ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 14 Second Order Transient Response

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Damped Harmonic Motion

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Drive voltagev S (t)Applied forcef(t) ChargeqDisplacementx CurrentiVelocityv InductanceLMass m Inductive EnergyU=(1/2)Li 2 Inertial Energy(1/2)mv 2 CapacitanceCSpring constant1/k Capac. EnergyU=(1/2)(1/C)q 2 Spring energyU=(1/2)kx 2 ResistanceRDampening constantb The Series RLC Circuit and its Mechanical Analog

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Second –Order Circuits Differentiating with respect to time:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Second –Order Circuits Dampening coefficient Undamped resonant frequency Define: Forcing function

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Solution of the Second-Order Equation

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Solution of the Complementary Equation

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Solution of the Complementary Equation Dampening ratio Roots of the characteristic equation:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. 1. Overdamped case (ζ > 1). If ζ > 1 (or equivalently, if α > ω 0 ), the roots of the characteristic equation are real and distinct. Then the complementary solution is: In this case, we say that the circuit is overdamped.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. 2. Critically damped case (ζ = 1). If ζ = 1 (or equivalently, if α = ω 0 ), the roots are real and equal. Then the complementary solution is In this case, we say that the circuit is critically damped.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. 3. Underdamped case (ζ < 1). Finally, if ζ < 1 (or equivalently, if α < ω 0 ), the roots are complex. (By the term complex, we mean that the roots involve the square root of –1.) In other words, the roots are of the form: in which j is the square root of -1 and the natural frequency is given by:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. In electrical engineering, we use j rather than i to stand for square root of -1,because we use i for current. For complex roots, the complementary solution is of the form: In this case, we say that the circuit is underdamped.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source Apply KVL around the loop:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source To find the particular solution, we consider the steady state. For DC steady state, we can replace the inductors by a short circuit and the capacitor by an open circuit:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source To find the homogeneous solution, we consider three cases; R=300 , 200  and 100 . These values will correspond to over-damped, critically damped and under-damped cases.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source To find K 1 and K 2 we use the initial conditions:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source To find K 1 and K 2 we use the initial conditions:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Analysis of a Second-Order Circuit with a DC Source To find K 1 and K 2 we use the initial conditions:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Overdamped

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Critically damped

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Underdamped

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Normalized Step Response of a Second Order System

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Example: Closing a switch at t=0 to apply a DC voltage A Normalized Step Response of a Second Order System

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Normalized Step Response of a Second Order System What value of  should be used to get to the steady state position quickly without overshooting?

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C We can replace the circuit with it’s Norton equivalent and then analyze the circuit by writing KCL at the top node:

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C This is the same equation as we found for the series LC circuit with the following changes for  :

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Example Parallel LC Circuit: RF-ID Tag C L

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C Find v(t) for R=25 , 50  and 250 

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C Find v(t) for R=25 , 50  and 250 

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C To find the particular solution v P (t) (steady state response) replace C with an open circuit and L with a short circuit. V p (t) = 0

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C Since the voltage across a capacitor and the current through an inductor cannot change instantaneously

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Circuits with Parallel L and C