Transient Circuit Analysis Cont’d. Dr. Holbert November 27, 2001 ECE201 Lect-24

Introduction In a circuit with energy storage elements, voltages and currents are the solutions to linear, constant coefficient differential equations. Real engineers almost never solve the differential equations directly. It is important to have a qualitative understanding of the solutions. ECE201 Lect-24

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-24

The Differential Equation Every voltage and current is the solution to a differential equation. In a circuit of order n, these differential equations have order n. The number and configuration of the energy storage elements determines the order of the circuit. n  # of energy storage elements ECE201 Lect-24

The Differential Equation Equations are linear, constant coefficient: The variable x(t) could be voltage or current. The coefficients an through a0 depend on the component values of circuit elements. The function f(t) depends on the circuit elements and on the sources in the circuit. ECE201 Lect-24

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-24

Differential Equation Solution The total solution to any differential equation consists of two parts: x(t) = xp(t) + xc(t) Particular (forced) solution is xp(t) Response particular to a given source Complementary (natural) solution is xc(t) Response common to all sources, that is, due to the “passive” circuit elements ECE201 Lect-24

The Forced Solution The forced (particular) solution is the solution to the non-homogeneous equation: The particular solution is usually has the form of a sum of f(t) and its derivatives. If f(t) is constant, then vp(t) is constant ECE201 Lect-24

The Natural Solution The natural (or complementary) solution is the solution to the homogeneous equation: Different “look” for 1st and 2nd order ODEs ECE201 Lect-24

First-Order Natural Solution The first-order ODE has a form of The natural solution is Tau (t) is the time constant For an RC circuit, t = RC For an RL circuit, t = L/R ECE201 Lect-24

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: s1 and s2. ECE201 Lect-24

Initial Conditions The particular and complementary solutions have constants that cannot be determined without knowledge of the initial conditions. The initial conditions are the initial value of the solution and the initial value of one or more of its derivatives. Initial conditions are determined by initial capacitor voltages, initial inductor currents, and initial source values. ECE201 Lect-24

Transients and Steady State The steady-state response of a circuit is the waveform after a long time has passed, and depends on the source(s) in the circuit. Constant sources give DC steady-state responses DC SS if response approaches a constant Sinusoidal sources give AC steady-state responses AC SS if response approaches a sinusoid The transient response is the circuit response minus the steady-state response. ECE201 Lect-24

Step-by-Step Approach Assume solution (only dc sources allowed): x(t) = K1 + K2 e-t/ x(t) = K1 + K2 es1t + K3 es2t At t=0–, draw circuit with C as open circuit and L as short circuit; find IL(0–) and/or VC(0–) 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) At t=, repeat step #2 to find x()=K1 ECE201 Lect-24

Step-by-Step Approach Find time constant (), or characteristic roots (s) 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 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-24

Class Examples Learning Extension E6.3 Learning Extension E6.4 ECE201 Lect-24