Tài liệu Circuits & Electronics P15 docx

6.002 Fall 2000 Lecture 1 15 6.002 CIRCUITS AND ELECTRONICS Second-Order Systems 6.002 Fall 2000 Lecture 2 15 Second-Order Systems C A B 5V + – 5V C GS large loop 2KΩ 50Ω 2KΩ Demo 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] S 6.002 Fall 2000 Lecture 3 15 Second-Order Systems C A B 5V + – 5V C GS large loop 2KΩ 50Ω 2KΩ Demo + – 5V C GS 2KΩ B L Relevant circuit: S 6.002 Fall 2000 Lecture 4 15 Now, let’s try to speed up our inverter by closing the switch S to lower the effective resistance t v A 5 0 v B 0 t v C 0 t Observed Output 2 kΩ 2kΩ 6.002 Fall 2000 Lecture 5 15 t v A 5 0 v B 0 t v C 0 t Observed Output ~50Ω 50 Ω Huh! 6.002 Fall 2000 Lecture 6 15 v , i state variables + – C L + – )( tv )(tv I )( ti Node method: dt dv Cti =)( dt dv Cdtvv L t I =− ∫ ∞− )( 1 2 2 )( 1 d t vd Cvv L I =− I vv d t vd LC =+ 2 2 time 2 dt di Lvv I =− idtvv L t I =− ∫ ∞− )( 1 Recall First, let’s analyze the LC network 6.002 Fall 2000 Lecture 7 15 Recall, the method of homogeneous and particular solutions: ( ) ( ) tvtvv HP += 1 2 3 Find the particular solution. Find the homogeneous solution. L 4 steps The total solution is the sum of the particular and homogeneous. Use initial conditions to solve for the remaining constants. Solving 6.002 Fall 2000 Lecture 8 15 And for initial conditions v(0) = 0 i(0) = 0 [ZSR] I v 0 V 0 t Let’s solve I vv d t vd LC =+ 2 2 For input 6.002 Fall 2000 Lecture 9 15 1 Particular solution 0 2 2 Vv d t vd LC P P =+ 0 Vv P = is a solution. 6.002 Fall 2000 Lecture 10 15 0 2 2 =+ H H v d t vd LC Solution to Homogeneous solution 2 Recall, v H : solution to homogeneous equation (drive set to zero) Four-step method: D tj 2 tj 1 H oo eAeAv ωω − += General solution, Roots C o js ω ±= LC 1 o = ω Assume solution of the form * A ?s,A,Aev st H == so , 0 2 =+ stst AeeLCAs * Differential equations are commonly solved by guessing solutions 1−=j LC js 1 ±= B LC s 1 2 −= characteristic equation [...]... 2 Total energy in the system is a constant, but it sloshes back and forth between the Capacitor and the inductor 6.002 Fall 2000 Lecture 15 18 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 6.002 Fall 2000 Lecture 15 19 ... 2 2 CV0ωo 0 2π ωo t i (t ) π π 2 3π 2 2π ωo t − CV0ωo 6.002 Fall 2000 Lecture 15 13 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 . 6.002 Fall 2000 Lecture 1 15 6.002 CIRCUITS AND ELECTRONICS Second-Order Systems 6.002 Fall 2000 Lecture 2 15 Second-Order. between the Capacitor and the inductor 6.002 Fall 2000 Lecture 19 15 RLC Circuits See A&L Section 12.2 add R no R + – C R L + – )( tv )(tv I )( ti )( tv