ECE 1 Circui Analysis Lesson 37 Chaper 8: Second Order Circuis Discuss Exam Daniel M. Liynski, Ph.D.
Exam CH 1-4: On Exam 1; Basis for work CH 5: Operaional Amplifiers CH 6: Capaciors and Inducor CH 7-8: Laer CH 9: Sinusoids and Phasors CH 1: Sinusoidal Seady Sae CH 11: AC Power Analysis
Pracice Exam Topics Sinusoid and Phasor Basics Impedance concep Operaional Amplifiers RLC AC circui conceps AC power ransfer and load maching Circui Theorems in AC circui analysis 3
Exam Prep Tex - Chaper summaries Tex Chaper review quesions HW Topics covered Class Noes commens and examples 4
Exam Forma (Tenaive) Approx 5 problems One shor answers OpAmp AC Circui Analysis AC power 5
ECE 1 Circui Analysis Chaper 8 Second-Order Circuis Copyrigh The McGraw-Hill Companies, Inc. Permission required for reproducion or display. 6
8.1 Examples of Second Order RLC circuis (1) Wha is a nd order circui? A second-order circui is characerized by a secondorder differenial equaion. I consiss of resisors and he equivalen of wo energy sorage elemens. RLC Series RLC Parallel RL T-config RC Pi-config 7
8. Finding Iniial and Final Values Finding Iniial and Final Values of v(), i(), dv()/, di()/, v( ), i( ) 8
8. Finding Iniial and Final Values Key Poins for Finding Iniial and Final Values: 1. Polariy of v() and i() using passive sign convenion (see Figs. 6.3 & 6.3). Capacior volage is always coninuous so v( + ) = v( - ) for swich a = 3. Inducor curren is always coninuous so i( + ) = i( - ) for swich a = 9
8. Finding Iniial and Final Values Example 8.1: The swich in Fig. 8. has been closed for a long ime. I is opened a =. Find: (a) v( + ), i( + ), (b) dv( + )/, di( + )/, (c) v( ), i( ) 1
8. Finding Iniial and Final Values Example 8.: In he circui of Fig. 8.5, calculae: (a) i L ( + ), v C ( + ), v R ( + ), (b) di L ( + )/, dv C ( + )/, dv R ( + )/, (c) i L ( ), v C ( ), v R ( ) 11
8.3 Source-Free Series RLC Circuis (1) The soluion of he source-free series RLC circui is called as he naural response of he circui. The circui is excied by he energy iniially sored in he capacior and inducor. The nd order of expression d i R L di i LC How o derive and how o solve? 1
8.3 Source-Free Series RLC Circuis () Mehod is illusraed on pp. 3-34 of exbook (review) 13
8.3 Source-Free Series RLC Circuis (3) There are hree possible soluions for he following nd order differenial equaion: d i R L di i LC d i di => i General nd order Form where R and 1 L LC The ypes of soluions for i() depend on he relaive values of and 14
8.3 Source-Free Series RLC Circuis (4) There are hree possible soluions for he following nd order differenial equaion: d i di i 1. If > o, over-damped case s1 s i( ) A1 e Ae where s 1,. If = o, criical damped case ( ) ( A A 1 e where s 1, i ) 3. If < o, under-damped case i( ) e ( B1 cos B sin ) where d 15
Example 1 8.3 Source-Free Series RLC Circuis (5) If R = 1 Ω, L = 5 H, and C = mf in Fig. 8.8, find α, ω, s 1 and s. Wha ype of naural response will he circui have? Pracice Problem 8.3 16
Example 8.3 Source-Free Series RLC Circuis (6) The circui shown below has reached seady sae a = -. If he make-before-break swich moves o posiion b a =, calculae i() for >. Pracice Problem 8.4. Please refer o lecure or exbook for more deailed elaboraion. Answer: i() = e.5 [5cos1.6583 7.538sin1.6583] A 17
8.4 Source-Free Parallel RLC Circuis (1) Le 1 v( ) i( ) I L v() = V Apply KCL o he op node: v R 1 L v C dv Taking he derivaive wih respec o and dividing by C The nd order of expression d v 1 RC dv 1 LC v 18
8.4 Source-Free Parallel RLC Circuis () There are hree possible soluions for he following nd order differenial equaion: d v dv 1 1 v where and RC LC 1. If > o, over-damped case s1 s v( ) A1 e A e where s 1,. If = o, criical damped case v ( ) ( A A 1 ) e where s 1, 3. If < o, under-damped case v( ) e ( B1 cos B sin ) where d 19
Example 3 8.4 Source-Free Parallel RLC Circuis (3) Refer o he circui shown below. Find v() for >. Pracice Problem 8.6 modified. Please refer o lecure or exbook for more deailed elaboraion. Answer: v() = 66.67(e 1 e.5 ) V
8.5 Sep-Response Series RLC Circuis (1) The sep response is obained by he sudden applicaion of a dc source. The nd order of expression d v R L dv v LC v s LC The above equaion has he same form as he equaion for source-free series RLC circui. The same coefficiens (imporan in deermining he frequency parameers). Differen circui variable in he equaion. 1
8.5 Sep-Response Series RLC Circuis () The soluion of he equaion should have wo componens: he ransien response v () & he seady-sae response v ss (): v( ) v ( ) v ( ) The ransien response v is he same as ha for source-free case v s1 s ( ) A1 e Ae (over-damped) v ( ) ( A1 A ) e (criically damped) v ( ) e ( A1 cos A sin ) ss (under-damped) The seady-sae response is he final value of v(). v ss () = v( ) The values of A 1 and A are obained from he iniial condiions: v() and dv()/.
8.5 Sep-Response Series RLC Circuis (3) Example 4 Having been in posiion for a long ime, he swich in he circui below is moved o posiion b a =. Find v() and v R () for >. Pracice Problem 8.7. Please refer o lecure or exbook for more deailed elaboraion. Answer: v() = {1 + [( cos3.464 1.1547sin3.464)e ]} V v R ()= [.31sin3.464]e V 3
8.6 Sep-Response Parallel RLC Circuis (1) The sep response is obained by he sudden applicaion of a dc source. The nd order of expression d i 1 RC di i LC Is LC I has he same form as he equaion for source-free parallel RLC circui. The same coefficiens (imporan in deermining he frequency parameers). Differen circui variable in he equaion. 4
8.6 Sep-Response Parallel RLC Circuis () The soluion of he equaion should have wo componens: he ransien response v () & he seady-sae response v ss (): i( ) i ( ) i ( ) The ransien response i is he same as ha for source-free case i s1 s ( ) A1 e Ae (over-damped) i ( ) ( A1 A ) e (criical damped) i ( ) e ( A1 cos A sin ) ss (under-damped) The seady-sae response is he final value of i(). i ss () = i( ) = I s The values of A 1 and A are obained from he iniial condiions: i() and di()/. 5
8.6 Sep-Response Parallel RLC Circuis (3) Example 5 Find i() and v() for > in he circui shown below: Pracice Problem 8.8 modified. Please refer o lecure or exbook for more deail elaboraion. Answer: v() = Ldi/ = 5xsin = 1sin V 6