The feedback theorem derivation step there are actually six steps that we need to do, and those six steps are going to be first of all finding or defining transfer functions when only input signal is present, that is going to define the system closed loop transfer function G. Then we are going to find the transfer function that is only the injection is pleasant, and that's going to give us the loop gain T. We already know that, but let's do it in the context of the feedback theorem. Then the next step is very important. We are going to apply superposition and express all output quantities in terms of the contributions that we find in steps one and two. So those are the three steps that I think we will be able to do with just the basic circuit analysis knowledge. The steps that follow are a little bit unusual, they rely on what is called null double injection. I'll stop with that and try to explain it carefully. You'll we'll see if actually have three such steps are first nulling u_y and finding expression for G_infinity, then nulling u_x and finding an expression for G_naught, and then finally nulling the output u_naught and finding expression for T_n. So let's go through these steps. All right. So first transfer functions when only input u_i is present. In the block diagram, the injection signal is set to zero, it's not null is just equal to zero. It's an independent source, we just set it to zero. In the particular derivation step, we have the current injection used as an example. You could do exactly the same steps with voltage injection. Let's see where the i_z signal is actually injected. It's injected in parallel with a current source that represents the propagation of the error signal through this block whatever it may be, and the output port of that block looks like a current source, like a control current source. That's an ideal injection point for the current injection signal because i_y signal is completely dependent on the error signal here. So this i_y is going to be directly proportional to the error, it's not going to be directly affected by the presence or not of IC. So that meets the condition of the ideal injection point. Set-up for the transfer functions when the input u_i is present, is that, i_z is simply equal to zero, there is no injection. Output over input under the condition that i_z is equal to zero is what? That's G itself. So this here is the transfer function we are looking for. This is the definition of the transfer function we are looking for. What can we say about the other two outputs and we have i_x and i_y here? What do you see for i_x and i_y? Well, i_x is equal to negative i_y. We can just read that from the diagram, and we can also say that i_x is some function of the input. So we can say i_x is equal to G_a times the input signal, where this G_a is, we don't know what it is. So here's the summary of what we have so far. So we have i_x is equal to negative i_y is equal to some G_a function where we don't know what G_a is right now. We have u-output under the condition that the injection is set to zero is equal to G times u-input, very simple. Next, we want to do the step number two is to find the transfer functions when only the injection u_z is present. We set u_i to zero, there is no input present. If it is a voltage source which are short, if it is current source it's an open, but we do have i_z present. The injection signal is in fact present. We can say that i_y over i_x under the condition that u_i is equal to zero is equal to what? As the loop gain, that's T, as is by definition T. T is i_y over i_x under the condition that the input signal is set to zero. What can we say about the i_x and i_y signals with respect to i_z? Well, do we have that I_x plus i_y is equal to i_z, I_y is equal to T times i_x. You can have I_x 1 plus t is equal to i_z. You can have that i_x is equal to i_z one over one plus T. Then of course i_y can be expressed in terms of i_z in a similar manner. To summarize this particular step in here is what we have. So we have i_y is equal to t times i_x. On the other hand, we can also express the output voltage as some other function, we don't know what it is right now, G_b times i_x. So i_naught is equal to G_b times i_x, or in terms of the independent input the injection signal that's going to be G_b over one plus T times i_z. So this is basically just introducing yet another unknown transfer function G_b into play. Next step, is going to be superposition, very simple. So we have two independent inputs, we have the independent inputs is u_i and i_z, and with respect to each one of them independently, we have expressions for the outputs i_x, i_y, and u_naught. Now, it is a linear time invariant system, so we can certainly write these contributions together and say that in the presence of both independent signals at the same time, we have these general expressions for i_x, I_y, and u_naught given as superposition of the results we had in steps one and two. Now of course this actually does look a little bit confusing, because we are setting up definitions for these various transfer functions, but in the process we had to introduce these two quantities G_a and G_b, and those G_a and G_b represents transfer functions in the systems that are generally not of interest. They are auxiliary transfer functions in the process of deriving the feedback theorem, but ultimately, rarely would care exactly what G_a or G_b really are. So the remaining steps have the main purpose of eliminating G_a and G_b and putting everything in terms of these quantities that we do care about. Those are the remaining steps of the feedback theorem derivation, and those are going to be done through what is called null double injection.
