Let's look at Z_D,old. Notice that this is now open. My extra element here is gone. This now means this is zero, it's not null to zero, it is actually zero. That means this current is going to be equal to zero, and you can say Z_D,old, hopefully, by inspection, is equal to what? It's just as L_1 plus L_2. This is good news. It's pretty easy. You just get it immediately. Very nice. Takes one second to do that. You don't even need to draw this. Let's do Z_D,z; i_test is equal to 0 and we are finding a transfer function in this circuit right here by injecting the test source at a point where the extra element impedance was connected, and finding the response voltage to that test source. So Z_D,z is going to be equal to v_a over i_a, and you tell me by inspection what that is going to be. There are two elements now in series right here. Like this, like that. One component to that is just L_2. What do we see when we look at this side of the transformer? We see L_1 scaled by the square of the turns ratio of the transformer, so this is going to be D over D prime square times L_1, and so the final result is going to be s times L_2 plus D over D prime square times L_1. Z_D,z, basically by inspection very quick. Z_N,z is going to be a little bit tricky. Unfortunately, this is not by inspection, but let's go for it. We're going to have now two independent sources. I'm going to place here the test source that we are going to call i_a. We are looking for a voltage, v_a, and we keep i_test present at the same time. So there are now these two independent sources acting together so that Z_N,z is equal to v_a over i_a under the condition that what is nult? V_test, exactly. So V_test is nult, that's Z_N,z. Now, that means this V_testing is nult, which really gives us a couple of different things we can do. i_a is the current that is going to go through here. This is going to be current, i_a. So i_a is the current that is going into the dot of the secondary side of the transformer. What is coming out of the dot of that secondary side of the transformer is going to be D over D prime times i_a. The current that is flowing through the inductor, L_1, is going to be that current minus i_test. So here we have D over D prime times i_a minus i_test, which will give us the voltage drop across L_1 equal to SL_1 times that current. Now, i_a current also comes in right here. This is i_a. If you look at this node right here, you'll see that that node, the i_test and i_a combined together. We have the current i_test plus i_a going through L_2, which means the voltage drop across L_2 is going to be equal to SL_2 times the sum of these two currents. Yet again, when you are working with Z_N in particular, do as much as you can on the diagram itself before writing equations. Yet another example of doing that. But at this point here we've exhausted what we can do without writing loop or nodal equations that are not trivial. What do we do? Eliminate i_test, and you can do that by exploiting this nulling condition right here. This voltage we've actually written down, that's equal to SL_1, D over D prime i_a minus i_test hat, and this voltage right here, we've just found that, that's going to be equal to SL_2 i_test hat plus i_a. Eliminate i_test from here, and we get an expression for i_test. Then the final thing is to find what v_a is. For example, we can go through the slope right here, go down here, down here, and back here, and for that loop, we have that v_a, since we have solved for i_test right here, we plug it into here, and we do need to do a couple of lines of [inaudible] for it. The end result for Z_N,z is going to look like this. The expression actually looks really simple, but it does take a few steps to get there. Now, we can actually put things together. You did now backwards. When we go to our second layer of Extra Element Theorem, put everything into Z_D. So Z_D is Z_D,old 1 plus Z_N,z over Z_z, 1 plus Z_D,z over Z_z. This is our extra element in the second layer of that Extra Element Theorem application, and these are the expressions we have for the old impedance, N impedance and the D impedance. Let's just write down, see how that goes. Z_D is going to be in front. We have s L_1 plus L_2, that's the old. I would really encourage you to do this, sit on the side and try to solve for Z_D by hand. It's not that bad. I'm not trying to torture you. But compared that work to what we have done right here in terms of applying the Extra Element Theorem, and you'll find that it's much less easy to make mistakes when you do Extra Element Theorem, and you'll also get results in the form that are really preferred. The end result of writing this down is basically from these expressions directly, and you'll see how it works. You'll see that is going to be some inductance interacting with this capacitance in the numerator, and some other inductance interacting with the capacitance in the denominator. Plus in front, it looks inductive overall. So we have as L_1 plus L_2. It is very revealing to actually do this through this second layer of Extra Element Theorem.