Boundary Conditions

Objective of LectureDemonstrate how to determine the boundary

conditions on the voltages and currents in a 2nd order circuit. These boundary conditions will be used when

calculating the transient response of the circuit.

2nd Order CircuitsA 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

Boundary ConditionsSteady state

For step response functions u(t- to) for all times betweent = +/- ∞ except for some time period after t = to

Capacitors are opens Inductors are short circuits

During the transition at the step t = to

Voltage across a capacitor is continuous vC(to

+) = vC (to -)

Current through an inductor is continuous iL(to

+) = iL(to -)

Initial ConditionRedraw the circuit at t < to

Determine the value of all voltage and current sources at t< to

Make the appropriate substitutions for the energy storage devices.Substitute an open circuit (∞ resistor) for all

capacitors. Note: IC(t < to ) = 0A.

Substitute an short circuit (0 resistor) for all inductors. Note: VL(t < to ) = 0V.

Calculate VC(t < to ) and IL(t < to ).

Final ConditionRedraw the circuit at t =∞ sDetermine the value of all voltage and current

sources at t =∞ sMake the appropriate substitutions for the

energy storage devices.Substitute an open circuit (∞ resistor) for all

capacitors. Note: iC(t =∞ s) = 0A.

Substitute an short circuit (0 resistor) for all inductors. Note: vL(t =∞ s) = 0V.

Calculate vC(t =∞ s) and iL(t =∞ s).

Example #1

Example #1 (con’t)Initial Condition: The circuit is:

Example #1 (con’t)

∞

iL (-∞) = iL (to-) = 0A vL (-∞) = vL (to

-) = 0V

iC (-∞) = iC (to-) = 0A vC (-∞) = vC (to

-) = [R2/(R1+R2)]V

∞

Example #1 (con’t)

Example #1 (con’t)Final Condition: The switch opens,

which removes V1 and R1 from the circuit.

Example #1 (con’t)The energy stored in the inductor and

capacitor will be dissipated through R2 and R3 as t increased from t= to.

Example 1 (con’t)At time t = ∞s, the energy stored in the

inductor and in the capacitor will be completely released to the circuit.

∞

Example #1 (con’t)

iL (∞s) = 0A vL (∞s) = 0V

iC (∞s) = 0A vC (∞s) = 0V

∞

Example #1 (con’t)

For to < t << ∞s

iL (t) ≠ 0 vL t) ≠ 0

iC (t) ≠ 0 vC (t) ≠ 0

Electronic ResponseDraw the circuits when t < to and t = ∞s for

the following circuit:

Example #2 (con’t)

Example #2 (con’t)

iL (-∞s) = 0.3mA vL (-∞s) = 0V

iC (-∞s) = 0A vC (-∞s) = 3.5V

Example #2 (con’t)

iL (∞s) = 0A vL (∞s) = 0V

iC (∞s) = 0A vC (∞s) = 5V

Example #3

Example #3 (con’t)

iL1 (-∞s) = -1mA vL1 (-∞s) = 0V

iL2 (-∞s) = 1mA vL1 (-∞s) = 0V

Example #3 (con’t)

Example #3 (con’t)

iL1 (∞s) = -1mA vL1 (∞s) = 0V

iL2 (∞s) = 1.4mA vL2(∞s) = 0V

Example #4

iL1 (-∞s) =- 1 mAvL1 (-∞s) = 0V

iC1 (-∞s) = viC2 (-∞s) = 0A

vC1 (-∞s) = vC2 (-∞s) = 4V

Example #4 (con’t)

Example # 4 (con’t)

iL1 (∞s) = 0mA vL1 (∞s) = 0V

vC1 (∞s) = vC2 (∞s) = 1V iC1 (∞s) = iC2 (∞s) = 0A

Summary