Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12EE201 Lecture 19 P. 1Step Response (continued)Step 1 : Write differential equations to model circuit.5+vL(t)-Example: Find iL(t) for t 1given vs = 10 u(t-1)V and iL(1-) = 1A. 2HiL(t)+-vsdiL(t)dt+RLiL(t)=10 u(t-1)LStep 2: Evaluate iL(t). iL(t) = iL() + [ iL(1+) – iL()]e-(t-1)/EE201 Lecture 19 P. 2Step 3: Calculate vL () and iL().iL(1+) = iL(1-) = 1 AvL(t) = LdiL(t)dt=-L[ iL (1+) – iL()]e-(t-1)/Lim vL (t) = 0 (short circuit)tiL() = 10 u(t-1) V5= 2 AStep 4: Solve for iL(t) for t 1. = LRTH= 2H5= 0.4s1 = 2.5 s-1EE201 Lecture 19 P. 3iL (t) = 2 + [ 1-2 ]e-2.5(t-1) AiL (t) = [ 2 - e-2.5(t-1)] AExample: Find vc(t) for t 0 if vc(0-) = -12V.+--+vc(t)+_t = 9s3018V0.6F206048u(t) VStep1: Find Thevenin equivalent for 0 t 9sEE201 Lecture 19 P. 4Step2: Evaluate vc(t) for 0 t 9s20 60 +48 V+- -12 V15Thevenin Equivalent+_12 V15 0.6 FEE201 Lecture 19 P. 5 + = d vc (t) 1 dt RTHCvc(t) = voc + [ vc(0+) – voc ]e-( t - to)/RThCvc(t)1RTHCvocFinal voltage = voc (capacitor is open)vc(0+) = vc(0-) = -12 VRTHC = (15 )(0.6 F) = 9 svc(t) = 12 + [ -12-12]e-(t-0) / 9 Vvc(t) = 12 – 24e-(t-0) / 9 VEE201 Lecture 19 P. 6Step3: Find Thevenin equivalent for t 9 s.+--+3018V0.6F206048VRTH = 30 || ( 60 || 20 )RTH = 30 || 15 RTH = 10 Use superposition to find voc.48V : voc1 = 20 || 3060+(20 || 30)48 =127248EE201 Lecture 19 P. 7voc1 = 8 V18V: voc2 = 20 || 60 30 + (20 || 60) 18 = 154518voc2 = 6 Vvoc = voc1 + voc2 = 14 VStep4: Solve for vc(t) when t 9 s.+_+vc(t)_0.6 F14 V10 EE201 Lecture 19 P. 8vc(t) = voc + [ vc(9+) – voc ]e-( t - to)/RthCvc(t) = 14 + [ vc(9+) – 14 ]e-( t - 9 )/RthCRTHC = (10)(0.6F) = 6 svc(9+) = vc(9-) = 12 - 24e-1 = 3.17 Vvc(t) = 14 + [ 3.17 – 14 ]e-( t - 9 )/6 Vvc(t) = 14 – 10.83e-( t - 9 )/6 VFinal solution is plot of results from steps 2 and 4.EE201 Lecture 19 P. 9+_+Example: First find vc(0-), then find vc(t) for t 0 given vs = 12 u(-t)+ 24u(t)V vs6 kt = 10 s3 k0.5 mF_+vc(t)Step1: Find vc(0-)Due to 12u(-t), the 12 V excitation has been applied for a long time. Thus, by the continuity property, vc(0-) = vc(0+) = 12 VEE201 Lecture 19 P. 10Step 2: Evaluate vc(t) for 0 t 10s (open switch)τ = RthC = (6 kΩ)(0.5 mF) = 3 svc(t) = voc + [ vc(0+) – voc ]e-( t – t0 )/RThCvoc = vc(∞) = 24 Vvc(t) = 24 + (12 – 24)e-t/3 = 24 – 12e-t/3 VStep 3: Compute initial condition for t < 10s, vc(10+)vc(10-) = vc(10+) = 24 – 12e-10/3 = 23.57 VStep 4: Find vc(t) for t > 10s (Find Thevenin Eq.)+_+voc = 8 VRth = 2 kΩ0.5 mF_+vc(t)EE201 Lecture 19 P. 11vc(t) = voc + [ vc(0+) – voc ]e-( t – t0 )/RThCUsing Step 2 equation (where voc is 8 V):vc(t) = 8 + [ 23.57 – 8 ]e-( t - 10 )/1vc(t) = 8 + 15.57e-(t-10) VStep 5: Find complete response using step functionsVc(t) = (24-12e-t/3)[u(t) – u(t-10)] + [8 + 15.57e-(t-10)]u(t-10) V-5 0 5 10 15 2081012141618202224Vc(t) = (24-12e-t/3)[u(t) – u(t-10)] + [8 + 15.57e-(t-10)]u(t-10) VTime (s)Vc(t)EE201 Lecture 19 P. 12Step 6: Plot
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