponds to iL 1 w1 2 A Vo V LC C 1 2 Vi Vi 1 R A L LC A 2 w2 3 w3 4 w4 7 none of the above 2 5 w5 6 y 6 003 Signals and Systems Lecture 9 October 8 2009 Second Order Systems Second Order Systems Analogous systems The effect of 0 is to scale time RLC circuit Vo Vi 1 1 2 LC A R A 1 A2 L LC 02 A2 1 Q0 A 02 A2 r 0 H 1 LC There is one 0 paired with each A Mass and spring let Ae 0 A Then the system is reduced to a function of one parameter Q K 2 MA K A2 1 M Y X 1 02 A2 0 2 2 Q A 0 A 02 A2 1 Q0 A 02 A2 r 0 K M H Ae2 1 Leaky tanks 1 2 R2 1 2 A R0 1 1 1 A 1 A2 1 2 1 2 1 02 A2 0 2 2 Q A 0 A 0 r Since A 1 1 2 Scaling 1 e e2 QA A Z t d Ae Z 0 t d 0 Second Order Systems Scaling details Z t Ax t x d Find the poles by factoring the denominator of the system functional e Ay r Z r y d H Want x t y 0 t It follows that Let 0 r 0 t d 0 d pe0 pe1 Z 0 t y 0 d 0 0 Z t 1 Q and pe0 pe1 1 Alternatively substitute Ae 1 and find roots of denominator e s 1 se2 se 1 0 Q Then e 0 t Ay Ae2 Ae2 1A e e Ae2 e pe1 A 1 Q 1 pe0 A 1 y 0 d 0 Ay 0 t e Ax t 0 Ax t Either way the result is s 1 1 2 se pe0 pe1 1 2Q 2Q Ae 0 A Second Order Systems Second Order Systems Map pole locations as a function of Q For small Q the poles are at Q and 1 1 If Q 0 then Q 1 se2 se 1 0 Q z se plane 1 Q Im se 1 Now if e s 1 the se2 term can be neglected and se Q 1 Or if e s 1 the 1 term can be neglected and se Q 3 1 Q Q Re se 6 003 Signals and Systems Lecture 9 October 8 2009 Second Order Systems Second Order Systems As Q 12 the two poles converge to a double pole at se 1 For Q 12 the poles have imaginary parts 1 se 2Q s 1 2 1 2Q 1 se j 2Q s 1 1 2 e j fd 2Q Im se se plane se plane Im se fd 1 2 Re se 1 Re se e f d The poles are complex conjugates and lie on the unit circle Second Order Systems Second Order Systems For Q 3 frequency of oscillation is nearly 1 0 986 fd 1 We have now found the locus of pole locations for the dimensionless Ae version of the system functional se plane Im se Ae2 1 1 1 e e2 QA A Re se 1 Ae2 e e 1 pe0 A 1 pe1 A se plane Im se 1 1 Re se 1 Q 3 is often called the high Q region 1 Check Yourself Second Order Systems The impulse response of the RLC circuit is then a weighted sum of the modes 02 A2 02 A A H 1 A 2 A2 p0 p1 1 p0 A 1 p1 A 1 Q 0 0 Find locus of pole locations for original system functional 02 A2 02 A2 1 p0 A 1 p1 A 1 Q0 A 02 A2 Im s s plane For 0 Q 21 the modes are real valued exponentials When x t t y t 02 …
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