ECE634S09_L11_ElectricalNetworks

ECE634S09_L11_ElectricalNetworks - ECE634 Signals and...

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ECE634 Signals and Systems II, Spring 2009 - Lecture 11, February 16 Daniel S. Brogan 1 4.4 Analysis of Electrical Networks in the Frequency Domain: Example 7.28 (G. E. Carlson, Signal and Linear System Analysis ) – Circuit to equation in the time domain Consider the electrical circuit shown below. The switch is closed for all 0 t < , so the circuit is in equilibrium at 0 t = . Find ( ) o vt for 0 t Solution: Since the voltage supply is disconnected for 0 t , there is no input signal and () 0 Xs = . Some definitions to recall about circuit component equations: L L di vL dt = and C C dv iC dt = . In the differential equation, any inductor will require an initial condition 0 L i and any capacitor will require an initial condition ( ) 0 C v . Steady-state definitions at time 0 t = specify that a capacitor is an open, and an inductor is a short. Thus, at time 0 t , the circuit is a simple resistor divider network such that 8 02 13 L V iA == Ω+ Ω and 3 08 6 C vV V Ω Now for some circuit equations… () () 2 L oL di vt vt dt () () () () 33 2 L CL LL di it vt it dt =+ CC LC dv dv C dt dt =− Converting with Laplace transforms…
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ECE634 Signals and Systems II, Spring 2009 - Lecture 11, February 16 Daniel S. Brogan 2 () () () () () () ()[ ] 32 0 2 2 3 4 CL L L L L L V s I s sI s i I s sI s I s s ⎡⎤ = +− = =+ ⎣⎦ () () () () 06 LC C C I s sV s v sV s =− + () () () () () () 20 2 2 2 2 4 oL L L L V s sI s i sI s sI s = −=− Solving… () 6 L C Is Vs s −+ = [] 6 234 L L Is s s () () 2 62 3 4 LL I sI s s s s = + 2 23 1 4 6 L I ss s s ++=+ 2 46 1 L s + = ++ ( ) ( ) 22 2 81 24 1 4 4 1 1 1 2 1 1 o s s s s s s + + = = = + + ( ) 4 1/2 1 1 1 o AB s s == + = + + + + + 1/2 44 t t o vt e e u t
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This note was uploaded on 03/05/2009 for the course ECE 651 taught by Professor Messner during the Spring '09 term at New Hampshire.

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ECE634S09_L11_ElectricalNetworks - ECE634 Signals and...

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