First-Order Circuits in Electrical Engineering

Lecture 12 First-order Circuits (2) Hung-yi Lee

Outline Non-constant Sources for First-Order Circuits (Chapter 5.3, 9.1) ( ) t ( ) t , v i oc sc sc i

Outline Examples 5.12 and 5.11 Solved by Differential Equation Solved by Superposition and State

Example 5.12 RL circuit R=4 , L=0.1H Find i(t), t>0 i(t)=0, if t<0 ( ) 0 0 + = i ( ) t = 400 sin 280 v t (t>0) ( ) dt ( ) dt di t di t ( ) t ( ) t ( ) ( ) t + = +4 = L Ri v 1 . 0 t i v

Example 5.12 Differential Equation ( ) ( ) ( ) t v t i dt iN(t): General solution (Natural Response) ( ) ( ) ( ) t i t i t i F N + = ( ) 0 0 di t ( ) t +4 = + = = 1 . 0 400 sin 280 i v t iF(t): Special solution (Forced Response) ( ) ( ) 0 4 N = + t i dt Natural Response: di t N 1 . 0 ( ) t = = = t 40 t i Ae Ae A determined by initial condition 0 Ae ( ) 0 = t i NOT N = 40= t 0 A + = t t 1 . 0 4 0 Ae Ae ( ) t = 0 i 40 N

Example 5.12 Differential Equation ( ) ( ) ( ) t v t i dt ( ) 0 0 di t ( ) t +4 = + = = 1 . 0 400 sin 280 i v t ( ) ( ) t ( ) t = + t i i i iN(t): General solution (Natural Response) N F iF(t): Special solution (Forced Response) Forced Response: ( ) t di ( ) t + = 1 . 0 4 400 sin 280 F i t F dt 0k k1 k2 k 3 K K + K2 K 3 Table 5.3 (P222) 0 t e cos K t e cos iF(t)= v(t)= 1 0 at at + + sin t k t sin t K t 4 4

Example 5.12 Differential Equation ( ) ( ) ( ) t v t i dt ( ) 0 0 di t ( ) t +4 = + = = 1 . 0 400 sin 280 i v t ( ) ( ) t ( ) t = + t i i i iN(t): General solution (Natural Response) N F iF(t): Special solution (Forced Response) Forced Response: ( ) t di ( ) t + = 1 . 0 4 400 sin 280 F i t F dt ( ) t ( 4 = + K cos K 1 280 sin K + sin 280K 280 i t t + K = 28 4 0 K F 1 2 2 1 ) 1 . 0 + 280 280 + cos280t + = t 28 4 400 K K 2 1 2 ( ) K cos 280 K sin 280 t t = = 1 2 K 14 , 2 K 1 2 = 400 sin 280 t

Example 5.12 Differential Equation ( ) ( ) ( ) t v t i dt ( ) 0 0 di t ( ) t +4 = + = = 1 . 0 400 sin 280 i v t ( ) ( ) t ( ) t = + t i i i iN(t): General solution (Natural Response) N F iF(t): Special solution (Forced Response) ( ) t t 280 cos -14 + = ( ) t = 2 sin 280 i t 40 t i Ae F N ( ) = + 40 t A ( ) t i - 14 cos 280 2 sin 280 t i e t t = = = 14 A 0 A - 14 0 ( ) = = 14+ = A - 14 V V A V t i x x x

Example 5.11 Differential Equation ( ) ( ) ( ) t v t i dt iN(t): General solution (Natural Response) ( ) ( ) ( ) t i t i t i F N + = ( ) t = 40 t i Ae N Change v(t) ( ) 0 0 di t ( ) t +4 = + = = 1 . 0 40 t i 10 v e iF(t): Special solution (Forced Response) ( ) ( ) 0 4 N = + t i dt Natural Response: di t N 1 . 0 ( ) t = 40 t i Ae Independent to the sources N This circuit always has this term.

Example 5.11 Differential Equation ( ) ( ) ( ) t v t i dt iN(t): General solution (Natural Response) ( ) ( ) ( ) t i t i t i F N + = ( ) t = 40 t i Ae N ( ) 0 0 di t ( ) t +4 = + = = 1 . 0 40 t i 10 v e iF(t): Special solution (Forced Response) Forced Response: ( ) t di ( ) t + = 40 t 1 . 0 4 10 F i e F dt ( ) t ( ) t = 40 t = 40 t i K te ? i K e 2 F 2 F If the form for iF(t) contains any term proportional to a component of the natural response, then that term must be multiplied by t. P224 - 225

Example 5.11 Differential Equation ( ) ( ) ( ) t v t i dt iN(t): General solution (Natural Response) ( ) ( ) ( ) t i t i t i F N + = ( ) t = 40 t i Ae N ( ) 0 0 di t ( ) t +4 = + = = 1 . 0 40 t i 10 v e iF(t): Special solution (Forced Response) ( ) ( ) e t i dt Forced Response: di t + = 40 t 1 . 0 4 10 F F ( ) t ( ( ) t = = 40 40 t t 100 i K te i te 2 F F ) + = 40 40 40 40 t t t t 1 . 0 40 4 10 K e K te K te e 2 2 2 2= 2= 1 . 0 10 K 100 K

Example 5.11 Differential Equation ( ) ( ) ( ) t v t i dt iN(t): General solution (Natural Response) ( ) ( ) ( ) t i t i t i F N + = ( ) t = 40 t i Ae N ( ) 0 0 di t ( ) t +4 = + = = 1 . 0 40 t i 10 v e iF(t): Special solution (Forced Response) ( ) t ( ) t = = 40 t i Ae 100 ( ) N = + 40 40 t t A 100 t i e te 40 t i te F ( ) 0 = A = 0 0 i

Differential Equation - Summary List differential equation and find initial condition (from the property of inductors and capacitors) 1. Find general solution (natural response) Exponential form: Ae- t Find 2. Find special solution (forced response) Form: Consult Table 5.3 (P222) If a term in special solution is proportional to general solution, multiplying the term by t Find the unknown constant 3. Add the general and special solution together, and then find A in the general solution by initial condition

Example 5.12 Superposition + State RL circuits R=4 , L=0.1H i(t)=0, if t<0 ( ) ( ) i t i state = ( ) t ( ) t + t i = 400 sin 280 v t input Find istate(t) Consider the circuit from t=0 ( ) 0 0 i 4 + = i State is zero No state term ( ) = Only input term 100 sin 280 t i t

Example 5.12 Superposition + State ( ) t iL Review: pulse response A ( ) t i R ( ) t iL D t t A ( ) t i 1 A e e e D A If D is small ex + 1 x (If x is small)

Example 5.12 Superposition + State ( ) t i 100 = sin 280 t 100 t 0 100 The sin wav is composed of infinite tiny pulse! Find the response of each tiny pulse and sum them together.

Example 5.12 Superposition + State Response of the pulse between time point t0- t and t0 ( ) t t i 280 sin 100 = A = 100 sin 280 t t t A 0 ( ) t 0 = t i e t 0 t t 0t t0 0t - t 0 t t 100sin280t 0 ( ) t = t 0 i e t 0 t is small

Example 5.12 Superposition + State ( ) t 100sin280t t i 0 2t t t 0t 1t 0 = 0 e t 0t 2t 1t The response of sin wave is the summation of all the pulse responses.

Example 5.12 Superposition + State Current Source (Input) Let s focus on the response of sin wave at time point a 2t a 0t 1t We do not have to care the pulse after time point a. Current on Inductor (Response) a The function is zero at point a

Example 5.12 Superposition + State ( ) t i = t t 100sin280t 0 t 0 e t 0 Current on Inductor (Response) 2t 0t 1t a 3t a t 100sin280t 2 Value at a: 100sin280t 2 e D a t a t 100sin280t 0 1 0 e D 1 e D

Example 5.12 Superposition + State ( ) t i = t t 100sin280t 0 t 0 e t 0 Current on Inductor (Response) 2t 0t 1t a 3t Value at a: = t a a t a t = 100 sin 280 100 sin 280 t a t t ( ) a = = = = t e d i e t 0 t 0 t

Example 5.12 Superposition + State = t a a t 100 sin 280 t ( ) a = = t i e d 0 t RC = t a 100 1 ( ) a ( ) = = 40 a t sin 280 40 L i t e dt = = 40 0 t R = t a ( ) = = 40 a t 4000 sin 280 t e dt RL circuit R=4 , L=0.1H 0 t = t a = = 40 40 a t 4000 sin 280 e t e dt 0 t

Example 5.12 Superposition + State at e ( ) = at sin sin cos bt e dt a bt b bt (P806) + 2 2 a b = t a ( ) a = = 40 40 a t 4000 sin 280 i e t e dt 0 t 40 t e + ( ) = 40 a a 0 4000 40 sin 280 280 cos 280 | e t t 2 2 40 280 4000 2 + ( ) = 40 sin 280 280 cos 280 a a 2 40 280 1 ( ) 40 a We can always replace a with t . 4000 280 e + 2 2 40 280 = + 40 a 1 4 - 14 cos 280 2 sin 280 e a a

Example 5.12 Superposition + State ( ) t i 4 1 = iN + 40 t - 14 cos 280 iF 2 sin 280 e ( ) t t t ( ) t

Example 5.12 Superposition + State From Differential Equation If we have initial condition i(0)=Vx ( ) ( t i V 4 1 + = ) + 40 t - 14 cos 280 2 sin 280 e t t x From Superposition State Superposition (no state) ( ) t i input = + 40 t 1 4 - 14 cos 280 2 sin 280 e t t State ( ) ( ) t ( ) t ( ) t = + = t i i i 40 t i V e state input state x

Example 5.11 Superposition + State RL circuit R=4 , L=0.1H Find i(t) i(t)=0, if t<0 i 4 ( ) t ( ) = = 40 40 t t 10 5 . 2 v e t i e

Example 5.11 Superposition + State Current Source (Input) Current on Inductor (Response) ( ) = 40 t 5 . 2 t i e a a 0 Response of the pulse between time point t0- t and t0 t t t t -40t A 2.5e 0 0 ( ) t = t = i e t e t 0 2.5e Its contribution at point a: a t -40a 0 ( ) a = t i e t 0

Example 5.11 Superposition + State For the response of the pulse between time point t0- t and t0 Its contribution at point a is: a t -40a 2.5e 0 ( ) a = t i e t 0 1 = t a a t a t 40 40 a a = 5 . 2 5 . 2 t a e e = ( ) a = = = i e t e dt 40 = 0 t 0 t = t a = t a 5 . 2 ( ) a ( )dt = ( )dt = = 40 40 t a t = 100 e e 4040 40 t a t i e e 0 t 0 t = t a a = 40 a 100 e a = 40 100 dt 1 e = 0 t

Homework 5.56 5.60 5.64

Thank you!

Answer 5.56: iF(t) =-10te^(-20t) - 3e^(-20t) 5.60: vF(t) = 2 50te^(-25t) 5.64: i(t) = 0.05e^(25t) + 0.02 0.07e^(-25t)