Second-Order Circuits Cont’d Dr. Holbert April 24, 2006 ECE201 Lect-22 1 Important Concepts • The differential equation for the circuit • Forced (particular) and natural (complementary) solutions • Transient and steady-state responses • 1st order circuits: the time constant () • 2nd order circuits: natural frequency (ω0) and the damping ratio (ζ) ECE201 Lect-22 2 Building Intuition • Even though there are 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 ECE201 Lect-22 3 Second-Order Natural Solution • The second-order ODE has a form of 2 d x(t ) dx(t ) 2 2 0 0 x (t ) 0 2 dt dt • To find the natural solution, we solve the characteristic equation: s 2 0 s 0 2 2 0 • Which has two roots: s1 and s2. ECE201 Lect-22 4 Step-by-Step Approach 1. Assume solution (only dc sources allowed): i. ii. x(t) = K1 + K2 e-t/ x(t) = K1 + K2 es t + K3 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)=K1+K2 (+K3) 4. At t=, repeat step #2 to find x()=K1 ECE201 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; =RThC 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 K2 & K3 (try repeating the circuit analysis of step #5 at t=0+, which basically means using the values obtained in step #3) ECE201 Lect-22 6 Class Examples • Learning Extension E7.10 • Learning Extension E7.11 ECE201 Lect-22 7