To the feedback theorem, we are in the middle of the derivation steps, so we did steps 1, 2, and 3 and we're going to do now double injections steps today. Step three was the superposition of hearing these two independent sources present at the same time, and contributing together to the three dependent outputs currents ix and iy and the output signal u naught. Those are the signals ux and ui right here and the output signal u naught. Those are three dependent outputs in feedback theorem. All right. So let's see how the null double injection that I tried to introduce last time actually works in the derivation steps. The first of the null double injection steps is to null uy, we have ui and iz present, together null iy signal. So you have two signals present at the same time, jointly together they act to null the iy signal and under that condition, we want to find out G infinity has the ratio of output over input under the condition that iy is nulled. So now, let's see what nullying iy actually implies. So having this signal here equal to zero, nulled to zero, not shorted to zero not open, it's nulled to zero then implies that the error signal must be nulled to zero as well because we know by the injection point discussion that the ideal injection point is such that iy signal is directly proportional to the error. So if yy signal is nulled to zero the error is nulled to zero as well. So if we null the error, what is the response that we have in this system, what is output over input when the error is nulled? That's the ideal response of the feedback system. So this is the ideal closed loop response. Now on the right-hand side you have several steps into what that G infinity in fact comes out to be. The starting point is this expression that you obtain from the superposition step. This is right here, so iy is negative GA, ui plus t over one plus t times iz. You put that right here, that's your starting point. Then you say these two signals together are in combination, are set up so that we null iy. So that means from this equation right here, we have the next one which says iy is set to zero, it's nulled to zero by the combined action of these two signals. So now we are going to have this particular value of iz that together with ui is going to be responsible for setting the left-hand side of this equation to zero and that is the term that we can then eliminate and express G infinity. So how does that go? Well, the u not, again from the superposition, this is this term right here, is repeated right here so u naught is Gui plus Gb over plus t times iz. But now we take this iz from this point right here and we plug it into right here. Why do we do that? Because we want to eliminate that the iz and have everything in terms of ui, we are interested in u naught over ui. That's why we use this step right here to eliminate iz and that particular iz that's used to null iy. So plugging iz from this equation to the third equation, you get the expression that is shown right here, and then you see that on the left-hand side we have u naught under the condition that iy is nulled to zero. On the right-hand side, we have an expression that multiplies ui. and that's exactly what we were looking for. So g infinity can be expressed in terms of g plus Ga, Gb over t. That G infinity expression is still by itself not particularly useful because it still contains these unknown quantities Ga and Gb. So we need further steps to be able to eliminate those unwanted quantities. The next step is going to be the null double injection to null ux. Nulling ux, that's right here, is done by having combined presence of ui and iz which together now ix. Then we know that by the definition of the ideal injection point, there is not going to be any transmission from the error signal to the output. The only way the output signal can be null zero in this case here is in fact this direct transmission through the feedback path to the output. That's why this g naught which is equal to u out over ui, under the condition that ux is nulled to zero is called direct transmission through the feedback path. On the right-hand side, let's see how that goes. So what is the starting point? You now take the expression for ix from the superposition of ui and iz, then the next step is to say all right the left hand side is going to be nulled and that gives you the capability of finding out what iz is necessary in order to null ix. So from this expression right here, we can solve for iz that is needed to null ix. What do we do with that iz? We plug it into the expression for u naught. Where is that expression coming from? That's another one of these superposition terms that we had found in earlier steps. So again, you take this value here sulfur it, plug it in the next expression right here, and finally we get u naught as a function of u i and you can see that the g naught, u naught over ui when ix is nullled to zero is going to be equal to g minus Ga, Gb. So this is the result that we were looking for, that's now the expression for g naught. Yet again, that by itself is not the end result because we still have this Ga, Gb terms inside that we need to eliminate in the end. The final nulling is going to be the case where we have ui and iz together to null the output itself and that gives us an opportunity to compute a particular version of the loop gain that is called null loop gain. So tn is going to be defined as iy over ix under the condition that output signal is nulled to zero. Yet again on the right-hand side, you start with the expression for u naught, then you say right there is a particular value of iz that is going to null that u naught, so we can say zero on the left-hand side and that particular value of iz on the right-hand side. Then, we have two other terms for ix and iy, where are these two expressions coming from? Again, from superposition. We're going to take this value of iz from here, solve for it, plug it into here and here and we are going to have ix as a function of ui and iy as a function of ui. So these two are going to become ix is something times ui and iy is something times ui and the last step is going to be to eliminate ui and express iy over ix, and that's going to be our tn. We obtain this last result right here that is going to give us the expression for tn. Yet again, notice that the tn does still include the Ga and Gb that finally we need to get rid of. From step four, we have an expression for G infinity, from step five we have an expression for G naught, and from step step six we have an expression for tn. That gives us an opportunity to now eliminate Ga and Gb in actually couple of different ways. Well first of all, the first one can be written in the form of G infinity T is equal to T times G plus Ga, Gb. You recognize this term here shows up right here. So you can say all right this Tn is going to be equal to G infinity times T. The second equation gives us G naught in the form of G minus Ga, Bb and you see that actually can be plugged in right here. So this is equal to G naught and see from that we have directly, that Tn is equal to G infinity times T over G naught and that is very commonly written in the form of Tn over T is equal to G infinity over G naught and that represents what is called the reciprocity relationship for this feedback system. Notice that that no longer involves these unknowns of Ga and Gb, we have successfully eliminated those entirely from the reciprocity relationship. The main result of course is the G itself. Yet again, we have these three relationships that are coming from null double injection and we can combine them to eliminate Ga, Gb and come up with expressions for the overall transfer function of the system. This is the actual transfer function of the complete closed loop system. So we can put it together in terms of these four quantities that we have derived in two different ways. If you combine steps 4 and 6, you get the first expression right here. If you combine steps 4 and 5, you get the second expression right here. So there is a little bit of algebra to do that, but it's very easy. The point to note here is really simple, take the Ga, Gb and eliminate that from these two equation right here and we get what we consider really the main result of the feedback theorem that gives us G in terms of G infinity, T, and G naught. Alternatively, you can express G in terms of G infinity, Tn and T. Those are two expressions that are entirely equivalent and they are equivalent because we have the reciprocity relationship between these four quantities. So the derivation itself is not so important by itself, but the process of deriving and being able to apply the null double injection is important because that's what we will be doing in the applications of the feedback theorem.