KVL in the frequency domain

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 going to basically show that Kirchhoff's voltage law works in the frequency domain.

What I have here is a circuit that has some voltage source, an AC voltage source. Let's put AC on it like that. It has three impedances connected. Inside each of these boxes is an R and L or a C, and we're not going to show which because we're going to carry these along just as general impedances.

In AC analysis, the voltages are all cosine waves. So ( V_{in} = V_m \cos(\Omega t + \phi_0) ), where ( V_m ) is some voltage amplitude and ( \phi_0 ) is some starting phase shift. This is our input signal. Now let me label the voltages on everything else. We'll call this ( V_1 ) and I'm going to label it this way here. This will be ( V_2 ) and I'll put it upside down like that: plus and this will be ( V_3 ) minus plus ( V_3 ).

Now that I have ( V_1 ) here, let's change the name of this to ( V_0 ) just so I don't get ( I ) and ( 1 ) mixed up. So we'll change that to ( V_0 ). The input source is ( V_0 ), 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 going to see how that applies to AC circuits here.

In the time domain, we say that ( V_0 + V_1 + V_2 + V_3 = 0 ). So let's talk about how this is going to turn out. Well, what do we know right now? Well, we know that ( V_0 ) is a cosine wave at some phase angle.

Now, what do we know about the other voltages? In this AC analysis, what we're doing is we're looking for a forced response. We've let the natural responses die out; there's no switch in this circuit, and we just assumed the circuit has been in this state forever. 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 know that all these voltages are going to resemble the input voltage. They're all going to be sinusoids. All the voltages here are going to be AC sinusoids because the forcing function is a sinusoid. The other thing we know is they're all going to have the same ( \Omega ); the frequency of this voltage and this voltage and this voltage is going to be identical to ( \Omega ).

Here, I'm going to 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—if we've done all the analysis—don't create new frequencies. They're all ( \Omega ).

Now, some other things we know: there's going to be phase shifts involved here. Remember when we do impedance? We are multiplying it by ( j ) and rotating things by 90 degrees. So we're going to have different ( \phi ) for each one. The other thing we're going to have is we're going to have different amplitudes. The amplitude of ( V_1 ) could be different than the amplitude of ( V_2 ).

So this is what an AC solution is going to look like. Let's move on a little further here. What I'm going to do now is we're going to take this input voltage plus these things that we know here, and we're going to see how Kirchhoff's voltage law works in the frequency domain when we work with these transformed Z's, these impedances. Okay, let's go ahead and do that.

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_0 \cos(\Omega t) + V_1 \cos(\Omega t + \phi_1) + V_2 \cos(\Omega t + \phi_2) + V_3 \cos(\Omega t + \phi_3) = 0 ).

Now, all these ( \Omega )s are the same exact number, the same radian frequency. All the ( \phi )s are different, and all the ( V_2 )s and ( V_3 )s are different.

Okay, now I'm going to switch to complex exponential notation. We're just changing notation here. We could represent this number as the real part of ( V_0 e^{j(\Omega t + \phi_0)} ). That's exactly the same as this cosine can be represented as the real part of a complex exponential with this frequency.

I can write out the rest of these: ( V_1 e^{j(\Omega t + \phi_1)} + V_2 e^{j(\Omega t + \phi_2)} + V_3 e^{j(\Omega t + \phi_3)} = 0 ).

One thing I can do next is we can start to factor this. We can start to take this apart a little bit. I know that if I have the expression ( e^{j(\Omega t + \phi)} ), just in general, I can change that by exponent properties to ( e^{j \phi} e^{j \Omega t} ).

So I'm going to do this transformation on all four of these terms here. Let's keep going. So we're still working on this. Let’s go the real part now. I'm going to take apart ( V_0 ) here, and I get ( V_0 e^{j\phi_0} e^{j\Omega t} + V_1 e^{j\phi_1} e^{j\Omega t} + V_2 e^{j\phi_2} e^{j\Omega t} + V_3 e^{j\phi_3} e^{j\Omega t} = 0 ).

Here's a nice simplification. We take out this common term; we factor out this common term across the entire equation. What do we come up with? We come up with the result: ( (V_0 e^{j\phi_0} + V_1 e^{j\phi_1} + V_2 e^{j\phi_2} + V_3 e^{j\phi_3}) e^{j\Omega t} = 0 ).

We're getting close! How do we make this equation zero? Does ( e^{j\Omega t} ) ever become zero? Well, ( e^{j\Omega t} ) is a rotating vector. It's never zero, so that's not going to do it.

That means that this other term here has to be equal to zero. So how am I going to do that? I'm going to make one more notational change. This kind of number here is ( e^{j\phi} ) is called a phasor; it's some amplitude times ( e^{j\text{angle}} ) and there's no time up here. There’s no time; the time is only over here. This is the only place the time appears in the equation, and this is the only place that ( \Omega ) appears in the equation, and these are just phase angles, starting phase angles.

So my notation for a phasor is going to be: this is going to be called ( \overline{V_0} ), and I'm going to put a line over it to indicate that it's a complex vector, and that equals ( V_0 e^{j\phi_0} ). So when you see the vector symbol and ( \overline{V} ), that's that right there.

Now we can write finally that ( \overline{V_0} + \overline{V_1} + \overline{V_2} + \overline{V_3} = 0 ). So this is KVL in the frequency domain. Fortunately, 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 that is equal to zero.