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1 1/3/14 hapter 8 Second order circuits A circuit with two energy storage elements: One inductor, one capacitor Two capacitors, or Two inductors i Such a circuit will be described with second order differential equation: Equialently: b; () V ; () V 1 d d b; dt dt d () V; () V1 dt We first learn how to sole nd order diff equ. Soling nd order differential equations onsider a second order differential equation b with initial condition: ()=V, ()=V 1 Need to find (t) for all t. In chapter 8, b= corresponds to source free case, b for step response. et = b/ et the roots to s s = (not =b) be s 1, s, s, s 1 For example: s 3s ( s 1)( s ) s1 1; s s s 4s 5 ( s j)( s j) s1, s 1s 5 ( s 5) s1 s 5 The solution will be constructed using the roots. Three cases: ase 1: >, two distinct real roots ase : =, two identical real roots ase 3: <, two distinct complex roots j 1

2 1/3/14 b, (1) () V, () V (I) 1 ase 1: >, two distinct real roots for s s = s, s 1 The solution to (1) is not unique. For any real numbers A 1, A, the following function satisfies (1), s1t st where A b/ t () A Ae Ae 1 This is called the general solution. To erify, t () sae 1 1 sae t () sae 1 1 sae s1t st s1t As e As e ( As e As e ) ( A Ae Ae ) st 1 st st 1 st s1t st st 1 st A1e ( s1 s1) Ae ( s s ) A A b b (t) satisfies (1) for any A 1, and A The solution will be uniquely determined by the initial conditions (I) s t b, (1) () V, () V (I) 1 The general solution t () A Ae Ae s1t st 1 where A b/ There is only one pair of A 1, A that satisfy the I 1 To find A 1, A, use I to form two equations. Recall: Ealuate (t) and (t) at t=: () A A A 1 () s A s A 1 1 To satisfy the I, we obtain A A A V sasa V A V 1 A s1 s A V 1 You may use other methods to sole for A 1 and A. t () sae sae s t s t A V A A s 1 s V 1

3 1/3/14 b, (To compare) Example: Sole 4 36, () 1; () 1 4; 3; b 6 ; 3; ; ase 1, two distinct real roots for. Step 1: Find s 1, s, s1, s 31, s1 1, s 3 A b/ 6/3 The general solution: t 3t t () Ae Ae 1 Step : Find A 1,A, use initial condition () A A 1 (*) 1 () A 3A 1 (**) 1 Sole (*) and (**) to obtain A 1 = ; A =1 s, s 1 Note:, are roots to: , 3 Note: 3 Finally, 3 t () e t e t b, (1) () V, () V (I) 1 An Example: 4 36, () 1; () 1 The general solution: Since, are negatie,, t () Ae Ae t 3t 1 lim A Another way to see /,, lim lim From 1, lim 3

4 1/3/14 ase : = Two identical roots, s 1 =s = to s s = The general solution: t () A ( A Ate ) t, A t Its deriatie: t () Ae ( A Ate ) Use I to find A 1, A. At t =, () A A V 1 t 1 1 () A A V 1 1 A V A 1 A V A 1 1 b/ ase 3: < Two complex roots to s s = et d 1, 1, The general solution: t t () A e ( Bcos tbsin t), 1 d Its deriatie: t t () e ( Bcos tbsin t) Use I to find B 1, B. () A B V 1 d t e ( B sin t B cos t) 1 () B db V 1 1 d 1 d d d d 1 d A B V A b/ B ( V1 B1)/ d 4

5 1/3/14 t t () A e ( Bcos tbsin t) 1 1 d d A t A e M A 5 d ase 3: < A t A e M Example: Sole 8 5,, 1 ompare with Solution:??,?? 4, 5,, case 3, two complex roots : Step 1: Form general solution using roots and. cos sin, Imaginary part of the roots: omplex roots: 43? 5 8?? 8 cos 3 sin 3, General solution Step : Find,, using initial conditions,, 1 4 cos 3 sin 3 3 sin 3 3 cos ; 16 3 Final solution: 8 4cos4t sin 3 5

6 1/3/14 R Practice 1: Sole 5 618, () ; () 4 Practice : Sole 4 48, () 1; () 1 Practice 3: Sole , () 1; () 1 Second order R circuits Example : Find, for t >. 1 General circuit: i 6 6 t = 3V.5H 6 1V c As always,, are continuous function of time. Use the circuit before switch to find,. For t>, satisfies dc dc c b; dt dt dc c() V; () V1 dt We also need Generally, at t=. How to find. is not continuous? 6

7 1/3/14 onerting a circuit problem to a math problem To sole a second order circuit, we need to 1. Derie the differential equation:. Find the initial condition d () (), () ( c ) b dt 8.. Finding initial alues d( ) di( ) We need to find c( ), i( ),, dt dt 1. apacitor oltage and inductor current don t jump Key points: () (), i () i () Use circuit before switch (t<) to find these two. d( ) di( ). For, dt dt onsider the circuit right after switch, i.e., at t = Apply KV to find ( ), then Apply K to find ( ), then di ( ) 1 d ( ) 1 ( ), i ( ), dt dt 1 Example: Find 4 1V d( ) di( ) ( ), i( ),, dt dt.5h t= Use circuit before switch to find At t = i 4 1V.1F ( ), i ( ), 1 ( ) A, 4 4 By continuity,, 4 To find 4 1V onsider t =,.5H.1F V di( ) ( ) A/ s dt ; 4 A 1 d( ) i( ) V / s dt.1 7

8 1/3/14 hapter 8 circuits 1). Source free series R circuits ). Source free parallel R circuits 3). Step response of series R circuits 4). Step response of parallel R circuits 5). Other second order circuits We will focus on 1) and 3). We study 3) first. 1) is a special case of 3) Step response of series R circuits 1 A general circuit Before switch (t<), and may not be in series. Determine, from circuit before switch by considering as short, as open. After switch ( t> ), and in series R By Theenin s theorem, the two terminal circuit can be replaced by a oltage source and a resistor 8

9 1/3/14 By Theenin s theorem, the two terminal circuit can be replaced by a oltage source and a resistor 1 R Rearranging the elements does not change loop current and capacitor Gien We call (, ) Theenin s equialent w.r.t ( ) V, d( ) V dt 1 The equialent circuit: V th R th By KV, R Vth Gien d d R th V dt dt (V th, R th ) : Theenin s equialent w.r.t ; th ( ) V, d( ) V dt hoose as key ariable. We derie diff. equation for Express other ariables in terms of : d i ; dt ; 1 1 Normalize: d Rth d 1 Vth ; dt dt 9

10 1/3/14 Step response of series R circuit: 1 V th R th If V th =, source free R, A special case d Rth d 1 Vth ; d( ) ( ) V, V1 dt dt dt d d As compared with b; dt dt general form Rth 1 Vth b,, b, A th V ( ) The solution: ase 1: ; s, s 1 t () ( ) Ae Ae s1t st 1 ase : ; s1 s t () ( ) ( A Ate ) t 1 ase 3: ;, s, s j d 1 t t () ( ) e ( Bcos tbsin t) 1 d A 1,A,or B 1,B will be determined by initial conditions d d ase 1: 1 ; s, s 1 t () ( ) Ae Ae s t 1 s t 1 Find A 1,A from () ( ) A1 A d( ) s1a1sa dt ase : ; s1 s t () ( ) ( A Ate ) t 1 Find A 1,A from () ( ) A1 d( ) A A dt 1 ase 3: ;, s, s j d 1 t t () ( ) e ( Bcos tbsin t) 1 d d d Find B 1,B from () ( ) B1 d( ) B B dt 1 d 1

11 1/3/14 Key points for step resp. of series R circuits: Obtain, from circuit before switch d Rth d 1 Vth ; dt dt Rth 1,, ( ) For circuit after switch, obtain Theenin s equialent (V th,r th ) w.r.t Assign by passie sign conention i ( ) i () Depending on how is assigned. Then V d( ) i ( ) th dt Then follow straightforward math computation on preious slide Key parameters to obtain: V th R th, from circuit before switch, from circuit after switch. 1 Example 1: Find (t) for t >. t = 4V 5 1H.5F Step 1: find (), () from circuit before switch 4V 5 =4/6=4A = 4V 1 Since, 1 Step : Find V th, R th, w.r.t from circuit after switch 4V 5 V th =4V, R th =5. 1H.5F ( ) = 4V 16/ 1 11

12 1/3/14 Step 3: Math computation R th.5,, ase 1.5 s, s s11, s4 1 1 The general solution: t () 4 Ae Ae t 4t 1 Note: t Find A 1, A from initial condition: () = 4V; d( )/dt= =16 64 t 4 4t Final solution: t () 4 e e V /3, 4/3 Practice 4: Find d( ) di( ) ( ), i( ),, dt dt R1 4 i 3u(t)A.5F V.6H 1

13 1/3/14 Practice 5: Find (t) for t >. R1 t = 6 3V 4 i.f 4.5H Practice 6: The switch is at position a for a long time before swinging to position b at t =. Find i (t), (t) for t >. R1 a b 6V 1 t = 1 16mF i.5h 13

14 1/3/14 Example : Find i(t), (t) for t >. 6 6 t =.5H 6 i 1/8F Step 1: find (), i() from circuit before switch i 3V 1V 3V 1V (3 1) i() 5A 6//(66) 4 A simple circuit in hapter. By oltage diision, 6 6 (3 1) 1 V 66 By KV, ()=V Step : After switch 6 6.5H 6 i 1V 5,? Since, 5 1/8F R th =(66)//6=4, V th =1V = ( ) i 5/1/8 4/ R th =?, V th =? Step 3: math?? Which case? R th 4; 4.5.5/8 ase. General solution: (t) = 1(A 1 A t)e 4t Find A 1,A using (), : () 1 A1 A 1 =1, A = d( ) 4A1 A 4 dt Final solution: = 1 1 e 4t V Find inductor current, = (1/8)*1( 4)e 4t = 5 e 4t A 14

15 1/3/14 For parallel R circuit i By Norton s Theorem I N i R R i i hoose i as key ariable. Derie diff equ. with K di di d i, ir, i dt R dt dt ir i i I di N di i I N Rdt dt di 1 di 1 I 1 1 I N Normalize: i,, b N dt R dt R Same math problem. No parallel R circuits in homework or Final Exam. Example 3: Find (t) for t >. 1V 3u(t)A 6 Recall:, t < ut () 1, t > 6.5H 1 16mF Thus for t<, 3u(t)A=A The current source is turned off. Replace it with open circuit for t< Key parameters to obtain: (), i () from circuit before switch (t<) V th, R th w.r.t. from circuit after switch (t>). Step 1: Find, from circuit for t< 15

16 1/3/14 Step 1: Find, from circuit for t< For t <, 3u(t)A=, current source open, open, short 1V 6 16mF 6 1 Since =, 1V, 6Ω 1Ω in the outer loop are in series By oltage diision: ,, Step : For t>, 3u(t)A=3A. Need to find, w.r.t 1V 3A 6 6.5H 1 16mF 16

17 1/3/14 Step : Fot t>, 3u(t)A=3A. Need to find, w.r.t 1V 3A 6 For 1V 3A 6 3A 6 6.5H 16mF 1 For, turn off 1V with short, turn off 3A with open 6 6 R 66//1 =1Ω 1 6 ( ) 1 By KV, R 3618, by ohms law , Step 3: Math Step : Another way to look at the circuit for t> 1V 6 3A.5H 16mF 6 1 To see connection better, moe 1 Ω to the left 6 1V R 3A 1.5H 16mF 6 61//61Ω Under D condition, 8, 18,

18 1/3/14 Step 3: math From step 1, step, found 1Ω, 8186, 8, 1.5 1; 1 15,.5.16, 3, The general solution: t 6e B cos 5t B sin 5t To satisfy initial condition: Soling equations to get 18; 36 Step response: t 6e 18 cos 5 36 sin 5 18