Lecture 15 Notes: Second-Order Systems

6.002 CIRCUITS AND ELECTRONICS Second-Order Systems 6.002 Lecture 15 Second-Order Systems 5V 5V Demo 2KΩ 50Ω S A + – 2KΩ C B large loop CGS Our old friend, the inverter, driving another. The parasitic inductance of the wire and the gate-to-source capacitance of the MOSFET are shown [Review complex algebra appendix for next class] Second-Order Systems 5V 5V Demo 50Ω 2KΩ 2KΩ S C A + – Relevant circuit: 5V + – B large loop 2KΩ CGS L B CGS Observed Output 2kΩ 5 vA t 0 vB 2kΩ 0 t vC 0 Now, let’s try to speed up our inverter by closing the switch S to lower the effective resistance t Observed Output ~50Ω 5 vA t 0 vB 0 50Ω t vC t 0 Huh! First, let’s analyze the LC network i (t ) L vI (t ) + – C + v(t ) – Node method: dv i (t ) = C dt t dv 1 v v dt C ( − ) = ∫ I L −∞ dt 1 (v I − v ) L d 2v =C 2 dt Recall di vI − v = L dt 1 t (vI − v) dt = i ∫ L −∞ d 2v LC 2 + v = vI dt time2 v, i state variables Solving Recall, the method of homogeneous and particular solutions: 1 Find the particular solution. 2 Find the homogeneous solution. L 4 steps 3 The total solution is the sum of the particular and homogeneous. Use initial conditions to solve for the remaining constants. v = vP (t ) + vH (t ) Let’s solve d 2v LC 2 + v = vI dt For input V0 vI 0 And for initial conditions v(0) = 0 i(0) = 0 [ZSR] t 1 Particular solution d 2 vP LC 2 + vP = V0 dt vP = V0 is a solution. 2 Homogeneous solution Solution to d 2 vH LC 2 + vH = 0 dt Recall, vH : solution to homogeneous equation (drive set to zero) Four-step method: A Assume solution of the form* vH = Ae st , A, s = ? so, B LCAs 2 e st + Ae st = 0 s2 = − characteristic equation 1 LC 1 s=±j LC C Roots j = −1 1 ωo = LC s = ± jω o General solution, D vH = A1e jωot + A2 e − jωot Differential equations are commonly solved by guessing solutions * 3 Total solution v(t ) = vP (t ) + vH (t ) v( t ) = V0 + A1e jωot + A2 e − jωot Find unknowns from initial conditions. v(0) = 0 0 = V0 + A1 + A2 i ( 0) = 0 dv i (t ) = C dt i( t ) = CA1 jωo e jωot − CA2 jωo e − jωot so, 0 = CA1 jωo − CA2 jωo or, A1 = A2 − V0 = 2 A V0 A1 = − 2 so, V0 jωot v( t ) = V0 − (e + e − jωot ) 2 3 Total solution Remember Euler relation e jx = cos x + j sin x (verify using Taylor’s expansion) e jx + e − jx = cos x 2 so, v( t ) = V0 − V0 cos ωot where i( t ) = CV0ωo sin ωot 1 ωo = LC The output looks sinusoidal v(t ) Plotting the Total Solution 2V0 V0 0 π π 3π 2 2 CV0ωo 0 i (t ) π 2 − CV0ωo 2π ωo t π 3π 2 2π ωo t Summary of Method 1 Write DE for circuit by applying node method. 2 Find particular solution vP by guessing and trial & error. 3 Find homogeneous solution vH A Assume solution of the form Aest . B Obtain characteristic equation. C Solve characteristic equation for roots si . D Form vH by summing Ai esit terms. 4 Total solution is vP + vH , solve for remaining constants using initial conditions. Example What if we have: L iC + C vC – vC (0) = V iC (0) = 0 We can obtain the answer directly from the homogeneous solution (V0 = 0). Example L iC + C vC – vC (0) = V iC (0) = 0 We can obtain the answer directly from the homogeneous solution (V0 = 0). vC ( t ) = A1e jωot + A2 e − jωot vC (0) = V V = A1 + A2 iC (0) = 0 0 = CA1 jωo − CA2 jωo or V or A1 = A2 = 2 V jω o t vC = (e + e − jωot ) 2 vC = V cos ωot iC = −CV ωo sin ωot Example vC V 2π ωo t CVωo iC 2π − CVωo ωo t Energy EC C: 1 CV 2 2 1 2 CvC 2 2π ωo t EL 1 2 L : LiC 2 1 CV 2 2 2π Notice 1 1 1 2 2 CvC + LiC = CV 2 2 2 2 Total energy in the system is a constant, but it sloshes back and forth between the Capacitor and the inductor ωo t RLC Circuits R L vI (t ) + – i (t ) C + v(t ) – v(t ) no R add R t Damped sinusoids with R – remember demo! See A&L Section 12.2