6.003: Signals and Systems Lecture 9 October 8, 200916.003: Signals and SystemsSecond-Order SystemsOctober 8, 2009Last TimeWe analyzed a mass and spring system.x(t)y(t)F = Kx(t) − y(t)= M¨y(t)+KMA A−1x(t) y(t)˙y(t)¨y(t)YX=KMA21 +KMA2Last TimeWe also analyzed a leaky tanks system.r0(t)r1(t)r2(t)h1(t)h2(t)τ1˙r1(t) = r0(t) − r1(t)τ2˙r2(t) = r1(t) − r2(t)+1τ1A+1τ2Ar0(t) r2(t)˙r1(t) r1(t) ˙r2(t)− −R2R0=A/τ11 + A/τ1×A/τ21 + A/τ2Second-Order SystemsToday: Look more carefully at growth and decay of oscillatory re-sponses by studying an analogous electrical circuit.vivoR LCBut First ...The canonical forms for CT and DT differ.+As0X Y+Delayz0X YH =YX=A1 − s0AH =YX=11 − z0Rh(t) = es0tu(t) h[n] = zn0u[n]A →1sR →1zs-plane s-planez-plane z-planeCheck YourselfWhat if we had used the DT canonical form for CT?+As0X YWhat is the impulse response of this system?1. s0es0tu(t)2. st0u(t)3. 1 + s0es0tu(t)4. δ(t) + s0es0tu(t)5. none of the above6.003: Signals and Systems Lecture 9 October 8, 20092Second-Order SystemsToday: Look more carefully at growth and decay of oscillatory re-sponses by studying an analogous electrical circuit.vivoR LCSecond-Order SystemsSolve with state variable approach.State variables represent the minimum knowledge of the past (t < t0)needed to propagate the output into the future (t > t0).Check YourselfvCiCvLiLiC= CdvCdtvL= LdiLdtWhich of the following can be state variables?1. vCand vL2. iCand vL3. iCand iL4. vCand iL5. iCand vCand iLand vL6. none of aboveSecond-Order SystemsState variable approach: determine expressions for derivatives ofstate variables in terms of (undifferentiated) state variables.vivC= vovLR LCiLiCdvCdt=1CiC=1CiL(KCL)diLdt=1LvL=1L(vi− RiL− vC) (KVL)Second-Order SystemsDetermine the system functional.vivC= vovLR LCiLiCdvCdt=1CiLdiLdt=1L(vi− RiL− vC)Use first equation to eliminate iLfrom the second equation:Cd2vCdt2=1L(vi− RCdvCdt− vC) .Integrate twice (ignoring initial conditions ... why?)CvC=1L(A2vi− RCAvC− A2vC)VoVi=VCVi=A2LC1 +RLA +1LCA2Check YourselfAlternatively, determine system functional from block diagram.dvCdt=1CiLdiLdt=1L(vi− RiL− vC)+ +1L1CRA A− −viyw1w2w3w4w5Which node corresponds to iL?1. w12. w23. w34. w45. w56. y7. none of the above6.003: Signals and Systems Lecture 9 October 8, 20093Second-Order SystemsAnalogous systems.RLC circuitVoVi=1LCA21 +RLA +1LCA2=ω20A21 +ω0QA + ω20A2ω0=r1LCMass and springYX=KMA21 +KMA2=ω20A21 +ω0QA + ω20A2ω0=rKMLeaky tanksR2R0=1τ1τ2A21 +1τ1+1τ2A +1τ1τ2A2=ω20A21 +ω0QA + ω20A2ω0=r1τ1τ2Second-Order SystemsThe effect of ω0is to scale time.H =ω20A21 +ω0QA + ω20A2There is one ω0paired with each A → leteA = ω0A.Then the system is reduced to a function of one parameter Q.H =eA21 +1QeA +eA2Since A =Zt−∞dτ,eA =Zω0t−∞d(ω0τ).ScalingScaling details.Ax(t) =Zt−∞x(τ)dτeAy(r) =Zr−∞y(ρ)dρWant x(t) = y(ω0t).Let ρ = ω0τr = ω0tdρ = ω0dτTheneAy(ω0t) =Zω0t−∞y(ω0τ)d(ω0τ) = ω0Zt−∞y(ω0τ)dτ = ω0Ay(ω0t)eAx(t) = ω0Ax(t)eA = ω0ASecond-Order SystemsFind the poles by factoring the denominator of the system functional.H =eA21 +1QeA +eA2=eA2(1 − ep0eA)(1 − ep1eA).It follows thatep0+ ep1= −1Qand ep0ep1= 1 .Alternatively, substituteeA →1esand find roots of denominator:es2+1Qes + 1 = 0Either way, the result ises = ep0, ep1= −12Q±s12Q2− 1 .Second-Order SystemsMap pole locations as a function of Q.If Q ≈ 0, then1Q 1.es2+1Q|{z}1es + 1 = 0Now if |es| 1, the es2term can be neglected and es ≈ −Q.Or if |es| 1, the +1 term can be neglected and es ≈ −1Q.Second-Order SystemsFor small Q, the poles are at −Q and −1Q.Re esIm eses-plane−Q−1Q6.003: Signals and Systems Lecture 9 October 8, 20094Second-Order SystemsAs Q →12the two poles converge to a double pole at es = −1.es = −12Q±s12Q2− 1 2Re esIm eses-plane−1Second-Order SystemsFor Q >12, the poles have imaginary parts.es = −12Q± js1 −12Q2≡ −eσ ± jfωdRe esIm eses-plane−1−eσfωd−fωdThe poles are complex conjugates, and lie on the unit circle.Second-Order SystemsFor Q > 3, frequency of oscillation is nearly 1 (0.986 < fωd< 1).Re esIm eses-plane−1−11Q > 3 is often called the “high-Q” region.Second-Order SystemsWe have now found the locus of pole locations for the dimensionlesseA version of the system functional.eA21 +1QeA +eA2=eA2(1 − ep0eA)(1 − ep1eA).Re esIm eses-plane−1−11Check YourselfFind locus of pole locations for original system functional.ω20A21 +ω0QA + ω20A2=ω20A2(1 − p0A)(1 − p1A).Re sIm ss-planeSecond-Order SystemsThe impulse response of the RLC circuit is then a weighted sum ofthe modes.H =ω20A21 +1Qω0A + ω20A2=ω20p0− p1A1 − p0A−A1 − p1A.For 0 < Q <12the modes are real-valued exponentials.When x(t) = δ(t),y(t) =ω20p0− p1ep0t− ep1t; t ≥ 0This is the case with the tanks, and for the RLC circuit if R is large.6.003: Signals and Systems Lecture 9 October 8, 20095Second-Order Systems: ResonanceFor Q >> 1, the impulse response is oscillatory.y(t) = vC(t) = ω0sin ω0t ; t > 0This is case for mass and spring, and for RLC circuit if R is small.tvC(t)tiL(t)Second-Order Systems: ResonanceFor Q >> 1, energy flows back and forth between the capacitor andinductor, a phenomenon we call resonance.tvC(t)tiL(t)Capacitor energy (12Cv2C) is large when |vC(t)| is large.Inductor energy (12Li2L) is large when |iL(t)| = |CdvCdt| is large.Second-Order SystemsIf R > 0 then the impulse response decays with time.Such a system is lossy.As the energy flows back and forth between the capacitor and induc-tor, current flows through R on each cycle, and energy is dissipated.vivoR LCy(t) =ω20ωde−σtsin ωdt ; t ≥ 0 where σ =ω02Q.The exponential term e−σtrepresents this dissipation.Second-Order SystemsIn the high-Q region, the number of cycles before the amplitudediminishes by a factor of e is approximatelyQπ.y(t) =ω20ωde−ω0t2Qsin ωdtAmplitude decays by factor of e in one time constant =2Qω0.Cycle time =2πωd≈2πω0.Number of cycles till amplitude down by e:2Q/ω02π/ω0=Qπ.Check YourselfWhich corresponds to a Q of 10?1t2t3tSummarySecond-order systems are characterized by two numbers: Q and ω0.High Q systems are resonant.If Q > 3, the impulse response of a second order system decaysexponentially with time constant2Qω0.If Q > 3, the number of cycles before the amplitude diminishes by afactor of e is ≈
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