The impedance model

Good morning. All right. Today we are going to take a fresh look at some of the stuff we covered in the last two lectures. And the graph I want you to keep in mind as we go through this lecture in terms of what to expect.
This was time. And last Tuesday's lecture we covered some stuff. I talked about a method for the sinusoidal response which was agony, I warned you it will be agony, and then towards the end I showed you another method that was quite a bit easier but still pretty hard.
And I promised you that today there will be a new method which is going to be so easy , actually almost trite. Just imagine. I am going to make a statement right now that I think you will all find hard to believe.
What I am going to say is just imagine your RLC circuit, your resistor, inductor and capacitor, a parallel form or series form. Imagine that you could write down the characteristic equation for that by observation in 30 seconds or less.
Just imagine that. By observation, boom, write down the characteristic equation for virtually any RLC circuit or RC circuit or whatever. And we all know that once you have the characteristic equation you could very easily go from there to the time domain response intuitively or to the sinusoidal steady-state response, too.
So just keep that thought in mind. Imagine 30 seconds. And that is what you should expect in today's lecture. Students often ask me, if this stuff is actually so easy why do you take us through this tortuous path? Are we just mean? Do we just want you show you how hard things are and then show the easy way? I have argued with myself every year as to whether to just go ahead and give the easy path and that's it.
But I think the reason we cover the basic foundations is that it gives you a level of insight that you would not have otherwise gotten if I directly jumped into the easy method. So you need to understand the foundations and you need to have seen that at least once.
And second, once you do something the hard way, you appreciate all the more the easy method. All right. Today we cover what is called "The Impedance Model." First let me do a review just because of the large amount of content in the last two lectures.
I did them using view graphs. I usually don't like to do that, but even then it was quite rushed. So let me quickly summarize for you kind of the main points. We have been looking at, on Tuesday, the sinusoidal -- --looking at the sinusoidal steady state response.
Also fondly denoted as SSS. And the readings for this were Chapters 14.1 and 14.2. what we said was if you took this example circuit and we fed as input cosine of omega t, we have an R and a C, and let's say we cared about the output response and we cared about the capacitor voltage.
What we talked about was focused on the sinusoidal steady-state response. And what that meant was first of all focus on steady-state. In other words, just to capture the steady-state behavior when t goes to infinity after a long period of time.
And for most of the circuits that we consider, because of the R or presence of any resistance, the homogenous response usually would die out because the homogenous response is usually of the form minus t by tau.
And as t goes to infinity this term tends to go to zero. We are just looking at the steady-state. And therefore, because of the circuits we looked at, we can ignore the homogenous response. All we are left to do is to find the particular response to sinusoids of this form.
And second was focus on sinusoids. We said the reason for this was that, let's say we did not care particularly What happened when I just turned on my amplifier. I just turned on my amplifier, often times you see some distorted sound coming out for a few seconds and then hear a much clearer sound.
And that initial part is due to the transient response. And let's say we don't care about that. We care about the steady state. Second we focus on sinusoids because based on the Fourier series experience that you had previously, we can represent repeated signals as a sum of sines.
And therefore it is important to understand the behavior of these circuits when the input is a sinusoid. And what was important was this introduced a new way of looking at circuits, and that was the frequency viewpoint.
When we looked at transient responses, we plotted response as a function of time. And when we look at sinusoidal steady-state, it becomes interesting to plot the response as a function of the frequency, a function of omega.
What I will do is draw a little chart for you to sort of visualize the various processes we have been going through. We can liken obtaining the sinusoidal steady-state response to following these steps.
Here is my input. What I did as a first step was fed my input to a usual circuit model. My elements were lumped elements, built the circuit and wrote down the VI relationship for the element. As a second step I set up the differential equation.
This was the first of four steps, set up a differential equation. And then the path that I took first was fraught with real nightmarish trig. By the end of the day it would still yield an answer. It could be a nightmare.
But I would get something cosine omega t plus something, some phase. I could grunge through the trig. And I gave up halfway in class here, but you could grunge through it if you would like. And you would get the answer to be some sinusoid with some amplitude and some phase.
So Vi cosine omega t would produce the response that was something cosine omega t plus some phase. We said this was too painful so let's punt this. Instead, what we said we would do is take a detour, take an easier path.
And the easier path looked like this. I said let's sneak in -- -- Vie^(j omega t) drive. That is just imagine, do the math as if you had fed in not a Vi cosine omega t but a Vie^(j omega t). And from Euler's relation you know that the real part is Vi cosine omega t.
So we said that I am going to sneak in this thing, find the response and just take the real part of that because the real part of the input gives me this. So this is my "sneaky path." And what I did there, as soon as we fed in the e^(j omega t), because of the property of exponentials, the e^(j omega t) cancelled out in my equation.
And what was left was some fairly simple complex algebra. And at the end of the day, after I grunged through some fairly simple complex algebra, I ended up with some response that looked like this.
Vpe^(j omega t). What I would find is that for the input Vie^(j omega t), I would get a response Vpe^(j omega t). And then what I said we would do is take the real part. Why take the real part? Because this is a fake, a sneaky input.
The input I really care about is the real part of the sneaky input. So this is my sneaky output. And what I care about is the real part of the sneaky output. That is sort of the inverse superposition argument that I made on Tuesday that if what I care about is the real part of this input, then I just take the real part and get the output that I care about.
So I take the real part. Notice that Vp here, in the examples we did, we did an RC example. The Vp here was a complex number. So I could represent that complex number as, in many ways. This is e^(j omega t).
I could represent Vp in an amplitude, as a phasor, actually polar coordinates. I can say that the equivalent to Vpe to the j angle Vp. Vp is a complex number. If you look at the complex appendix in your course notes, I can represent a complex number as an amplitude multiplied by e raised to j times some phase.
It's simple complex algebra. And then what I could do here is take the real part of that. And when I took the real part of that what came about was that this was simply Vp. Notice that the angle Vp goes in here so it becomes j times omega t plus angle Vp.
It is Vp amplitude times e raised to j omega t plus j angle Vp. And the real part of that is simply Vp cosine of that stuff. What I end up getting here is Vp cosine omega t plus Vp. The cool thing to notice was that once I found out this response here, I could immediately write down the output based on Vp.
In other words, once I had Vp, I could stop right there in my math. I got Vp very quickly here. This step produced Vp very quickly, after two algebraic steps. And then from here I could directly write down the answer as homogenous of Vp cosine omega t plus angle Vp.
Boom, right there. So this was a much shorter path. And here I just described to you how this yields an expression for Vp and angle Vp. And for our example Vp was 1/(1+j omega RC). And we often times write a shorthand notation 1+sRC, where S is simply j omega.
We commonly jump back and forth between the shorthand notation S and j omega. S has some other fundamental, has another fundamental significance you will learn about in future courses, but for now S is simply a short form for j omega.
This was the path that we took. There is a hard path and an easier path. Today I am going to claim that even this was too hard. There is an even easier path. And today what I am going to show you is that from here we are going to take one step and get here.
I am going to show you today that we won't do this, we won't do this, not this, not this, none of this. One step and then we are going to get the answer. So let's do that. Before we jump into the impedance method and get into doing that, I just would like to plot for you this function here just so we can understand a little bit better exactly what is going on.
As I mentioned to you, the output vO for our circuit there was simply Vp cosine of omega T plus angle Vp. Oh, that's Vp so this one should be Vi here. I am showing you Vp so there is a Vi in there.
Vp/Vi=1/(1+j omega RC). This is a complex number, and it is simply a number that when multiplied with Vi gives me the output. This is also called a transfer function and represented as H(j omega). This guy is a transfer function, much like the gain of my amplifier.
Which when multiplied by the input to get me the output. This guy is a complex multiplier which when multiplied by Vi gives me Vp. And as such we call it a transfer function H(j omega). And we can plot this function.
Notice that this a function of omega. Remember we are taking the frequency domain view, so where has time vanished? Remember that we are taking the steady state view. So we are saying in the steady state, if I wait long enough this is how my circuit is going to behave, this is how a circuit is going to behave.
And the transient responses have died away and I have time in my output here so my output is a cosine. But that in itself is not very interesting. It is a cosine of some amplitude and has some phase.
What we will plot is we are going to plot this property here, Vp as a function of the frequency. Vp is frequency dependent. As an example, I could plot the absolute value of Vp/Vi, the modulus of that versus omega.
And notice that when omega is zero again intuitive ways of plotting this is to look at the value at zero and look at the value at large omega. For small omega, omega goes to zero this is one, so it starts off here.
And when omega is very large then it is much bigger than one here, so this goes down. Far away this one looks like 1/omega RC. And this function, assuming I have linear scales on my X and Y axes looks like this.
We also commonly plot this using log-log scales. And when you do log-log scales you get a straight line here, and then you actually get a straight line of slope minus one because the log of this gives you a line with a constant slope, it's a slope of negative one so it becomes a straight line going down.
The other interesting thing to realize is that this magnitude is simply one by one plus omega squared R squared C squared, the square root of this. That's the magnitude here. And notice when omega equals 1/RC, this thing, the denominator becomes one by square root of 2.
Somewhere here when omega equals 1/RC The output is one by square root 2 times the input. It's an interesting point. And this is called the "break frequency." You can view it as a frequency where I am getting this transition from one to a lower value, and it is where the output is one by square root two times the value of the input.
Now you can think back on the demo we showed you earlier. And in the demo remember that as I increased the frequency of my input sinusoid my output kept becoming smaller and smaller and smaller. And you notice that you can see this dying out or decaying of the amplitude as I increase my omega.
Let me go back. What you have done is that, we're going to apply a bunch of sinusoids to the same circuit and plot the frequency response, the ratio of the output versus input as a function of frequency.
And kept applying a variety of frequencies. So you can listen to the frequencies as they go by, and we will plot the amplitude up on the screen for you. Just for fun we are going to play frequencies between, say, 10 hertz and 20 kilohertz.
It will be fun for you to figure out at what point you stop hearing the frequencies. We are going to play from 10 hertz to 20 kilohertz. And figure out where your ears cut out. That will tell you what the break frequency of your ear is.
You can see the amplitude being articulated. The bottom figure is the phase. This is the frequency axis. This is the amplitude, log-log scales. I am not sure about you but I cannot hear anymore.
If you bring your canine friends to class it is quite possible that they would go berserk somewhere here. As I promised you, when I plot this on a log-log scale I get a straight line here and a straight line out there as well and the bottom line gives you the phase.
Now, what you can also do is you can also go to Websim. Websim is now linked on your course homepage. You can go to Websim and you can play with various L and C and R values. And if you plot frequency response, if you click on the frequency response button, boom, it will give you frequency responses for your circuit that look exactly like that.
You can go and play around with that. Thank you. All right. As the next step I promised to show you an easier path. And let's build some insight. Is there a simpler way to get where we would like to get? In particular, is there a simpler way to get Vp? Let's focus on Vp.
Why Vp? Because remember Vp was the complex amplitude of e to the j omega t. And once I know Vp then I know this expression here. Also notice that this here, the denominator is simply the characteristic equation for, I wonder how many of you noticed it, is simply the characteristic equation for the RC circuit.
If I can write down Vp, I can write down the characteristic equation, it will be in the denominator. I can also write down the frequency response very easily by taking the magnitude and phase of Vp.
So Vp has all the information humankind needs for those circuits. Is there a simpler way to get Vp? To bring some insight, let's go ahead and write down -- Let's stare at this for a while longer and see if light bulbs go off in our minds.
Of course, I could write this as Vi/(1+sRC). I just replaced the shorthand notation for a j omega. And I simply divide by SC throughout. So I get Vi times, I simply divide by SC throughout. Here is Vi.
I have one by SC, one by SC plus R. Light bulbs beginning to go off? The form we have here is 1/SC, some function of my capacitance divided by something connected to my capacitance plus R. This is Vi multiplied by something connected to capacitance divided by something connected to capacitance plus R.
And remember your circuit. What is that reminiscent of? What does that remind you of? Voltage divider? Hmm. There is some voltage divider thing going on here. I just cannot quite pin it. It is something about the capacitor, capacitor plus booster, some voltage divider thingamajig happening here.
We will try to figure that out. What I will do is replace those terms with something called Zc. Zc plus Zr. If I can find out the Zr and Zc somehow, I can write down the Vp by inspection by the voltage divider action, by some generalization of the good old Ohm's law that I know about.
Let's proceed further and see if we can make some kind of a connection between this and this. If I can make the connection then boom, I'm done. I will just use voltage dividers and I am home. OK, so let's play around and see.
There is something in there. By now you should know that we are very close. There is something going on in there. I just need to get that spark. I just need to make that spark so I can bridge the gap between something that is really easy versus where I am.
Let's take a look at the resistor. I have my resistor with the voltage vR across it and a current iR. Remember to get to any sort of steady state you are going to be dealing with the drives of the form vI e to the j omega t, exponential drives.
And by taking the real part, I know I get the input, and the real part of the output gives me the actual output. Let's say my iR is simply Ire^st and my vR is Vre^st. The S is, again, a shorthand notation for j omega.
If my current Ire^st of the exponential form shown there and here is Vr, I need to find out what relates Vr and Ir for the element relationship for the resistor to hold. In general, Ir and Vr are complex numbers.
For the resistor, I know that Vr=RIr. And I substitute using my complex drives here. So it is Vre^st=RIre^st. I am just substituting for these drives, Ohm's law should apply, and I cancel off e^st.
And so I get Vr=RIr. Interesting. For the resistor I find that, based on the fundamental principles of resistor action, the complex amplitude of the voltage simply relates to the complex amplitude of the input by the proportionality factor R.
In other words, for the resistor -- Just as the time domain V and I were related by the proportionality constant R, the complex amplitudes Vr and Ir are also related in the same way. That's interesting.
Now let's look at the capacitor. Some current ic flowing through it and a voltage vc. Let's say the current is Ice^st and the voltage is Vce^st. Let's plug these into the element law for the capacitor and see if we can find out a way of relating vc and ic.
I know that ic is simply Cdvc/dt. So I replace this with Ice^st=Cd/dt(vce^st), which is simply Ice^st=CsVce^st. So I can cancel this out again. Interesting. Ic=CsVc. Very interesting. What is interesting here? Notice that in the time domain Ic=Cdvc/dt, the element law for the capacitor.
So I said let's use exponential drives, Ice^st, Vce^st, that's an exponential drive, and try to find out what the relationship between the complex amplitudes are. I plug them and what do I find? I find that if my input is Vce^st, and Vc is the amplitude of the input, then the current is simply given by something multiplied Vc.
It's very similar in form to what I saw here. The resistor, Vr=RIr. For the capacitor, Vc=Ic/sc. 1/sc kind of plays the role of R. In other words, the complex amplitudes around the capacitor are related by Vc equals some constant times Ic.
Almost like a funny Ohm's law kind of relationship where Vc and IC are complex amplitudes. For the inductor it is the same way, iL, vL and L. Let's say iL=Ile^st and vL=Vle^st. Substitute the values for the inductor into its element relationship as well.
I know that vL=LdiL/dt. Therefore, substituting the complex amplitudes is L. And diL/dt will simply be Ilse^st. So I cancel out the exponentials. The reason we're able to do all of this is simply the remarkable beauty of exponentials.
Exponentials are absolutely stunningly beautiful. The reason is that when I differentiate them what I get back is the exponential times some constant, and the constant was in its numerator multiplying t.
And that's the beauty of exponentials. If this was a sine then I would get cosine and a sine. With exponentials these cancel out and what I am left with is something that is LsIl. Again, for the inductor, the voltage across the inductor relates to some constant Ls here times Il.
This is absolutely stunning and almost looks like a form of Ohm's law here. What I am going to do is let's give this the name Zr. Let's give this 1/sC the name Zc. And let's give this the name ZL.
It kind of behaves like a resistor, so the resistor simply becomes Zr. And 1/sC behaved like a resistor so I called it Zc. And this is a ZL. These are called "impedances." In other words, for a capacitor, as far as complex inputs and outputs are concerned, if Vc and Ic is fed to it, the capacitor can be replaced by an impedance Zc where I can write the relationship between Vc and Ic as Vc=ZcIc.
Where Zc is simply one by sc. Similarly, for an inductor -- -- I can write its impedance ZL as sL and I get Vl=ZLIl. And finally for a resistor it is pretty simple. What I am saying is that if I am in the region of the playground, if I constrain myself in the region of the playground where my inputs are something Vi e to the j omega t or exponentials, in that little region of the playground now, I am focusing more and more on small parts of the playground so I am kind of boxed in right now.
In that region of the playground this applies. In that region of the playground, I can replace resistors by impedances, capacitors with impedances of value 1/sC. And within that playground the beauty of analysis there is that in that region of the playground where the inputs are of the form Vie^st, it turns out that the element laws are simply generalizations of Ohm's law.
That is absolutely stunning. It is one of the biggest hallelujah moments in learning circuits. This is really big. And I think this is almost as big as the realization that you can take a nonlinear circuit, operate it at a given operating point, and you can sit around doing Zen things, looking at small perturbations in there, those are going to be linearly related.
This is one of the big hallelujah moments in 6.002. And this is of the same magnitude as the small signal response being linear. It is something that is completely non-intuitive. It is something that you just would not have known until you had seen it happen.
The same way here. This is very important so I will repeat it again. I have boxed myself into this small region of the playground where all I care about are sinusoidal inputs and steady-state responses.
So there I focus on complex inputs, Vi e to the j omega t. And I have just shown you that I can replace inductors, capacitors, resistors with their impedances. And the amplitudes of the corresponding signals around them are related by just a simple Ohm's law like relationship using impedances.
I am sort of boxed into this playground, right? In my playground it is all about e to the ij omega t. e to the ij omega t is implicit everywhere. I just don't show it. If I want to talk to somebody else outside but within MIT in this small region, it's all e to the ij omega t in there.
If I want to talk to somebody outside, get out of MIT, get out of this playground, what else do I have to do? I have to take the real part. Don't forget that. Remember that, take for example Vc here, so Vc is this, so implicit in all of this is that if I measure Vc at some place it is really going to be Vce to the j omega t.
And if we the cosine, the real part, then I have to take a real part of this. And the real part of that would Vc cosine of omega t angle Vc. This piece here kind of goes unsaid. We will agree that we have to do it, but we just skip that step because it is obvious.
We just deal with Vcs and Vls now. So a new notation certainly sneaked by you, and that notation looks like a big letter and a small letter. Remember you have seen vL, this is the total behavior, you have seen vl, that's a small signal behavior, and now you see this, Vl, capital V small l.
And we also have DC, we have labeled operating point values as VL, capital V, capital L. We have one thing left so nobody go out there inventing something new because we would be in trouble. This is capital V, small l, and this is simply "complex amplitude" in the small boxed region of my playground where good things happen and exponentials fly.
Whenever someone gives you a variable, capital V, small l, remember it's a complex amplitude, a complex number, and you know how to get to the time domain from there. You take that number, take the real part, multiple the number by e to the j omega t and take the real part, which is tantamount to magnitude cosine omega t plus angle of that number.
Actually, you know what? Let's send this up. Back to an example. Oh, I'm sorry. I'm sorry. This is not good. This is my time domain circuit. Remember this was my time domain circuit. A vI input.
A vC output. I wanted to analyze this. What I am telling you now is let's box ourselves in this impedance playground. And in the impedance playground the input becomes the complex amplitude of the input, my resistance gets replaced by a box Zr, my capacitor gets replaced by a box Zc.
And the voltage I care about here is Vc. Zr = R and Zc=1/sC. Now, there we go. I can write down Vc using a voltage divider action as Vc is simply Zc/(Zc+Zr), done, times Vi of course. And that gives me 1/sC divided by 1/sC+R and multiplying throughout by sC I get 1/1+sCR where S is j omega.
Just cannot get any simpler. How long did I take to do this? 30 seconds. Where I spent a whole lecture on Tuesday grinding through first trig, giving up halfway and collapsing, and then showing you the sneaky path which was still pretty painful, but 30 seconds, boom.
This stuff is spectacularly beautiful. The really cool thing here is that in this impedance domain for linear circuits all your good old tricks apply. Your Thevenin, your Norton, your superposition, name it and it applies for this linear circuit.
If you close your eyes and make believe that Zr is like an R and simply apply all the techniques you have learned so far in this linear playground. Just a little hack at the end where this is the complex amplitude.
And if you want to go to the time domain part then you do the usual thing. Modulus Vc cosine omega t plus angle Vc. Just remember that. That's the jump to get back to the time domain. Just to show you that this not just works for one little rinky-dink circuit here, let me take a more complicated circuit.
If I believe in my own BS, I should be able to apply this theory to my series RLC, the big painful circuit that we did differential equations for about a week ago. Let's do it. I have an inductor, a capacitor and a resistor.
What I am going to do is replace this with the impedance model. Input Vi. Let's say this was vI. Let's say I cared about vR. L, C and R. The impedance model would simply be Vi. What's the impedance of an inductor? SL.
And for the capacitor it is 1/sC. And for a resistor it is simply R. And just remember, if I can find out VR then for an input cosine of the form Vi cosine omega t the output will given by Vr cosine of omega t plus angle Vr.
Just remember this last step. But Vr itself is trivially determined. It is the voltage divider action again times Vi. And the voltage divider action is in the denominator I sum these thingamajigs, so ZL+ZC+ZR, ZR in the numerator.
And Zr is simply R. ZL is sL. Zc is 1/sC. And R is R. Vi. And I multiply through by, in this particular situation, by s/L. I want to get it into the same form as you've seen before. Multiply throughout, the numerator and denominator by s/L, what do I get? I get RS/L and out here I end up getting S squared plus 1/LC, and I get plus R/L S.
I am done. Look at that. Well, a little more than 30 seconds. Maybe a minute. What is this? Where have you seen this before? The denominator of this expression here? Ah, characteristic equation for the RLC.
Remember I promised you in the beginning that when we come to the end of the day using a simple one-minute expression I am going to write down the characteristic equation? Boom, here is what I get. Did somebody hear an echo in there? Notice that just by doing a simple voltage divider thingamajig, I got this expression.
And now I can write down the frequency response by replacing s is equal to j omega. Even more beautiful and what is even more stunningly pretty here is that remember the intuitive method I taught you about? The characteristic equation gives you alpha, omega nought, omega d and Q.
And based on those we can sketch even the time domain response. Guess what? RLC circuits are passé now. You can just write this thing down and you're done, 30 seconds or less. No DEs, no trig, no nothing.

Sinusoidal steady state

OK. Good morning. Let's get going. As always, I will start with a review. And today we embark on another major milestone in our analysis of lumped circuits. And it is called the "Sinusoidal Steady-state."
Again, I believe this will be the second and the last lecture for which I will be using view graphs. And the idea here is that, just like in the last lecture, there is a bunch of mathematical grunge that needs to be gone through.
And I want to show you a sequence in a little chart today that talks about the effort level in doing some problems based on time domain differential equations, as in the last lecture, something slightly different today and something much better on Thursday.
And so, in some sense, Thursday's lecture and today's lecture involve talking about the foundations of the behavior of certain types of circuits. And it is good for you to have this foundation as background, but when you actually go to solve circuits you don't quite use these methods.
You use much easier techniques, which I will talk about next Thursday. Let's start with a quick review, and then we will go into sinusoidal steady state. The last lecture we talked about this circuit and we did the same two lectures ago on Tuesday.
We had one inverter driving another inverter. And we said that the wire over ground had some inductance. CGS is the capacitor of the gate and R is the resistance at the drain of the first inverter.
And if you look at this circuit, that circuit formed an RLC pattern. And what we did was we said let's drive this with a one to zero transition at the input. And the one to zero transition at the input would cause this transistor to switch off, and this node would then go from a very low value to a high value.
So it as if a 5 volt step was applied at this input. We also saw that using time domain differential equations that by applying a step input here the output looked like this. The output would show some oscillatory behavior when we have a capacitor and inductor.
I also gave you some insight as to why it oscillates like this. And you also heard in recitation that the reason for this oscillation was because of these two storage elements. Each of these storage elements tries to hold onto its state variable.
For example, the capacitor tries to maintain its voltage while the inductor tries to maintain its current. And, much like a pendulum which oscillates back and forth, it swaps potential energy versus kinetic energy down here and swings back and forth.
In the same way, in an LC circuit like this, energy swaps back and forth between a potential energy and a kinetic energy equivalent, which swaps back and forth between energy stored in the inductor and energy stored in the capacitor and sloshes back and forth.
And because of this resistor the energy eventually dissipates and you end up getting a final value which corresponds to the 5 volts appearing here. And why is that? That is because remember the capacitor is a long-term open for DC.
It is a DC voltage. After a long time this capacitor looks like an open circuit and the inductor looks like a complete short circuit, an ideal inductor as a complete short circuit for DC. And so therefore in the long-term it is as if this guy is a short, this guy is an open, so 5 volts simply appears here.
And this is the transient behavior. Then we just switch the first transistor off. In the last lecture, I left off with intuitive analysis. Let me quickly cover that and redo the intuitive analysis for you.
I left it the last time by having you think about whether the transient response would begin by going down or begin by going up. And I will cover that today. This was the circuit that we analyzed.
A VI input with a step and an RLC out here. And we were plotting the capacitor voltage. And intuitively we can plot this in the following way. I have also marked for you the section number in the course notes which has a discussion of this intuitive analysis, (Section 13.8).
Let's do the easy stuff first. Notice that the capacitor wants to hold its voltage. And so given that we don't have any infinite impulse here, I am going to start out with the capacitor voltage being where it is.
And the initial conditions are given to you. You are given that the capacitor voltage starts out being positive at v zero and the current starts out being negative at time zero. So I am telling you that there is a voltage v across the capacitor at time t=0 and there is a current that is flowing.
Since i is negative there is a current initially that is flowing in the opposite direction to this arrow here. The i zero is negative. In light of that, I can start plotting my curve here by intuition.
I start by saying at time t=0 I am told that the initial voltage of the capacitor is at zero. This is about as simple as it gets. Completely intuitive. I also know that after a long time, can someone tell me after a long time what the voltage will be at the end of the capacitor? You should be able to get his by observation? VI.
And why is it VI? It is vI because this is a DC value VI. And after a long time this guy behaves like an open circuit to DC. This guy behaves like a short circuit to DC. So since this is an open circuit, VI will appear here after a long time.
And so therefore, after a long time, the capacitor voltage is going to be at VI. And I just showed you that. There you go. You already have the two endpoints of your curve completely by observation, intuition.
No DEs, no nothing. Just by staring at it and understanding the fundamentals of how simple primitive circuit elements work. Absolutely simple stuff. So you've nailed the two ends now and you cannot go wrong with the stuff in the middle.
Let's see. As a next step what I do is I need to understand what the dynamics of the circuit looks like here. What I do is I develop the characteristic equation. And initially you will write the differential equation and then substitute e^st and get this characteristic equation.
But you could also remember it as a pattern. For a series RLC circuit you always get an equation of this form, always. If this were R, L and C. And whether you are looking at L up here or C up here, as long as you're looking at the capacitor voltage, the capacitor voltage is going to behave the same.
And for this circuit the characteristic equation remains the same as well for a series RLC. It is exactly this. And I write the standard canonic form as s squared plus two alpha s + omega nought squared.
And omega nought is simply one by square root of LC and alpha is simply R divided by L and I have two in the denominator as well. And then I get omega d which is my damped frequency given by omega nought squared minus alpha squared.
And Q is simply called the quality factor. And we will learn about Q in a lot more detail in about a couple of lectures from today. That is given where omega nought divided by two alpha. These parameters, alpha, omega nought, Q and omega d pretty much characterize everything else that I need to know about the circuit.
First of all, omega d is the frequency of oscillation. And so since omega d is a frequency of oscillation then I know that the time period of oscillation is two pi by omega d. Omega is in radians. Notice that for typical values of circuits like this when R is pretty small, alpha squared is going to be very small.
It's a common case for underdamped circuit that omega d will more or less be equal to omega nought. Commonly that is going to be the case. This frequency is governed by LC. And if R is large it is governed by this omega d here.
So I have the frequency of oscillation. I also know that Q is related to the frequency of oscillation divided by alpha. It is a ratio of the frequency divided by how badly I am being damped. So it is a fight between the frequency of oscillation and now heavily I damp.
And the ratio of that is an indication of how many cycles I ring. So Q tells me that the ringing stops approximately after Q cycles. These four values, omega d, Q, alpha and omega nought are telling me more and more now.
So I have got these two factors. So I know now, based on omega d and Q, that it is going to look something like this. Some ringing here and then I stop at this point. The last thing that is left to do here for me for now is to figure out whether I start out going down or I start out going up.
I start out going down or I start out going up? I don't know that yet. And I stopped at this point in the last lecture and let you think about how you can stare at the circuit and intuitively figure out whether this goes down or that goes up.
So here is the insight. Let's stare at this for a minute purely through intuition. The capacitor has a voltage v across it, right? And that is because I am telling you that it has an initial voltage v.
Now, I want to find out at prime t equals zero plus, in which direction does a capacitor voltage go? Increase or decrease? What do I do? Let me take a look at the inductor current. I am told that the inductor current is negative which means I am told that the inductor current is going in this direction initially.
The inductor current is pushing in this direction. Now, remember, just as the capacitor is one stubborn nut trying to hold its voltage, the inductor is as stubborn. It is trying to hold its current.
It is trying to maintain its current. And its initial current i(0) is in this direction. Capacitor has a voltage here, that is v(0), and the inductor is yanking the current in that direction. So what should happen to the capacitor voltage initially? If I am at v(0) and someone is yanking current out, at least initially in that direction, what should the initial dynamics of the capacitor voltage look like? Pardon? It should drop, which means that if the initial current is being pulled in that direction the capacitor voltage must droop to begin with.
Completely through intuition. No math here. This means that i(0) is negative. It is as if i(0) is being pulled out in this manner, so the capacitor voltage must drop to begin life. Therefore, the dynamics look somewhat like this.
Notice that this is very reminiscent of the ringing that we saw at the gate node of the second inverter. Let's stop here in terms of time domain analysis of RLC, and today let's take another big step forward.
Today marks another transition in life here. This is actually a huge transition so I want to just pause and take like ten seconds of a breather just to clearly demarcate the fact that we have a huge transition coming up.
The key to this transition is that I want to look at today the steady-state response of networks to a sinusoidal drive. We are going to do two things differently starting today on this new journey of ours.
In the past, we looked at time domain behavior of circuits. For RLC, for example, we looked at the transient response. And what happened the instant I turn something on? The transient response. Today we are going to look at a steady-state response.
Steady-state response says that if I wait long enough, for whatever the circuit wants to do in the beginning of life to die out. If I wait long enough, how is the circuit going to behave after a long time? I will tell you why that is important in a second.
I look at the steady-state behavior. Second, I am going to look today at sinusoidal drive. Two things that are different from, say for example, what I covered in the past ten minutes. In the past ten minutes I covered two things which were different.
One is that I looked at the transient response and then steady-state. And remember for a DC input, for a DC voltage the steady-state was a DC voltage across the capacitor, correct? So the steady-state was pretty boring, it was steady-state DC.
But what we are going to do today is instead of having DC inputs or step inputs which settle to a DC value after some time, we are going to drive a circuit through the sinusoidal input. So you may ask me, Agarwal, are you nuts? Why do you want to drive something with a sinusoidal input? Why not just steps in DC and so on? That was painful enough.
Why sinusoidal? Why not do triangular or why not do some other exponentially decaying stuff or something really cool like a whacko music signal? What is special about sinusoidal stuff? The key thing to realize is that, well, let me ask you a question first.
How many people here know about Fourier series? Good. It looks like some of you have taken the prerequisites. Good. Need I say more as to why this is important? Just that question should give you the answer, right? You've learned about Fourier series.
And when you learned about Fourier series you were wondering why on earth are we learning about Fourier series? Who cares that you can represent the periodic signals as a summation of a series of sine waves? Why is that interesting? Why are you telling me that I can take a square wave and represent that as a summation of periodic square waves and represent that as a summation of sines? Who cares that I can take a sequence of pulses with a fixed period and represent that as a sum of sines? Who cares that I can take a triangular wave and represent that as a sum of sines? I am not sure what answer your math professors gave you when they taught you Fourier series.
But in math they are purists. They don't care about applications. The answer could well have been because it is aesthetically pleasing. I mean isn't it cool that you can represent a sequence of pulses as a sum of sines? That is good enough for mathematicians.
But I am an engineer. If it I cannot see how it helps humanity in the short-term then I probably don't care too much about it. Let me give you some practical significance of this. So it turns out that, since we know that we can represent periodic signals with sums of sines.
What this means is that if I can figure out the behavior of networks to a sinusoidal input, if I can understand how to analyze a network for a sinusoidal input that means that if the circuit is linear I can then compute the response of the circuit to any periodic waveform.
Here is the argument. I can represent any periodic waveform as a sum of sines. The Fourier series, remember? If I just figure out the response of my network for a sine wave, then if this is a linear network, I can compute the response to any sequence of scaled sum of sines.
A some sine, B sine two, omega whatever, C sine something or the other. I can simply take the response of the one sine. And from there I can go ahead, and knowing the response of a sine wave I can compute the response to a sum of sines.
That is pretty cool. Therefore, doing it for sinusoidal drives is really important. Why steady-state now? Hopefully, I have convinced you that looking at the response of circuits to sinusoidal drive is important and interesting because we can long ways from there.
What about steady-state? Well, it turns out that, and I am going to show you, when you listen to music, you have an amplifier and listen to music, what you are observing by and large is the steady-state behavior of the amplifier.
You are listening to something over many seconds or many hours. And the transients used for most of our common circuits, the transients die out pretty quickly. And so the transients are quite complicated and they die out quickly.
We say we are engineers. Let's focus on what is practically important. And we focus on the steady-state behavior as a large part of our analysis and just completely ignore the transient response when we care about the response to sinusoidal input.
The transient response will die away, and I will show that mathematically to you in a few seconds. And let's focus on the steady-state because that what I am going to hear most of the time. I am going to listen to an average over long periods of time.
That's the motivation behind this. And let me put this in perspective for you. By now this should bring memories to your mind. This is the playground that we are in. This is the lumped circuit playground here.
Remember we came from the playground of nature to the playground of EECS where we made the big leap from Maxwell's equations to lumped circuits, that's lumped circuit abstraction. And within there we spent a large part of the last couple of months doing linear circuits.
We also analyzed nonlinear circuits. Remember the amplifier circuit of the MOSFET large signal analysis was nonlinear? Well, there is linear and nonlinear. Within linear we also showed that if you take a digital circuit, at least as we understood them, and draw the subcircuit for a given set of switch settings, if I set the switches in a given way what I was left with was another linear circuit for a given value of all the switch settings.
My small signal analysis was also linear. If I focused on small signals I also had linear analysis. Today what we are going to do is this. We are going to articulate a different part of the playground.
This was a big linear playground. We've done this. We've done this. We are going to explore this territory. This is that territory of the playground in which we have sinusoidal inputs to circuits.
Furthermore, we are going to look at a subcircuit of that region which is steady-state. We will look at sinusoidal input and in the steady-state. So that is going to be our focus for the next two or three lectures just to give you a perspective of where we are.
To motivate this, what I would like to do is consider your amplifier. This is our friend the amplifier circuit, this part here. And remember, even though this is an amplifier, I am using a MOSFET here.
And a MOSFET, as you know, has this gate capacitance CGS. I am explicitly drawing it out for you here. And I am going to drive this with a bias voltage plus some small signal vI, the usual template for amplifiers.
And there is some Thevenin resistance attached to that source. I am going to model my source as a bias voltage, a small signal plus some source resistance. And I want to apply a sine wave here and I am going to look at what this looks like.
You may think, look, this is a linear amplifier. And so if I apply a sine wave here I should see some response here, and that should be the same amplitude if I feed the same amplitude here over any frequency.
But let's see what happens. Keep a look at, switch over to that view graph while I show you a little demonstration here. What you see here are three sine waves, a yellow sine wave which is the input here, you see a green sine wave which is the input vC at the gate of the MOSFET, and then you see the blue which is the output here.
For now simply focus on the yellow and the blue. The yellow is the input and the blue is the output. So I apply some input and I get an output that looks more or less some linear function of this input here.
It is a small signal. What I am going to do is I am going to change the frequency of the input. Remember, I want to look at the response of the circuit to a sinusoid. I am feeding a sinusoid here.
I look at the response. I am going to change the frequency. That is a big shift that we are making in that the curve that we drew in the last lecture had to do with varying time. Now I am going to focus on sinusoids and vary their frequency.
I am going to change the frequency. Stare at the blue curve as I increase the frequency and just think of what you might expect. Based on the knowledge you have so far you would expect that look, as I change the frequency, the frequency should change but I should see the same amplitude.
OK but take a look. Let me increase the frequency of the input. What do you see at the output? I am increasing the frequency. What do you see happening there? If you don't see anything changing there you will need to see an optometrist.
What do we see here? As I changed the frequency, as I increased the frequency what happened to the blue? The blue kept decreasing in amplitude. And you are saying whoa, what is happening here? We don't have the tools to deal with this.
I expected that when I changed my frequency, my frequency here should change of course, but why is the amplitude changing? What is happening here? That is weird. I noticed that this amplitude became smaller because the amplitude of the green became smaller.
And remember the green was the voltage across the capacitor. So this is your RC. And here is my input. My input has the amplitude, which I am holding more or less constant. And notice that vC decreased in value as I increased my frequency.
Just hold that thought. As I increased the frequency of my input the amplitude of the output kept diminishing. In other words, the gain of the system seemed to have decreased as I increased by frequency.
And today we will look at why that is so and how we can analyze that. The other thing that is not so obvious here is there is a phase shift. What I am going to do is try to see if we can observe the phase shift here.
Notice here. What we have been used to is for the amplifier we get a complete inversion at the output. Inversion means a phase difference of 180 degrees for a sine wave, right? This peak should have been here, but notice that there is a shifting of the peak.
In other words, if the yellow was my input my output should have had its minimum when the input had its maximum. But notice there is a shifting of the signal such that the output is a maximum, not quite at the point where the input is a minimum but a little bit after that.
Also weird. Not only has this little circuit here lost its gain somehow, but also it seems to have shifted the signal in phase. That is weird. And today we will look at why that is so and try to understand the frequency behavior of this little subcomponent here.
Notice that vC is exactly 180 out of phase with vO. So vO is faithfully an inverted amplified form of the input vC. However, vC itself should have been the same as vI but it looks quite different. So let's understand why that is so.
The subcircuit to model is the subcircuit comprising the source, resistor and the capacitor. And I am just showing that to you here. I have a vI, a resistor and capacitor. And I am going to understand how this behaves.
And it is a first order circuit, single capacitor. My input is a vI cosine of omega t. vI is a real number for t greater than zero. And I am telling you that the capacitor voltage starts out being zero.
And my vI is a sinusoid. It's not a step this time. It's a sinusoid. So vI is a sinusoid and I want to find out what vC looks like. The behavior here tells me, I will give you the answer, that when I feed a sinusoidal input as the frequency increases, vC should be decreasing somehow and also be shifting in phase.
Let's do the derivation for that and see what happens. To give you some insight as to how to go about analyzing this let me draw a little graph as to the effort level of doing this. To determine vC of t on the y-axis here is our effort.
How hard do we have to work to solve this circuit for a sinusoidal input? And on this graph, down here is easy and up here is pure agony. Sheer agony up here. So it's the scale of effort level ranging from easy to complete agony.
And this is the time axis. And the time axis starts out at 11 o'clock, the early part of today's lecture, and ends at roughly 12, that is today's lecture and this is next lecture. What I am going to show you today is a method that uses a usual differential equation approach, and this is going to be pure agony.
If you thought last Thursday was agony watch today. This is going to be sheer, sheer, sheer hell. So I am going to grunge through that, and I think I will sort of give up halfway because it's just too painful even for me here.
Then what I am going to do is at the end of this lecture I am going to show you an approach that I give a cutesy name. I call it the "sneaky approach." We are going to sneak something in there it is going to be a lot easier.
And then in the next lecture I am going to show you an even sneakier approach that is just going to be absolute bliss. So let's start here. Indulge me as I go through the agony part. I am going to blast through it because that is not of how we are going to do things, but you just need to know that that is agony.
Let's do a usual differential equation approach. Steps one, two, three and four. Set up differential equation, find the particular solution, find the homogenous solution, add up the two and solve for the unknowns.
It's a mantra. The four-step manta. Let's do it. Step one, write the DE. That's easy. We have done this before the RC circuit. It's RC dvc/dt+vc=vI. This is no different from what you got from what you got from your RC circuit with a step input, just that my input is VI cosine of omega t in this case.
It is not just a DC voltage VI. Stare at that. Enjoy it while the going is easy. It's like traversing rapids, and before you come to a class five, you have calm and raging waters there, you kind of sit there saying oh, the scenery around here looks really good and so on.
All you are doing is stalling before you have dive down to class five. We will get to class five rapids in a few seconds here, so just enjoy this. RC dvc/dt+vC=vI. You've seen this before. Nothing fancy.
Good old stuff. VI cosine of omega t. You want to hold onto your seatbelts? OK. Let's find the particular solution to the cosine input. Let's use our standard method. What I will do is just so, there is going to be so much crapola up there, so that I draw your attention to vP, which is what we are trying to get, I am just going to put a box around vP in red.
If you see like all sorts of garbage appear, just look for the red square. That is what we are trying to get at. That's the equation. Let's try. First try, A worked before. A constant value A worked before for DC inputs.
Let's try that again. If it worked then it may work now. If I use A, a constant value, and I substitute it here, I get dA/dt goes to zero, vP is A, but on the right-hand side I have VI cosine omega t.
So clearly A doesn't work. Sorry. I struck out. Well, cosine omega t here, let's try A cosine omega T as my particular solution. Things are getting a little harder now, a little more painful. So substitute A cosine omega t here.
So I do get A cosine omega T for vP, but out here I get RCA sine omega t times omega times minus one. So I have a sine and a cosine, and I have a cosine on the right-hand side. Sorry, it doesn't work.
Nope, doesn't work either. Well, let's try A cosine omega t plus phi. We are now embarking into the rapids here. You can begin feeling the pressure. Just to refresh your memories of sines and cosines.
A is the amplitude of the cosine. Omega is the frequency. Phi is the phase. Remember the signal I showed you earlier? If something happens to the amplitude of the sine, something happens to the phase.
A cosine omega t plus phi. Let me plug it in here and go by standard practice. Here is what I get. I plug in A cosine omega t to this equation, and this is what I get. I differentiate it. I get omega out minus sine, sine of negative d plus phi, A cosine omega t plus phi equals VI cosine omega t.
That might work. Now we get into the class six part of the class five. All class fives have a little bit of class six rapids. Remember, the rapids go up on an exponential scale so it like earthquakes.
What I do now is expand out sine omega t plus phi, blah, blah, blah, it goes on and on. I could go on and on, but this is even tiring me. This can be made to work, but I am not sure I want to put all of us through this trig nightmare here.
If I am really, really nasty I could give this to you as a homework assignment or something, but I am not that nasty so you won't see that. But if I go down this path it will get me to the answer, but I would have to soon negotiate class six, class seven rapids to get to where I want.
So let me punt on it, let me start from scratch. I am at step two, let me start from scratch. Instead what I would like to do is let's get sneaky here. Rather than negotiating the class five rapids, what we can say is ah-ha, we can take our canoes and jump onto shore and run down and then get back onto the river.
Let's do that. That is called the sneaky approach. So that all our friends who are behind us think we are negotiating the rapids, but what we are going to do is get sneaky and take the shore path. Let's get sneaky, just walk down the shore and see what is there.
I want to do something completely weird here. I want to look at solving this equation through the shore method. S stands for shore or S stands for sneaky, whatever you want. What I am going to do is rather than trying to solve for VI cosine omega t.
I am going to say let's try a different input all together. And you will understand why in a second. It's like I am the captain of my canoe and I tell my teammates, hey, let's not negotiate the rapids, let's go and explore the shore.
Maybe down the shore we can find a path that gets us across to the other side more easily. So here is me and my colleagues carrying our canoe and getting onto shore and taking a sneaky path. This is not what I set out to solve.
I don't know where this will lead me. But let's see where the shore path leads us. I want to try solving this equation Vie^st. S stands for shore or sneaky or whatever you want. I don't know where I am going, but let's see where this leads us.
Let's explore. Make believe you are Columbus or something. I don't know. Let's use the usual techniques and see how this works out. Let's try a particular solution, Vpe^st. My input is Vie^st. I am trying to solve the circuit for a different input.
And let me try solution Vpe^st and see if that works out nicely. I substitute Vpe^st into my equation here, so RCVpe^st blah blah blah. What I get here is Vie^st, Vpe^st stays the same, but here vP comes out, s comes out and e^st stays the same.
That is nice property of exponentials. This is what I get. A really cool property of exponentials is that when I differentiate it I get the exponential back. Unlike a cosine I got a sine, and for a sine I got a cosine.
Exponentials are very plain and simple, are straightforward. What you see is what you get. You differentiate it. You get the same thing, you get scaling vP, S and so on, and some scaling here. You get S scaling here, but the basic form stays the same.
This is quite nice. I have e^st in all three places, so I can cancel those out and I get this expression. And I get this. Wow. So if I go down the shore I get some place fast. I don't know where I am yet, but whatever I did, it was easy.
I am just exploring this path, down the shore path. I am making progress. I don't know where I have gotten yet. We will see where we got to in a second, but I got some place quickly, fast. I was able to solve for this input Vie^st and get this solution very quickly.
So what happened here? I assumed the solution of the form Vpe^st, substituted it back, and found that if vP were equal to Vi/(1+sRC) then Vpe^st is a solution. What I have done here is that Vi/(1+sRC) is a particular solution to this equation for the input Vie^st.
I put a box around it because this is important. This was easy. We went down the shore, and said let's try this other input. We made rapid progress on shore and I got some place. I don't know where I am yet.
I got this. Let me pause here and let me give you the final answer. I am going to show you over the next five minutes that this is our answer. You are staring at the answer already. I am a party, I have taken a shore path and we have gotten some place.
We see the river there, so it turns out we are exactly where we want to be, just after the rapids. All I have to do now is get my colleagues into the river with myself in the canoe and we are done.
You don't know that yet. My colleagues and I are sitting on the shore looking at the river. We've gotten some place. There are no rapids there. We have gotten some place. We don't quite know is this just after the rapids or not.
We don't know yet, but I got there very quickly. And I will tell you right now, that is the place we wanted to go to. The next five view graphs I am going to blast through. There is going to be more pain and agony to show you why that is the case.
It's me thinking I am Columbus and proving to my colleagues that this is where we want to be. And pulling out my sextant, and the compasses and so on that cartographers and people use to prove to my colleagues that this is where I want to be.
This is the answer. The next five view graphs will be demonstrating that this is indeed the answer, or close enough to the answer that we will be satisfied. Isn't this spectacular? I am going to show you in about five minutes that this gives us all the information we need to know to compute the sinusoidal steady-state response to this differential equation.
Let me write that down here just so you know. Just so you remember. I am going to put a marker on the shore that says this is where we are right now. Now let me prove to you. As I just said, vPS is Vi, it's this stuff here multiplied by e^st is the solution to Vie^st.
This guy here is a solution for Vie^st and vP is simply the coefficient that multiplies e^st. Similarly, if I substitute S equals j omega. I told you five view graphs of more hell, but I am just going to prove to you that this is it.
I am substituting S equals j omega. This is Columbus giving a big speech at the end convincing his colleagues that we are where we want to be. I substitute j omega for S and this is what I get. This is a solution for e to the st, and so this is a solution for e to the j omega t.
And let me mark this for you as something to remember. Vi/(1+j omega RC). In terms of that, I am substituting j omega for S. And this is a complex number. It is a complex amplitude. It is a complex number because of j here, and it multiplies e to the j omega t.
Just keep this in mind. So that was easy. The steps were easy. I am still proving to you that this is where we want to be. Now let me draw the connection back to vP. And the first fact was that finding the response to Vie^(j omega t) was easy.
We know that. Second was the following observation that Vi cosine omega t is simply the real part of this number here. So Vi cosine omega t is simply the real part of Vie^(j omega t) from the Euler relation.
So cosine omega t is simply the real part of this guy. Light bulbs beginning to go off? The first fast was that finding the response to Vie^(j omega t) was easy. And the response was this, right? Times e to the j omega t.
That was easy. All right. And the second part is that the real part of this is Vi cosine omega t was our input. Draw the connection between two steps. Finding the response to Vie^(j omega t) was easy.
The real part of that was the input we cared about. Are light bulbs going off? Let me draw you a little picture here to show you what is happening. Response to vI is vP. It's the particular response we are looking for.
Remember the red square? But we threw in a sneaky input vIS and we formed the response vPS to that. This step was easy. This step was hard. vI to vP was hard, trig nightmare, remember? But vIS to vPS was easy.
It was a simple Vpe^st thing. We also know that the real part of vIS is vI. The real part of this is simply vI. If you have a real circuit, if you have a real linear circuit, for a linear circuit, if the real part of this gives me vI then the real part of the solution should give me vP.
So computing vPS was easy. If I take the real part of this, I take the corresponding real part of this. This is sort of an inverse superposition argument. Superposition, I said, take the response for A, take the response for B, add them up and you get the response for A plus B.
Here what we are saying is that get the response to A plus B, or to A plus jB and the real part of the input will produce the response given by the real part of the output. So this is an inverse superposition argument.
If it is a linear circuit, then if vI is the real part of this sneaky input then I find the response to the sneaky input and take its real part I should get vP. Here I am, Columbus, staring down at the entrance to this part of the river.
I just proved to my colleagues that all we have to do is take the real part of what we have. We can just jump right back into the river and get back to vP. And what I am going to do next is just grind through the math and just show you that.
I will just blast through it. It is not important, but you have it in your notes. I am telling you that vP is simply the real part of the sneaky output. And I take the real part of vP e to the j omega t.
And I take the real part. And just a bunch of math here. I am just taking the real part and doing a bunch of complex math. Remember vP was given by this quantity here. And I take the real part and I end up with vP is simply this quantity multiplied by cosine omega t plus phi, where phi is given by is given by tan inverse of omega RC, and this is the coefficient multiplying the cosine.
So by taking the sneaky path and then taking the real part of that output answer, I was able to very quickly get to where I wanted to be. So from here to here it is only math. Recall, that vP, the thing in the red was what we set out to find out, which was the particular response to VI cosine of omega t.
And remember that two grunge is all of this stuff. I am going to blast through two or three more view graphs that just give you more insight and more math, nothing particular. And remember to solve the equation we have to find a homogenous solution, too.
But recall that the homogenous solution for an RC circuit is of the form Ae^-t/RC. This means that as time becomes very large this part goes to zero. As time becomes large in the steady state, remember I care about the steady state? This goes to zero.
I don't care about the homogenous solution. Isn't that fantastic? Most the circuits we will deal with, except for purely oscillatory ones, the homogenous part dies away. You have something like e to the -t whatever.
It just dies away. It's gone. So the total solution has vH going away. And what I end up with is just vP. My total solution in the steady state is simply vP. And A is given by this that we just calculated.
I just have a bunch more insight that I talk about that you can look through in your notes. And I just want to show you a very quick summary. In summary, what we have is we computed vP. It was a complex coefficient.
And all these steps, 2 grunge, 3 and 4 were a waste of time. And what I showed you was that for the input VI the coefficient vP was complex. And I can take the ratio and represent it in this manner as well.
And from vP, I can then compute the multiplier for the cosine as follows. I divide by vP here. Remember the cosine was multiplied by, in the mathematical step that I did, VI divided one plus, this stuff here, so I could get the magnitude and phase of the transfer function of this circuit in the following manner.
And to wrap up very quickly, I am going to cover this again the next time and show you a magnitude plot. Notice here that if I plot Vp/Vi. Remember this was Vp here. That's the answer. The magnitude looks like this.
On a log scale Vp/Vi for small frequencies omega is at one, but as omega increases Vp/Vi keeps decreasing. That is the output. Remember Vp was the amplitude of the output? That keeps decreasing. And this is the reason why.
As I increase the frequency, the amplitude of my output cosine kept decreasing. I could also plot the phase for you. And the phase, in the same manner as omega increased, my phase also kept shifting from zero initially to pi/2 finally.
Let me stop here and start with this the next time and revisit this. Unfortunately, I won't have time for the demo. I will show it to you next time.

Second-order systems - part 2

Before I begin today, I thought I would take the first five minutes and show you some fun stuff I have been hacking on for the past three years. This has to do with 6.002 and circuits and all that stuff, but this is completely optional, this is for fun, this is to go build your intuition, this is to check your answers, whatever you want.
This is not a required part of the course. Just for fun. There is this URL out here that I put down here. I have been hacking on this system for the past three years, and for the first time this year and very tentatively and gingerly introducing it to students.
The idea here is that it is a, that is kind of defocused. Any chance of focusing that a little bit better? The idea of this is that it is a Web-based interactive simulation package that I have pulled together.
And what you can do is you can pull up a bunch of circuits. Notice that the URL is up here. It is euryale.lcs.mit.edu/websim. And there is the pointer to it. So you have a bunch of fun things you can play with.
And we have gone through all of these things in lecture. Let's pick the MOSFET amplifier. You come to this page. This is something you have seen in class. And let's play with this little circuit.
And you see the mouse? Good. You can set up a bunch of parameters. You can set up the MOSFET parameters VT and K. You can set up the value of R for your resistor, you can establish a bias voltage, and you can have an input voltage vIN.
So you can apply a bunch of input voltages. You can apply a zero input, unit in pulse, unit step, sine wave, square waves. Or this was the part that took me the longest to get right. You can also input a bunch of music.
And so far I just have two clips, so you are going to get bored listening to them. Good. So you can also input music. And what you can do is you can watch the waveforms, you can listen to the output and do a bunch of fun stuff.
One experiment I would love for you guys to try out. Again, remember, this is completely optional. Just for fun. You can apply some input. Step input, for example, to an RLC circuit and spend 30 seconds thinking about what should the output look like.
I divine that the output should look like this and then do this and see if what you thought was correct. And it's fun to kind of play around with it. Let me start with, just as an example, let's say I input classical music.
And let us say I would like to listen to the output here that is the voltage at the drain terminal of the MOSFET. For listening it sets up a default timeframe to listen to, so you go ahead and do it.
This shows you the time domain waveform of a clip of the music and then you can listen to it. Lot's of distortion, right? As you can see, there is a bunch of distortion. And that is as you expect because the peak-to-peak voltage is 1 volt, the bias is 2.5, and so this is clipping at the lower end, plus the MOSFET is nonlinear.
You can play around with a bunch of things and you can have a lot of fun. And the reason I created this is that MIT is putting a bunch of its courses on the Web. And one of the hottest things about courses like this is the lab component.
If you are beaming a course to, say, a Third World country or something, how do you get people to set up the massive lab infrastructure? I know you hate your oscilloscopes, I know you hate your wires, I know you hate the clips, but the fact is you have them.
I know a lot of places those are way too expensive to pull together, which is why I have been creating this Web-based kind of interactive laboratory so that people can learn this stuff over the Web.
Let's go do another example very quickly. Let's say you learned about, well, let's do RC circuits. Here is the parallel RC circuit. And you can set up capacitor values, resistor values, you can set up input.
Here, let me look at the time domain waveform for the voltage across the capacitor. And this time around let me play a unit step. And let's see what the output is going to look like. You can think in your minds what should the output look like, and then you can go and plot it.
There you go. That's what the output looks like. So you can play around with it and have fun. That's all the good news. The bad news is that so far I just have one Pentium III machine behind us.
It is a Linux box, so don't all of you try it at once. However, what I have also done, and that took me another six months of hacking in the small amount of time professors have to hack on stuff, I've hacked an incredibly elaborate cashing system so that once anyone in class tries out some combination of parameters it goes and squirrels away all the outputs.
If anybody else types in the same sets of parameters it will just get all the output and play it back to you. So if enough of you play with over time, we may end up cashing all the important waveforms and music clips and all of that stuff.
I have allocated a few gigabytes of storage, so I am hoping that it may work. Go forth. Play with it. And this is completely my fault, so if there are any bugs or anything simply email them to me.
This is the first time this is coming alive so bear with it. Now let me switch back to the scheduled presentation for today. All right, hope and pray that this works. Yes. Good. I am going to do today's lecture using view graphs.
And the reason I am going to do that and not do my usual blackboard presentation which I way, way, way prefer to a view graph presentation. The only reason I am going to do this for today, and maybe one more lecture, is that there is just a huge amount of math grunge in this lecture.
What I want to do is kind of blast through that, but you will have it all in the notes that you have, so that you don't waste time in class as you watch me stumbling over twiddles and tildes and all that stuff.
The key thing here is that the insight is actually very simple. The beginning and the end are connected very tightly and very simple. There is a bunch of math grunge in the middle that we are going to work through and, again, follows a complete old established pattern.
So, in that sense, there is really nothing dramatically new in there. Let me spend the next five minutes reviewing for you how we got here, what have we covered so far and set up the presentation. The first ten view graphs I am going to blast through and just tell you where we are in terms of LC and RLC circuits.
I began by showing you this little demo, two inverters, one driving. I can model the inductance here with a little inductor, the capacitor of the gate here. And recall that when I wanted to speed this up by introducing a 50 ohm smaller resistance, I got some really strange behavior.
Just to remind you, for Tuesday's lecture it would help if you quickly reviewed the appendix on complex algebra in the course notes. Remember all the real and imaginary j and omega stuff? It would be good to very quickly skim through that.
It is a couple of pages. Remember this demo? And the relevant circuit that is of interest to us is this one here. It is the resistor, there is the inductor and there is a capacitor. This is Page 3.
I am just going to blast through the first ten view graphs. It is all old stuff. Then we observed the following output. We applied this input at VA and we got this output, a very slowly rising waveform because of the RC transient.
And because of that you saw a delay. Notice that this delay was because of the slowly rising transient. This waveform took some time to hit the threshold of the neighboring transistor. So we say ah-ha, let's try to speed this sucker up by reducing the resistance in the collector of the first inverter.
And so I had this input. Now, to my surprise, instead of seeing a nice little much higher and much faster transitioning circuit, well, I did see a much faster transitioning circuit but I got all this strange behavior on the output that I was interested in.
And because of that, if these excursions were low enough, I could actually trigger the output and get a whole bunch of false ones here because of these negative excursions which should not really be there.
That was kind of strange. In the last lecture we said let's take this one step at a time. Let's not jump into an RLC circuit. Let's go step by step. Let's start with an LC, understand the behavior.
We started off with an LC circuit of this sort, and using the node equation we showed that this was the equation that governed the behavior of the circuit. And then we said that for a step input and for zero initial conditions, that is the zero state response, let's find out what the output, the voltage across the capacitor looks like.
And so we obtained the total solution to be this. And there was a sinusoidal term in there. And the omega nought which was one by square root of LC. And this was the circuit. And so for this step input notice that the output looked like this.
So far an input step I had an output that went like this. Notice that it is indeed possible for the output voltage to actually go above the input value VI. This is kind of non-intuitive but this can happen.
So this waveform jumps up and down. But the steady state value, on average if you will, is VI. On the other hand, it does have sinusoidal excursions and this kind of goes on because there is nothing to dissipate the energy inside that circuit.
By the way, the fact that the capacitor voltage shoots above the input voltage is actually a very important property. We won't dwell on it in 6.002, but just squirrel that away in your brain somewhere.
I promise you that some time in your life you will have to create a little design somewhere that will need a higher voltage than your DC input. And you can use this primitive fact to actually produce a DC voltage higher than you are given, and then use that somehow.
In fact, there is a whole research area of what are called DC to DC converters, voltage converters. Let's say you have 1.5 volt battery, a AA battery, but let's say a circuit needs 1.8 volts. The Pentium IIIs, for example, needed 1.8 volts.
And the strong arm is another chip that required 1.8 volts a few years ago, but the AA cell was 1.5 volts. How do get 1.8 from 1.5? Well, you have to step it up somehow. And this basic principle where the voltage can jump up above the input is actually used, of course with additional circuitry, to kind of get higher voltages.
It is a really key point that you can squirrel away. This was pretty much where we got to in the last lecture. This starts off the material for today. What we are going to do is take that same circuit, but instead we are going to put in this little resistor here.
This is what we set out to analyze. And for details you can read the course notes Section 13.6. The green curve here was the behavior of the LC circuit. And what we are going to show today is that the moment we introduce R this sinusoid here gets damp.
It kind of loses energy. And I am going to show you that the behavior is going to look like this. By introducing R this guy doesn't keep oscillating forever. Rather it begins to oscillate and then kind of loses energy and kind of gets tired and settles down at VI.
And remember the demo. This is exactly what you saw in the demo. You had a step input and you had this funny behavior. And for the RLC that is exactly what it was. So today's lecture will close the loop on what you saw in the demo and the weird behavior, and I am going to show you the mathematics foundations for that today.
Let's go ahead and analyze the RLC circuit. I purposely created the entire presentation to follow as closely as possible both the discussion of the RC networks and the LC networks so that the math is all the same.
Exactly the same steps in the mathematics are in the exposition of the analysis. What's different are the results because the circuit is different. So don't get bogged down or whatever in the mathematics.
Just remember it is the same set of steps that you are going to be applying. We start by writing down the element rules for our elements. Nothing new here. For the inductor V is Ldi/dt. The integral form which is simply 1/L integral vLdt=i.
We saw this the last time. And for the capacitor, the current through the capacitor is simply Cdv/dt. Those are the two element rules for the capacitor and inductor. The element rule for the resistor, of course, is V=iR.
You know that. And for the voltage source we know that, too, the voltage is a constant. Just follow the same established pattern. By the way, just so you are aware, I have booby trapped the presentation a little bit to prevent you from falling asleep.
You see the dash lines here? Whenever you see a dash line, that stuff needs to be copied down. Don't trip over that. Don't say I didn't warn you. We start by using the usual node method. And I have two nodes in this case.
Unlike the LC circuits, I have two unknown nodes. One is this node here with the node voltage vA and the second node is the node with voltage vT. Let me start with vA and write the node equation for that.
It is simply 1/L, the node equation for this is the current going in this direction with is vI-vA integral and that equals the current going this way which is vA-v/R, node equation. I then write the node equation for the node v, for this node here, and that is simply (vA-v)/R=Cdvdt.
And that is what I have here, two node equations. Let me summarize the results for you and then show you a view graph where I grind through the math as to how I got the result. Here is the result I am going to get.
If I take these two node equations and I massage some of the mathematics, I am going to get this result. And I will show you that in a second. By grinding through some math and solving these two equations and expressing this in terms of v, I get a second order differential equation, d^2v blah, blah, blah.
Notice that this is different from the LC in this term. Every step of the way you can check to see if I am lying or I am correct. I will indulge you, indulge myself rather with a little story here.
Richard Fineman was a known smart guy. And one of the reasons that he was that was in the middle of talks he was known to get up and ask some of the darndest, hardest questions and say ah-ha, you have a bug in this proof here or a bug in this equation that is not right.
And usually he would be correct. So his trick in doing this and which is one reason how he became a known smart guy. What he would do is, as the speaker went on talking he would kind of follow along and think of a simple initial primitive case.
In this case, I have an RLC circuit. So think of a simpler case of this. A simpler case of this is R=0. Whenever you set R to be zero, you should get exactly what we got in the last lecture, correct? That is what Fineman would do.
He would boil this down to a simpler case, make some assumptions and just follow along. And whenever he found a discrepancy between the math here and his simple case he would say oh, there is a bug there.
If you want you can catch me that way. Here, what Fineman would do is replace R being zero, and notice then this equation here is exactly what we got the last time with R being zero. Just remember that Fineman trick.
This is the equation we get, the second-order differential equation with an R term in there. And let me just grind through the math and show you how I got this from this. So the two node equations again.
And what I do is I start by taking these two equations and differentiating this with respect to t and this is what I get. And, at the same time, I have replaced (vA-v)/R here by this term. I replace this with this and differentiate.
Then I simply divide the whole thing by C. Then I take this expression here and write down vA is equal to this stuff here. Next I am going to substitute this back for vA and eliminate vA. So I take this vA, stick the sucker in here, and thereby eliminate vA and get this.
And then I simplify it and here is what I get. That is what I get. I just grind through the two equations and get that result. So like a stuck record I will repeat our mantra here, which is here is how we solve the equations that we run across in this course, the same three steps.
Find the particular solution. Find the homogenous solution. Find the total solution and then find the constants using the initial conditions. Same steps. You could recite this in your sleep. And the homogenous solution is obtained using a further four steps.
Let's just go through and apply this method to our equation and get the results. vP is a particular solution and vH is the homogenous solution. With a particular solution, oh. Before I go on to do that, let me set up my inputs and my state variables.
My input is going to be a step. Remember, I am trying to take you to the point where the demo left off. The demo had a step input, so I am going to use a step input rising to vI. And I am going to with the initial conditions being all zeros.
So the capacitor voltage is zero, inductor current, another state variable is also zero, and therefore this is also fondly called the ZSR or the zero state response because there is only an input but zero state.
Again, remember the dashed lines here. Don't say I didn't warn you. Let's start with a particular solution. This is as simple as it gets. I simply write down the particular equation and stick my specific input.
And remember the solution to the particular equation is any old solution, it doesn't have to be a general solution, any old solution that satisfies it. And I am going to find a simple solution here.
And V particular is a constant VI. It works. Because remember this has been working all along. And I am going to keep pushing this and see if this works until the end of the course. Guess what? It will.
So this is a solution. I'm done. That is my particular solution. Simple. Second, I go and do my homogenous solution. And the homogenous equation, remember, is the same old differential equation with the drive set to zero.
Remember that sometimes this equation with the drive set to zero is the entire equation you have to deal with in situations where you have zero input, for example. Or in other situations in which you have an impulse at the input.
And the impulse simply sets up the initial conditions like a charge in the capacitor or something like that. So we are going to blast through this four-step method. The method simply says that four steps, I am going to assume a solution of the form Ae^st.
And if you think you've seen that before, yes, you have seen it many times before. And you will see it again, again and again. And we need to find A and s. We want to form the characteristic equation, find the roots of the equation and then write down the general solution to the homogenous equation as this.
Same old same old. Let me just walk through the steps here. Step A, assume a solution to the form Ae^st. And so I substitute Ae^st as my tentative solution to the equation. Again, let me remind you that the differential equations that we solve here are really easy because the way you solve them is you begin by assuming you know the solution and stick it in and find out what makes it work.
I am going to stick Ae^st into this differential equation, and A comes out here. Differentiate this d squared, I get s squared down here, A s here and this simply gets stuck down here with the 1/LC coefficient.
The next step I begin eliminating what I can, so I eliminate the A's, then eliminate the e^st's, and I end up with this equation here. I end up with this equation. This is my characteristic equation.
It is an equation in s. Do people remember the characteristic equation we got for the LC circuit? Remember the Fineman trick? That's right, LC. S^2+1/LC=0. This thing wasn't there. All you do is simply follow the R.
Just follow the R. Just imagine this is a dollar sign and kind of follow it. And you will see what the differences are between the LC and the RLC. So this is the characteristic equation. What I am going to do, iss much as I wrote the characteristic equation for the LC circuit, by substituting omega nought squared for 1/LC.
Let me do the same thing here but introduce something for R and L as well. What I will do is let me give you this canonic form. The very first second-order equation I learned about when I was a kid was this one, S^2+2AS+B^2 or something like that.
Let me write it in that form where I get 2 alpha s plus omega nought squared. Again, remember the alpha comes about because of R. So omega nought squared is 1/LC and alpha is RL/2. Omega nought squared is 1/LC and R/L is equal to two alpha.
I am just writing this in a simpler form so that from now on going forward I am just going to deal with alphas and omega noughts. Once I get to this characteristic equation, after that I can give you one generic way of solving it.
And depending on the kind of circuit you have, a series RLC, which is what we have, or a parallel RLC we will simply get different coefficients for the alpha term. This is going to stay the same but this term will look different, alpha is going to look different.
There is a real pattern here. And what I am doing is simply focusing on what is important, what the differences are between the pattern. You learned the LC situation and the RLC situation. Given this I can now write down, I am just simply replacing this as my characteristic equation in dealing with alphas and omegas.
I will give you a physical significance of alpha in a little bit. Do you remember the physical significance of omega nought? That was the oscillation frequency. In other words, given an inductor and capacitor, you put some charge on the capacitor and you watch it, it will oscillate.
And its oscillation frequency will be one by a square root of LC. The magnitude of the initial conditions will determine how high are the oscillations or what the phase is in terms of when it starts, but the frequency is going to be the same.
Step three, to solve the homogenous equation, is find the roots of the equation, s1 and s2, and here are my roots. Good old roots for a second-order, little s squared equation here. Finally, given that I have the roots, I can write down the general homogenous solution.
So general solution is simply A1e^s1t, A2e^s2t. That's it. That's the solution. This looks big and corny, but we are going to make some simplifications as we go along and show that it ends up boiling down to something cos omega t.
The math is kind of involved but we get down to something very simple, a cosine. Hold this general solution. From that, as a step three of the differential equation solution, I write the total solution down.
And my total solution is the sum of the particular and the homogenous, so therefore I get this. VI was my particular and this term here is my homogenous solution. Now, if I wasn't doing circuits and simply trying to solve this mathematically here is what I would do.
I would find the unknown from the initial conditions, so I know that v(0)=0. And so therefore if I substitute zero for V(0) I get this. If I substitute zero here, t is 0, t is 0, so I simply get V1+A1+A2.
And let me just blast through because I am going to redo this differently. i=Cdv/dt. And so that's what I get. I substitute zero and this is what I would get. I hurried through this. Don't worry.
I'm going to do it again. If you just do it mathematically, you can solve this equation here and these two simultaneous equations in a1 and a2 and get the coefficients and you are done. But it doesn't give us a whole lot of insight into the behavior of these terms here.
What I am going to do for now is kind of ignore that. Ignore I did that and instead try to go down a path that is a little bit more intuitive. Let's stare at this expression we got for the total solution.
That is the expression we got. All I did is, I had alpha in there, I simply pulled out the alpha outside. So this is my total solution, V1-A1e^(-alpha t) something else and something else. Three cases to consider depending on the relative values of alpha and omega nought.
If alpha is greater than omega nought then I get a real quantity here. The square root of a positive number, I get a real number, and that number will add up to the minus alpha and I am going to get a solution that will look like, oh, I'm sorry.
Let me just do it a little differently. There are three situations here. One is alpha greater than omega nought. Alpha equal to omega nought. Alpha less than omega nought. Alpha is greater, alpha is less, alpha is equal to this term inside the square root sign.
For reasons you will understand shortly, we call this "overdamped" case, the "underdamped" case and the "critically damped" case. When alpha is greater than omega nought this term gives me a real number, and I get something as simple as this.
Remember, for the series RLC circuit, alpha was R/2L. So if R is big, in other words, if in my RLC circuit R is huge then I am going to get this situation. My output voltage on the capacitor is going to look like this, the sum of two exponentials.
And if I were to plot it very quickly for you, for a VI step, V would look like this. So v would simply look like this because it is the sum of a couple of exponentials. All right. Now, alpha is positive here.
Remember alpha1 and alpha2 are both positive. These two added up, because of this constant VI, give rise to something that increases in the following manner. Let's look at the situation where alpha is less than omega nought, where the term inside the square root sign is negative.
What I can do is pull the negative sign out and express it this way. What I am going to do is since alpha is less than omega nought, I am going to reverse these two and pull out square root of minus one to the outside.
This is what I get. I am just playing around with this so that whatever is under the square root sign ends up giving me a positive real number. So I pull the j outside and this is what I get. Now, let me blast through a bunch of math and end up with something very, very simple for this underdamped case.
Let me define a few other terms. I am going to call omega nought minus alpha squared the square root of that. I am going to call it omega d. And here is what I get. So I have defined three things for you now, alpha, omega nought and omega d.
And I get this equation in terms of alpha and omega d. And then, remember from your good-old Euler relationship? e to the j omega d is simply cosine plus a j sine. I am just going to blast through a bunch of math rather quickly.
Once I replace this in terms of a cosine and sine, cosine and a j sine and then collect all the coefficients together, I get an equation of the form VI plus some constant e to the minus alpha t, cosine, the sum of the constant e to the minus alpha t, sine.
Remember the sines and cosines are coming out, but because of my R I am getting this funny alpha here, e to the minus alpha here. So I am getting sums of sine and cosine. And K1 and K2 are some constants which I will need to determine for my initial conditions.
I am going to continue on with this and keep on simplifying it because, as I promised you, I want to get to something that is just a cosine. I want to go down this path. I am not going to cover this case, the critically damped case.
And I will touch upon it later but not dwell on it. Let me continue down the path of the underdamped case, and this is what we have. Continuing with the math, let's start with the initial conditions, v nought equals zero, and that gives me K1 is simply -VI.
So at v(0)=0 t is zero, so this terms goes away, the cosine becomes a 1, e^(alpha t) goes away, and K1=-VI. Then I know that i(0) and i is simply Cdv/dt. And I get this nasty expression. I substitute t=0 and I get something that looks like this.
I know what K1 is, and so therefore K2 is simply -V1alpha divided by omega nought. I have taken this expression where the unknowns K1 and K2 are to be found. I set the initial conditions down at t=0 and I get K1 and K2 as follows, which gives me the following solution.
This is the solution I get where I do not have any unknowns anymore. Remember that omega d and alpha are directly related to circuit parameters. Alpha was R/2L and omega d was square root of alpha squared minus omega nought squared.
** omega d = sqrt(alpha^2 - omega_0^2) ** And omega nought squared was 1 by square root of LC. So I know it all now. I still have sines and cosines here, so I am going to simplify this a little further.
Oh, before I go on to do that, let's do the Fineman trick again and notice if I am still true to the LC circuit I did the last time. Remember when R goes to zero alpha goes to zero. Because alpha is R divided by 2L.
If alpha was zero what happens? If alpha was zero, this guy goes to one, this whole term goes to zero and omega dt now ends up becoming omega nought, and I get this term here. I get VI-VIcosine(omega t), which is exactly what I expected in my equation.
This is the same as the LC case that I got. Let's go back to this situation and simply if further. If you look at Appendix B.7 in your course notes, Appendix B.7 is a quick tutorial on trig. And in that trig tutorial you will see that, and you have probably seen this before, too, multiple times, the scaled sum of sines are also sines.
This is an incredibly cool fact of sinusoids. If you take two sinusoids of the same frequency and you scale them up in any which way and add them up you also end up with a sinusoid. It is hard to believe but it is true.
It is an incredible property of sinusoids. Take any two sinusoids, scale them in any way you like, same frequency, add them up, you will get a sinusoid. What that is saying is that, look, here is a sinusoid, here is a sinusoidal function, and I am scaling them up in some manner.
So I should be able to add them up and be able to express that as single sine. And to be sure you can, look at the Appendix, and there is an expression for a1 sinX plus a2 cosX is equal to a cosine of blah, blah, blah.
This is what you get. No magic here. Just math. From here I directly get this. And look at what I have. It is absolutely unbelievable. v(t) is simply VI, there is a constant here, this an e to the minus alpha term and there is a cosine.
Again, to pull the Fineman trick, if this alpha were to go to zero here then you would end up with the expression you had for the LC situation. Let's stare at this a little while longer. There is a constant plus a minus, a cosine term, so there is a sinusoid at the output, and there is an e to the minus alpha which ends up giving you the decay you have seen before.
In other words, to a step input, the LC circuit would give you a sinusoid. That is what the LC circuit would do if alpha was zero. But because of this alpha term here, e to the minus alpha t, that gives rise to a damping effect, so this causes this thing to become smaller and smaller as time goes by until this term goes to zero at t equals infinity.
This guy damps down and so therefore you end up getting the curve that you saw like this. Twenty minutes of juggling math solving a second-order differential equation, but what ends up is the same sinusoid but it is damped in the following manner such that the frequency, rather the amplitude keeps decaying until it starts off at zero and then settles down at vI.
This is exactly what you saw in the demo that we showed you earlier. The critically damped case, I am not going to do it here. I am going to point you to the following insight. The underdamped case looked like this.
It was a sinusoid that kind of decayed out. That is the underdamped case. And then I showed you the overdamped case. The overdamped case looked like this. And, as you might expect, the critically damped case is kind of in the middle and looks like this.
So the overdamped case would look like this, underdamped like this, and the critically damped case kind of goes up and kind of settles down almost immediately. This is when alpha equals omega nought.
I won't do that case here, but I will simply point you to Section 13.2.3. Just to tie things together, recall this demo here that we showed you in class yesterday. This is exactly the kind of form of the sinusoid you saw because of that input step.
If you want to see a complete analysis of inverter pairs and look at the delays and so on because of that, you can look at Page 170 and example 898. In the next five or six minutes, what I would like to do is stare at the RLC circuit.
And much like I showed you some intuitive methods to get the RC response, what we are going to do is do the same thing for the RLC. In the RLC situation, much like the RC situation, experts don't go around writing 15 pages of differential equations and solving them each time they see an RLC circuit.
They stare at it and boom, the response pops out, the sketch pops out. This one is going to be another one like our Bend it Like Beckham series here. And this one is in honor of Leslie Kolodziejski.
And I call it "Konquer it like Kolodziejski." Again, as I said, experts don't go around solving long differential equations and spending ten pages of notes trying to get a sinusoid. They look at a circuit and sketch response.
I am going to show you how to do that, too. And what you can do is, to practice, go to Websim and try out various combinations of inputs and initial conditions and sketch it, time yourself, give yourself 30 seconds or a minute if you like, and sketch it and check it against the Websim response.
If it doesn't match either you are wrong or there is a bug in Websim. What I am going to do is, the response to the critically damped and underdamped case was very easy to sketch out. You started with an initial condition, you settled at VI and just kind of drew it like that.
The interesting case is the underdamped case, and that is what I am going to dwell on. Before we go on and I show you the intuitive method, as a first step I would like to build some intuition. Let's stare at this response here and try to understand what is going on.
This is the response that we saw. And this fact that you see an oscillation happening is also called "ringing." You say that your circuit is ringing. All right. You see some interesting facts. You see that frequency of the ringing is given by omega d.
This cosine omega d, so that is the frequency omega d. So the time is 2 pi divided by omega d. The oscillation frequency is omega d, but omega d is simply omega nought squared minus alpha squared.
Once you have a big value of R alpha becomes very small and omega d is very commonly equal to, very close to omega nought. So omega d and omega nought very commonly are very close together. And when that happens this frequency is directly omega nought.
Alpha governs how quickly your sinusoid decays. e to the alpha t here is the envelope that governs how quickly my sinusoid decays. And notice that each of these terms, alpha and omega nought, comes directly from my characteristic equation.
Which means that once you get your characteristic equation you really don't have to do much else. And up until now you still have to write the differential equation to get the characteristic equation, so you still have to do some differential equation stuff, but in two lectures I am going to show you a way that you can even write down the characteristic equation by inspection.
Look at your circuit and boom, in 15 seconds or less write down the characteristic equation. It is absolutely unbelievable. What are the other factors that are interesting here? Of course I need to find out initial values.
I start off at zero. This is my capacitor voltage. If I don't have an infinite spike or an impulse my capacitor voltage tries to stay where it is and starts off at zero. And the final value is given by VI, the capacitor is a long-term open so therefore VI appears across the capacitor.
In the long-term my final value is going to be VI. There is one other interesting parameter, which I will simply define today but dwell on about a week from today, and that is called the Q. Some of you may have heard the term oh, that's a high Q circuit.
Q is an indication of how ringy the circuit is. And Q is defined as omega nought by 2 alpha. It is called the "quality factor." And it turns out that Q is approximately the number of cycles of ringing.
So if you have a high Q you ring for a long time and if you have a low Q you ring for a very short time. That is called the quality factor defined by omega nought by 2 alpha. Notice that Q, omega nought, alpha, omega d, all of these come from the terms in the characteristic equation.
We will spend more time on Q later. With this insight here is how I can go about very quickly sketching out the form of the response. Here is my circuit. I want to sketch the form of the response for a step input at vI.
Zero to vI step input here, I want to find out what happens at this point. This is described to you in a lot more detail in Section 13.8 in your course notes. Let's go through the steps. Let's do the really simple situation first.
Let's also assume for fun that you are given that v(0) starts out being some positive value. Some v(0) which is a positive number. And, to make it harder on ourselves, let's say i(0) starts out being some negative number.
So i(0) is some negative current. The first thing I know is v(0), the capacitor voltage starts out here, which can change suddenly. And I also know that in the long-term this is an open circuit. So that this voltage vI will appear directly across the capacitor in the long-term.
So I get starting out at v(0), ending at vI, I am also half the way there. I know the initial and ending point of the curve. And then I know that somewhere in here there must be some funny gyrations here, because remember I am dealing with the underdamped case.
And you can determine that from alpha and omega nought. If alpha is less than omega nought, you know that you are in the underdamped case and this is what you get. Let's compute and write the characteristic equation down.
A week from today you will write it by inspection, but for now you will do it by writing down a differential equation. And from the characteristic equation you will get omega d, you will get alpha, omega nought and Q.
So omega d gives you the frequency of oscillations. My frequency of oscillation is now known. From Q I know how long it rings, because I know it rings for about Q cycles. I know that ringing stops approximately here.
And then I know that between that the start and end point my curve kind of looks like this, something like this. Right there we are 95% of the way there. The only question is I do not know if it goes like this or it goes like this.
I am not quite sure yet if it starts off going high or starts off going low. Not quite clear. I also do not know what the maximum amplitude is. It turns out this is rather complicated to determine so we won't deal with that.
Just simply so you can draw a rough sketch. The questions is which way does it start? I could leave it for you to think about. Yeah, let me do that. It is given on this page so don't look at it.
Think about it, and think about how you can determine whether it goes up or down. It turns out that in this case it is going to down and then ring. See if you can figure it out for yourselves and then we will talk about it next week.