1.7 Transfer Function Example

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Do curso por Georgia Institute of Technology

Linear Circuits 2: AC Analysis

45 classificações

Georgia Institute of Technology

Linear Circuits 2: AC Analysis

45 classificações

This course explains how to analyze circuits that have alternating current (AC) voltage or current sources. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

Na lição

Module 1: AC Circuit Analysis

This module describes how to analyze circuits with sinusoidal inputs

1.0 Introduction to Linear Circuits 2: AC Analysis3:11

1.1 AC Circuits7:50

1.2 Phasors11:06

1.3 Impedance5:29

1.4 Circuit Analysis with AC Impedances10:16

1.5 AC Circuit Analysis Example7:35

1.6 Transfer Functions13:28

1.7 Transfer Function Example9:04

1.8 Module Summary3:50

Conheça os instrutores

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

[MUSIC]

Welcome back to linear circuits.

This is Dr. Ferri.

This lesson is an example of using the transfer function.

Our objective is to use the transfer function to find the output for

inputs of different frequencies.

We're going to be building upon our last lesson

where we introduced the transfer function and

showed it to be a way of relating the input phasor to the output phasor.

And the transfer function was defined again as being the output phasor

divided by the input phasor.

And they key here was that we can look at the change in amplitudes between

the input and the output as being the magnitude of that transfer function and

the angle as being the difference between those angles.

And we use it to find the output angle and amplitude using these formulas right here.

Let's take a look at this example, we're going to be using the same impedance

method that we outlined before when we were trying to solve for

a particular output.

In this case, our output is going to be iout and our input is the voltage.

So our transfer function is going to between a current and a voltage.

The first step is to convert this into an impedance equivalent impedance circuit.

So we've got the Vin, the phasor of this,

all our components get converted into their

impedance form, and I'll just call them Zc and Zl for now.

And we're trying to solve for the phasor of Iout.

Z sub L, I'll keep this in sort of a generic form.

J omega L, Z sub C is equal to 1 over j omega C.

And I'm not going to substitute in specific values for omega.

That's the key difference between the transfer function and

explicit solution methods that we've done before.

Okay, this particular thing I want to look at this circuit say,

I want to solve for i out.

So if I wanted to do that I'm going to find the equivalent impedance for

this entire circuit, and then just use Ohm's law.

So the equivalent impedance is equal to R + Z sub L in parallel with Z sub C.

And that is equal to R + Z sub L,

the product over the sum,

and I can combine my terms here.

Common denominator, and I get this term here.

R times that denominator.

Okay.

Now, my transfer function is what I'm trying to solve for.

So, I want an expression for Iout,

it's going to be Vin/Zeq.

In other words, Iout is equal to 1 over Z equivalent times Vin.

And this right here is my transfer function.

That's because if I divide the output by the input, this is what I'm left over.

Another way of looking at it and saying the output is equal to something

which is the transfer function times the input and

whatever that something is, again, is a transfer function.

So I've solved for, Z equivalent,

I have to invert that to get this expression right here.

Z sub L plus Z sub C over this to get that transfer function.

If I then plug in these quantities right here,

these expressions right here, I can simplify this.

I've got this expression right here and

I want to notice that j squared is equal to minus one.

So I'm going to have a minus one times this term.

Let me rewrite is as 1- omega squared LC, and

in the denominator, I get R(1- omega squared LC) + j omega L,

and that is my transfer function right there.

Let's continue on with this example, but with actual numbers in there.

So, we've defined the capacitor, conductor, and resistor

values here and just to remind you, this is what we solved for in our last slide.

I just want to substitute in for there, and

wanting to keep omega as a variable because this is a function of omega, so

if I substitute in I get the following expression.

I've got 1 times 10 to the minus 5 as my coefficient of omega squared.

And then in the denominator I get this expression here.

And that is my transfer function.

Now, the thing is, I want to be able to use this transfer function to found i out,

and this is a general form for various values of omega.

So, if I have a cosine of a particular frequency, or

different amplitudes, different phases,

I have now an expression to find that output In a generic form like a formula.

And the formula that I use, again, is A out is equal to

the amplitude at that frequency times A in.

And theta out is equal to the angle at that particular frequency

plus the input angle.

So, let's suppose that if omega = 100.

And, let's say, let's find H at that frequency.

Let's go ahead and say,

Ai = 10, and theta i = 0.

So in other words V in = 10

cos(100t) + 0.

Okay, so what I need to do is substitute in from omega.

I have H(100)=,

if I substitute in for omega,

I can simplify this all out to .9 over 9+j.

If I find what the magnitude is,

that is equal to 0.9 over the magnitude of 9 + j.

Which is 0.099 and the angle.

Well the angle of the numerator is zero because it's just a real number.

And so I've just got the angle of the denominator so

since it's in the denominator I've got a minus sign here.

Minus arc tan of the imaginary over the real.

Imaginary part of the real is one-ninth, and that's the inverse tangent.

And that is equal to a degrees- 6.3 degrees.

So using these formulas up here with what

you just saw for the magnitude in angle,

then I have iout as my output is equal

to the amplitude of H times 10.

So it would be 0.99 cosine

same frequency 100 t and

then this angle, -6.3 degrees.

And that to shows how to use the transfer function to solve a particular problem

at a particular frequency.

If I wanted to go back and solve for a different input frequency,

different input amplitude or frequency, I don't have to resolve this whole problem.

I now have this relationship between the input and the output that I can use

very easily if I'm going to change the angles,

the angle or the amplitude, or even the frequency.

To summarize, in this lesson, we've looked at the transfer function and

we've used that to solve input, output problems.

We looked at particular circuit and found its transfer function and

then moreover we used this relationship right here between the input and

output amplitudes and input and output phases to be able to solve for

a particular output for a given input.

All right, thank you very much.

[MUSIC]

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