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 v t for 0 t Solution: Since the voltage supply is disconnected for 0 t , there is no input signal and ( ) 0 X s = . Some definitions to recall about circuit component equations: L L di v L dt = and C C dv i C 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 0 2 1 3 L V i A = = Ω + Ω and ( ) 3 0 8 6 1 3 C v V V Ω = = Ω + Ω Now for some circuit equations… ( ) ( ) 2 L o L di v t v t dt = = ( ) ( ) ( ) ( ) 3 3 2 L C L L L di v t i t v t i t dt = + = + ( ) ( ) C C L C dv dv i t i t C dt dt = − = − = − Converting with Laplace transforms…
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