Analysis of Second-Order Circuits Using Differential Equations

Lecture 13 Second-order Circuits (1) Hung-yi Lee

Second-order Circuits A second order-circuit contains two independent energy-storage elements (capacitors and inductors). Capacitor + inductor 2 Capacitors 2 inductors

Second-order Circuits Steps for solving by differential equation (Chapter 9.3, 9.4) 1. List the differential equation (Chapter 9.3) 2. Find natural response (Chapter 9.3) There is some unknown variables in the natural response. 3. Find forced response (Chapter 9.4) 4. Find initial conditions (Chapter 9.4) 5. Complete response = natural response + forced response (Chapter 9.4) Find the unknown variables in the natural response by the initial conditions

Solving by differential equation Step 1 Step 1: List Differential Equation

Systematic Analysis 1 = = v i dt i C v C C C C C 1 i = = L i v dt v i L L L L L Mesh Analysis 1 = + + = + + v v v v vs i L i R i s L R C C 1 1 R 1 i + i + = + = + v i vs i L Ri idt s L L LC C

Systematic Analysis 1 = = v i dt i C v C C C C C 1 i = = L i v dt v i L L L L L = i C cv Mesh Analysis Find vC: 1 1 R 1 1 R + = + C C C i + i + v v v v = v i s C C C s L 1 L R LC L L LC 1 = + + v v v v i = L i Find iL: s C C C C L L LC

Systematic Analysis v 1 = = v i dt i C v C C C C C 1 = = L i v dt v i L L L L L Node Analysis 1 1 + = + + = + i i i i is v v C v s L R C R 1 L 1 v 1 1 = + + + + = is vdt C v i v v v s L R RC C LC

Systematic Analysis v Systematic Analysis 1 = = v i dt i C v C C C C C 1 = = L i v dt v i L L L L L = L v Li Node Analysis Find iL: 1 1 1 1 1 1 i + L i + L + + = = L i i i v v v s L L L s RC C LC 1 LC RC C LC 1 1 i + = + i i i vC=v Find vC: L s L L RC LC

v1 v2 Example 9.6 Find i2 =1 + is i i v1: 1 L 1 1 1 v v = + = + is v v v 1 2 is v dt 1 1 2 1 R R L R 1 1 = x+ v 2 i i 2i v2: + R 1 R R v v R = + = + 1 2 x v dt v v v 2 1 2 2 R L R L 2 2 x x

v1 v2 Example 9.6 Find i2 1 L R 1 1 1 = + = is v v v i v dt Target: 1 1 2 2 2 R R L 1 + R R = + x v v v Find v2from the left equations 1 2 2 R L 2 x Then we can find i2 Equations for v1and v2

v1 v2 Example 9.6 Find i2 + R R R 1 L R 1 1 = + x v v v dt = + is v v v 1 2 2 R L 1 1 2 R R 2 x 1 + 1 L R R R + R R = + x i v v dt = + x v v v 2 2 s R L 1 2 2 R L 1 2 x 2 x + 1 1 R R R + + Find v2 x v v v 2 2 2 R R R L 2 x

v1 v2 Example 9.6 Find i2 + 1 L R R R = + x i v v dt 1 2 2 = s i v dt R L 2 2 1 2 x L Li + 1 1 R R R = v 1 R 1 L + + x v v v 2 2 2 2 2 R R R L 2 x Replace 2v with 2i + R R R i = + + + x v v v 2 2 2 s L R L L 1 2 1 2 x x

Example 9.7 Please refer to the appendix

Summary List Differential Equations 1 1 1 1 1 R i + + L= + + = i i i v v v v L L s C C C s RC LC LC C L LC L

Solving by differential equation Step 2 Step 2: Find Natural Response

Natural Response The differential equation of the second-order circuits: ( ) t y t y + 2 ( ) ( ) t ( ) t + = 2 0 y f y(t): current or voltage of an element = damping coefficient 0 = resonant frequency

Natural Response The differential equation of the second-order circuits: ( ) t y t y + 2 ( ) ( ) ( ) t y t y t y F N + = ( ) 2 N N + y t y ( ) y t y + F F 2 ( ) ( ) t ( ) t + = 2 0 y f ( ) t ( ) t ( ) t + = 2 0 0 ( ) t y N ( ) t + = 2 0 y f F Focus on yN(t) in this lecture

Natural Response ( ) N t y ( ) t ( ) t N + + ( ) N = 2 0 2 0 y y N = t A y t e yN(t) looks like: Characteristic equation + + = 2 2 0 + + = 2 2 0 ( t t t 2 0 2 0 Ae Ae Ae ) 2 2 0 2 2 4 = = 2 2 0 2 2 = + ( ) t N 2 0 = 2 2 0 1 2 + = t t A A y e e 1 2 1 2

Natural Response = + = 2 2 0 2 2 0 1 2 Overdamped 1 2 Real 2 2 0 = = Critical damped 1 2 2 2 0 1, 2is 0 Underdamped Complex 2 2 0 = 0 Undamped

Solving by differential equation Step 2 Step 2: Find Natural Response Overdamped Response

Overdamped Response = + = 2 2 0 2 2 0 1 2 1, 2are both real numbers 1 2 2 2 0 ( ) t = t A y e yN(t) looks like N ( ) t + = t t A A y e e 1 2 N 1 2

Overdamped Response ( ) t + = t t A A y e e 1 2 N 1 2 2 A 1 A 0 0 t A 2 e 0 1 2 0 2 t A e 1 1

Solving by differential equation Step 2 Step 2: Find Natural Response Underdamped Response

Underdamped = + = 2 2 0 2 2 0 1 2 = 1 j 2 2 0 = 2 0 2 = + 1 2 0 2 1 2 1 = = + 2 0 2 2 0 2 j j 2 1 = + j = j 1 d 2 d = 2 0 2 d

Euler's formula: = cos + Underdamped ejx sin x j x = j = + j 2 d 1 d ( ) t + ( ( A ( A = = = t t A A y e e 1 2 N 1 2 ) t ( )t + ) + + + j j A A e e d d 1 2 j t j t t A e e e d d 1 2 t d ( ) t ) ( ) = + t A cos A A sin y e t j N 1 2 1 2 d A = A yN(t) should be real. 1 2

Euler's formula: = cos + Underdamped ejx sin x j x = j = ( A + j 2 d 1 d ) t d ( ) t ( ) = + + t A cos A A sin y e t j N 1 2 1 2 d jB (no real part) yN(t) should be real. * = A A 1 2 1 1 1 1 = + = A2 a jb A1 a j b 2 2 2 2 = + = A A A A jb a 1 2 1 2

Underdamped = j = ( A + j 2 d 1 d ) t d ( ) t ( ) = + + t A cos A A sin y e t j N 1 2 1 2 d * = + = = A A A A jb a A A 1 2 1 2 1 2 t d ( ) t = + t cos sin y e a t b a and b will be determined by initial conditions N d Memorize this!

Underdamped ( ) a e t cos = t d + t sin y t b N d a b ( ) t = + + 2 2 t cos sin y e a b t t N d d + b L + 2 2 2 2 a b a b a L e ( ) t = + t L cos sin y e t t = a + 2 2 L b L N d d t d ( ) t ( ) t = ( + t cos cos sin sin y t N d ) = t L cos y e t L and will be determined by initial conditions N d

Underdamped t d ( ) t = + t cos sin y e a t b N d ( ) t ( ) = t L cos y e t N d L t L e Lcos

Solving by differential equation Step 2 Step 2: Find Natural Response Undamped Response

Undamped = j = + j 2 d 1 d Undamped is a special case of underdamped. = 1 j = j = 0 d 2 d t d ( ) t = + t cos sin y e a t b N d ( ) t = + cos sin y a t b t N d d ( ) t ( ) = t L cos y e t N d ( ) t ( ) = d L cos y t N

Solving by differential equation Step 2 Step 2: Find Natural Response Critical Damped Response

Critical Damped ( ) t ( ) t ( ) ? = t + a = = t t A A y e e 2 2 0 Overdamped 1 2 N 1 2 t d + t 2 2 0 cos sin y e t b Underdamped N d = yN 2 2 0 Critical damped = = = 2 2 0 1, 2 1 2 ( ) t ( ) t = Not complete = + t t t A ? A A t y e y e e 1 2 N N

Critical Damped (Problem 9.44) ( ) t ( ) t ( ) t ( ) t = + t t = t A2 t y A e e A2 A t ( ( 2 h e 1 N ) ) 2 = t 1 t h e 2 A = - t t h A e 2 e

Solving by differential equation Step 2 Step 2: Find Natural Response Summary

Summary ( ) t ( ) t ( ) t ( ) t + + = 2 0 2 y y y f Fix 0, decrease ( is positive): Critical damped Underdamped Undamped Overdamped 1 1 1 1 1 R i + = + i i i = + + v v v v L s L L s C C C RC LC LC C L L LC Decrease , smaller R Decrease , increase R

Fix 0, decrease ( is positive) The position of the two roots 1and 2. = 2 2 0 1, 2 =0 Undamped

Homework 9.30 9.33 9.36 9.38

Thank You!

Answer 9.30: v1 + 3 v1 + 10 v1 = 0 9.33: yN=a e^(-0.5t) + b te^(-0.5t) 9.36: yN=a e^(4t) + b e(-6t) 9.38: yN=2Ae^(3t) cos (6t+ ) or yN=2e^(3t) (acos6t + bsin6t) In 33, 36 and 38, we are not able to know the values of the unknown variables.

Appendix: Example 9.7

Example 9.7 Mesh current: i1and ic = i i 1i L c ( ) = = = v Kv KRi di KR i 1i ) out x di L c L di di ( ) ( = + 1 c i i i = + R 1 c i L R i i 1 1 c 1 1 c R dt dt dt ) 1 i dt 1 C di di ( ( ) = + + KR 1 c i i dt L R i i 1 c c c dt dt 2 2 di d i di d i ( ) = 1 1 1 c c i RC K LC c 2 2 dt dt dt dt

Example 9.7 L di di ( ) = + 1 c i i i (1): 1 1 c R dt dt 2 2 di d i di d i ( ) = di 1 1 1 c c i RC K LC (2): c 2 2 dt dt dt dt 2 2 di d i d i ( ) = 1 1 1 c c i i RC K LC (2) (1): 1 c 2 2 dt dt dt dt di L di ( ) 1 c i i 1 c R dt dt

Example 9.7 2 2 d i di d i L di ( ) ( ) 0 = + + 1 2 1 1 c c LC RC K i i 1 c 2 2 dt dt R dt dt 2 2 1 2 d i di d i R di ( ) ( ) 0 = + + 1 1 1 c c K i i 1 c 2 2 dt dt RC L dt dt LC 2 1 2 d v dv R ( ) + + = 1 0 out out K v out 2 dt RC L dt LC

Appendix: Figures from Other Textbooks

Undamped

Acknowledgement (b02) Equation (b02) Equation