ECE634 Signals and Systems II, Spring 2009 - Lecture 11, February 16 Daniel S. Brogan 14.4 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 0t<, so the circuit is in equilibrium at 0t=. Find ( )ovtfor 0t≥Solution: Since the voltage supply is disconnected for 0t≥, there is no input signal and ( )0Xs=. Some definitions to recall about circuit component equations: LLdivLdt=and CCdviCdt=. In the differential equation, any inductor will require an initial condition ()0Li−and any capacitor will require an initial condition ()0Cv−. Steady-state definitions at time 0t−=specify that a capacitor is an open, and an inductor is a short. Thus, at time 0t≥, the circuit is a simple resistor divider network such that ()80213LViA−==Ω + Ωand ()308613CvVV−Ω==Ω + ΩNow for some circuit equations… ( )( )2LoLdivtvtdt==( )( )( )( )332LCLLLdivtitvtitdt=+=+( )( )CCLCdvdvititCdtdt= −= −= −Converting with Laplace transforms…
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