1.3 Impedance

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来自 Georgia Institute of Technology 的课程

Linear Circuits 2: AC Analysis

41 个评分

Georgia Institute of Technology

Linear Circuits 2: AC Analysis

41 个评分

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.

从本节课中

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

与讲师见面

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 on impedance.

Remember, the objective of this particular module is to be able to analyze circuits

like this when we've got a sinusoidal input to it.

Now we've already gone through in the last lesson and

shared how we can represent the sinusoidal input as a phasor.

Now the goal in this particular lesson is to try to represent these circuit

components as what we call impedances.

This lesson builds upon phasors, which we've already mentioned,

as well as the IV characteristics of R, L, and C components.

Remember that a capacitor has this relationship between the voltage and

the current, i is equal to C dv dt.

Now an inductor has this relationship, which is v is equal to L di dt.

So we want to introduce impedance as a way of substituting for a circuit element.

So we want to be able to look at it in terms of the IV characteristics of it.

So we're going to define impedance as a ratio of phasors of voltage and current.

So, the V over I in terms of the phasors.

So, this is the polar form in the phasors, right here.

So, then if I take I times Z to give me V,

that looks very much like Ohm's Law, V = ZI.

Except that all these terms are complex or they have imaginary components to it.

So what we look like in terms of impedances is that they look like we

are going to treat them like complex resistors.

So we're going to be able to do circuit analysis with impedances,

as if they were complex resistors.

We need to be able to find expressions for the impedance of a capacitor and

an inductor.

Let's look at a capacitor first.

So we want to replace this circuit element with this one right here.

Where Z sub c indicates the capacitance, the impedance of a capacitor.

We're going to assume a sinusoidal steady-state across that capacitor.

This is in terms to voltage, where we've defined the phase

of the voltage as theta sub v.

The impedance of a capacitor has its form 1 over j omega C,

where omega is that frequency and C is the capacitance.

And remember j is c square root of minus 1.

Where this comes from is from this derivation,

where we've got this form comes from directly from here,

that's what we said and then I take the derivative of it and I get this.

So C dv dt gives me i, so this is now i.

And this is a sine wave.

I have to go from a sine wave to a cosine and so that's where I subtract off 90 and

I have to get rid of this minus sign right there and that's why I have to add 180.

This is just trig identities, trig formulas there.

Now, phasor form, because impedance, remember, is defined as the ratio

of the output phasor over the input phasor, the v over i, in other words.

So, this is the phasor for v, the phasor for i.

I take the ratio and I get 1 over j mega c.

We can do a similar derivation for the impedance of an inductor.

I want to replace this inductive element with and

impedance z sub L and In this case we are sinusoidal steady-state of the current.

And we can derive an expression for the impedant Z sub L is j omega L.

Let me go through that derivation over here.

We start out with the i right here, and then we have L di dt,

we take the derivative of that expression, we get this for V.

And again, I have to go from this sine to a cosine, so I subtract off minus 90, and

then I have to get rid of that minus sign right there.

And that's where I add 180 degrees in here.

I have to go to the phasor form.

So the phasor of this LI omega is c amplitude.

And the angle is theta i plus 90.

And the phasor for the current is I at theta i.

I form the ratio V over I is equal to Z.

And this is where I get L omega j.

So the key concepts on this lesson are, first of all, we defined impedance.

And I'm showing it here in terms of the Ohm's Law form of it.

But it's really the ratio of the V over I is gives me impedance.

And I showed it here because I think it's important for

you to realize that we treat it as if it's a complex resistor.

So then I have to go and replace these components and we didn't show this

explicitly, but it falls directly out of this impedance formula right here.

That if this was a resistor,

this is the V and the I for resistor, then the Z is simply R.

So the impedance of a resistor is simply R.

And we derive the formulas for a capacitor, 1 over j omega C.

And the impedance for an inductor is j times omega L.

All right. Thank you very much.

[MUSIC]

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