ECE 201, Section 4Lecture 18Prof. Peter BermelFebruary 22, 2012Returning Homeworks• HW #13: N=91, µ=83%, σ=20%• HW #14: N=97, µ=89%, σ=15%2/22/2012 ECE 201-4, Prof. BermelExams• Posted Exam #1 solution on Blackboard – is it useful as is?• Exam #2 review session tentatively scheduled for Wed., March 7 at normal class time2/22/2012 ECE 201-4, Prof. BermelSolving Differential Equations• Using KVL, RL circuits obey: = + • RC circuits can be written as: =1+• General form decomposes into homogeneous and inhomogeneous equations:0 = + = +2/22/2012 ECE 201-4, Prof. BermelSolving Differential Equations• Can solve each part separately– Homogeneous part:= −= −– Inhomogeneous part: = += , constant2/22/2012 ECE 201-4, Prof. BermelSolving Differential Equations• Use separation of variables and the following rules:= ! −"! #+1$$%%= ln$ −ln$"ln'−ln( = ln'('ln( = ln()*)+,-= ()2/22/2012 ECE 201-4, Prof. BermelSolving Differential Equations• We can integrate our homogeneous ODE:= −• To obtain:./.0= −/0ln1= −1−1= *2(/20)/62/22/2012 ECE 201-4, Prof. BermelSolving Differential Equations• Overall solution is given by: = +• For RC or RL circuits, we obtain: = *2(20)/6+, constant• Where the time constant τ is given by:– RL circuits: τ=L/R– RC circuits: τ=RC2/22/2012 ECE 201-4, Prof. BermelExample 1• What is the voltage response of this circuit as a function of time if V(0)=1 mV?2/22/201210 mFVx/25 Ω+Vx-ECE 201-4, Prof. BermelSolution• Calculate Thèvenin equivalent for left-hand side:– No independent sources, so Voc=0– Thèvenin resistance obtained from KCL: 7+)-/2 = )-/5 7= −3)-/10= −10/32/22/201210 mFVx/25 Ω+Vx-ECE 201-4, Prof. BermelSolution• Applying homogeneous solution from before: = "*2/6Since τ=RC=-(1/30) s, Vo=1 mV, Qo=10 µC: =(10;)*<=Charge exceeds that of known universe in: = <=ln =>?∙ =AB =AC= 5.06s!2/22/201210 mF-10/3 ΩECE 201-4, Prof. BermelExample 2What is the current flow in this circuit before and after the switch is flipped at time t=0?2/22/2012+-9 V3 Ω9 HI+-9 V3 Ω9 Ht=0I6 Ω6 ΩECE 201-4, Prof. BermelSolutionFor t<0: I = (9 V) / (9 Ω) = 1 AFor t>0: KVL yields 9 = (3 Ω)I + (9 H) dI/dtSince L/R=3, assume solution takes the form:I = 3 - Be-t/3 2/22/2012+-9 V3 Ω9 HI+-9 V3 Ω9 Ht=0I6 Ω6 ΩECE 201-4, Prof. BermelSolutionFor t>0: 9 = 3 3−G*2HI+9J<*2HIB is set by continuity condition for current: 0 = 0 = 3−G*2/<Thus, B=3, and =31−*2/<,>02/22/2012+-9 V3 Ω9 HI+-9 V3 Ω9 Ht=0I6 Ω6 ΩECE 201-4, Prof. BermelHomework• HW #16 solution posted today• HW #17 due today by 4 pm in EE 325B• HW #18 due Fri.: DeCarlo & Lin, Chapter 8:– Problem 4– Problem 5– Problem 9(b)2/22/2012 ECE 201-4, Prof.
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