Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6EE201 Lecture 21 P. 1Waveform GenerationUse RL or RC circuit to obtain a specific waveform. Response is controlled by a switch.Recall general solution for RL or RC circuits (with sources).x(t) = x() + [x(to+) - x()]e-(t-to)/Suppose two excitation levels are required to activate (turn off or turn on) a switch:x(t1) = X1(e.g. turn on @ iL(t1) = 3 A)x(t2) = X2(e.g. turn off @ iL(t2) = 1 A)EE201 Lecture 21 P. 2The time required for circuit to go from one excitation level to the other is, X1 - x() X2 - x()t2 – t1 = ln [ ](21.1)This is the elapsed time formula.Example: Plot vc(t) for the circuit below when the switch is activated three times (closed, opened, closed). 0.5 mF 12 u(t) V3 k6 k+_vC(t)+_EE201 Lecture 21 P. 3Initially, switch is open as shown, and it closes when vc(t) equals 9 V. Switch opens again when vc(t) equals 5 V. vc(0+) = 0 V.General solution for RC circuit:vC(t) = vC() + [vC(to+) - vC()]e-(t-to)/t = ta: switch closest = tb: switch openst = tc: switch closesStep 1: Find waveform for 0 t ta when open.vC(0+) = 0 V; vC() = 12 VRTH = 6 k; RTHC = 3 svC(t) = 12(1 - e-t/3) VEE201 Lecture 21 P. 4How long is this solution valid? Use elapsed time formula (21.1).0 - 12 9 - 12ta – 0 = 3ln [ ] = 4.16 sec.Step 2: Find waveform for ta t tb when closed.vC(ta-) = vC(ta+) = 9 V; vC() = 4 VRTH = 2 k; RTHC = 1 svC(t) = 4 + 5e-[(t – ta)/1] VvC(t) = vC() + [vC(to+) - vC()]e-(t-to)/How long to go from 9 V to 5 V? Use elapsed time formula (21.1).9 - 4 5 - 4tb – ta = 1ln [ ] = 1.61 sec.EE201 Lecture 21 P. 5Step 3: Find waveform for tb t tc when reopens.vC(tb-) = vC(tb+) = 5 V; vC() = 12 VRTH = 6 k; RTHC = 3 svC(t) = 12 - 7e-[(t – tb)/3] VvC(t) = vC() + [vC(to+) - vC()]e-(t-to)/How long to go from 5 V to 9 V? 5 - 12 9 - 12tc – tb = 3ln [ ] = 2.54 sec.EE201 Lecture 21 P. 6Step 4: Plot solutions to obtain complete waveform.t (s)vC(t)369ta24 68tbtcPeriod: time interval for waveform to repeat itself (tc – ta = 4.15 sec).Waveform frequency = 1/4.15 s = 0.24 Hz5 V9
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