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Published byMagdalene Thompson Modified over 9 years ago
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SECOND ORDER CIRCUIT
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Revision of 1 st order circuit Second order circuit Natural response (source-free) Forced response SECOND ORDER CIRCUIT
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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
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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
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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
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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
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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
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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(∞)
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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 (∞)
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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).
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Second order circuit Assuming Sincecannot become zero, This is known as the CHARACTERISTIC EQUATION of the diff. equation
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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
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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
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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,
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Second order circuit Overdamped response Case 1 100 0.05H 0.5mF +vc+vc Initial condition v c (0) =100V
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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:
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Second order circuit Critically damped response Case 2 20 0.05H 0.5mF +vc+vc Initial condition v c (0) =100V
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Second order circuit Underdamped response Case 3 Roots to the characteristic equation are complex - known as damped natural frequency
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Second order circuit Underdamped response Case 3 Using Euler’s identity: e j = cos + jsin where
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Second order circuit Underdamped response Case 3 (i) At t = 0, (ii) At t = 0,
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Second order circuit Underdamped response Case 3 10 0.05H 0.5mF +vc+vc Initial condition v c (0) =100V
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Second order circuit Underdamped, overdamped and critically damped responses
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