AC analysis superposition

So in the last video, we talked about Oilers formula, and then we showed the expressions for how to extract a cosine and a sine from Oilers formula. We have a powerful set of expressions there for relating exponentials to sine waves.

Now, I want to show you an example, just a preview of when we get to the formal AC analysis. How are we going to exploit these expressions? How are we going to exploit these formulas? This is a real-world signal. This cosine term, let's suppose we build something that has a cosine to it. That could be something like a microphone that's hearing sounds that look like sine waves, and we would model those sine waves as a cosine wave that comes into some electronic system.

So let me draw a sketch of the clever approach that we're going to use. What I'm going to do is I'm going to build here. Here's a circuit I'm imagining. There are resistors, capacitors, and inductors—linear elements—and we can have sources and stuff like that. So there's something in here, and we have something going in and something coming out.

Okay, voltages and currents are coming out, voltage and current are coming in. I'm going to drive my circuit with some sort of sinusoid. I'll call that sinusoid—I'll give it an amplitude—and I'll call it cosine of Omega T, where Omega is the frequency, T is time, and A is the amplitude of the signal coming in here.

Now that signal is going to go into this circuit. Something's going to happen, and there's going to be a voltage coming out of here. We'll get V out over here of some sort, and it's going to be some amplitude. It's going to be some sort of a cosine wave of Omega T plus some phase angle—some angle—that's our job to discover. That's the circuit analysis problem for AC analysis.

We put in an AC signal, and we're going to get out another AC signal. This is the forced response. Remember, it's going to look like the input; it's going to be at the same frequency, but it's going to be at some different phase angle.

So, in order to do this, there's a fair amount of hard trigonometry we have to do. There's going to be a lot of cosines and sines and angles and things inside this, so that's pretty challenging analysis. Now, what we do with Oilers formula is we turn it into exponentials, and we already know how to solve exponentials.

So, we take the same circuit—it has the same stuff inside it as resistors and capacitors, the same exact circuit. Let's draw the same exact circuit there—that's identical—and it has the same output port right here. This time, what we're going to do is we're going to basically take this cosine, and we're going to cast this into an exponential. The way we do that is we use that formula; we use Oilers formula.

We basically create two sources—two separate sources—and they're exponential sources. It's going to be A over 2 e to the J Omega T; that's this source here. This source here is A over 2 e to the minus J Omega T; that's this waveform. If I add those together like that, remember the previous equation we just looked at? Oilers formula says that cosine equals that. The voltage here is exactly the same, and all we've done is describe the same exact cosine waveform as these two imaginary exponentials.

Now, I can't actually on my workbench build one of these things; these don't exist in real life. But they can exist mathematically; they can exist in my head. And I know that if I add these two voltages together, I do get a cosine. So, in our heads and on paper, we can actually drive circuits with these things. We can't actually build it, but on paper, we can do it.

Now that I have two sources, I can use the principle of superposition. This is another use of the very powerful idea of superposition. So, using the idea of superposition, I'm going to apply each of these two inputs one at a time and then add the results together.

So, over here, I'm going to get two outputs. I'm going to get a V out1, let's call it V out plus, which is what happens when I put in this plus source and suppress this one, which means I short it out. I'm going to get a V out plus. How do you solve a differential equation when you have an exponential going in? Well, we know this; it's going to be an exponential answer. It's going to be some constant times e to the J Omega T plus some constant, some angle.

Then I'm going to solve it again. I'm going to add to that V out minus, and I do that by suppressing this input and turning this one back on using superposition. V out for the plus—or, sorry, V out for this source here—is going to equal some other K, let's call that K plus plus K minus e to the minus J Omega T.

So, I'm going to put in two exponentials. I'm sure I'm going to get out two exponentials. Now, using Oilers formula, we know how to combine these. We can use that same expression, and we can recover our cosine. We can recover the real signal, and this will have some magnitude B.

We don't know the magnitudes yet, but we know what the shape of the waveform is. If we look at this, this is the same thing as this here. We did it by decomposing our cosine into exponentials, putting each exponential through this, and then recombining to get cosine. We do all those steps because solving differential equations with exponentials is the easiest way to do it.

So, just so you know, it's not twice as much work. What's going to happen is it's going to turn out that this whole solution down here, using the negative exponential, the answer is going to be exactly the same as the positive exponential except for there's just this conjugate—this complex conjugate—in here.

So, for real signals going in, the answer goes through, and it's always—these answers are always complex conjugates of each other as long as we start with a real input.

Okay, so that was a review of Oilers equation and a little preview, a little sneak preview of how we're going to do AC analysis using this really powerful tool.