This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.
1.6 Review of Frequency Response Plots
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Do curso por Georgia Institute of Technology
Introdução à Eletrônica
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Introduction and Review
Learning Objectives: 1. Review syllabus and procedures of this course. 2. Review concepts from linear circuit theory to aid in understanding material covered in this course.
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Dr. Bonnie H. FerriProfessor
Electrical and Computer Engineering
Dr. Robert Allen Robinson, Jr.Academic Professional
School of Electrical and Computer Engineering
Welcome back to electronics.
This is Dr. Ferri.
In this lesson,
we will do a review of frequency response plots including Bode Plots.
In our last lesson, we did a review of transfer functions.
Well, we need transfer functions for us to be able to do a frequency response.
So this lesson will define frequency response for transfer function.
Actually I'm going to go ahead and
define it right here, because this, this is really all I want to say for it.
A frequency response is the magnitude plot
of, taking the transfer function, H of omega.
We want to take the magnitude of H versus omega.
Plot it that way.
And the angle is the angle of H versus omega.
H is a complex variable.
And then we want to show the linear plots and
Bode Plots corresponding to the frequency response.
Lets look at a particular circuit.
This is an RC circuit that we've looked at a couple of times in
the impedance lesson as well as in the transfer function lesson.
We found that the transfer function is equal to this H right here.
And what we want to define is the magnitude and the angle.
Well lets look at, lets take a step back and think about.
A magnitude of A, complex number.
A plus jb.
The magnitude of 1 over a plus jb, is going to be 1 over.
The square root of the real part squared,
which is a squared plus the imaginary part squared, which is b squared.
So, in this particular case, b is equal to RC omega,
that is what multiplies J, and I've got that squared.
And then the real number in the denominator is one, so
that's squared right there.
So the magnitude is this.
Now the angle 1 over a plus jb
rather the, the angle of that, is minus arc tan.
And the minus sign comes because this is.
The complex number's actually in the denominator.
And then it's the imaginary part over the real part.
So, in this case, it's b over a.
And A is equal to 1, and B is equal to RC omega.
If I want to plot this, what I'm plotting.
And the magnitude is H of omega and then this magnitude.
So I'm applying this function right here.
And it's the function of omega, and
one thing you can see is that as omega gets very large, this gets small.
And then this is the angle is the arc tangent of RC times omega.
And in this case as omega gets large, this approach is minus 90 degrees.
And as omega is equaled 0, it's 0 degrees.
What's it mean, the transfer function here?
It's really how a circuit processes signals of different frequencies.
So in this case, suppose I've got an input that is a sum of two sign waves.
And the sum of two sign waves, I've got an amplitude of one and an amplitude of one.
But different frequencies.
50 and 800.
You can see the 800 as being the high frequency here and
the 50 is being kind of the low frequency there.
So, I have got this summation of two frequencies and one of them at 50.
So, what I do is go to this plot at 50.
This is plot of my frequency response.
At 50, I look at this value here, that's about .95.
And at 800, I look at the value there, that's at about .13.
So my output.
Amplitude is equal to the input amplitude times the corresponding
H at that frequency, the corresponding magnitude.
So 1 times .95 is right here.
The input amplitude times H to give me that output amplitude.
And remember just to write it up here, the output amplitude.
Is equal to the input amplitude times H at that frequency.
That's what we're using.
And the other thing is the output angle is equal
to the input angle plus H at that frequency.
We'll in these cases I have no angle, so the input angle is 0.
So looking at the course trying to figure out what these angles are, at 50.
That's about minus 20 degrees.
So that's where I get the minus 20 degrees in there.
And at 800, that's at about, minus 85 degrees.
So that's where I get minus 85 degrees.
So the transfer function tells us what the output amplitude is and
what the output phase is.
And what we see here is that this particular case, it rolls off at we,
as we get to high frequency.
So that means that higher frequency part of this signal.
Gets attenuated.
And that you can see it here,
the high frequency is much smaller than it was before,
whereas its a lot more dominated by the low frequency then it was in this case.
So it let the low frequency through without much change and
it attenuated the high frequency.
And we can see that by, by seeing what happens to this plot here.
What we've looked at so far was linear plots.
I want to show you a Bode Plot, which is the same values,
the same information, but this time on log scales.
So looking at the magnitude, now the magnitude is actually written in
decibels and that's what the dB stands for.
When I take 20 times the log of the magnitude.
And I plot it versus frequency.
Either radians per second or hertz.
On a log scale.
So I'm going to define what a decade is.
I look at this value, it's a frequency and ten times it's frequency.
That's a decade.
So it's omega to ten times omega.
Or in terms of hertz it's, frequency at hertz to ten times that frequency.
And it doesn't matter.
It doesn't have to be ten and 100, it could be somewhere in between.
That's also a decade right there, so it's that distance right there.
And the angle, we're just plotting the angle, on the frequency scale,
not doing anything special to it.
It's just on a log scale on the frequency part.
So what's the difference, in, the looks of these the linear plot versus the bode,
bode plot [SOUND] this is a linear plot and its corresponding bode Plot.
So at every point, so example right here I would take 20
times the log of 1 and I get 0 DB, which is what I get over here.
This is where this goes to 0 DB.
Over here I'm at 0.1.
So 20 times the log of 0.1 is equal to minus 20 decibels.
And that's where we get minus 20 decibels over here.
So this is at 1,000 radians per second.
This is 10 to the 3rd-1,000 radians per second.
And the angle's the same information.
I'm just plotting out a log scale over here.
Characteristics of the, of a Bode Plot.
At high frequency this has a slope of minus 20 db per decade.
And this, I what is plotted specifically here is this rc circuit.
The transfer function of this RC circuit.
And we can, draw this at high frequence, higher frequency,
is this being kind of an asymptotic line there?
And where these two lines meet up,
that's what we call the corner frequency, omega 0.
Is a corner frequency.
For this particular circuit, the point where that,
they cross, they intersect, is, 1 over RC.
Now what happens in the angle, it goes from 0 degrees
to minus 90 degrees as you get higher in frequency.
This, so this levels off at minus 90.
So that's the characteristics of a first order circuit, and
what the Bode Plot looks like.
This is an RLC circuit, it's a second order circuit.
And there are two cases that we're going to look at.
One is called overdamped, and the other's called underdamped.
So this is a transfer function with, we've seen this before.
And at high frequency, I can figure out what this slope is.
Because this is second order circuit I've got an L and a C.
This is minus 40 db per decade.
And I can figure out what this corner frequency is by drawing these
asymptotic lines right there.
So that corner frequency down here.
Is 1 over the square root of LC.
That's the corner frequency.
[SOUND] It's kind of where the plot changes.
Now, the angle goes from 0 degrees to minus 180 degrees.
As we go from low frequency to high frequency.
Then, underdamped case, we're doing the same transfer function.
And we can come up with these same.
Asymtitic curves.
And I still have the value, omega zero,
where omega 0 is equal to 1 over the square root of LC.
What we've found here is that.
We didn't have any dependence on R here.
But as R gets small, as R decreases, we get a resonant peak.
For large R, what we saw in our previous case, it overdamped.
There was no resonant peak.
So this is the resonant frequency range, and
this, actually, this peak value, is a resonant frequency.
Frequency at which that happens, resonant frequency.
And that. At that point, that means the output.
Is greater than the input.
Output amplitude, is greater
than the input amplitude, in resonance.
In the resonance region In terms of and the smaller R is, the higher the peak.
The only difference with it, as R gets smaller, this becomes sharper.
In terms of the phase transition.
Sharper transition.
And again we're going from 0 degrees to minus 180 degrees on the angle.
So that's the under damp case whenever we've got this, this peak there.
So in summary, we have defined a frequency response.
It's a plot of the transfer function versus frequency.
In particular, we are looking at the magnitude and the angle.
We can use it to determine the steady-state sinusoidal response
of a circuit.
At different frequencies.
And the whole point of the, the transfer function that the frequency response is to
understand how the circuit processes signals of different frequencies.
Does it attenuate frequencies in certain ranges?
Or does it amplify signals?
Or make them larger?
Like that last case with the resonant peak we had the output greater than the input.
If we take a look at the transfer function and plot it versus, in other words,
look at the frequency plots, we can understand what's going on.
Is it amplifying or is it attenuating?
And if so, in what frequency range?
So this ends our series of lessons that are review.
And in the next module, we will start the, the regular electronics part .Thank you.
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