EE201 Lecture 19 P 1 Step Response continued Example Find iL t for t 1given vs 10 u t 1 V and iL 1 1A vs iL t 5 2H vL t Step 1 Write differential equations to model circuit 10 u t 1 R diL t iL t dt L L Step 2 Evaluate iL t iL t iL iL 1 iL e t 1 EE201 Lecture 19 P 2 iL 1 iL 1 1 A Step 3 Calculate vL and iL L diL t vL t L dt iL 1 iL e t 1 Lim vL t 0 short circuit t 10 u t 1 V 2A iL 5 Step 4 Solve for iL t for t 1 2H L 5 RTH 1 2 5 s 1 0 4s EE201 Lecture 19 P 3 iL t 2 1 2 e 2 5 t 1 A iL t 2 e 2 5 t 1 A Example Find vc t for t 0 if vc 0 12V t 9s 60 20 48u t V 0 6F vc t 30 18V EE201 Lecture 19 P 4 Step1 Find Thevenin equivalent for 0 t 9s 15 60 20 48 V 12 V Thevenin Equivalent Step2 Evaluate vc t for 0 t 9s 15 12 V 0 6 F EE201 Lecture 19 1 d vc t 1 voc vc t dt RTHC RTHC vc t voc vc 0 voc e t to RThC Final voltage voc capacitor is open vc 0 vc 0 12 V RTHC 15 0 6 F 9 s vc t 12 12 12 e t 0 9 V vc t 12 24e t 0 9 V P 5 EE201 Lecture 19 P 6 Step3 Find Thevenin equivalent for t 9 s 60 48V 20 30 0 6F 18V RTH 30 60 20 RTH 30 15 RTH 10 Use superposition to find voc 12 20 30 48 48 48V voc1 72 60 20 30 EE201 Lecture 19 P 7 voc1 8 V 18V voc2 20 60 30 20 60 18 15 18 45 voc2 6 V voc voc1 voc2 14 V Step4 Solve for vc t when t 9 s 14 V 10 0 6 F vc t EE201 Lecture 19 vc t voc vc 9 voc e t to RthC vc t 14 vc 9 14 e t 9 RthC RTHC 10 0 6F 6 s vc 9 vc 9 12 24e 1 3 17 V vc t 14 3 17 14 e t 9 6 V vc t 14 10 83e t 9 6 V Final solution is plot of results from steps 2 and 4 P 8 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 vs t 10 s 6 k 3 k 0 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 V EE201 Lecture 19 P 10 Step 2 Evaluate vc t for 0 t 10s open switch RthC 6 k 0 5 mF 3 s vc t voc vc 0 voc e t t0 RThC voc vc 24 V vc t 24 12 24 e t 3 24 12e t 3 V Step 3 Compute initial condition for t 10s vc 10 vc 10 vc 10 24 12e 10 3 23 57 V Step 4 Find vc t for t 10s Find Thevenin Eq Rth 2 k voc 8 V 0 5 mF vc t EE201 Lecture 19 P 11 Using Step 2 equation where voc is 8 V vc t voc vc 0 voc e t t0 RThC vc t 8 23 57 8 e t 10 1 vc t 8 15 57e t 10 V Step 5 Find complete response using step functions Vc t 24 12e t 3 u t u t 10 8 15 57e t 10 u t 10 V EE201 Lecture 19 P 12 Step 6 Plot vc t 24 Vc t 24 12e t 3 u t u t 10 8 15 57e t 10 u t 10 V 22 20 Vc t 18 16 14 12 10 8 5 0 5 Time s 10 15 20
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