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Transient Circuit Analysis Cont’d. Dr. Holbert November 27, 2001 ECE201 Lect-24 1 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 2 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 3 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 4 The Differential Equation • Equations are linear, constant coefficient: d n x(t ) d n 1 x(t ) an an 1 ... a0 x(t ) f (t ) n n 1 dt dt • 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 5 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 6 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 7 The Forced Solution • The forced (particular) solution is the solution to the non-homogeneous equation: d n x(t ) d n 1 x(t ) an an 1 ... a0 x(t ) f (t ) n n 1 dt dt • 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 8 The Natural Solution • The natural (or complementary) solution is the solution to the homogeneous equation: d n x(t ) d n 1 x(t ) an an 1 ... a0 x(t ) 0 n n 1 dt dt • Different “look” for 1st and 2nd order ODEs ECE201 Lect-24 9 First-Order Natural Solution • The first-order ODE has a form of dxc (t ) 1 xc (t ) 0 dt • The natural solution is xc (t ) Ke t / • Tau () is the time constant • For an RC circuit, = RC • For an RL circuit, = L/R ECE201 Lect-24 10 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-24 11 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 12 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 13 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-24 14 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-24 15 Class Examples • Learning Extension E6.3 • Learning Extension E6.4 • Learning Extension E6.11 ECE201 Lect-24 16