7.7 (OPTIONAL): Obtaining the Differential Equation - Operator Method.

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Linear Circuits 1: DC Analysis

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Georgia Institute of Technology

Linear Circuits 1: DC Analysis

124 notes

This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

À partir de la leçon

Module 7

This module demonstrates useful applications of capacitors and inductors in real engineering systems and introduces the behavior of second-order reactive circuits.

7.5 Lab Demo:RLC Circuit.8:05

7.6 (OPTIONAL): Obtaining the Differential Equation - Direct Method.6:50

7.7 (OPTIONAL): Obtaining the Differential Equation - Operator Method.8:43

7.8 (OPTIONAL): Finding the Initial Conditions7:00

Rencontrer les enseignants

Dr. Bonnie H. Ferri
Professor
Electrical and Computer Engineering
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering
Dr Mary Ann Weitnauer

[MUSIC]

Welcome back to Linear Circuits.

This is Doctor Weitnauer.

This lesson is the operator method.

Our objective is, for a second order circuit, learn how to apply the operator

method to determine the differential equation for the desired variable.

This builds on a second order circuit is modeled by a second

order differential equation I V relationships for capacitors and

inductors, node voltage analysis, mesh current analysis.

The motivation for this method is that allows one to

use a standard circuit analysis approach, in contrast to

the direct method where we are aiming at trying to use the derivatives.

In this approach we can use integrals and it's okay.

There may be increased algebraic complexity however,

compared to the direct method.

The summary of the operator method is first,

decide the best standard circuit analysis method for your circuit.

Write the circuit equations for that method, using either the derivative or

the integral V I relationships for each storage element.

For integrals, assume zero initial conditions.

Even though this may not be the case, do it because the correct initial

conditions will be applied later when we get the combined response.

This lesson is only about getting the differential equation.

Next we'll substitute the variable s for each derivative operator and

1 over s for each integral operator.

We will obtain a quadratic equation in s,

in terms of the variable of interest, and have to clear the fractions.

Substitute each factor of s with d by dt.

So we'll do the first example.

We're going to do two examples, so the same ones from the direct method lesson.

This one is the parallel RLC and

were finding i the current through the inductor.

Write a Node Voltage Analysis equation.

Why?

Because all of these elements run parallel and

besides the ground node there's only one other Node.

So in the Node Voltage Analysis we're going to be adding up all the currents

flowing out of the top node so we will have this equation at the top and notice

that the current through the inductor is expressed as 1 over L integral of VDT.

We have to do that because with node voltage analysis the variable is V.

But we're aiming at a differential equation in terms of i, so

we're going to need the iv characteristic for the adductor to relate i to v.

Now we convert to the s domain.

And what that means is that we take each term that has either an integral or

a derivative in it, and we replace it by the appropriate expression of s.

So for example, for the integral we're going to replace it with one over s,

and for the derivative, we're going to replace it with just s.

And we also do that for

the other equation because we're going to need to eliminate v.

So we do that.

We substitute v = Lsi in for every occasion of v here.

And now our goal will be to put this quadratic expression in standard form.

Which mean, and we're going to do that in to steps.

The first step, we will move the IS term, that's the forcing function term,

over to the right hand side.

And then we're going to divide through by LC every term.

And then finally, standard form has the S squared term first,

followed by the S term, follow by the 0 order s term.

All of these terms on this side involve i, and

all the other terms are on the other side.

Finally, we will exchange the s, each factor of s, with d by dt.

So you see the s squared becomes a second order derivative.

And the s just becomes d by dt.

And we end up with the differential equation that we seek and

you can compare this to the one that we got with the direct method.

Now, here's the other example, in this case,

we're going to find the differential equation for I2.

And we want to apply mesh current analysis because

we have just two meshes in this circuit.

So here I've defined the mesh currents.

Because I'm defining mesh currents, the currents I want in i are going to

be a little bit different in this example than they were in the direct method.

But you can still compare these two.

It's the same example and we're seeking the same differential equation.

In Mesh 1, we write the KVL equation.

So going around we have minus VS + R1 x I1 + L1 d by dt,

and I have to do the current through the branch flowing downwards.

So I want I1- I2, right here.

Mesh two,

I'm coming around and I'm leading notice I'm leading with the I2 here.

So it's I1- I1.

Then when I can catch this voltage, that's going to be L2 d by dt i2,

and then finally, r2 times i2.

Now, we map to the S domain, so every d by dt is going to be replaced by S.

There's the first mesh equation, there's the second mesh equation.

Now, since we want to end up with a differential equation in i2,

we have to do algebra to eliminate the i1.

So, we look at which equation has the fewest i1 terms, and we'll use that one.

The second equation here has only one i1 term, so we can solve that for i1.

And then we'll substitute that in to the top equation here.

Now this gets messy and I'm showing you all this for a point.

I want you to see that while this method is very straightforward it can get pretty

pretty messy algebraically.

So now we, I need to make a little room and

we substitute this expression for i1 into the top equation.

And the substitution is wherever you see the square brackets.

Now, I need to clear the fractions.

I multiply every term in the equation by L1s.

And what I'm doing here is, I'm distributing all these terms,

so now I have everything multiplied out, and it happens that

this term is the negative of this term, so we can cancel those out.

Then what I want to do is start grouping terms according to s.

This is an s squared term, and then I have three.

Single powers of S.

So I've got this one, this one, and this one, excuse me,

this one over here, I can write and show you those.

Those are all three going to get grouped together into this coefficient.

And then, to put this quadratic expression into standard form,

I need to divide every term by L1 over L2.

Finally, I replace all the factors of s with d by dt, and I have what

I was seeking, which is the differential equation for i2 in that circuit.

And you can compare this to what we did with the direct method.

Okay, to summarize this lesson, when using the operator method,

choose the most convenient circuit analysis technique for your circuit.

Substitute d by dt with s, and the integral dt with 1 over s.

Then eliminate the unwanted variable through substitution and

put the equation into standard quadratic form.

This can be tedious.

Finally, substitute s with d by dt and

s squared by d squared by dt squared to get back into the time domain.

Thank you.

[NOISE]

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