This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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From the course by Georgia Institute of Technology

Introduction to Electronics

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This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

From the lesson

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.

- 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.

Â This is Dr. Ferri.

Â In this lesson,

Â we will do a review of frequency response plots including Bode Plots.

Â In our last lesson, we did a review of transfer functions.

Â Well, we need transfer functions for us to be able to do a frequency response.

Â So this lesson will define frequency response for transfer function.

Â Actually I'm going to go ahead and

Â define it right here, because this, this is really all I want to say for it.

Â A frequency response is the magnitude plot

Â of, taking the transfer function, H of omega.

Â We want to take the magnitude of H versus omega.

Â Plot it that way.

Â And the angle is the angle of H versus omega.

Â H is a complex variable.

Â And then we want to show the linear plots and

Â Bode Plots corresponding to the frequency response.

Â Lets look at a particular circuit.

Â This is an RC circuit that we've looked at a couple of times in

Â the impedance lesson as well as in the transfer function lesson.

Â We found that the transfer function is equal to this H right here.

Â And what we want to define is the magnitude and the angle.

Â Well lets look at, lets take a step back and think about.

Â A magnitude of A, complex number.

Â A plus jb.

Â The magnitude of 1 over a plus jb, is going to be 1 over.

Â The square root of the real part squared,

Â which is a squared plus the imaginary part squared, which is b squared.

Â So, in this particular case, b is equal to RC omega,

Â that is what multiplies J, and I've got that squared.

Â And then the real number in the denominator is one, so

Â that's squared right there.

Â So the magnitude is this.

Â Now the angle 1 over a plus jb

Â rather the, the angle of that, is minus arc tan.

Â And the minus sign comes because this is.

Â The complex number's actually in the denominator.

Â And then it's the imaginary part over the real part.

Â So, in this case, it's b over a.

Â And A is equal to 1, and B is equal to RC omega.

Â If I want to plot this, what I'm plotting.

Â And the magnitude is H of omega and then this magnitude.

Â So I'm applying this function right here.

Â And it's the function of omega, and

Â one thing you can see is that as omega gets very large, this gets small.

Â And then this is the angle is the arc tangent of RC times omega.

Â And in this case as omega gets large, this approach is minus 90 degrees.

Â And as omega is equaled 0, it's 0 degrees.

Â What's it mean, the transfer function here?

Â It's really how a circuit processes signals of different frequencies.

Â So in this case, suppose I've got an input that is a sum of two sign waves.

Â And the sum of two sign waves, I've got an amplitude of one and an amplitude of one.

Â But different frequencies.

Â 50 and 800.

Â You can see the 800 as being the high frequency here and

Â the 50 is being kind of the low frequency there.

Â So, I have got this summation of two frequencies and one of them at 50.

Â So, what I do is go to this plot at 50.

Â This is plot of my frequency response.

Â At 50, I look at this value here, that's about .95.

Â And at 800, I look at the value there, that's at about .13.

Â So my output.

Â Amplitude is equal to the input amplitude times the corresponding

Â H at that frequency, the corresponding magnitude.

Â So 1 times .95 is right here.

Â The input amplitude times H to give me that output amplitude.

Â And remember just to write it up here, the output amplitude.

Â Is equal to the input amplitude times H at that frequency.

Â That's what we're using.

Â And the other thing is the output angle is equal

Â to the input angle plus H at that frequency.

Â We'll in these cases I have no angle, so the input angle is 0.

Â So looking at the course trying to figure out what these angles are, at 50.

Â That's about minus 20 degrees.

Â So that's where I get the minus 20 degrees in there.

Â And at 800, that's at about, minus 85 degrees.

Â So that's where I get minus 85 degrees.

Â So the transfer function tells us what the output amplitude is and

Â what the output phase is.

Â And what we see here is that this particular case, it rolls off at we,

Â as we get to high frequency.

Â So that means that higher frequency part of this signal.

Â Gets attenuated.

Â And that you can see it here,

Â the high frequency is much smaller than it was before,

Â whereas its a lot more dominated by the low frequency then it was in this case.

Â So it let the low frequency through without much change and

Â it attenuated the high frequency.

Â And we can see that by, by seeing what happens to this plot here.

Â What we've looked at so far was linear plots.

Â I want to show you a Bode Plot, which is the same values,

Â the same information, but this time on log scales.

Â So looking at the magnitude, now the magnitude is actually written in

Â decibels and that's what the dB stands for.

Â When I take 20 times the log of the magnitude.

Â And I plot it versus frequency.

Â Either radians per second or hertz.

Â On a log scale.

Â So I'm going to define what a decade is.

Â I look at this value, it's a frequency and ten times it's frequency.

Â That's a decade.

Â So it's omega to ten times omega.

Â Or in terms of hertz it's, frequency at hertz to ten times that frequency.

Â And it doesn't matter.

Â It doesn't have to be ten and 100, it could be somewhere in between.

Â That's also a decade right there, so it's that distance right there.

Â And the angle, we're just plotting the angle, on the frequency scale,

Â not doing anything special to it.

Â It's just on a log scale on the frequency part.

Â So what's the difference, in, the looks of these the linear plot versus the bode,

Â bode plot [SOUND] this is a linear plot and its corresponding bode Plot.

Â So at every point, so example right here I would take 20

Â times the log of 1 and I get 0 DB, which is what I get over here.

Â This is where this goes to 0 DB.

Â Over here I'm at 0.1.

Â So 20 times the log of 0.1 is equal to minus 20 decibels.

Â And that's where we get minus 20 decibels over here.

Â So this is at 1,000 radians per second.

Â This is 10 to the 3rd-1,000 radians per second.

Â And the angle's the same information.

Â I'm just plotting out a log scale over here.

Â Characteristics of the, of a Bode Plot.

Â At high frequency this has a slope of minus 20 db per decade.

Â And this, I what is plotted specifically here is this rc circuit.

Â The transfer function of this RC circuit.

Â And we can, draw this at high frequence, higher frequency,

Â is this being kind of an asymptotic line there?

Â And where these two lines meet up,

Â that's what we call the corner frequency, omega 0.

Â Is a corner frequency.

Â For this particular circuit, the point where that,

Â they cross, they intersect, is, 1 over RC.

Â Now what happens in the angle, it goes from 0 degrees

Â to minus 90 degrees as you get higher in frequency.

Â This, so this levels off at minus 90.

Â So that's the characteristics of a first order circuit, and

Â what the Bode Plot looks like.

Â This is an RLC circuit, it's a second order circuit.

Â And there are two cases that we're going to look at.

Â One is called overdamped, and the other's called underdamped.

Â So this is a transfer function with, we've seen this before.

Â And at high frequency, I can figure out what this slope is.

Â Because this is second order circuit I've got an L and a C.

Â This is minus 40 db per decade.

Â And I can figure out what this corner frequency is by drawing these

Â asymptotic lines right there.

Â So that corner frequency down here.

Â Is 1 over the square root of LC.

Â That's the corner frequency.

Â [SOUND] It's kind of where the plot changes.

Â Now, the angle goes from 0 degrees to minus 180 degrees.

Â As we go from low frequency to high frequency.

Â Then, underdamped case, we're doing the same transfer function.

Â And we can come up with these same.

Â Asymtitic curves.

Â And I still have the value, omega zero,

Â where omega 0 is equal to 1 over the square root of LC.

Â What we've found here is that.

Â We didn't have any dependence on R here.

Â But as R gets small, as R decreases, we get a resonant peak.

Â For large R, what we saw in our previous case, it overdamped.

Â There was no resonant peak.

Â So this is the resonant frequency range, and

Â this, actually, this peak value, is a resonant frequency.

Â Frequency at which that happens, resonant frequency.

Â And that. At that point, that means the output.

Â Is greater than the input.

Â Output amplitude, is greater

Â than the input amplitude, in resonance.

Â In the resonance region In terms of and the smaller R is, the higher the peak.

Â The only difference with it, as R gets smaller, this becomes sharper.

Â In terms of the phase transition.

Â Sharper transition.

Â And again we're going from 0 degrees to minus 180 degrees on the angle.

Â So that's the under damp case whenever we've got this, this peak there.

Â So in summary, we have defined a frequency response.

Â It's a plot of the transfer function versus frequency.

Â In particular, we are looking at the magnitude and the angle.

Â We can use it to determine the steady-state sinusoidal response

Â of a circuit.

Â At different frequencies.

Â And the whole point of the, the transfer function that the frequency response is to

Â understand how the circuit processes signals of different frequencies.

Â Does it attenuate frequencies in certain ranges?

Â Or does it amplify signals?

Â Or make them larger?

Â Like that last case with the resonant peak we had the output greater than the input.

Â If we take a look at the transfer function and plot it versus, in other words,

Â look at the frequency plots, we can understand what's going on.

Â Is it amplifying or is it attenuating?

Â And if so, in what frequency range?

Â So this ends our series of lessons that are review.

Â And in the next module, we will start the, the regular electronics part .Thank you.

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