The derivation follows really the same steps that we had in the feedback theorem. In fact, the derivation is perhaps a little bit simpler. So, let's go through it. So, we're going to look at the general representation of a linear circuit, with a general input u(s), general output y(s) and the port where the impedance may or may not be connected. And we are going to put the test source in the form of a current source, to test for these impedances are looking into the port. So, when you look at the system defined in this manner here, we have two inputs and two outputs. What are the two inputs? The two inputs are u(s) and i(s). And the two outputs are y(s), the output of the system, and the voltage across that source i(s). So, we are going to treat this system as a linear system. So, superposition is going to apply. First, you write down transfer functions when only input u(s) is present and input i(s) is not present. In that case, if only u(s) is present what is y over u when i is equal to zero? Well, that is what we call G_old with Z open. So, that's really giving immediately a definition for G_old when Z is open. Then, if you look at the transfer functions when only input i(s) is present, and u(s) is set to zero at port where the impedance is to be connected, the response v over i with u set to 0, is nothing else but the definition of the impedance Z_D seen at the port. Then we take these two and apply superposition. So, we're going to have transfer functions of these two outputs with respect to two inputs at the same time. And then, we want to eliminate, out of those transfer functions, non-essential transfer functions that we are not interested in. And we do that elimination by using Null double injection in the last step. In that last step, we narrow the output. You now find an expression for Z_N and use that expression to eliminate the non-essential transfer functions. That's the outline of the derivation. So, we have two outputs, so we can say y=G_old times u plus, now we're going to give it another transfer function, G_i times i. This, we don't know what it is. It's a transfer function from i(s) to y(s). And the second expression is that v is equal to something times u, and that something we are just going to give it the name, G_v times u plus something times i. That something being Z_D. Two expressions, these are complete by superposition transfer functions from two inputs to two outputs. Now, when Z is present- this is important thing that's going to link things. When Z is present, the relationship between v and i is not arbitrary. So, with Z present at that port, looking at just the Z itself, we have v = -Zi. So, let's use that because we know, we do want to have the final expression with Z present. Let's use that to eliminate i from these two expressions. So, v is going to be equal to negative Z times i, and that's equal to G_v u plus Z_D i. You say minus G_v u is equal to Z plus Z_D times i. And then we can solve for i, that's equal to minus G_v over Z plus Z_D times u. Plug that back into here and we have y is equal to G_old minus G_i G_v over Z plus Z_D times u. This is with Z present. What is this? This is y is equal to something times u. Output is equal to something times input when Z is actually present. What does this mean? Or is this of interest or not? >>Transfer function of your system. >>That's the transfer function we are looking for, right? This is exactly what we're looking for in the end. This is the y over u with Z present. This here is G. This is what we are after, right? We want this. Okay. So, here's an expression of that G in terms of G_old. That's great. We want to keep it G_old inside. We kept Z and Z_D inside. But we also have this G_i G_v thing that we don't care about and we don't know what they are. So, the next step is going to be the Null double injection that is going to eliminate G_i G_v. So, let's do the Null double injection. So, we now place both u and i into the linear circuit and we go back to this expression for y. We say, right, y by superposition is equal to G_old times u plus G_i times i and by null double injection, now we are going to set this to zero. Not to zero but now to zero. So, by simultaneous action of the u and i sources together, we null the output. Remember again, nulling does not mean shorting or opening or anything like that. You're not changing the circuit. You're simply nulling the output. Now, our second equation was v = G_v u + Z_D i. Now, from the first one, you're going to eliminate u. So, we find out from here that u is going to be equal to negative G_i over G_old times i. And we plug that u into here and we get that v is going to be equal to negative G_i G_v over G_old plus Z_D and all that times i. And this is under the condition that y has been nulled. Now, what is this? That's Z_N. By definition, that's Z_N. Z_N is v over i when the output is nulled. And so this is really that expression that we have down here. So, now we have these two. Put them together. You have G is equal to G_old(s) minus G_v(s) G_i(s) over Z(s) plus Z_D(s). You have that Z_N(s) is equal to Z_D(s) minus G_v(s) G_i(s) over G_old(s). And you can see now from these two, we can eliminate that product that we don't like from here and here and the end result is going to be the one version of the Extra Element Theorem. So, let's do that. So, from here we have G_v G_i. That's going to be equal to G_old times Z_N minus Z_D and with minus sign. So, plug that into G. G is equal to G_old minus. That's going to be plus G_old. How did I get the minus sign wrong here? Z_N minus Z_D over Z plus Z_D. That's equal to G_old one plus Z_N minus Z_D over Z plus Z_D. That's G_old Z plus Z_D plus Z_N minus Z_D over Z plus Z_D. This goes away and you have G is equal to G_old times the correction factor in the form of one plus Z_N over Z over one plus Z_D over Z. And that's the end of the derivation of the Extra Element Theorem for the case where we consider the Extra Element Theorem as being an open circuit in the old transfer function. You take advantage of the Null double injection to eliminate quantities that are not of interest and express everything in terms of these quantities that do have intuitive interpretation, that are useful in the interpretation or design process. So, summary of the elements is right here. So, yet again, we have two versions of that. Depending on whether G_old is considered with the Z open or G_old is considered with Z short. And those two expressions both have the same format. The G_old at front, with a correction factor that follow. The correction factor have two different forms, depending on whether G_old considers Z open or G_old considers Z short. And then finally on the side, there is a reciprocity relationship as well. So, notice that we have in the derivation, derived only the version when Z is considered open. The derivation steps for the other version are exactly the same.