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Boundary Conditions

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Objective of Lecture Demonstrate how to determine the boundary conditions on the voltages and currents in a 2 nd order circuit. These boundary conditions will be used when calculating the transient response of the circuit.

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2 nd Order Circuits A second order differential equation is required to solve for the voltage across or the current flowing through a component. The circuit will contain at least one resistor and the equivalent of two energy storage elements 2 capacitors, 2 inductors, or a capacitor and an inductor

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Boundary Conditions Steady state For step response functions u(t- t o ) for all times between t = +/- except for some time period after t = t o Capacitors are opens Inductors are short circuits During the transition at the step t = t o Voltage across a capacitor is continuous v C (t o + ) = v C (t o - ) Current through an inductor is continuous i L (t o + ) = i L (t o - )

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Initial Condition Redraw the circuit at t < t o Determine the value of all voltage and current sources at t< t o Make the appropriate substitutions for the energy storage devices. Substitute an open circuit ( resistor) for all capacitors. Note: I C (t < t o ) = 0A. Substitute an short circuit (0 resistor) for all inductors. Note: V L (t < t o ) = 0V. Calculate V C (t < t o ) and I L (t < t o ).

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Final Condition Redraw the circuit at t = s Determine the value of all voltage and current sources at t = s Make the appropriate substitutions for the energy storage devices. Substitute an open circuit ( resistor) for all capacitors. Note: i C (t = s) = 0A. Substitute an short circuit (0 resistor) for all inductors. Note: v L (t = s) = 0V. Calculate v C (t = s) and i L (t = s).

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Example #1

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Example #1 (cont) Initial Condition: The circuit is:

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Example #1 (cont)

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i L (-) = i L (t o - ) = 0Av L (-) = v L (t o - ) = 0V i C (-) = i C (t o - ) = 0A v C (-) = v C (t o - ) = [R 2 /(R 1 +R 2 )]V Example #1 (cont)

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Final Condition: The switch opens, which removes V1 and R1 from the circuit.

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Example #1 (cont) The energy stored in the inductor and capacitor will be dissipated through R2 and R3 as t increased from t= t o.

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Example 1 (cont) At time t = s, the energy stored in the inductor and in the capacitor will be completely released to the circuit.

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Example #1 (cont) i L (s) = 0Av L (s) = 0V i C (s) = 0Av C (s) = 0V

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Example #1 (cont) For t o < t << s i L (t) 0v L t) 0 i C (t) 0 v C (t) 0

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Electronic Response Draw the circuits when t < t o and t = s for the following circuit:

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Example #2 (cont)

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i L (-s) = 0.3mAv L (-s) = 0V i C (-s) = 0Av C (-s) = 3.5V

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Example #2 (cont) i L (s) = 0Av L (s) = 0V i C (s) = 0Av C (s) = 5V

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Example #3

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Example #3 (cont) i L1 (-s) = -1mAv L1 (-s) = 0V i L2 (-s) = 1mA v L1 (-s) = 0V

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Example #3 (cont)

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i L1 (s) = -1mAv L1 (s) = 0V i L2 (s) = 1.4mA v L2 (s) = 0V

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Example #4

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i L1 (-s) =- 1 mA v L1 (-s) = 0V i C1 (-s) = vi C2 (-s) = 0A v C1 (-s) = v C2 (-s) = 4V Example #4 (cont)

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i L1 (s) = 0mAv L1 (s) = 0V v C1 (s) = v C2 (s) = 1Vi C1 (s) = i C2 (s) = 0A

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Summary Calculation of the initial and final conditions for 2 nd order circuits requires: Knowledge of the magnitude of the voltage and/or current sources in the circuit before and after a step function transition. In steady state (t < t o and t = s), replace energy storage devices. Capacitors are opens circuits => i C = oA Inductors are short circuits => v L = o A Calculate the voltage across the capacitor and the current through the inductor. During the transition at the step t = t o Voltage across a capacitor is continuous v C (t o + ) = v C (t o - ) Current through an inductor is continuous i L (t o + ) = i L (t o - )

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