ECE2260F07_HW1p3soln

Matching this to the symbolic solution for t 0 we have

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Unformatted text preview: nd A2 by matching our symbolic solution to dv ( t ) + . We have already found v(0+) to € circuit values for v(0 ) and dt + t =0 be 0 V. Matching this to the symbolic solution for t = 0+, we have the following: € A1 + A2 = 0V or A1 = –A2 For the value of the derivative in the circuit, we always try to first write our variable, (v in this case), in terms of energy (or state) variables, iL and/or vC. Here, this is a simple matter: v ( t ) = vC ( t ) € Then we differentiate, and use the component equations involving d/dt for L and/or C: d d i (0+ ) −2.3 mA v(t) = vC ( t ) =C = + + dt dt C 5 µF t =0 t =0 We equate this to the symbolic derivative: d −2.3 mA v(t) = A1s1 + A2 s2 = A1 ( s1 − s2 ) = A1 2 α 2 − ω 2 = o dt 5 µF t =0+ € or € A1 = −2.3 mA 2 α 2 − ω 2 ⋅ 5 µF o −2.3 mA = 2 2 Mr/s ⋅ 5 µF 5 = −2.3 mV ≈ −257 µV 45 Plugging in values gives the solution for v(t > 0): € € −1+ v ( t ) = −257µV ⋅ e 2 Mr/s⋅t 5 −1− + 257µV ⋅ e 2 Mr/s⋅t 5...
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This document was uploaded on 03/17/2014 for the course ECE 2260 at University of Utah.

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