Chapter 8 Secondorder circuits A circuit with two

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Chapter 8 Second-order circuits A circuit with two energy storage elements: One inductor, one

Chapter 8 Second-order circuits A circuit with two energy storage elements: One inductor, one capacitor Two capacitors, or Such a circuit will be described with Two inductors second-order differential equation: + - C + - i. L Equivalently: L We first learn how to solve 2 nd-order diff equ. L 20

 L 20 Solving 2 nd-order differential equations Consider a second-order differential equation Need

L 20 Solving 2 nd-order differential equations Consider a second-order differential equation Need to find v(t) for all t. In chapter 8, b=0 corresponds to source free case, b≠ 0 for step response. Let the roots to s 2+2 s+ 02= 0 (not =b) be s 1, s 2, For example: The solution will be constructed using the roots. Three cases: Case 1: > 0, two distinct real roots Case 2: = 0, two identical real roots Case 3: < 0, two distinct complex roots

 L 20 Case 1: > 0, two distinct real roots for s 2+2

L 20 Case 1: > 0, two distinct real roots for s 2+2 s+ 02 = 0 The solution to (1) is not unique. For any real numbers A 1, A 2, the following function satisfies (1), This is called the general solution. To verify, v(t) satisfies (1) for any A 1, and A 2 The solution will be uniquely determined by the initial conditions (IC)

 L 20 The general solution There is only one pair of A 1,

L 20 The general solution There is only one pair of A 1, A 2 that satisfy the IC To find A 1, A 2, use IC to form two equations. Recall: To satisfy the IC, we obtain You may use other methods to solve for A 1 and A 2.

 L 20 Example: Solve The general solution: Step 2: Find A 1, A

L 20 Example: Solve The general solution: Step 2: Find A 1, A 2, use initial condition Solve (*) and (**) to obtain A 1=-2; A 2=1 Finally,

 L 20 An Example: The general solution:

L 20 An Example: The general solution:

Case 2: = 0 L 20 Two identical roots, s 1=s 2= - to

Case 2: = 0 L 20 Two identical roots, s 1=s 2= - to s 2+2 s+ 02=0 The general solution: Its derivative: Use IC to find A 1, A 2. At t = 0,

L 20 Case 3: < 0 Two complex roots to s 2+2 s+ 02=0

L 20 Case 3: < 0 Two complex roots to s 2+2 s+ 02=0 The general solution: Its derivative: Use IC to find B 1, B 2.

L 20 Case 3: < 0

L 20 Case 3: < 0

 Solution: Imaginary part of the roots: Final solution:

Solution: Imaginary part of the roots: Final solution:

R 20 Practice 1: Solve Practice 2: Solve Practice 3: Solve

R 20 Practice 1: Solve Practice 2: Solve Practice 3: Solve

Second-order RLC circuits L 21 General circuit: + + - 6 6 C t

Second-order RLC circuits L 21 General circuit: + + - 6 6 C t = 0 i. L + - 30 V L 6 0. 5 H + 10 V -

Converting a circuit problem to a math problem L 21 To solve a second

Converting a circuit problem to a math problem L 21 To solve a second order circuit, we need to 2. Find the initial condition 1. Derive the differential equation: § 8. 2. Finding initial values We need to find 1. Capacitor voltage and inductor current don’t jump Key points: Use circuit before switch (t<0) to find these two. Recall: Consider the circuit right after switch, i. e. , at t = 0+

4 + - L 21 Example: Find 0. 25 H 12 V t=0 +

4 + - L 21 Example: Find 0. 25 H 12 V t=0 + - 12 V Consider t = 0+ 2 0. 25 H + - 0. 1 F Use circuit before switch to find At t = 04 4 2 0. 1 F

Chapter 8 circuits 1). Source free series RLC circuits 2). Source free parallel RLC

Chapter 8 circuits 1). Source free series RLC circuits 2). Source free parallel RLC circuits 3). Step response of series RLC circuits 4). Step response of parallel RLC circuits 5). Other second order circuits We will focus on 1) and 3). We study 3) first. 1) is a special case of 3). L 21

§ 8. 5 Step response of series RLC circuits L 21 A general circuit

§ 8. 5 Step response of series RLC circuits L 21 A general circuit + + - Before switch (t<0), L and C may not be in series. C L After switch ( t>0 ), L and C in series + - R 0 + - L C By Thevenin’s theorem, the two terminal circuit can be replaced by a voltage source and a resistor

By Thevenin’s theorem, the two terminal circuit can be replaced by a voltage source

By Thevenin’s theorem, the two terminal circuit can be replaced by a voltage source and a resistor L + - R 0 C + - L C Given L 21

L 21 The equivalent circuit: + - V th (Vth, Rth) : Thevenin’s equivalent

L 21 The equivalent circuit: + - V th (Vth, Rth) : Thevenin’s equivalent w. r. t LC Rth L Given C By KVL, Normalize:

L 21 Step response of series RLC circuit: + - V th Rth L

L 21 Step response of series RLC circuit: + - V th Rth L C If Vth= 0, source free RLC, A special case As compared with general form The solution: Case 1: Case 2: Case 3: A 1, A 2, or B 1, B 2 will be determined by initial conditions

L 21 Case 1: Find A 1, A 2 from Case 2: Find A

L 21 Case 1: Find A 1, A 2 from Case 2: Find A 1, A 2 from Case 3: Find B 1, B 2 from

L 21 Key points for step resp. of series RLC circuits: C + +

L 21 Key points for step resp. of series RLC circuits: C + + - For circuit after switch, obtain Thevenin’s equivalent (Vth, Rth) w. r. t LC + - L Vth Rth L C Then follow straightforward math computation on previous slide Key parameters to obtain:

L 21 Example 1: Find v(t) for t > 0. t = 0 1

L 21 Example 1: Find v(t) for t > 0. t = 0 1 H 5 + - 24 V 1 0. 25 F Step 2: Find Vth, Rth, w. r. t LC from circuit after switch + - 24 V 5 1 5 + - 24 V Vth=24 V, Rth=5 1 H 0. 25 F

L 21 Step 3: Math computation The general solution: Find A 1, A 2

L 21 Step 3: Math computation The general solution: Find A 1, A 2 from initial condition: v(0) = 4 V; dv(0+)/dt=16 Final solution:

R 21 Practice 4: Find i. L 4 2 3 u(t)A 0. 5 F

R 21 Practice 4: Find i. L 4 2 3 u(t)A 0. 5 F + - 20 V 0. 6 H

R 21 Practice 5: Find v(t) for t > 0. 6 t = 0

R 21 Practice 5: Find v(t) for t > 0. 6 t = 0 i C 4 + - 30 V 0. 02 F 0. 5 H 4

Practice 6: The switch is at position a for a long time before swinging

Practice 6: The switch is at position a for a long time before swinging to position b at t = 0. Find i. L(t), v. C(t) for t > 0. a 10 60 V b t = 0 10 16 m. F 0. 5 H i L R 21 + -

Example 2: Find i(t), v(t) for t > 0. 6 i + - 30

Example 2: Find i(t), v(t) for t > 0. 6 i + - 30 V 6 6 6 t = 0 Step 1: find v(0), i(0) from circuit L 22 before switch 6 6 1/8 F i 0. 5 H + 10 V - + - 30 V + 10 V - A simple circuit in Chapter 2. By voltage division, By KVL, v(0)=20 V

Step 3: math Step 2: After switch L 22 6 6 6 i. C

Step 3: math Step 2: After switch L 22 6 6 6 i. C Case 2. i 0. 5 H + 10 V - 1/8 F General solution: v(t) = 10+(A 1+A 2 t)e-4 t A 1=10, A 2=0 Final solution: Find inductor current, Rth=? , Vth=?

For parallel RLC circuit L 22 i. R i. L + + - L

For parallel RLC circuit L 22 i. R i. L + + - L C By Norton’s IN Theorem R i. L L Choose i. L as key variable. Derive diff equ. with KCL Normalize: Same math problem. No parallel RLC circuits in homework or Final Exam. i. C C

L 22 Example 3: Find v. C(t) for t > 0. + 12 V

L 22 Example 3: Find v. C(t) for t > 0. + 12 V - 6 Recall: 3 u(t)A 6 0. 5 H 16 m. F 12 Thus for t<0, 3 u(t)A=0 A The current source is turned off. Replace it with open circuit for t<0 Key parameters to obtain:

L 22 For t < 0, 3 u(t)A=0, current source open, C open ,

L 22 For t < 0, 3 u(t)A=0, current source open, C open , L short + 12 V - 6 6 12 16 m. F By voltage division:

L 22 + 12 V - 6 3 A 6 0. 5 H 16

L 22 + 12 V - 6 3 A 6 0. 5 H 16 m. F 12

L 22 12 V + 3 A 12 V + 0. 5 H 6

L 22 12 V + 3 A 12 V + 0. 5 H 6 12 6 16 m. F 6 6 12 Step 3: Math 3 A 6 - 3 A

L 22 Step 2: Another way to look at the circuit for t>0 12

L 22 Step 2: Another way to look at the circuit for t>0 12 V + 3 A 6 - 6 0. 5 H + 12 V 12 3 A 6 12 16 m. F 6 0. 5 H 16 m. F

Step 3: math From step 1, step 2, found The general solution:

Step 3: math From step 1, step 2, found The general solution:

R 22 Practice 7: Find i. L(t), v. C(t), v. R(t) for t >

R 22 Practice 7: Find i. L(t), v. C(t), v. R(t) for t > 0. 20 m. F 4 6 + - 60 V 1 H 2 A i L t = 0 2 4

Practice 8: Find v. C(t) and v. R(t) for t > 0. R 22

Practice 8: Find v. C(t) and v. R(t) for t > 0. R 22 2 u(-t)A 2 2 H 3 A 250 m. F 4 4

Chapter 7, Chapter 8 Review Final exam: 12/16/19 (Monday), 8 -11 am, BL 210

Chapter 7, Chapter 8 Review Final exam: 12/16/19 (Monday), 8 -11 am, BL 210 One page one-sided note allowed, Calculators allowed Don’t include the solution to any circuit problem in the note!

Step response of general RC circuit One or more switches changes i structure of

Step response of general RC circuit One or more switches changes i structure of circuit at t = 0. + + - C Three key parameters: 1. Initial condition v(0)=V 0, Established from circuit before switch. Obtained by solving a DC circuit, t < 0 (capacitor = open circuit) 3. Equivalent resistance Rth, or Req with respect to capacitor for circuit after switch L 23

Step response of RL circuits + + - L 23 Step response – For

Step response of RL circuits + + - L 23 Step response – For t > 0, i L Three key parameters: (IN, RN) together as Norton’s equivalent, w. r. t inductor, for t > 0.

L 23 Key points for step resp. of series RLC circuits: + + -

L 23 Key points for step resp. of series RLC circuits: + + - For circuit after switch, obtain Thevenin’s equivalent (Vth, Rth) w. r. t LC C i. L + - L Obtain v(0), i. L(0) from circuit before switch Vth Rth i. L L i C C Assign i. C by passive sign convention Depending on how i. L is assigned. Then follow straightforward math computation on next slide Key parameters to obtain:

Case 1: Find A 1, A 2 from Case 2: Find A 1, A

Case 1: Find A 1, A 2 from Case 2: Find A 1, A 2 from Case 3: Find B 1, B 2 from L 23

 2 + - 10 V 4 H t =0 6 1 A

2 + - 10 V 4 H t =0 6 1 A

 t =0 24 V 8 + - 4 0. 05 F 10 24

t =0 24 V 8 + - 4 0. 05 F 10 24

 25 m. F 6 - 12 V + 4 t =0 1. 5

25 m. F 6 - 12 V + 4 t =0 1. 5 A

 1 A 15 6 2 A 4 3 H a b t =0

1 A 15 6 2 A 4 3 H a b t =0

 10 1 H 20 3 70 8 10 A 8 t =0 5

10 1 H 20 3 70 8 10 A 8 t =0 5 m. F

 -+ 20 4 V -+ 32 V 30 30 2 H 4

-+ 20 4 V -+ 32 V 30 30 2 H 4

 Step 2): For t > 0 (After switch): 4 H 2 + -

Step 2): For t > 0 (After switch): 4 H 2 + - 10 V t =0 6 1 A + - 10 V 2 2 + - 10 V 6 1 A Use source transformation: 6 2 + - 10 V 1 A 6 6 V 1 A

 t =0 24 V + - 8 4 0. 05 F 24 V

t =0 24 V + - 8 4 0. 05 F 24 V 24 10 + - 8 By voltage division: 4 10 24 24 V 8 + - 4 10 4 24 8 10 24

 25 m. F 6 - 12 V + 4 Step 2: For t

25 m. F 6 - 12 V + 4 Step 2: For t > 0, switch is open 1. 5 A 6 t =0 + 12 V 4 6 + 12 V 4 1. 5 A

 Step 2: For t > 0 6 1 A 15 2 A 4

Step 2: For t > 0 6 1 A 15 2 A 4 6 1 A 3 H a b 15 t =0 6 1 A 2 A 4 1 A 15 6 1 A 4 b a b t =0 6 4 4 15 a b 6 3 H 2 A 4

 10 1 H 20 3 70 8 t =0 10 20 70 General

10 1 H 20 3 70 8 t =0 10 20 70 General solution: 8 + - 80 V 8 5 m. F 8 3 10 A 5 m. F 1 H 70 20 3 10 A 8 10 Step 2: For t > 0 1 H

 32 V 30 30 4 2 H 4 V 30 4 30 4

32 V 30 30 4 2 H 4 V 30 4 30 4 V -+ 30 -+ 20 2 H 32 V -+ -+ -+ Step 2: For t > 0 4 V -+ 32 V 20 4 The general solution? Use initial conditions to find coefficients