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Â Welcome back to linear circuits, this is Doctor Weitnauer.

Â This lesson is finding initial conditions in second order circuits.

Â Our objective are to find out what type of initial conditions are needed,

Â and apply circuit analysis to calculate them.

Â This builds on many things.

Â A second order circuit is modeled by a second order differential equation.

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The current inductor is continuous in time.

Â The voltage across a capacitor is continuous in time.

Â And finally, KCL.

Â Our motivation is, we need these initial conditions in the last step of finding

Â the complete response of a circuit with reactive elements.

Â What initial conditions are needed?

Â If we want to find y(t) for t greater than or equal to 0,

Â when y(t) satisfies a differential equation such as this.

Â Because this is a second order differential equation,

Â we need y(0+) and its derivative at 0+.

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Or if we're not given those things, then we can derive them

Â by DC analysis prior to t=0, and use their continuity properties.

Â But even if you have these, we are not done,

Â we have to find that derivative and this usually requires circuit analysis.

Â Here's a simple example, the series RLC circuit.

Â Find the initial conditions for the solution for v(t).

Â Suppose we were given v(0+) and i,

Â i being the current through the inductor at times 0+.

Â However, because we are trying to find the initial conditions for

Â the solution for v(t), this is not all we need.

Â We have one part of it, but we need the derivative of v(0+).

Â So we can use i = C dv dt, and

Â then solve for dv dt at 0+.

Â And that will give us 1 over C times i(0+) which we are given.

Â This is a harder example.

Â We will determine the initial conditions needed for v(t),

Â t greater than or equal to 0, assuming the circuit is at rest,

Â that is, DC steady state, prior to t=0.

Â To do this, we should consider what this circuit looks like prior to t=0.

Â There are four things that we need to consider.

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This resistor will not play a role because the switch is open.

Â The capacitor will behave as an open circuit,

Â the inductor will behave as a short circuit.

Â And the voltage, because u(t) will be 0 for

Â negative time, the voltage is only 7 volts.

Â So this is what our simplified circuit looks like for t<0.

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First we can notice that the resistor is in parallel with the voltage source.

Â So that means that v(0-) is 7 volts because of the continuity

Â of voltage in capacitors, that is also v(0+).

Â There's one of the initial conditions that we need.

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The current in the inductor, we can find that by using Ohm's law.

Â And because of the continuity of current as a function of time

Â in an inductor, that is also the current at i(0+).

Â But that is not the second initial condition we need,

Â we need this derivative at t=0+.

Â Now, I want to emphasize this is at 0+.

Â The circuit is different for positive time, so

Â we need to redraw the circuit for t>0.

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Here is that circuit, and we see that we can do a little more simplification by

Â combining these parallel resistors into one effective resistance of R over 2.

Â Now I want to make a little room on the screen and we'll continue.

Â Note that we have these initial values of the voltage v and

Â the current i through the inductor, and

Â our goal is to find the derivative of the voltage at t = 0+.

Â It's going to be helpful for us to define two currents.

Â i sub r through that branch, and

Â i sub c through this branch, and

Â we'll be using KCL, which is, i = ir + ic.

Â This is that circuit analysis I was telling you about.

Â Now we'll going to use the ic = c dv dt.

Â That is why I've chosen to use KCL.

Â It allows me to bring in that derivative that we're seeking.

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I'm also going to note that ir, because it's in parallel

Â with the capacitor, we can find it by saying it's V

Â over the value of the resistance, which is R over 2.

Â Next I will solve the KCL equation for

Â ic, and then divide by c,

Â giving me dv dt = 1 over c [i- ir].

Â And this is going to be true for anytime, but

Â in particular we want it for t = 0+.

Â That's the other initial condition we need,

Â and we will be able to just plug in the current through the inductor at i+ and

Â then we can use the V(0+) divided by (R/2).

Â And that gives us the other initial condition that we need.

Â Okay, to summarize the key concepts in this lesson.

Â Finding the initial conditions for a second order circuit requires more

Â than just finding the inductor currents and the capacitor voltages.

Â In general, you're going to have to do some circuit analysis to

Â find the derivative of the desired variable at t = 0+.

Â And it's important to redraw the circuit for

Â t>0 when you're trying to find that derivative.

Â Thank you.

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