Elec 3202 Chap 7

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UAE University Department of Electrical Engineering Dr.Hazem N.NounouChapter (7)Second Order Transient CircuitsHere , we consider circuits with both capacitor , inductor , andresistors (RLC)We expect the circuit to be descried by a second order differential equation.Consider the parallel RLC circuit iS (t)iR (t) iL (t) iC (t)LCR+-VC(t)UAE University Department of Electrical Engineering Dr.Hazem N.NounouTake first derivative ( ) (t) idt(t) dvC d vL1) (t iR(t) vscttc 0 Lc0= + + + dt(t) didt(t) v dC (t) vL1dt(t) dvR1s2c2cc= + +Lets assume that the voltage across the capacitor is VC(t)(t) i (t) i (t) i (t) iC L R S + + = KCL :UAE University Department of Electrical Engineering Dr.Hazem N.Nounoudt(t) diC1(t) vLC1dt(t) dvC R1dt(t) v dscc2c2= + +0dt(t) diS=0 (t) vLC1dt(t) dvC R1dt(t) v dcc2c2= + +If IS(t) = constant (DC) Consider the series RLC :VS(t)+-RLCiUAE University Department of Electrical Engineering Dr.Hazem N.NounouKVL :( ) (t) V d iC1) (t Vdtdi(t)L i(t) R0 V (t) V (t) V (t) VStt0 CC L R S0= + + += + + + dt(t) dV(t) iC1dti(t) dLdtdi(t)RS22= + +0dt(t) dVS=Take first derivative :If VS= constant 0 (t) iC L1dtdi(t)LRdti(t) d22= + + Series RLCUAE University Department of Electrical Engineering Dr.Hazem N.NounouHence, lets assume that the differential equation we wish to solve is A (t) x adtdx(t)adt(t) d2 12X2= + +It is known that the solution x (t) can be expressed as (t) x (t) x x(t)C P + =xP(t) : particular ( forced solution )xC(t) : complementary ( natural solution)To find xP(t)The only solution is the constant xP(t) = k0UAE University Department of Electrical Engineering Dr.Hazem N.Nounou( ) ( )2P200 2 0 1 022aA(t) xaAkA k a kdtda kdtd= == + +0 (t) x adt(t) dxadt(t) dc 2c12cX2= + +To find xC(t) : ( natural solution )For simplicity lets define :20 20 1a 2 a ==UAE University Department of Electrical Engineering Dr.Hazem N.NounouWhere : : damping ratio.w0 : underdamped natural frequency0 (t) xdt(t) dx 2dt(t) dc20c02cX2= + + Assume xc(t) = k e s t| | | | | || | 0 s 2 s e k0 e k e s k 2 e s k0 e k e kdtd 2 e kdtd20 02 t st s 20t s0t s 2t s 20t s0t s22= + + = + += + + UAE University Department of Electrical Engineering Dr.Hazem N.Nounouk : non-zeroest :non-zero 0 s 2 s20 02= + + 1 s24 4 2s20 0202020 = = Characteristic polynomial The roots :1 s1 s20 0 220 0 1 = + = Complex frequencies[ rad / sec]UAE University Department of Electrical Engineering Dr.Hazem N.Nounout s2t s1 c2 1e k e k (t) x + = t s2t s1 0C P2 1e k e k k x(t)(t) x (t) x x(t)+ + = + = Q2 1 0k k k x(0) + + =k1, k2 can be found from initial conditions From x(0): From dx(0) / dt:2 2 1 1t s2 2t s1 1s k s kdtdx(0)e s k e s kdtdx(t)21+ =+ =UAE University Department of Electrical Engineering Dr.Hazem N.NounouNote :There are 3 types of responses based on the values of (S)1. Overdamped response ( > 1)S1and S2are real and distinct .20t s2t s1 0aAke k e k k x(t)2 1=+ + =k1and k2can be found from initial conditions 2 2 1 12 1 0s k s kdtdx(0)k k k x(0)+ =+ + =UAE University Department of Electrical Engineering Dr.Hazem N.Nounou2. Underdamped response ( < 1 )S1& S2are complex conjugate 20 0 1,220 0 1,220 0 1,2 - 1 j S1 - 1 S1 Smmm = = = 20 d0 1 LetDamped radian frequency[ rad / sec ] d 1,2 j S m = UAE University Department of Electrical Engineering Dr.Hazem N.Nounou( ) ( )( )( ) ( ) sin j cos ee k e k e ke k e k ke k e k k x(t) jt j2t j1t 0t j 2t j 1 0t s2t s1 0d dd d2 1 = + + = + + = + + = + Q( ) ( ) ( ) ( ) ( ) ( ) | |( ) ( ) ( ) | | t sin k j k j t cos ) k (k e k sin cos k sin cos k e k x(t)d 2 1 d 2 1t 0d d 2 d d 1t 0 + + + = + + + =t j t t j t( )2 1 22 1 1k k j Ak k A Let = + =UAE University Department of Electrical Engineering Dr.Hazem N.Nounou( ) ( )20dt 2 dt 1 0aAkt sin e t cos e A k x(t)=+ + = AA1& A2can be found from initial conditions :( ) ( )( ) ( )2 d 1t -2 d dt d 2t 1t d d dt 11 0A A dtdx(0) e A t sin e t cos Ae A e t cos t sin e Adtdx(0)A k x(0)+ = + =+ = UAE University Department of Electrical Engineering Dr.Hazem N.Nounou+ =+ ==2 d 11 020A A dtdx(0)A k x(0)aAk0 2 120 0 1,2 S S1 S Since = = = m3. Critical damped response ( = 1 )* 0 (t) x dt(t) dx 2dt(t) x d Let0 (t) x dt(t) dx 2dt(t) x dc20c2c20c20c02c2= + + = + + UAE University Department of Electrical Engineering Dr.Hazem N.NounouWhich means thatt 2t 1 ce k e k (t) x + =t 2t 1 ce k e k (t) x + = tIt is impossible for this solution to satisfy two initial conditions :It can be shown that the following solution also satisfies the differential equation:t 2t 1 0c Pe t k e k k x(t)(t) x (t) x x(t) + + = + = UAE University Department of Electrical Engineering Dr.Hazem N.Nounou2 1t 2t 2t 11 020k kdtdx(0)e k e t k e kdtdx(t)k k x(0)aAk+ =+ =+ == + =+ ==2 11 020k kdtdx(0)k k x(0)aAkUAE University Department of Electrical Engineering Dr.Hazem N.Nounou V (0)+-R LCExample :A 3 (0) i , V 12 v(0) AssumeF 1 CH 0.5 L31RL = ====Find v (t) ?UAE University Department of Electrical Engineering Dr.Hazem N.Nounou0C L1sC R1s0 (t) vLC1dt(t) dvC R1dt(t) v d2cc2c2=((

+ += + +Characteristic polynomial is :1 1.062 23 233 22 2 0C L1SC R1S000202> = = == = == + +OverdampedFrom before we knowUAE University Department of Electrical Engineering Dr.Hazem N.NounouA 363112(0) i 12 v(0) Since(1) k k 12 v(0)e k e k v(t)e k e k v(t)2 s , 1 sR2 1t 2 -2t -1t S2t S12 12 1= = =+ = = + = + = = =L L39dtdv(0)0dtdv(0)1 390dtdv(0)C 3 360 (0) i (0) i (0) iC L R == += + += + +UAE University Department of Electrical Engineering Dr.Hazem N.Nounout 2 t212 12 1t 22t1e 27 e 15 - v(t)27 k15 - k(2) 39 k 2 k39 k 2 kdtdv(0)e k 2 e kdtdv(t) + === = + = = =K KUAE University Department of Electrical Engineering Dr.Hazem N.Nounou24 V+-R LCt = 0F 4 . 0 CH 0.1 L 80 2 R ===Find vc(t) , t > 0 ?Example :If vc(0) = 0iL(0) = 0 UAE University Department of Electrical Engineering Dr.Hazem N.NounouKVL :( ) d iC1(0) V i(t) Rdtdi(t)L 24V (t) V (t) V 24ttCC R L0+ + + =+ + =0 (t) i 10 * 25dtdi(t)2800dti(t) d0 (t) iC L1dtdi(t)LRdti(t) d(t) iC1dtdi(t)Rdti(t) dL 06222222= + += + ++ + =UAE University Department of Electrical Engineering Dr.Hazem N.NounouChar. Poly. is 1 28 . 0100002800 22800800 2 2rad/sec 000 5 10 * 5 2 0 10 * 5 2 s 800 2 s0006206 2< = = = = = = = + +Underdamped 4800 j 1400 j - 1 w j w sd20 0 2 1,mmm = = =UAE University Department of Electrical Engineering Dr.Hazem N.Nounou( ) ( )0 kt sin e A t cos e A k i(t)0dt 2 dt 1 0= + + = ( ) ( ) t 4800 sin e t 4800 cos e A i(t)t 14002t 14001 + = AA1& A2can be found from initial conditions4 2 0 R (0) i (0) V4 2 (0) V (0) V (0) VLC R L= + + = + +i(0) = A1= 0Since homogenousUAE University Department of Electrical Engineering Dr.Hazem N.Nounout) sin(4800 e 0.05 i(t)0.05 AA 4800 A 40 24800 A A 40 2dtdi(0)4 2dtdi(0)L (0) vt 1400 -22 2 dd2 d 1L== = = =+ = == =dtdi(t)0.1 i(t) 280 4 2 (t) v(t) V i(t) Rdtdi(t)L 24V (t) V (t) V 24cCC R L =+ + =+ + =From KVL UAE University Department of Electrical Engineering Dr.Hazem N.Nounou( ) | |( )( ) ( ) (((

+ =t 1400t 1400t 1400ce 1400 - 0.05 * t 4800 sin4800 t 4800 cos e 0.050.1t 4800 sin e 0.05 280 4 2 (t) v( )( ) ( )( ) ( ) t 4800 cos e 4 2 t 4800 sin e 7 4 2 (t) vt 4800 sin e 7 t 4800 cos e 4 2t 4800 sin e 14 4 2 (t) vt 1400 t 1400ct 1400 t 1400t 1400c = + =UAE University Department of Electrical Engineering Dr.Hazem N.Nounou+-RLCiL(0)vc(0)Example :F91CH 1 L 6 R===Find vc(t)?vc(0) = 1 V , iL(0) = 0 0 (t) i 9dtdi(t)6dti(t) d0 (t) iC L1dtdi(t)LRdti(t) d2222= + += + +UAE University Department of Electrical Engineering Dr.Hazem N.Nounou = = = = = + = + +2 12 122s s3 s s 0 3) (s0 9 s 6 s0 k i(0)e t k e k i(t)1t 32t 31= = + = 1dtdi(0)0 1 0dtdi(0)0 1 R (0) idtdi(0)L0 (0) V (0) V (0) VC R L == + += + += + +From KVL :UAE University Department of Electrical Engineering Dr.Hazem N.Nounout 322 1e t i(t)1 k1 k kdtdi(0) = = = + = dtdi(t)L (t) i R (t) V0 (t) V (t) V (t) VCC R L == + +From KVL :Now, we need to find Vc(t):UAE University Department of Electrical Engineering Dr.Hazem N.Nounou| | ( ) ( ) | |t 3 t 3ct 3 t 3 t 3t 3 t 3 t 3ce e t 3 (t) ve e t 3 e t 61 e 3 e t - e t - 6 (t) v + = + = + =

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