SecondOrder Circuits Contd Dr Holbert April 24 2006

Second-Order Circuits Cont’d Dr. Holbert April 24, 2006 ECE 201 Lect-22 1

Important Concepts • The differential equation for the circuit • Forced (particular) and natural (complementary) solutions • Transient and steady-state responses • 1 st order circuits: the time constant ( ) • 2 nd order circuits: natural frequency (ω0) and the damping ratio (ζ) ECE 201 Lect-22 2

Building Intuition • Even though there an infinite number of differential equations, they all share common characteristics that allow intuition to be developed: – Particular and complementary solutions – Effects of initial conditions – Roots of the characteristic equation ECE 201 Lect-22 3

Second-Order Natural Solution • The second-order ODE has a form of • To find the natural solution, we solve the characteristic equation: • Which has two roots: s 1 and s 2. ECE 201 Lect-22 4

Step-by-Step Approach 1. Assume solution (only dc sources allowed): i. ii. x(t) = K 1 + K 2 e-t/ x(t) = K 1 + K 2 es t + K 3 es t 1 2 2. At t=0–, draw circuit with C as open circuit and L as short circuit; find IL(0–) and/or VC(0–) 3. At t=0+, redraw circuit and replace C and/or L with appropriate source of value obtained in step #2, and find x(0)=K 1+K 2 (+K 3) 4. At t= , repeat step #2 to find x( )=K 1 ECE 201 Lect-22 5

Step-by-Step Approach 5. Find time constant ( ), or characteristic roots (s) i. ii. 6. Looking across the terminals of the C or L element, form Thevenin equivalent circuit; =RTh. C or =L/RTh Write ODE at t>0; find s from characteristic equation Finish up i. ii. Simply put the answer together. Typically have to use dx(t)/dt│t=0 to generate another algebraic equation to solve for K 2 & K 3 (try repeating the circuit analysis of step #5 at t=0+, which basically means using the values obtained in step #3) ECE 201 Lect-22 6

Class Examples • Learning Extension E 7. 10 • Learning Extension E 7. 11 ECE 201 Lect-22 7