So in a ZD set up, that's the classical driving point impedance seen at the port where the extra element was connected. So we have I test source connected at the point where the extra element, the capacitor C1 was connected. And we want to find out ZD is equal to V test hat over I test hat. What do we do with other independent sources when we are looking for the D impedance? They are all set to 0, so D hat is equal to 0, no other independent sources. What do we know about ZD? As shown right here, ZD is a transfer function of which system? Of the quote mark old system, right? So by that we also know that ZD will have to have denominator that matches the denominator of Gvd-bb, the old transfer function. Right, but we don't know anything beyond that. We do need to find what exactly ZD actually is. We can certainly go ahead and start putting equations together. That's perfectly valid approach, in fact, it's not that difficult, really. And I'd encourage you to, in fact, do that. Do a little bit of a side exercise and see where you can derive the same result we are going to derive in class. But it will take this example to do this in a little bit different manner. What do you think would be an alternative possibility to solving for ZD? So ZD is a transfer function of a system. That system happens to be our old system, but it is a system, you're just finding the transfer function of a system. What technique could we apply to finding ZD? What technique are we talking about? It's too much of a hint. >> The extra element theorem. >> There we go, so we're going to go actually apply extra element theorem to finding ZD. So we have ZD as part of the extra element theorem calculation. Then we say, right, and I see this is a difficult circuit. I'm just going to resort to what I know how to do well and that's going to to be extra element theorem. So I'm going to apply extra element theorem to solve ZD, how about that? Okay, an extra exercise. An example within an example. So can we apply extra element theorem to find ZD? The answer in general, yes, whenever you're looking for any transfer function you like, you can apply extra element theorem. ZD happens to be transfer function of the old system. We can apply extra element theorem yet again to solving for ZD. So to do that, we need to decide which element are we going to consider an extra element, it's completely arbitrary, right? You can choose whatever you like. In fact, there may be better choices than what I'm going to do. But I'm going to make a suggestion here. If I could eliminate this stuff right here, let's say, suppose we consider this parallel R and C2 as an extra element, as an open extra element. It looks like the circuit becomes much, much simpler. So we are going to say, let's call this ZZ, we'll be careful about notations. So you don't get confused between these two levels of extra limit theorem application. So we're going to say ZZ is equal to the parallel combination of R11 over SC2. And we're going to consider this extra element as an open. And then we're going to write the transfer function ZD as ZD old times 1 + ZN, I'm going to call this ZNZ, right just to not confuse these ZN's and ZD's, over ZZ and 1 + ZDZ over ZZ. Okay, just transfer function, just apply the extra element theorem, find the transfer function of a system in terms of the ZD old, this is going to be with ZZ open. And then the correction factor in the usual form for the open circuited extra element. You don't have to do this, right? I'm doing this just as another exercise in extra element theorem calculations. And it's actually not that bad way to do it, I think it's actually easier than solving the circuit brute force wise. But I would still encourage you to do brute force type calculations and see where you can work through that and get the same result. Now we need to find a couple of things, we need to find this. We need to find this guy right here and ZDZ, so we're having to find three things. You will see in all of these extra element theorems we have this approach of looking at a complicated problem and splitting this into multiple, hopefully simpler, problems. So instead of solving for ZD all at once, we solve for these three different impedances in three steps that are hopefully simpler and less prone to error.