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- [Voiceover] As we do AC analysis, and we do operations in the frequency domain, we need to bring along Kirchhoff's Laws so that we can make sense of circuits. So, in this video, I'm gonna basically show that Kirchhoff's Voltage Law works in the frequency domain. And what I have here is a circuit that has some voltage source, an AC voltage source, let's put AC on it like that, and it has three impedances connected. Inside each of these boxes is an R, an L, or a C, and we're not gonna show which because we're gonna carry these along just as general impedances. So, in AC analysis, the voltages are all cosine waves. So, v-in equals some voltage amplitude times cos of omega-t plus we'll call it phi-zero, some starting phase shift. So, this is our input signal. Now let me label the voltages on everything else. We'll call this v-one, and I'm gonna label it this way, here, and this will be v-two, and I'll put it upside-down like that, the plus, and this will be v-three, minus, plus, v-three. And, oh, now that I have v-one here, let's change the name of this to v-zero just so I don't get i and one mixed up. So, we'll change that to zero. So, the input source is v-zero, and that's that voltage. Now, when we apply KVL to this, Kirchhoff's Voltage Law, what it says is if we start in a corner, if we start somewhere in the circuit, let's start right here, and go around the loop, it should add up to zero volts. That's KVL for normal DC circuits, and we're gonna see how that applies to AC circuits, here. So in time domain, we say that v-zero plus v-one plus v-two plus v-three equals zero. So, let's talk about how this is gonna turn out. Well, what do we know right now? Well, we know that v-zero is a cosine wave at some phase angle. Now, what do we know about the other voltages in this? In AC analysis, what we're doing is we're looking for a forced response. So, we've let the natural responses to die out. There's no switch in this circuit, and we just assume this circuit has been in this state forever. So, the natural response, the natural response has died out, and that means we're looking for the forced response. So, what we know is, we have three voltages. We have three voltages. We know that all these voltages are gonna resemble the input voltage. So, they're all gonna be sinusoids. All the voltages here are gonna be AC sinusoids because the forcing function is a sinusoid. The other thing we know, they're gonna all have the same omega. The frequency of this voltage, and this voltage, and this voltage is gonna be identical to omega, here. I'm gonna put a big bang there. That's really important. In an AC circuit, when you're driving it from a frequency, every other frequency in the system is the same frequency. This is a linear system and linear components. All the analysis we've done, linear components don't create new frequencies. They're all omega. Now, some other things we know. There's gonna be phase shifts involved here. Remember when we do impedance, we are multiplying by j and rotating things by 90 degrees. So, we're gonna have different, different phi, for each one. And, the other thing we're gonna have, is we're gonna have different... The amplitude of our sinusoids are gonna be different. The amplitude of v-one could be different than the amplitude of v-two. So, this is what an AC solution is going to look like. Let's move on a little farther here. What I'm gonna do now is we're gonna take this input voltage plus these things that we know, here, and we're gonna see how Kirchhoff's Voltage Law works in the frequency domain, when we worked with these transformed z's, these impedances. Okay, let's go ahead and do that. Okay, let's do a little more in the time domain. And we'll write out our KVL equation again. So, the KVL equation was v-naught, cosine omega-t plus phi-zero plus v-one, that's the amplitude of v-one, cosine omega-t plus some different phase angle. We don't know what that is yet. Plus, v-two, amplitude of v-two, cosine omega-t plus phi-three plus v-three, the amplitude of v-three, cosine omega-t plus phi-three all equals zero. And omega, all these omegas, are the same exact number, the same radian frequency. All the phi's are different, and all the v-twos and v-threes are different. Okay, now I'm gonna switch to complex exponential notation. We just changing notation here. We can represent this number here as this is the real part of v-naught, v-zero, e to the j times omega-t plus phi-zero. That's exactly the same as this. This cosine can be represented as the real part of a complex exponential with this frequency. And, I can write out the rest of these. V-one, e to the j omega-t plus phi-one plus the real part of v-two, e to the j-omega-t, plus phi-two, parentheses, plus v-three, oops, real part of v-three, e to the j-omega-t plus phi-three equals zero. All right, now, one thing I can do next, we can start to factor this. We can start to take this apart a little bit. So, I know that if I have the expression, if I have the expression, e to the j-omega-t plus phi, just in general, I can change that just by exponent properties to e to the j-phi, let's get the parentheses in there, like that, e to the j-phi times e to the j-omega-t. So, I'm gonna do this transformation on all four of these terms here. Let's keep going. So, we're still working on this. Let's go real part... Now I'm gonna take apart omega-t and phi-zero, here, and I get v-naught e to the j-phi-zero, e to the j-omega-t, plus real part, v-one, e to the j-phi-one, e to the j-omega-t, plus real part, v-two, e to the j-phi-two, times e to the j-omega-t, plus real part v-three, e to the j-phi-three, e to the j-omega-t all equals zero. And here's a nice simplification, we take out this common term. We're gonna factor out this common term across the entire equation. And what do we come up with? We come up with... The result is the real part of v-zero, e to the j-phi-zero, plus v-one e to the j-phi-one plus v-two, e to the j-phi-two. See the pattern. All that times e to the j-omega-t. And we close that and that equals zero. Now, we're getting close. We're getting close. All right, how do we make this equation zero? Does e to the j-omega-t ever become zero? Well, e to the j-omega-t e to the j-omega-t is a rotating vector. It's never zero. So, that's not gonna do it. So, how do we get it? Well, that means that this other term, here, has to be equal to zero. So, how am I gonna do that? I'm gonna make one more notational change. This kind of a number, here, is called a phasor. It's some amplitude times e to a complex one angle, and there's no time up here. There's no time. The time is only over here. This is the only place that time appears in the equation, and this is the only place that omega appears in the equation, and these are just phase angles, these are starting phase angles. So, my notation for a phasor is gonna be... This gonna be called... I'm gonna call it v-zero and I'm gonna put a line over it to indicate that it's a complex vector, and that equals v-naught, e to the j-phi-naught. So, when you see the vector symbol and the aught, that's that right there. And we can write now, finally, the real part of V-naught plus V-one phasor plus V-two phasor plus V-three phasor equals zero. So, this is KVL in the frequency domain. And fortunately, it looks like... It looks exactly like KVL that we remember from our DC analysis. The sum of the voltages, going around the loop, is equal to zero, and, in this case, it's the sum of the phasors going around the loop is equal to zero. Let's try to give a graphical interpretation to this. Here's our real and imaginary plane, the complex plane, and what it says is that these phasors... So, let's say that v-naught, let's say that v-naught look like that. That was our voltage source, okay. This represents a vector spinning around at the frequency, omega, and it's offset phase is this angle right here. So, this is phi-zero. Each of these other components is gonna have an AC voltage, a sinusoidal voltage on it, with some phase and some magnitude, and what KVL tells us, it puts a constraint on what those voltages can be. So, we have three impedances here. I don't know what they are because we didn't fill in the circuit, but there's gonna be some vector associated with each one of those. Let's say that's vector-one, and let's say that this is vector-two, and what it says is that vector-three, by the time we get done, vector-three has to sum back to zero. So, this kind of a constraint where voltage is going around in a circle, have to come back to zero, that's KVL and the frequency domain. That's what that means. So, we've shown that KVL works in the frequency domain. I could do a similar analysis and show that KCL, Kirchhoff's Current Law, also works in the frequency domain, and that means, fantastically, that all the tools that we developed for DC analysis of just resistor circuits, all those tools work for AC analysis as well. Thanks for listening.