5--4 EECS Electric Circuit Chapter 6 R Circuit Natural and Step Repone Objective Determine the repone form of the circuit Natural repone parallel R circuit Natural repone erie R circuit Step repone of parallel and erie R circuit
5--4 EECS Electric Circuit Natural Repone of Parallel R Circuit Step in Solving R Circuit n The firt tep i to write either KVL or KCL for the circuit. n Take the derivative to remove any integration n Solve the reulting differential equation
5--4 Natural Repone of Parallel R Circuit The problem given initial energy tored in the inductor and/or capacitor, find v( for t. Activity It i convenient to calculate v( for thi circuit becaue: A. The voltage mut be continuou for all time. B. The voltage i the ame for all three component. C. Once we have the voltage, it i pretty eay to calculate the branch current. D. All of the above. 3
5--4 Natural Repone of Parallel R Circuit KCL: C dv( + t L v(x)dx + I + v( R = Differentiate both ide to remove the integral: C d v( + L v(+ R dv( = Divide both ide by C to place in tandard form: d v( + v(+ dv( = RC Natural Repone of Parallel R Circuit The problem given ini al energy tored in the inductor and/or capacitor, find v( for t. Decribing equation: d v( + v( + RC dv( = Thi equation i ü Second order ü Homogeneou ü Ordinary differential equation ü With contant coefficient 4
5--4 Natural Repone of Parallel R Circuit Decribing equation: d v( dv( + v( + = RC The circuit ha two initial condition that mut be atified, o the olution for v( mut have two contant. Ue v( = A e t + A e t V; Subtitute: ( A e t + A e t )+ RC ( A e t + A e t )+ (A e t + A e t ) = [ + ( RC) + ( )]A e t +[ + ( RC) + ( )]A e t = For olution, either A = and A = or the quadratic part i. The quadratic part i called Characteritic equation: + ( RC) + ( ) = Natural Repone of Parallel R Circuit The two olution to the characteritic equation and can be calculated uing the quadratic formula:! + $! # & + $ # & = ; " RS % " %! $, = # &± " RC % where α = RC and ω =! $! # & $ # & = α ± α ω " RC % " % (the neper frequency in rad/) (the reonant radian frequency in rad/) 5
5--4 Parallel R Circuit,, α = = RC RC = α ± ± α ω rad/, ω = RC rad/,, ω = = ± RC = ξω ± ω, RC ξ ξ = R L C Neper frequency radian frequency Reonant Damping ratio to Parallel R Circuit v( = A e t + A e t n The olution of the differential equation depend on the value of and, A and A. n and can be found from the characteritic equation, conider three cae 6
5--4 Cae - Overdamping α> ω o, ζ> à real root,, α = = RC RC = α ± ± α ω rad/, ω = RC rad/,, ω = = ± RC = ξω ± ω, ξ ξ = R RC L C v( = A e t + A e t Cae : Underdamping α< ω o, ζ< à a pair of complex root,, α = = RC RC = α ± ± α ω rad/, ω = RC rad/,, ω = = ± RC = ξω ± ω, ξ ξ = R RC L C = α ± jω d ω d = ω α v( = B e αt co(ω d + B e αt in(ω d 7
5--4 Cae 3: Critical Damping α=ω o, ζ= à equal root,, α = = RC RC = α ± ± α ω rad/, ω = RC rad/,, ω = = ± RC = ξω ± ω, ξ ξ = R RC L C = = α v( = D te αt + D e αt to Parallel R Circuit v( = A e t + A e t v( = B e αt coω d t + B e αt inω d t v( = D te αt + D e αt Overdamped Underdamped Critically damped 8
5--4 Determine the Coefficient n We can calculate them from the initial condition n Keep in mind voltage acro a capacitor and current in an inductor can not change intantaneouly v i C L ( ( ) = v ) = i L C () = v () = i L C ( ( + ) + ) Activity Given v( + )=V and i L ( + )=3mA, Find v( for t>=. 9
5--4 Step : determine the olution form. Given R=Ω,C=.µF, L=5mH α = RC =. 6 =.5 6 rad / ω = = 5 3. 6 =4 rad / α > ω overdamped, = α ± α ω =.5 4 ±.565 4 = 5 + 75 = 5rad / = 5 75 = rad / v( = A e 5t + A e t V Step, find A and A Given v( + )= V, i L ( + )=3 ma 5,t,t and v( = A e + Ae v( + ) = = A + A () KCL: i C ( + ) = i R ( + ) i L ( + ), alo i C ( + ) = C dv(+ ) C dv(+ ) = V R i L (+ ) dv( + ) = V CR i L (+ ) C = 3 3. 6. 6 =.3 6.5 6 =.45 6 V / dv( + ) = 45, = 5A, A ()
5--4 Solve for equation () and (), A =-4V and A =6V v( = 4e 5,t + 6e,t V, t Plot v( from to 5m
5--4 Activity 3 Given V =V, I =-.5mA, Find v( for t>=. Step : determine the olution form. Given R=kΩ,C=.5µF, L=8H α = RC = = rad / 3 6.5 ω = = 8.5 6 =3 rad / α < ω underdamped ω d = ω α = 6 4 4 = 979.8rad / = α + jω d = + j979.8rad / = α jω d = j979.8rad / v( = B e t co(979.8+ B e t in(979.8
5--4 Step, find B and B Given v =v( + )= V, i R =à i C ( + )=-I =.5mA and v( = B e t co(979.8+ B e t in(979.8 v( + ) = B = () KCL: i C ( + ) = i R ( + ) i L ( + ), alo i C ( + ) = C dv(+ ) dv( + ) B = dv(+ ) = i L (+ ) C =.5 3 = 98V / 6.5 ω d = 98 V () 979.8 Solve for equation () and (), B =V and B =V v( =e t in(979.8v t 3
5--4 Plot v( from to m v( = e t in(979.8t ) Activity 4 In Activity 4, what i the value of R that reult in a critically damped voltage repone? Find v( for t>=. 4
5--4 Step : determine the olution form. Given C=.5µF, L=8H ω = = 8.5 6 =3 rad / α = ω = 3 = RC R = = 4kΩ 3 6.5 v( = D te αt + D e αt Step, find D and D Given v =v( + )= V, dv( + )/=98V/ and v( = D te αt + D e αt v( + ) = D = () dv( = D e αt + D t e αt α dv( + ) = D = 98V () v( = 98te t V t 5
5--4 v( = 98te t V 3 Cae Overdamped: Longer to ettle Underdamped: ocillate Critically damped: Fatet to ettle without ocillation 6
5--4 EECS Electric Circuit Natural Repone of Serie R Circuit Serie and Parallel R Circuit The difference() between the analyi of erie R circuit and the parallel R circuit i/are: A. The variable we calculate. B. The decribing differential equation. C. The equation for atifying the initial condition 7
5--4 Natural Repone of Serie R Circuit The problem given initial energy tored in the inductor and/or capacitor, find i( for t. KVL: L di( + t C i(x)dx +V + Ri( = Differentiate both ide to remove the integral: L d i( + di( i(+ R = C Divide both ide by L to place in tandard form: d i( + R L di( + i( = Activity 5 The decribing differential equation for the erie R circuit i d i( R di( + + i( = L Therefore, the characteritic equation i A. + (/RC) + / = B. + (R/L) + / = C. + (/) + /RC = 8
5--4 Natural Repone of Serie R Circuit The two olution to the characteritic equation can be calculated uing the quadratic formula: + (R L) + ( ) =, = α ± α ω α = R L ω = (the neper frequency in rad/) (the reonant radian frequency in rad/) Natural Repone of Serie R Circuit The olution are in the ame form a in the parallel R circuit: i( = A e t + A e t i( = B e αt coω d t + B e αt inω d t i( = D te αt + D e αt Overdamped Underdamped Critically damped 9
5--4 Activity 6 The capacitor i charged to V and at t =, the witch cloe. Find i( for t. R α = = L ω = α < ω ω = d o i( = B e = 56 (.) ω α 8t = 8rad/ =,rad/ (.)(.µ) o thi i the underdamped cae! = 96rad/ co96t + B e 8t in 96t A, t
5--4 i( = B e 8t Now wemut ue the coefficientin the equation to atify the initialconditioninthe circuit : i( t= di( inthe equation = i( t= co96t + B e 8t t= di( inthe equation = in 96t A, t inthe circuit t= inthe circuit i( = B e Equation: Circuit : 8t co96t + B e i() = B i() = I B = = 8t in 96t A, t (ame a the parallel cae!)
5--4 Equation: di( + ) Circuit: L di(+ ) = V di(+ ) B = di(+ ) i( =.4e 8t in96t A, t = αb +ω d B (ame a the parallel cae!) = V L = =A / 3 = ω d B = 96B = B =.4 EECS Electric Circuit Step Repone of R Circuit
5--4 Step Repone of R Circuit v L + v C + v R = V di L + vc + ir = V dvc i = C d vc dvc + vc + RC = V d vc R dvc vc I + + = L i L + i C + i R = I dv v il + C + = I R dil v = L d il L dil + + il = I R d il dil il I + + = RC Step Repone of R Circuit n A topic for a coure in Math. n Generally peaking, the olution of a econdorder DE with a contant driving force equal the forced repone plu the a repone function identical to the natural repone. function of the ame form i = I f + a natural repone function of the ame form v = V f + a natural repone n I f or V f i the non-zero final value. 3
5--4 EECS Electric Circuit Step Repone for Parallel R Circuit Step Repone of a Parallel R Circuit i L ( = I f + A e t + A e t i L ( = I f + B e αt coω d t + B e αt inω d t i L ( = I f + D te αt + D e αt Overdamped Underdamped Critically damped 4
5--4 Step Repone of R Circuit A t : The only component whoe final value i NOT zero i the inductor, whoe final current i the current upplied by the ource. Activity 7 There i no initial energy tored in thi circuit; find i( for t. 5
5--4 The problem there i no initial energy tored in thi circuit; find i( for t. To begin, find the initial condition and the final value. The initial condition for thi problem are both zero; the final value i found by analyzing the circuit a t. The problem there i no ini al energy tored in thi circuit; find i( for t. t : I F = 4mA 6
5--4 Next, calculate the value of α and ω and determine the form of the repone: α = RC = = 5, rad/ (4)(5n) ω = = (5m)(5n) = 4, rad/ Overdamped Since the repone form i overdamped, calculate the value of and :, = α ± = 5, ± 3,rad/ α ω = 5, ± =,rad/ and 5, 4, = 8,rad/,t 8,t il( =.4 + A e + Ae A, t 7
5--4,t 8,t il( =.4 + Ae + A e A, t Next, et the value of i() and di()/ from the equation equal to the value of i() and di()/ from the circuit. From the equation: i () =.4 + A + A From the circuit : i () = I L L = From the From the equation : circuit : dil() =,A 8,A dil() vl() V = = = L L Solve:.4 + A + A = and, A 8, A = A = 3 ma; A = 8 ma i L ( = 4 3e,t +8e 8,t ma, t 8
5--4 Activity 8 If the reitor value i changed to 65Ω, find i( for t. R = 65Ω,C = 5nF, α = RC = = 3, rad/ (65)(5n) ω = = (5m)(5n) = 4, rad/ α < ω Underdamped ω d = ω α = (6.4) 8 = 4rad/ = 3 + j4rad / = 3 j4rad / i L ( = I f + B e 3t co(4+ B e 3t in(4 9
5--4 Next, et the value of i() and di()/ from the equation equal to the value of i() and di()/ from the circuit. i L () =.4 + B = di L () = ω d B αb = B = 4mA, B = 3mA i L ( = 4 4e 3t co(4 3e 3t in(4ma t Activity 9 If the reitor value in i changed to 5Ω, find i( for t. 3
5--4 R = 5Ω,C = 5nF, α = RC = = 4, rad/ (5)(5n) ω = = (5m)(5n) = 4, rad/ α = ω Critically damped = = α = 4rad / i L ( = I f + D te 4t + D e 4t Next, et the value of i() and di()/ from the equation equal to the value of i() and di()/ from the circuit. i L () =.4 + D = di L () = D αd = D = 96mA /, D = 4mA i L ( = 4 96te 4t 4e 4t ma t 3
5--4 Plot Repone of 3 Cae The overdamped, underdamped, and critically damped repone of Activitie 7-9 are given below: Overdamped: i L ( = 4 3e,t +8e 8,t ma, t Underdamped: i L ( = 4 4e 3t co(4 3e 3t in(4ma, t Critically damped: i L ( = 4 96te 4t 4e 4t ma, t The current plot 3
5--4 EECS Electric Circuit Step Repone for Serie R Circuit Step Repone of a Serie R Circuit v c = V f + A e t + A e t v c = V f + B e αt coω d t + B e αt inω d t v c = V f + D te αt + D e αt Overdamped Underdamped Critically damped 33
5--4 Step Repone for Serie R Circuit The problem find v C ( for t. Find the initial condition by analyzing the circuit for t < : V I 5k = (8 V) 9k + 5k = 5 V = A Step Repone for Serie R Circuit Find the final value of the capacitor voltage by analyzing the circuit a t : V F = V 34
5--4 Step Repone for Serie R Circuit Ue the circuit for t to find the value of α and ω : α = R L = 8 (.5) = 8rad/ ω = = (.5)(µ) =,rad/ α < ω ω = d ω α = 6rad/ underdamped =, 8 Step Repone for Serie R Circuit Write the equation for the repone and olve for the unknown coefficient: 8t 8t v ( = + B e co6t + B e in 6t V, t v () = V dvc () = αb + ωd B B = 5 V, v C C C F + B ( = 5e = V 8t + B I = C B = 66.67 V co6t + 66.67e = 5 8B 8t + 6B = in 6t V, t 35