ECE201 Lect-221 Second-Order Circuits Cont’d Dr. Holbert April 24, 2006

ECE201 Lect-222 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 (ω 0 ) and the damping ratio (ζ)

ECE201 Lect-223 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-224 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. Second-Order Natural Solution

ECE201 Lect-225 Step-by-Step Approach 1.Assume solution (only dc sources allowed): i. x(t) = K 1 + K 2 e -t/  ii. x(t) = K 1 + K 2 e s 1 t + K 3 e s 2 t 2.At t=0 –, draw circuit with C as open circuit and L as short circuit; find I L (0 – ) and/or V C (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

ECE201 Lect-226 Step-by-Step Approach 5.Find time constant (  ), or characteristic roots (s) i.Looking across the terminals of the C or L element, form Thevenin equivalent circuit;  =R Th C or  =L/R Th ii.Write ODE at t>0; find s from characteristic equation 6.Finish up i.Simply put the answer together. ii.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)

ECE201 Lect-227 Class Examples Learning Extension E7.10 Learning Extension E7.11