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ECE201_30_Jung - ECE 201 Lecture 30 Complex Forcing...

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ECE 201 Lecture 30 Complex Forcing Function Announcements Last Lecture #29: RC Op Amp Circuits This Lecture #30: Continue RC Op-Amp circuit analysis Complex forcing function
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Oscillator Application KCL: KVL: + -
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0 1 2 3 4 5 6 7 8 -1.5 -1 -0.5 0 0.5 1 1.5 Solutions (R 1 =R 2 ) Characteristic equation Assume R 1 =R 2 =R Solutions are
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Solutions (R 1 <R 2 ) Characteristic equation Assume R 1 <R 2 Two roots are on the right half of the complex plane 0 1 2 3 4 5 6 7 8 -5 -4 -3 -2 -1 0 1 2 3 4 5 Solutions are
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0 1 2 3 4 5 6 7 8 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Solutions (R 1 >R 2 ) Characteristic equation Assume R 1 >R 2 Two roots are on the left half of the complex plane Solutions are
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Oscillator Application 2 3 f Out Out f f R V R R υ υ + = = + 2 3 Out V V υ υ - + = = = f G Advanced f G LPF HPF + = f G BPF 1 1 1 f G f G LPF HPF + = f G BPF 1 1 1 f G f G LPF HPF + = f G BPF 1 1 1
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Most Commonly Used Topologies + + + + + + + + + + +
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Motivating Example Initial condition Input Output Homogeneous equation : Characteristic equation with Particular solution to the inhomogeneous eq : General solution: Guess: Differential equation becomes
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Math Math Math…. Do we really need to solve differential equations every time to solve RLC network problems?
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