SECOND ORDER CIRCUIT

Revision of 1 st order circuit Second order circuit Natural response (source-free) Forced response SECOND ORDER CIRCUIT

Revision of 1 st order circuit NATURAL RESPONSE (SOURCE-FREE) -initial energy in capacitor Solution: -i.e. v C (0) = V o KCL   Solving this first order differential equation gives: +vC+vC iRiR iCiC +vR+vR C R

Solution: KCL   Solving this first order differential equation gives: +vC+vC V s u(t) R C ++ FORCED RESPONSE -no initial energy in capacitor -i.e. v C (0) = 0  Revision of 1 st order circuit

COMPLETE RESPONSE Complete response = natural response + forced response v(t) = v n (t) + v f (t) Complete response = Steady state response + transient response v(t) = v ss (t) + v t (t) Revision of 1 st order circuit

COMPLETE RESPONSE In general, this can be written as: - can be applied to voltage or current - x(  ) : steady state value - x(0) : initial value For the 2 nd order circuit, we are going to adopt the same approach Revision of 1 st order circuit

To successfully solve 2 nd order equation, need to know how to get the initial condition and final values CORRECTLY In 1 st order circuit need to find initial value of inductor current (RL circuit) OR capacitor voltage (RC circuit): i L (0) or v C (0) Need to find final value of inductor current OR capacitor voltage: i L (∞) or v C (∞) Before we begin ….. INCORRECT initial conditions /final values will result in a wrong solution In 2 nd order circuit need to find initial values of i L and/or v C : i L (0) or v C (0) Need to find final values of inductor current and/or capacitor voltage: i L (∞), v C (∞) Need to find the initial values of first derivative of i L or v C : di L (0)/dt dv C (0)/dt Section 8.2 of Alexander/Sadiku

Finding initial and final values Example 8.1 Switch closed for a long time and open at t=0. Find: i(0 + ), v(0 + ), di(0 + )/dt, dv(0 + )/dt, i(∞), v(∞)

Finding initial and final values PP 8.2 Find: i L (0 + ), v C (0 + ), v R (0 + ) di L (0 + )/dt, dv C (0 + )/dt, dv R (0 + )/dt, i L (∞), v C (∞), v R (∞)

Natural Response of Series RLC Circuit (Source-Free Series RLC Circuit) Second order circuit RL C i Applying KVL, Differentiate once,  This is a second order differential equation with constant coefficients We want to solve for i(t).

Second order circuit Assuming Sincecannot become zero, This is known as the CHARACTERISTIC EQUATION of the diff. equation

Second order circuit Solving for s, Which can also be written as where s 1, s 2 – known as natural frequencies (nepers/s)  – known as neper frequency,  o – known as resonant frequency

Second order circuit Case 1 A 1 and A 2 are determined from initial conditions Overdamped solution Case 2 Critically damped solution Case 3 Underdamped solution

Second order circuit Overdamped response Case 1 Roots to the characteristic equation are real and negative A 1 and A 2 are determined from initial conditions: (i) At t = 0, (ii) At t = 0,

Second order circuit Overdamped response Case  0.05H 0.5mF +vc+vc Initial condition v c (0) =100V

Second order circuit Critically damped response Case 2 2 A 3 is determined from 2 initial conditions: NOT POSSIBLE (i) At t = 0, (ii) At t = 0,  solution should be in different form: A 1 and A 2 are determined from initial conditions:

Second order circuit Critically damped response Case 2 20  0.05H 0.5mF +vc+vc Initial condition v c (0) =100V

Second order circuit Underdamped response Case 3 Roots to the characteristic equation are complex - known as damped natural frequency

Second order circuit Underdamped response Case 3 Using Euler’s identity: e j  = cos  + jsin  where

Second order circuit Underdamped response Case 3 (i) At t = 0, (ii) At t = 0,

Second order circuit Underdamped response Case 3 10  0.05H 0.5mF +vc+vc Initial condition v c (0) =100V

Second order circuit Underdamped, overdamped and critically damped responses