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Kevin D. Donohue, University of Kentucky1 Transient Response for Second- Order Circuits Characteristics Equations, Overdamped-, Underdamped-, and Critically Damped Circuits

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Kevin D. Donohue, University of Kentucky2 Second-Order Circuits: In previous work, circuits were limited to one energy storage element, which resulted in first-order differential equations. Now, a second independent energy storage element will be added to the circuits to result in second order differential equations:

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Kevin D. Donohue, University of Kentucky3 Example Find the differential equation for the circuit below in terms of v c and also terms of i L Show:

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Kevin D. Donohue, University of Kentucky4 Example Find the differential equation for the circuit below in terms of v c and also terms of i L Show: is(t)is(t) RLC +vc(t)_+vc(t)_ iL(t)iL(t)

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Kevin D. Donohue, University of Kentucky5 Solving Second-Order Systems: The method for determining the forced solution is the same for both first and second order circuits. The new aspects in solving a second order circuit are the possible forms of natural solutions and the requirement for two independent initial conditions to resolve the unknown coefficients. In general the natural response of a second-order system will be of the form:

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Kevin D. Donohue, University of Kentucky6 Natural Solutions Find characteristic equation from homogeneous equation: Convert to polynomial by the following substitution: to obtain Based on the roots of the characteristic equation, the natural solution will take on one of three particular forms. Roots given by:

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Kevin D. Donohue, University of Kentucky7 Overdamped If roots are real and distinct ( ), natural solution becomes a 1 =-3 a 1 =2

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Kevin D. Donohue, University of Kentucky8 Critically Damped If roots are real and repeated ( ), natural solution becomes a 1 =-3 a 1 =2

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Kevin D. Donohue, University of Kentucky9 Underdamped If roots are complex ( ), natural solution becomes: or

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Kevin D. Donohue, University of Kentucky10 Example Find the unit step response for v c and i L for the circuit below when: a) R=16, L=2H, C=1/24 F b) R=10, L=1/4H, C=1/100 F c) R=2, L=1/3H, C=1/6 F Show: a) b) c)

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