2.5 Frequency Response: Bode

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Linear Circuits 2: AC Analysis

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Georgia Institute of Technology

Linear Circuits 2: AC Analysis

54 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.

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Module 2: Frequency Reponse

The description goes here

2.1 Frequency Response Module Overview3:19

2.2 Frequency Spectrum9:01

2.3 Frequency Spectra of Real Signals11:42

2.4 Frequency Response: Linear7:14

2.5 Frequency Response: Bode9:46

2.6 Bode Plot of RC Circuits7:34

2.7 Bode Plot of RLC Circuits6:04

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 on the frequency response using Bode plots.

We would like to introduce the Bode plot as a way of showing

the frequency response.

In our last lesson,

we showed the frequency response on linear scales like this.

And we're going to be representing the same data but using logarithmic scales.

The other thing that we were going to be building on in this lesson is the input-

output behavior using transfer functions.

Remember this relationship between the amplitudes and

the phases of the input signal and the output signal.

The Bode plot is simply defined as being the plot of the transfer function

on logarithmic scales.

So if I look here, the frequency is on a logarithmic scale.

You see the spacing between these things are right here.

Goes up by powers of ten.

And then along this over here,

we define this as not being just the magnitude of H, but 20 times the log,

base 10 of the magnitude of H, and we're going to call that in decibels.

The angle is plotted, this is straight, directly the angle, but again it's plotted

on this axis as being with the logarithmic scale.

So, the magnitude actually has a log scale here, and along this way.

Now we're going to define a decade as being the distance between a frequency and

ten times it's frequency.

So right here between 10 and 1000, we've got 1 decade.

And that distance is true no matter where we start from.

So if we start from over here that distance is the same from

frequency to ten times its frequency is a decade.

Now let's compare the linear frequency response plot to the Bode plot.

If I look at the linear plot, for example here.

It starts out a magnitude of 1 at omega equals 0.

But on log plot we never have omega equals 0 because we keep going down by a factor

of ten every time we go over a decade, but you can see that it's asymptotic to 0.

And that corresponds to the fact that if I look at this, this is a magnitude,

it's 20 times the log of whatever these values are.

20 times the log of 1 is 0.

So that's why we have a level of 0.

And that's again in decibels.

Over here for example, at omega equals 1000, we have a value of 0.1.

So if I take 20log(.1).

I get approximately -20dB.

So that point right there, goes over to this point right here.

And if we look at another point, say, this point right here,

this will be a special point that we'll talk about some time in the future.

But this point right here is at, say .707.

How does that get?

And that occurs at about this point right

here at about 100 radiance per second,

20 times a log of .707 is -3dB.

So this point goes over to, and this is 10, this is a 100.

This point here is 200.

So at 100 we're at minus 3dB.

So, that's where this point gets moved to.

And the other thing is this on the angle plot,

it's just plotted on different scales.

I'm not changing the angle here at all, so

it still goes from 0 to -90, and this goes from 0 to -90.

Now let's look at example of how to use the Bode plot in order to find

an output for a given input.

So, this circuit has this transfer function, this frequency response.

We're looking at a DC component.

Omega = 100, and

omega = 3000.

So if I go over here at omega = 100, that's 10 squared is 100,

is right there, that's 0 decibels.

And at 3000, that's 1000, 2000, 3000, follow-up here is right there.

And the same in terms of the angle goes down about right there,

and at 100 I get this.

Now, DC value is not on here, but what we see is it looks constant at 0 decibels.

So it's going to be 0 decibels and the angles going to be 0.

So following that, what we need to be able to do is

we need to have a relationship that is A out for

example, is equal to A in times H at that frequency.

What we have is this, in decibels.

20 times the log of that.

So if I have 20 times a log of H,

and I'll call that H in dB.

So this for example is 10.

That would be the value in decibels.

I want to come back to what this is.

So I have to divide this by 20, and then raise it to the power of 10.

So H is equal to 10 raised to this value right here.

H of omega and dB all over 20.

And that's just basically the inverse of that formula right there.

So I want to find what H is at 100, it is 0 decibels corresponds to magnitude of 1.

So at omega = 0, I've

got 0 db, H(0) is 1.

At omega = 100, again I've got 0 decibels.

H of 100 is equal to 1, the angle is 0.

Omega is equal to 3000.

I've got 10dB,

And this magnitude is equal to, following this formula, I take 10 raised to

this value right there which turns out to be 10 decibels divided by 20.

And that's equal to 3.16.

So now I'm ready to solve it.

V0, the output is equal to the DC

value times 1, plus at 100,

it's the amplitude of 1 times 1.

And the angle here was 0.

So, I just keep the original angle, so it's the input angle plus the transfer

functioning angle or frequency responds angle, and so it stays the same.

And then on this one, the amplitude was 1.

I multiply it by 3.16, cosine of 3000

Okay, and, this original angle was 0, and

I add to it the angle here, which looks like it's about -90.

And, that's the output.

So what we've done here, is shown how we can use the Bode plot.

To find the output of a circuit, given this particular input.

To summarize our key concepts, we've introduced the Bode plot,

which shows way of representing the frequency response on logarithmic scales.

So we use logarithmic scales on the frequency access, and for

the magnitude plot, we also use it over here on the scale for the magnitude.

Now, the units right here for the Bode plot magnitude is in decibels.

The other thing we looked at was our input-output behavior.

Now this relationship we've used in a lot of different lessons.

The relationship between the input and output amplitudes and input and

output phases.

Now, this particular relationship requires a linear magnitude,

and up here, oftentimes we call it a magnitude in decibels.

So we have to convert anything that we see, any value that we pull of this plot,

we have to convert it back through this formula to the magnitude

in the linear scales to use that formula.

And so we use this relationship, so we pick something off of the plot right here

divide it by 20, and raise it 10 to that power to find this value here,

and that's the value we plug into that formula.

All right, thank you.

[MUSIC]

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