 Welcome to this lecture. We are going to continue about norms, norms of signals we have already seen. We are going to see in more detail about systems which we are going to consider as operators. So, we are going to see in more detail about this. In particular, we are going to speak about finite gain L2 stable, a very important class of stable systems. This is what we will see in more detail today. We had just begun in the previous lecture. So, we saw that if this is a system, input-output system, there are various notions of stability and we are speaking of input-output stability in which use and wise live in some space of signals. We said a good space of signals is L2E. L2E is a rich space of signals. For example, there are e to the power plus t, e to the power any pi t, etc. These are all there inside L2E are they are all growing. You see these signals are all growing and they are not in L2 for that reason. Why? Simply because they are all growing and they are not going to be square integrable. But for each chopped version at whatever tau one chops, the chopped version is in L2. So, we recall that the definition of L2E was you take a function f and put it into L2E if f tau is in L2 for every chopped f tau. So, now one can also speak of L2E 0 to infinity to Rn. Here we are considering functions that are defined over the domain 0 to plus infinity not from minus infinity to plus infinity like in this, but from 0 to infinity like in this. So, we define a system h with input u output y to be finite gain L2 stable. So, h is called finite gain L2 stable. L2 plays a key role in the norm that we are taking if there exist there exist constants gamma and beta. So, this one is not y, but gamma and this one is beta such that such that what inequality satisfied y. So, notice this which one we are putting at various places is less than or equal to u. The inequality that I am writing in now is not yet correct. We will correct it very soon gamma, this one is gamma for every u in L2E. So, notice that if u is in L2E then its L2 norm need not be defined. Of course, L2 space of functions is contained inside L2E, but you take arbitrary u in L2E its L2 norm is not defined. L2 norm of which functions are defined after chopping by tau. So, notice that you have to chop by tau, chop at value tau then it comes into L2 space and then the L2 norm is defined for every tau greater than or equal to 0. So, you chop and then it is in L2, but then the right hand side you might chop at different different values of tau and if you chop very late then the right hand side might become large. If you chop very early the right hand side might become very small. So, notice that you should also be chopping at the same value here. So, here also y will be in L2 after this chop. So, now is when this equation has lot of meaning. So, how much energy comes? So, we are going to chop at tau and how much energy comes in the output? Why we are measuring the output energy in the output and energy so far this tau you chop by tau and then take the L2 norm means you look at the energy that has come out so far that energy cannot exceed the energy that went in times a constant. The energy that went in so far times a constant and perhaps an offset. This offset beta is not very important. You should understand this offset beta as suppose you do not give any energy inside, you do not give any energy input u at all. If 0 is the input then this particular term becomes 0 then the output can still be non-zero but it has to be bounded. For every tau whatever comes out is bounded by some number beta. So, the energy that comes out for every tau is bounded by some fixed constant beta. So, gamma and beta are some fixed constants that are not allowed to be changed depending on the particular u that you take from L2 e and the particular tau at which you decide to chop. The independent of which u and which tau the same gamma and beta has to work for this inequality to hold that is the significance of this particular result. So, we will see some consequences of this inequality. First consequence is when u is equal to 0, u is identically equal to 0. This should be understood as identically not just equal to 0 at some time instant but equal to 0 at every time instant. u, u of t equal to 0 for all. For all t, all our signals are defined only for t greater than or equal to 0 and hence u of t equal to 0 for all t greater than or equal to 0. Then what this says is y tau L2 less than or equal to beta for there is no u anyway for every tau. So, in particular you can make this tau very large for every tau for every positive tau you can make this tau tend to infinity and still it will be bounded. So, this implies that y itself is in L2. Why? Because for every tau the chopped version is bounded by some number beta independent beta itself is not allowed to depend on beta. Beta itself is not allowed to depend on the tau and hence this L2 norm of the chopped y is independent of the tau and hence y itself has to be in L2. This is not very hard to prove but what this says is that when you give 0 input then the energy in the output is finite. The energy in the output is finite not just when you take the energy so far until now but also when you integrate up to plus infinity. So, what does this say in terms of LTI system? That is what we will see in much detail very soon now. Also let us see some more consequences. So, beta should be understood as some offset in output energy due to non-zero initial condition for example. There might be an initial condition inside the system. This might cause the output to be non-zero even though the input is identically 0. So, the significance of beta you already understood when the input is 0 that time the output can still have some energy but the output energy cannot exceed beta for any tau and hence y itself will be in L2 for the special case that 0 is the input. Now, let us understand in more detail. Let us look at this inequality again of course plus beta. So, if you give some finite amount of energy into the system from the input then the output will also have some energy but the output energy is expected to be more if input more energy is given to the system from the input. So, for linear system also if you give more energy into the system of course output should be allowed to give output should be allowed to have more energy in its output in the output signal should be allowed to have more energy but the output energy cannot exceed some constant times the input energy. This kind of quantifies that all the energy that comes in the output has to have gone in through the input plus some constant which is perhaps because of non-zero initial conditions. This kind of also quantifies that the system H itself does not have a source of energy inside. Why this rules out a source of energy within? Because if there is a source of energy within even when you give 0 input the output can have lot of energy outside that is being ruled out by this fact that when you give 0, when you give input identically 0 then the output energy is bounded from above by a value beta. Of course what remained to say was that gamma and beta are greater than 0 but that easily follows. You see left hand side is some positive number L2 norm because it is a norm it is non-negative quantity and hence the right hand side is also non-negative. Both gamma and beta have to be positive numbers because this U itself is also positive. Beta is positive follows by taking the fact that U itself might be identically 0 and hence the left hand side is non-negative hence right hand side is also non-negative. So, this is how we should understand this finite gain L2 stable notion of stability. So, what is finite gain about it? This gamma is like a gain of the system. Now, the way we have defined it here we have called the system H finite gain L2 stable L2 played a role because of the particular norm L2 norm that we took here. Hence, U and Y were taken from L2E and gamma is a gain of the system. Is it okay to say the gain? Is it unique or gamma and beta unique? We already noted in the previous lecture that if one gamma and beta pair works anything gamma larger than if gamma 1, beta 1 works. When we say works, it should say works for all, for all U in L2E and for all choppings tau then gamma 2, beta 2 will also work. Also works as long as you take gamma 2 greater than or equal to gamma 1 and beta 2 greater than or equal to beta 1. This pair will also work. So, clearly this is not unique. One might try to say what is the minimum gamma for which this inequality holds. But the minimum gamma for which this inequality holds for all U and for all tau that is reasonable to be called the gain of the system, the infimum such gamma. But then we do not need that concept in more detail. Hence, we are going to see the L2 norm, the gain of the system only for LTI systems. CISO. Consider some CISO transfer function. For example, take S plus 1 over S plus 2. So, this particular system, is it stable? It has its poles in the left half complex plane. This is a 0, this is a pole, minus 2, minus 1. So, in this particular, for this transfer function, this is stable. So, this is, does there exist gamma and beta such that, such that for this system, S plus 1 over S plus 2 for this system, this, as I said, this beta depends on the initial condition. Let us assume that the initial condition is equal to 0. If the initial condition is more, of course, more energy will come out in the output even when 0 is, when the input is identically 0 and hence beta is playing a role because the initial condition also could be sitting inside and could be non-zero. But we are going to consider the case that the initial condition is equal to 0 and hence we will not be considering beta, the role that beta plays. That is a relatively easier issue anyway. So, we are going to see what is the minimum gamma that allows us to call that minimum gamma as the so-called L2 induced norm of the system. So, this L2 induced norm has a very important significance. It is also the peak value of the Bode magnitude plot. That is what we will see in detail now. So, it is well known. Consider again this transfer function s plus 1 over s plus 2. This is how the Bode plot looks. Magnitude of g of j omega in the log scale, in log scale. Also, this is in log scale. It starts increasing at the 0, 1 radians per second. It stops increasing at 2 radians per second at a pole. So, these are the cutoff frequencies. So, notice that the peak value is equal to 1 which is 0 dB and this corresponds to 1 by 2. Now, for s equal to 0, s equal to 0 which is a DC gain, this norm, this gain is 1 by 2 and as s tends to infinity which is the gain at very high frequencies, this is equal to 1. As s tends to infinity, you see this constant terms do not play a role because the leading coefficients are equal and hence it becomes tends to 1. The degrees are also same. It is bi-proper. The numerator and denominator degrees are equal to each other. So, this is tends to 1. So, we can ask what is the question? The L2 induced norm. What is induced about it? So, we take the system g is defined as induced norm. This is g is not a signal. We define L2 norm of signals so far but now we are going to define a so-called induced norm. This is not the L2 norm that you saw so far. We are defining norm of an operator of a system that takes input u, output y and has a transfer function like we had h so far. Now, we have an LTI system with a transfer function g. Hence, we are defining its L2 norm with g here as outputs L2 norm divided by inputs L2 norm. Of course, this ratio, it is reasonable to call this as the gain of the system but this ratio might be different for different inputs that you gave. Even though it is being scaled by the inputs, it is being divided by the energy in the input but still this ratio might depend on what frequency you have concentrated your energy in. So, we are going to look at the supremum over all u in L2 and of course, u should not be equal to 0 because otherwise you will have 0 in the denominator. So, this supremum only means that it is like the maximum but we are not looking at the maximum value over a finite set. L2 is a very large set, not a compact set and hence we are replacing max by sup. Sup is just supremum. So, this ratio indicates the maximum amplification that the system g can cause to an input u and this amplification is being measured in terms of the L2 norms. Notice that these L2, this L2 and this L2 here stand for signals, for systems, for signals that you have already seen and hence this is what you already understood unlike here where this is the L2 norm being used for a system g. So, by using the ratio of L2 norms of output divided by input and then the taking supremum over all such signals u, we have used that to induce a notion of norm for an operator g. So, these induced norms also have those properties that we indicated so far like triangle inequality and the norm being at least greater than or equal to 0 and norm is equal to 0 only when the operator itself is 0. Further, if you scale the operator by some number and the L2 norms get scaled by absolute value of the number. All these properties are satisfied for norm in addition to some more properties called sub multiplicative property that we do not need in much detail in this course. When we need it, we will see. So, the induced norm is defined like this. It turns out that this norm will meet the purpose of the gamma written there. So, let gamma be defined as the induced norm, L2 induced norm. Since we have used L2 for an operator, it is understood that this L2 is not of a signal, but the induced norm that we used and induced refers to taking the ratio of the outputs L2 norm divided by the inputs L2 norm. So, since this one was defined as supremum over all u in L2 of outputs L2 norm divided by inputs L2 norm, since we took this as a ratio, we are now going to quickly see that this induced norm will serve the purpose of the gamma. So, define gamma as this L2 induced norm. When this L2 has been used in the context of a system of an operator, then we are going to understand it as obviously being the induced norm. From the context, it is clear because G is not a signal, it is a system. Hence, this L2 here refers to the supremum that we just now defined of the outputs L2 norm divided by the inputs L2 norm. Of course, we have not yet seen whether this is finite, whether it exists or not. But you take the supremum of this ratio, supremum over what? Supremum over all u in L2 means you vary this u for different different values in different different signals in L2 and look at this ratio and ensure that the denominator is not equal to 0, that is ensure by taking u unequal to 0, unequal to the signal that is identically 0. And this we will define as gamma. What this means is that the supremum of y L2 is less than or equal to is equal to gamma times u L2, L2 norm of the input. If the maximum, supremum is as good as the maximum, you see as I said almost reaches this, it is not attained that is the only concern. So, if the supremum is equal to this, then any other is less than or equal to gamma times u, L2 norm of u. Why this inequality has come? Because this gamma was defined as a supremum, what all we did is we took this denominator from this side to that side and we wrote this inequality, gamma equal to supremum of y L2 norm divided by u L2 norm, that is what we have multiplied and written here because we know that the denominator is unequal to 0. And if the supremum of the left hand side is equal to this, you if you get rid of the supremum, then this inequality ends up coming. Now, this is true for all u in L2. That is how we have obtained back that inequality we started with, except that one might say that what happened to the E, the extended. That extended has not come because we have already assumed that this ratio exists. This ratio could easily have been unbounded, it might not have existed, it might be very large and infinity, infinite. So, this ratio, this is finite if and only if G is stable, that is what we are going to see in detail now. So, when G has no poles in the right of plane nor does it have poles on the imaginary axis, that is when this gamma will exist. So, G L2 is defined as supremum overall u in L2 of y L2 norm, L2 norm and u are equal to 0. This sup exists. Sup meaning supremum. Sup is the short form for supremum. Supremum exists if and only if and important result if and only if G has no poles in closed right half plane, right half complex plane. If G has no poles in the closed right half complex plane, what is right half? Of course, that is clear. What is closed about it? No poles on the imaginary axis either. All the poles are in the open left half complex plane. What is open about it? The boundary of the left half complex plane is the imaginary axis. The boundary has not been included. That is what makes it open left half complex plane. So, to say all the poles are in the open left half complex plane is same as saying there are no poles in the closed right half complex plane. The right half complex plane is where I put my fingers. To say closed means it includes its boundary. The imaginary axis is its boundary. If it has no poles in the closed right half complex plane, that is when we will call G also as herwits. This property G has no poles in the right half complex plane is also called herwits. Herwits also depends on the context and in this context it means when G is rational, when G is ratio of two polynomials, G having no poles in the closed right half complex plane is also called as herwits. That is precisely the case when this supremum that we have defined here is a finite value. The supremum could easily become infinity. Why? Because the numerator might become unbounded even though the denominator is finite. That is when this ratio need not exist. It might become very large and infinite. That is when we will say L2 norm does not exist. When will the L2 norm exist? When is this supremum finite? When is it bounded? It is bounded precisely when G has no poles in the right half complex plane nor does it have poles on the imaginary axis. This is precisely the case that the L2 norm exists. In that case, you give a u in L2. The output will also be in L2 and hence this ratio will exist and that justifies. Let us go back to our previous slide. For that case, this gamma is finite and hence this inequality holds for all u in L2. The L2e we did not have to write the e because this holds for all u in L2 itself. So, now let us look at the notion of gain for some more systems. What is the meaning of this maximum? So, well known. There are many, it is also called R H infinity. Well known result, if G has no poles in closed right half plane which means G has all poles in open left half complex plane and if G is proper, this proper is another condition I forgot in the previous slide. That is, proper is also required if you do not want, if you want G to have a finite L2 norm. If you want the induced L2 norm of G to be finite, then G has to have all its poles in the left half complex plane, in the open left half complex plane, further G also has to be proper. It is proper is same as saying denominator degree greater than or equal to numerator degree. This means that G is proper which means that as S tends to infinity G, S exists. G infinity is a finite number. What is well known is that the L in L2 induced norm that we defined is equal to, is attained, is in fact attained on the imaginary axis. The supremum value, supremum over omega in R of G of j omega consider for the case that G is so and G is single input single output that and this L2 induced norm is nothing but supremum overall omega in R of G of j omega. This is a very important well known result and what is R H infinity that I have written here? R stands for real rational, those transfer functions G which is a ratio of two real polynomials. Polynomials whose coefficients are all real, that is what R stands for and H infinity stands for those transfer functions which are proper and have all their poles in the open left half complex plane. H is in memory of a person called Hardy who worked a lot in such spaces and many others and is a close associate of well known Ramanujan. So, this class of transfer functions G which are real rational and which have all their poles in the open left half complex plane. It is well known that the L2 induced norm that we defined is equal to just the supremum of this particular value here. Take the absolute value of G at different different points on the imaginary axis and look at the supremum over all these points on the imaginary axis. So, that is what brings us to the Bode plot, the Bode magnitude plot to be precise. So, supremum over all omega in R of G of J omega. So, this is a complex quantity we take the absolute value. This to do this is nothing but suppose this is the Bode magnitude plot then this supremum this value here suppose this value is equal to 8 then we will say this is a Bode magnitude plot. Why will this be finite? Because G has no poles on the imaginary axis here because G has no poles on the imaginary axis nor is G proper nor is G improper. Because G is proper as omega tends to infinity it either slopes down which is the case when G is strictly proper the denominator degree is strictly more than the numerator degree or it saturates. It can saturate to a value different from the DC gain if G is by proper. To say it saturates means as omega tends to infinity it is a non-zero finite value and to say it is finite means G is proper the numerator degree does not exceed the denominator degree and to say that it is non-zero means the two degrees are equal. So, it will either come down like this with some roll off, roll off that depends on the difference in the degrees of the numerator denominator or it will saturate to some value for the case that the degrees are equal. So, this peak value is finite as soon as these two cases are satisfied G has no poles on the imaginary axis and G is proper. What is the value of the peak that is equal to 8? This peak will indeed be equal to the L2 induced norm precisely when G has no poles in the right half complex plane also. Supremum of G of G omega, supremum overall omega in R exists when G has no poles on the imaginary axis and two conditions are required and G is proper. So, this supremum when will it exist it could easily become infinity it could become infinity for it will become infinity for example if G has poles on the imaginary axis in which case this goes off this become very large it becomes unbounded the body magnitude plot. So, if G has no poles on the imaginary axis at no finite value omega does it become unbounded and if G is proper then as omega tends to infinity also it does not become unbounded. So, when these two conditions are satisfied that time the supremum exists to say it exists means it is a finite value, but just because it exists does not make it does not make the supremum equal to the L2 induced norm. When is it equal to the L2 induced norm? When supremum G j omega exists and if G has no poles in the open left half complex plane open left half complex plane has poles have not got open right half no poles in the open right half complex plane this has got ruled out. Imaginary axis already got ruled out because supremum exists then this supremum is equal to the L2 induced norm this is a frequency domain condition and what I write here is a time domain. So, you look at the system U the system G the input U output y equal to we are going to say is just GU G acting on U for this system you look at the outputs L2 norm divided by the inputs L2 norm this L2 norm is what we define in the time domain for a signal. So, this right hand side quantity is a supremum over all L2 signals and U not equal to 0 that ratio we have taken in the time domain that is equal to this frequency domain condition the peak value of the Bode magnitude plot for these two to be equal we want G to have no poles in the open right half complex plane. The left hand side exists as soon as G has no poles on the imaginary axis nor no G has no poles on the imaginary axis and G is proper in that case the left hand side exists and it will become equal to the right hand side when G has no poles in the open right half plane also. So, here are some exercises. So, find L2 norm of a system means it is the induced norm of G equal to S over S plus 1 S plus 1 over S S plus 1 and S plus 3 over S minus 3. What are these values equal to? Please take your time and calculate to get for this the value is equal to 1, G of S equal to this is these, L2 induced norm of this is equal to 1 this does not exist. Here also does not exist. Please take your time to find out why they do not exist. These do not exist. These all three do not exist. This is because there is a pole on the imaginary axis. This is because it is improper here because there is a pole in the right half complex plane. But for each of these four cases also please check for which case does supremum of G of j omega as I said this supremum could exist under milder conditions. Under what condition does this exist? This supremum need not be equal to the L2 induced norm of G. It will be equal only under certain conditions for which of these four cases does this exist. Of course it will be equal to this for this case but for the other three cases is what requires thorough investigation. So, we have seen some examples of transfer functions and when they have a finite L2 induced norm what remained to be told is a very important word L2 induced norm has another name. It is only a different name that is also called as H infinity norm. What is H infinity norm? It is nothing but the L2 induced norm. For single input single output systems it is nothing but peak value of Bode magnitude plot. For single input single output systems it is very easy to calculate it. It is equal to the peak value of the Bode magnitude plot. This peak value exists if G has no poles on the imaginary axis. It is equal to the L2 induced norm when G has no poles on the imaginary axis when G is proper and G has no poles in the right half complex plane. Under that condition is when the H infinity norm equivalently the L2 induced norm exists and is finite. That time it will be equal to the peak value. The peak value is what is equal to the H infinity norm precisely when G is stable. When G has all its poles in the left half complex plane and also when G is proper. So what is this peak value? It has another significance. For the case that G is stable, one can also look at its Nyquist plot. So let us suppose this is the Nyquist plot. Of course, it is symmetric about the imaginary axis. It is some orientation. Let us say this is a complex plane Nyquist plot. At any omega this is where G of j omega is. The Nyquist plot is nothing but a map of the imaginary axis under the action of G. This is the j omega the imaginary axis. It gets mapped under G to this particular contour with a orientation. Of course, this closed contour gets mapped to this closed contour in some orientation, important orientation and there is this point minus 1. And we can ask at each point what is the distance of that g of j omega from the origin. The peak value is nothing but the maximum distance. This is absolute value of g of j omega. It is just a complex number. At any omega, g of j omega is a complex number and absolute value is its distance from the origin. So supremum is nothing but the maximum distance as you go as you traverse along the Nyquist plot and you have never gone to infinity. This point is not gone to infinity precisely because G has no poles on the imaginary axis and also because G is proper. In that case, this contour will be a closed contour and that time one can look at the peak distance from the origin. So, the supremum overall omega in R is nothing but farthest point of Nyquist plot, farthest from what? Farthest from the origin. This is the significance with the Nyquist plot. Of course, we will see this in little more detail very soon. So, we are going to, after this development of L2 induced norm, we are close to seeing an example called the small gain theorem. So, for that purpose, we need to see the notion of feedback interconnection being stable. Let us think of one of them with a negative sign. For the small gain theorem, it does not matter whether this is plus or minus. Suppose, H1 and H2 are finite gain L2 stable. Finite gain L2 stable means there exists gamma. This particular symbol with a E reversed means there exists. There exists gamma 1, beta 1, this is for H1 and gamma 2, beta 2 for H2 such that what inequality is satisfied? This inequality is what we have seen several times today already. Y1 chopped L2 norm is less than or equal to gamma 1 times E1 chopped L2 norm plus beta 1. Similarly, the same inequality for H2 which puts an inequality between the L2 norm of Y2 chopped and L2 norm of E2 chopped related by gamma 2 instead of gamma 1. So, we can ask the question, is the feedback interconnection stable? That is the question that we are trying to answer. So, before we come to that question, look at this feedback interconnection. We have already assumed that H1 is finite gain L2 stable. We have assumed that H1 is finite gain L2 stable. We have assumed that H2 is also finite gain L2 stable and we want to ask, is the interconnection also finite gain L2 stable? Now, the question arises, what does it mean for the interconnection to be finite gain L2 stable? To say that the interconnection is finite gain L2 stable means that the map with U1, U2 as the input to Y1, Y2 as the output, we can think of this interconnected system as a map, as a system which takes U1, U2 as the input and gives you output Y1, Y2. Of course, you might say, should the output be Y1, Y2 or should the output be E1, E2? Both are okay. Or you can also consider U1, U2 as the input and all four signals as the output. But if this map from U1, U2 to all these signals is also finite gain L2 stable, then we will say this feedback interconnection is finite gain L2 stable. This is an important concept. So, let me just recap this concept. Feedback interconnection is called stable. Here, by this word stable, here we mean finite gain L2 stable. If what is satisfied? U1, U2, Y1, Y2, map is finite gain stable. It is finite gain L2 stable. This particular map from the input U1, U2 to the outputs Y1, Y2, if this map is also finite gain L2 stable, then we will say that this feedback interconnection is stable, is finite gain L2 stable. Which feedback interconnection? This feedback interconnection. You might ask why take outputs Y1, Y2? It can be shown that if the map from U1, U2 to Y1, Y2 is finite gain L2 stable, then the map from U1, U2 to E1, E2 is also finite gain L2 stable. These are equivalent. So, between Y1, Y2 and E1, E2, you can take any one pair as the output. If you want to take all four of them, both pairs that is that is also okay. But it is not okay to take this E1, Y1. One should take cross Y1, Y2 or E1, E2. So, this is a meaning that feedback interconnection is stable. Now, we want to ask the question under what conditions on H1, H2 will we have stability of the feedback interconnection. Stability of the feedback interconnection as I said here is finite gain L2 stable from these inputs U1, U2. These are like being injected at the interconnection points. There is also a notion of well-posedness of interconnection. Well-posedness of interconnection means that U1, U2 are genuinely are inputs and as soon as U1, U2 inputs are given, rest all outputs are determined. Rest all variables are outputs means that they all get determined as soon as U1, U2 get determined. So, to say that this feedback interconnection is well-defined. To say that it is well-posed means for every U1, U2 input there is a unique output E1, Y1, E2, Y2. All these four are unique. So, we have already assumed that this particular feedback interconnection is well-defined. After it is well-defined, we have introduced a notion of feedback interconnection being stable. Feedback interconnection being stable means that this interconnection finite gain L2 stable, the map from U1, U2 to Y1, Y2 is feedback is finite gain L2 stable. So, for that for this feedback interconnection to be finite gain L2 stable, we are going to see a sufficient condition. So, for that sufficient condition, we have assumed that both h1, h2 are finite gain L2 stable. To say that h1, h2 are finite gain L2 stable is that there exists gamma1, beta1, gamma2, beta2 such that these two inequalities are satisfied. We are seeing an example one sufficient condition for the feedback interconnection to be stable. Of course, the feedback integration can be stable in when several other conditions are also satisfied not necessarily this. This is just an example of one sufficient condition. What is that sufficient condition? Suppose gamma1, gamma2 product is strictly less than 1, product gamma1, gamma2 is strictly less than 1, then if this is satisfied, then feedback interconnection is finite gain L2 stable. So, this is called the small gain theorem. What is the small gain theorem? It is an example, it is a sufficient condition for the feedback interconnection to be stable. That sufficient condition is as follows that we consider this feedback interconnection. Suppose it turns out that h1, h2 are themselves finite gain L2 stable. To say that they are themselves finite gain L2 stable means that these two inequalities are true. Of course, for all e in L2 e and for all e1 for all choppings greater than or equal to 0 and here for all e2 in L2 e, this particular symbol is for all. It is currently commonly used to say for all for every. Under the conditions of h1, h2 are both finite gain L2 stable and if the product is strictly less than 1, then one can also guarantee that the feedback interconnection is also finite gain L2 stable. This is a sufficient condition for the finite gain for the feedback interconnection to be finite gain L2 stable and this theorem is called the small gain theorem. One can understand the small gain theorem like this. Consider this feedback interconnection again. Remove u1, u2, start from e1, go to y1. The maximum gain that can happen is gamma 1. Then you go from here to here, the maximum gain is gamma 2. So, when, forget the minus sign, multiplication by minus sign does not really change the gain. Gain of the minus 1 is nothing but plus 1. When you go round this loop, how much amplification has occurred? From here to here by gamma 1, there is this plus u2, there is no u2. From here to here it is gamma 2, but the product of gamma 1, gamma 2 is strictly less than 1. So, when one traverses the loop completely and comes back here, the net magnification is strictly less than 1. The extremely important point to note that you traverse the loop and come back to the same point, the maximum amplification that can occur is strictly less than 1. If somebody assures us that the maximum amplification that can occur is strictly less than 1, then the small gain theorem assures us that this feedback interconnection is finite gain L2 stable. And for that maximum amplification to be strictly less than 1, notice that this plus sign or minus sign does not matter. Why? Because multiplication by minus 1 does not change the maximum amplification at all because the gain of the operator minus 1 is again equal to plus 1 due to that reason. So, this completes this topic about norms of various signals, norms of an operator. We have seen what is the meaning of the finite gain L2 stable in the context of LTI systems. For that case, we have seen that it is nothing but all poles in the left half complex plane, in the open left half complex plane and G is proper. For that case, we have seen that the L2 induced norm, which is also called H infinity norm is nothing but the peak value of the Bode magnitude plot. So, this and this peak value of the Bode magnitude plot is nothing but the distance of the farthest point of the Nyquist plot farthest from the origin, the distance from the origin. So, this completes this important topic and this is required for various other topics. We will continue with the next topic in the next lecture. Thank you.