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ELECTRIC CIRCUITS ECSE-2010 Spring 2003 Class 15. ASSIGNMENTS DUE. Today (Tuesday): Homework #5 Due Computer Project #1 Report Due Activities 15-1, 15-2 (In Class) No Regular Class on Wednesday: Section 2 will meet to review Exam I Strictly Optional Thursday: Experiment #5 Report Due
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ASSIGNMENTS DUE • Today (Tuesday): • Homework #5 Due • Computer Project #1 Report Due • Activities 15-1, 15-2 (In Class) • No Regular Class on Wednesday: • Section 2 will meet to review Exam I • Strictly Optional • Thursday: • Experiment #5 Report Due • Activities 16-1, 16-2 (In Class)
EXAM I - STATISTICS • Class Average = 75.1 • Class Median = 77.1 • Section 1 Average = 74.1 • Section 1 Median = 76.5 • Section 2 Average = 75.7 • Section 3 Median = 77.0 • Section 3 Average = 75.3 • Section 3 Median = 78.0
REVIEW Output Input Circuit y(t) = yN + yF x(t) = Switched DC
REVIEW • 1ST Order Switched DC Circuit: • yo = y(t = to+) • yss = y(t ) • to = Switching Time • = Time Constant = Req C for Ckt with 1 C = L / Req for Ckt with 1 L • Req = Equivalent Resistance Of Dead Network Seen at Terminals of C or L
REVIEW • Simply Write Down Output: • y(t) = yss + (yo - yss) e - (t - to)/ • Find yss, yo , directly from circuit • Follow 4 Step Procedure • Draw Ckt at t = t0-: Find vC(t0-) = vC(t0+); iL(t0-) = iL(t0+) • Draw Ckt at t = t0+: Find yo • Draw Ckt at t : Find yss • DC Steady State, C => Open Ckt; L => Short Ckt • Draw Dead Network: Find Req => Calculate • Practiced in Activity 14-1, Observed in Experiment 6; Now do Activity 15-1
ACTIVITY 15-1 • Two Switching Times: • Upper Position for t < 0 • Lower Position at t = 0 • Upper Position again at t = .3 sec • Follow 4 Step Procedure for Each: • For 0 < t < .3s: y(t) = yss1 + (yo1 - yss1) e - t / • => Find yss1, yo1, • For t > .3: y(t) = yss2 + (yo2 - yss2) e - (t - .3) / • => Find yss2, yo2, • Will have 2 different time constants
ACTIVITY 15-1 • Part a); Initial Values: • Step 1: Draw Ckt at t = 0-; Find iL(0-) • L will be a Short Circuit (DC Steady State) • 32 V Input to this Circuit for t < 0 • Source Convert to 8 A in parallel with 4 ohms • 4//12 = 3 ohms ( now in parallel with 3 ohms) • => iL(0-) = 4 A • => v(0-) = 3 x 4 = 12 V
ACTIVITY 15-1 • Part a); Initial Values: • Step 2: Draw Ckt at t = 0+; Find i01; v01 • iL Cannot Change Instantaneously • => i(0+) = i(0-) = 4 A = i01 • 12 // 12 = 6 ohms, i01 flows UP thru 6 ohms • => v(0+) = - i01 x 6 = - 24 V = v01
ACTIVITY 15-1 • Part b); Find i(t) and v(t) for 0 < t < .3 msec: • Step 3: Draw Ckt in DC SS: Find iss1, vss1 • Ckt will head for Steady State with vL = 0 • => iss1 = 0 A • => vss1 = 0 V • No Source!
ACTIVITY 15-1 • Part b); i(t) and v(t) for 0 < t < .3 msec: • Step 4: Draw Dead Network: Find Req1 • Req1 = 6 + 3 = 9 ohms
ACTIVITY 15-1 • Part b); i(t) and v(t) for 0 < t < .3 msec: • For 0 < t < .3 sec: • => i(t) = 0 + (4 - 0) e – t /.2 = 4e – t /.2; t in sec • => v(t) = 0 + (-24 - 0) e – t/.2 = -24e – t /.2 ; t in sec • At t = .3 sec: • => i(.3-) = 4 e –1.5 = .893 A = i(.3+) = i02 • => v(.3-) = - 24 e –1.5 = - 5.36 V
ACTIVITY 15-1 • Part c); t > .3 sec: • Step 1: Draw Ckt at t = .3- sec: • => i(.3-) = 4 e –1.5 = .893 A • => v(.3-) = - 24 e –.1.5 = - 5.36 V
ACTIVITY 15-1 • Part c); t > .3 sec: • Step 2: Draw Ckt at t = .3+ sec: • i02 = i(.3+) = i(.3-) = .893 A • v02 = v(.3+) = 4//12 (8 - i02) = 21.3 V
ACTIVITY 15-1 • Part c); t > .3 sec: • Step 3: Draw Ckt as t : • Circuit again heads for DC Steady State • L is a Short Circuit • => iss2 = 4 A • => vss2 = 12 V (as in Part a)
ACTIVITY 15-1 • Part c); t > .3 sec: • Step 4: Draw Dead Network: • Req2 = 3 + 3 = 6 ohms • For t > .3 sec: • i(t) = 4 + (.893– 4) e – (t-.3)/.3 = 4 – 3.11e -(t-.3)/.3 • v(t) = 12 + (21.3 - 12) e – (t-.3)/3 =12 + 9.3 e -(t-.3)/.3
COMPLETE TIME RESPONSE • Can Always Relate y(t) to x(t) by a Differential Equation: • Obtain Differential Equation from Circuit using Techniques we have developed • Order of Differential Equation determined by the number of Energy Storage elements • For 1st Order Circuits, We Solved the Differential Equation:
COMPLETE TIME RESPONSE • For Higher Order Circuits => Want to Find a Better Way! • Solving Second Order or Higher Differential Equations can be a Real Pain • Would Prefer Algebraic Equations, not Differential Equations
COMPLETE TIME RESPONSE • Use Method of Laplace Transforms: • Powerful Mathematical Tool • Will transform Differential Equations into Algebraic Equations • Can then use all the Circuit Analysis Techniques we developed for Resistive Circuits!!
LAPLACE TRANSFORMS • Very Important Topic in Course: • Will Allow Us to Find the Complete Time Response, y(t) = yN(t) + yF(t) for Any Circuit with Any Input: • Very Powerful Technique: • Will Use in Other Courses • Will Introduce Laplace Transforms before looking at 2nd Order Circuits
LAPLACE TRANSFORMS • Concept :
LAPLACE TRANSFORMS • Definitions:
LAPLACE TRANSFORMS • Definitions:
