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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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Circuit Elements Electric circuit is the interconnection of circuit elements ActivePassive Not capable of generating energy e.g. resistor, inductor, capacitor.

Circuit Elements Electric circuit is the interconnection of circuit elements ActivePassive Not capable of generating energy e.g. resistor, inductor, capacitor.

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