 So, last class we have if you recollect the last lecture we have covered the what is called internal stability of linear time invariant system. What is this? You have a plant, you have a controller then in addition to this we have a input disturbance, output disturbance and measurement noise. Generally measurement noise is high frequency signal and disturbance is the what is called low frequency signal and how to check the stability of this system in presence of all disturbances and noises we have discusses. Since there are four inputs are there 1, 2, 3, 4 inputs are there and we have all internal signal must be stable and we have consider the internal signal the summer output E, the summer output UP and summer output Y and even all internal signal must be stable. Then we have this problem then we have seen that there are four inputs and four outputs. So, we will get in all together the 16 transfer functions. So, we have a 16 transfer function and each transfer function must be stable for internal stability of the systems that we have derived. Next we have seen it that if you see the detail derivation of that one output is due to the reference input due to the disturbance, due to the output disturbance this is the input disturbance and this is the measurement noise that output this expression. So, then we have consider how disturbances effective disturbances the output can be reduced by properly designing a controller that C of s. Then design objective is what there are four design objective is there good output disturbance rejection. In order to achieve this one loop gain that G s and C s must be high at low frequency range and in turn it will reject the all disturbances. Then for good set point for achieving good set point then we have to make the transfer function or the transfer function t 0 is nearly equal to 0. This we have proved it this one next is noise suppression. For noise suppression is noise means high frequency signal at high frequency signal the loop gain is very small tends to 0. So, if you can make it this controller design in such a way. So, t s 0 at in presence of noise is equal to 0 then noise will be suppressed. Then another design objective is there for less control effort naturally when we design a controller control effort should be less because we cannot apply the what is called input magnitude very large to the physical systems. So, in order to achieve this one that what is called loop gain frequency response should be like that low frequency range you see the loop gain is large and high frequency range loop gain is in this region loop gain is low and the mid frequency range that determines the stability and bandwidth speed of response or bandwidth. Next we have considered that is when the plant that plant which is equivalent to model by g m agree and the controller is designed based on the plant model. If there is an uncertainty in the plant model means if there is a discrepancy between the plant model and the actual plant then whether the design controller based on the plant model will stabilize the system or not that we are going to study in this lecture. So, we have a two types of uncertainty involved in the plant model parameters one is structure uncertainty another is the unstructured uncertainty. We have discussed last class the case one the additive uncertainty that we have a controller this is the plant model and we have assumed there is a discrepancy between the plant model and the actual plant and that is compensated with a additive perturbation delta a m which is parallel connected to g m and this transfunction of that one g m of s plus delta of m s delta a of s a stand for additive perturbations. So, now we will detail we will study what are the different types of perturbation we can represent the discrepancy between the plant and actual plant and the plant model. So, let us call we are talking about the case one. So, if you see the case one now we are writing case one that case one is additive uncertainty or perturbations. So, we have a controller c of s we have the plant model g m of s and output is y of s and we have a simple unity feedback systems. This is the reference input this is our controller and this is the error signal and this is the input u to the plant and we assume there is a discrepancy between the actual plant and the plant model. So, that we are compensating the discrepancy is compensated by assuming that it is a additive perturbation delta a of s. So, that is sum here. So, now this block if you see this block is nothing but a g m of s plus delta a of s. Now, our problem is this we have to study the stability of these systems. So, now we assume that this is our one point is there u and this is our a point v look the controller is design the controller is design based on the plant model. And we have design the controller in such a way the nominal plant is stable. Now, we have a perturbation additive perturbation that the discrepancy between the plant model and the actual model the discrepancy and that is compensated through delta f. Now, we have to ensure that same controller whether it will be able to stabilize this system or not if it is stabilized how big that perturbation can withstand before the system becomes unstable. So, that is our objective. So, the perturbation block which is connected between the point u and v that you take out if you take out that one and equivalent circuit you can redraw this one like this way v look at this one v then our output is y or y of t then from there there is a feedback c s and g s then this. So, I can write it c s then g s g m of s then this is minus plus then which one is our v the output of the controller is our u this is our u. So, you have to find out this is u we have to find out the trans function between this point this is input and this is the output this transfer and since we have taken you have to in order to analyze the system that the controller based on the plan model whether it will be able to stabilize the perturb model or not that one to remove the perturb model and we have denoted the way it connected denoted by this two point v and u find out the trans function between v and u this which we can write it equivalently this is nothing but a trans function is one by c one forward path is one by plus g m of s into c of s. Since, this is a single input it does not matter the pre multiplied or post multiplied the product of this two trans function, but in case of multi input multi output it matters. So, this is the trans function then I need about the this this signal u of s that multiplied by is multiplied by c of s and what is the c of s if you see this one minus sign is there again minus sign c of s is the out of this. So, it will be minus one into c of s and you can put it in state of this one minus sign is here it does not matter minus sign is there effectively it is a negative feedback. So, this is the trans function which is minus c of s one plus g m of s into c of s. So, this is the trans function between the point this is the input the part of model output is the input to the systems and that part of model input from the controller output is acting as input to the model. So, you have to study the stability of this one. So, now see our u is this is our delta a of s this is our u and this is our v and we have found out the trans function between them is our that one that you consider m of s agree. So, it is m of s. So, now this whole block whole block is equivalent to you can write it equivalent to this block the whole block from this whole block is equivalent to in order to study the stability of the part of model is same as study the stability of this system that is it is m delta structure systems. So, this means the and also you have to find what is the how big this perturbation delta a delta a of s. So, that the design controller c of s based on the plant model will stabilize this one. So, in that case I call the design controller what we have designed is robust in presence of parameter perturbations. So, that is our problem to be stable. So, in order to study this one we assume that delta a of s is stable and this is naturally this is stable this m a process is a stable one because controller is designed based on the plant model. So, you want by s this is stable c s is the controller which is stable. So, m of s is also stable. So, now we have to use that is called a theorem which is called that what is small gain theorem we have to use next is small gain theorem to study the stability of that systems. So, what is the small gain theorem we will just tell you this one and that concept we will use to study the stability of this block. So, this problem is now here we have designed a controller c of s based on the plant model then we want to see whether the same controller will be able to stabilize the perturbation or not and if it is able to stabilize what is the how big that perturbation is there that how big that perturbation is there that system will remain stable before that what is the how big the perturbation delta a is there agree. So, that the system remains stable. So, this is our delta m of s or g s let us call g s this is our h of s feedback transaction and this is our negative feedback this is y of s and this is e of s and this is r of s the small gain theorem tells if g s is stable and h s is stable assume that is what we assume that g s and h s both are stable. Then the closed loop system then the closed loop systems will remain stable will remain stable if mod of g of s h of s is less than 1 how it is just a Nyquist stability criteria for all frequency range if it is a b since g s h is stable all frequency range that Nyquist plot should not encircle the minus 1 magnitude that means magnitude of this one will be less than 1 note due to this note due to the following inequalities due to the following inequalities that means g of s h of s is less than equal to g of s individual mod h of s is less than 1 because product of this one absolute value for all frequency is always less than equal to individual block mod for all frequency and if this quantity is less than 1 obviously the product of this one absolute value also will be less than 1 but this results will give you the conservative results. So, this implies so this implies this implies that this implies if mod of s h of s is less than 1 guarantees the closed loop system closed loop system internally stable system stability and internally this is stable because I told when I will talk about the stability that means internally the system must be stable. So, this thing so let us see the so this is our objective. So, now refer to the back to our plan so now we have to show it that m into delta s absolute value of that m into delta s must be less than 1 or individual absolute value product must be less than 1. So, our objective is here mainly objective for a given nominal stable nominal stable transfer stable closed loop systems this means what for a given nominal the nominal plant is there we have designed the controller based on the nominal plant and if it is a stable then we will call nominal closed loop nominal stable closed loop systems. So, now our problem is how big can be the can be the uncertainty how big can be the uncertainty delta a of s before the part of system before the part of system becomes unstable that means what is the how big that perturbation is there. So, that how big that perturbation is there so that the overall system becomes unstable. So, that perturbation upper limit we have to find out then the design controller based on the plant model will stabilize that one. So, keeping this thing in this one so using now we can write it if you apply the small gain theorem apply small gain theorem small gain theorem. If you apply the small gain theorem here according to small gain theorem using small gain theorem what you can write it that individual delta a of j omega into c j omega c j omega into what is the trans function of that one m of s actually this should be a if you see this one delta a g j omega mod and m of s mod m of s means j omega when s is all frequency this should be less than 1. What is m? m we have calculated this one agree c s 1 by g s 1 by 1 by 1 plus g m c h is the sensitivity function for nominal plant. So, we can write it this one that c s minus sign will go because m s where you can write it if you write it you are aware m s m of s is equal to minus c of s divided by 1 plus g m of s c of s. So, if you take the absolute value of that one that will be a this divided by 1 by this quantity that means this quantity is our sensitivity function s 0 of s this is the sensitivity function. So, you can write it this into mod s 0 of j omega agree sorry this that whole thing whole this will be less than 1 what is m? m is that quantity c s into s of 0 of this one is less than this. So, name therefore, the upper bound of this perturbation delta a g j omega mod should be less than equal to 1 by c j s 0 j omega mod then which one this perturbation which one you will take because for all frequencies you can write it this is for all frequency omega now which frequency for all frequency which one agree the supremum value supremum value of this mod and supremum maximum the largest possible value of that one. So, you will write it supremum sup of w. So, we will change the frequency we will take this magnitude next we change the frequency increasing at what value of frequency this will be the maximum that value will take in. So, that will be the maximum perturbation of before the system becomes unstable. So, you can find out the perturbation bound by using the small gain theorem case 2. So, naturally this results will give you the conservative results actually what is the condition for becoming in acoustic stability category delta a into mind it absolute value of delta a into m m j omega whole mod less than 1 then we have written product individual product and that is why it is giving a small gain theorem gives a what is called a conservative results provides conservative results. So, the way next our perturbation is multiplicative perturbations another is case is multiplicative perturbations. The perturbation the discrepancy between the actual plant and the plant model can be compensated by additive perturbation. Now, we will discuss the multiplicative perturbations the case 2 is the multiplicative uncertainty or perturbations uncertainty. So, once again we draw the systems this is our controller and this is our plant model and we compensate the discrepancy between the plant model and this by using what is called a multiplicative perturbations. So, this is delta m stands for multiplicative this coming here and this is the thing. So, that block if you see this block is nothing but a 1 plus delta if you see this block is nothing but a 1 plus delta m of s. The whole block now from here to here this is the actual plant g of s is nothing but a g m of s 1 plus delta m of s. If there is no perturbation the plant model an actual model actual plant is the same. So, this and this is our output y of s the feedback this is r of s this is e of s and this is y of s. Again we will do in similar manner the perturbation block you take out again. So, perturbation this is our u this is u and this signal is our v. You take out find out the transfer function between the v and u input and this then push this block across the perturbation multiplicative position. So, if you read all you remove this one and read all this one it is a like this v then g of m s this see this one v the g m of s. Then this is the output and it is coming back and it is where it is going this then c of s c of s then this minus this this if you read out this one this is the case. So, what is this circuit what is the if you just simplify this block you will get it this one is how much this block g m of s 1 plus g m of s plus c of s agree. Now, what is the transfer function between v and u v is your if you see v and u is what here your u is at this point. So, your the transfer function between the v and u will be that is m of s is the transfer function between g m a v will be this is the transfer function between this and this I have to find out the transfer function between v and u here u then this multiplied by c s into minus. So, minus g m of s divided by 1 plus g m of s c of s into c of s this is the transfer function between the u of s divided by v of s this is the transfer function. So, that so now this block you place it here between the this terminal the simplified transfer function you place it. So, if you place it there that one what will be this one you will get it you will get it this is our v u this is our m of s this is our v and this is our del time this is v this is equivalent to you write u. So, again applying the small using the small game theorem small gain theorem small gain theorem what you can write it without any details I can write delta m of j omega mod is less than equal to 1 by c that one what is this one from here to here it is nothing, but a transfer function t of 0 complementary transfer function agree that we denoted by t 0 of s based on the plant model. So, that is nothing, but equal to that t 0 of s what just see this one t 0 this t 0 into c of s. So, t m into so it is a t 0 j of omega mod supreme of w see once again what is the transfer function between this one first find out the transfer function of that one from here to here that is the transfer function I am writing it from y s to v s g m 1 plus g h s then I am finding out the transfer function between d s and so it is already that there from here you multiply it by c s with minus sign. So, this will be your t s this is your loop transfer function agree the loop transfer function is g m g m in c h that g s into c s plus this and this is that loop transfer functions. So, you have to this you have to minimize I again this quantity yields conservative results. So, now look this one additive perturbations and multiplicative perturbations. So, here c s into s of 0 mod of this you have to mod of this one you have to make supremum value of that one highest possible value or all frequency rates which are the highest value of this absolute value of this reciprocal of that one you have to take we have to find out the additive perturbation model. Similarly, here that similarly, here that you see this is t s t 0 s c s. So, I have just missed this here c s c j omega. So, this one that will give you the maximum perturbation of multiplicative perturbations we have designed the controller look here we have designed the controller based on the plant model c s. Then if you consider the discrepancy between the plant model and the actual plant is delta m to compensate that discrepancy between the actual plant and the plant model we have used the multiplicative perturbations because g m is multiplied by this. And now question is we have to our objective is what is the perturbation bound of delta m. So, that the system will become remain unstable that means how big the perturbation delta m will be required for system become unstable before going to the unstable region. So, I found out the transformation between these two points that is your this one is nothing but a if you see this is nothing but a and if you see this is nothing but a t 0 of s into c of s. Then we replace this one with a that block transformation between b and w m s you got it we kept it then we have to apply the small gain theorem. And we have found out that t s c s absolute value 1 by reciprocal of this one for over the all frequency then which frequency will consider that numerator part the denominator part for what frequency will give you the maximum value that supremum of that one for w varies from 0 to infinity that quantity you have to take it in a and that will give you the upper bound of multiplicative perturbations. So, again this results will give you the conservative results of that one. So, we have now discussed to compensate the discrepancy between the plant model and the actual plant we have used the tooth model one is additive perturbation model another is multiplicative perturbation. For stability analysis for stability analysis the Nyquist plot is useful for stability for stability analysis of analysis of additive for uncertainty why you see g of s is equal to additive perturbation g m of s plus delta a of s. So, what is this one if you do the Nyquist plot magnitude versus the phase angle plot then you can just nominal plant transformation you plot it then perturbation maximum bound you got it let us call nominal plant plot is like this way this r of s real part of this is the imaginary part of this and this is our g m of j omega agree this direction. And now if you know the perturbation of the upper bound of perturbations then you just make it that a circle of that magnitude and along the g j omega it will give you you can plot it that one that means ultimately if it is a delta a of j omega if it is a delta a of j omega then it is nothing but a boundary of like this way you will get it. So, if it is additive perturbation is there then plot Nyquist plot to study the stability of the part of plant if it is a and this Nyquist plot all these things we have discussed in detail in lecture 44 Nyquist plot we have lecture note refer to the lecture note 44 then if it is a multiplicative perturbation is there then it is useful if you use the Bode plot for stability analysis Bode plot is useful for stability analysis multiplicative perturbations or uncertainty because g of s is equal to 1 plus delta m of s into g m of s. So, if you take the log both side log of m plus this and you know the upper bound of delta m of s then it is if it is a product form then Bode plot will be useful for stability analysis of uncertain systems. So, now question is next question is we know that if the plant model is if you know it if there is a discrepancy between the plant model and the actual plant then how to compensate this one that we have seen by additive perturbation and multiplicative perturbations. And if you use the additive perturbations then we know what is the how big that perturbation can withstand just before becoming an unstable. Similarly, multiplicative perturbations also we can find out the what is called how big that perturbation will exist before the system become unstable. So, these are the condition what we have got it by using the small gain theorem and that gives the I told you that conservative results of this one means in the sense if the if your system with this small gain theorem if you find the system is unstable actually the system may not be unstable or may be unstable. But if the condition small gain based on small gain condition whether if you use it if the condition is satisfied it indicates the system will be stable. But if it is not satisfied connect say whether it will be actual system will be stable or not with under that perturbations. So, next is we will see how the perturbation involved in the system can be taken out from the plant model from the plant model. So, next is general control configurations with uncertainty. So, let us take a simple example mass spring and damper systems a mechanical system damper system. So, we have a reference frame is there we have a mass m 0 and we have a damper then we have a spring and we have applied a force f of t here from the reference when a force is applied from the reference equilibrium position this is displaced by x of t. Now, write the dynamic equation of the systems this is b 0 nominal spring constant is k 0 mass nominal from mass is m 0. So, mass may be change with time damper damping coefficient may change spring coefficient may change with time. So, we write the part of model of this one is dynamic equation m 0 plus delta a m dynamic equation when you force this one how you reach to the steady state value then it has a some displacement from the equilibrium point it has a some at any instant of time the displacement x of t. So, we will write it now at of t d t square plus b plus b 0 plus delta a m delta a b b stands for damping this d x d t plus k 0 plus delta a k x of t is equal to forcing function this f of t is the input or forcing input function. So, this first initially the spring mass system was equilibrium position then I applied a force at any instant of time displacement is x of t then we can write the dynamic equation. So, m 0 b 0 and your k 0 all are nominal parameters nominal parameters and this is the forcing function let us call this equation number 1. Now, we will see that how we can take out the part of parameters delta a delta a b delta a k k stands for spring b stands for damping m square m perturbation how we can take out from the plant model. So, define the state variables form define state variables the x is defined by x 1 x is the displacement of the body is x t and x 2 is the velocity of the mass or x 1 dot let us call this is nothing but a x 1 dot x dot is in x 1 dot. Then from 1 you look at this expression from 1 we can write it this is x 1 this is x 2 and this is x 2 dot. So, x 2 dot is equal to this is nothing but our x 1 this is x 2 and this is x 2 dot. So, we can write it x 2 dot is nothing but a equal to 1 by m 0 plus delta a m. So, I divided by both side by this quantity then you see this one minus k 0 plus delta a k agree into x 1 that quantity into x 1 take that side right inside then this into x 2 minus b 0 plus delta a b x 2 plus the forcing function all are divided by m 0 plus this. So, let us call this is equation 2 one can easily draw the block diagram of this one keeping in mind that x t is your forcing function that input and x 1 x 2 x 1 is the position of the mass x 2 is the x 1 dot or x dot or velocity of the mass. So, we can if you just represent the block diagram form block diagram representation. So, if you do the block diagram representation you see this one what you can write it. So, our input is we can write f of t then your I am just drawing from this one again there are two states are there x 1 and x 2 now I am drawing this one that 1 by m 0 delta m delta a m plus then we have a integrated output is there and then we have a another integrated now we tell this one f of t f of t and what is this block this is the negative feedback what is this block we will get it this block 1 by if you see 1 by m 0 divided by 1 plus 1 by m 0 into delta a m. So, what will come ultimately it is nothing but a 1 by m 0 plus delta a m this block is nothing but a if you see 1 by m 0 delta a m this block. So, f t into 1 by f block you see f t 1 by this you got it again then what you need it I need the information how the x 2 and x 1 is related x 2 is nothing but a if you integrate differentiation of that x 2 is what x 2 is the derivative of x 1 that means if it x 2 dot it is if you integrate this one you will get it x 1 again this x 1 dot sorry which is equal to x 2 which is equal to your x 2 you see our x 2 is if you recollect this our x 2 is what x 1 dot. So, x 1 dot is equal to x 2 if you integrate this one you will get. So, this is our x 2. So, x 2 multiplied by again this block is your b 0 and from this one delta a m a b and this what is this block you will get it and this is your plus positive feedback this is the positive way. So, this is nothing but a b 0 plus delta a b. So, all dotted block is the part of block similarly x 1 if you do this one this is k 0 delta k a delta a k plus. So, what is this block k 0 plus delta a k now you see the x 2 multiplied by b 0 a delta k x 0 minus x 2 b 0 minus minus sign is there. So, this now you see this signal x 2 multiplied by b 0 this coming here and it is a minus sign and then multiplied by 1 by a 0. So, I am getting the that term similarly x 1 this block k 0 by delta 1 this plus this plus minus sign is there multiplied by this block I am getting that one. So, this is the block this equation representation of block form is that one. So, tomorrow I will discuss in details about this one and our main job will be how to extract delta a delta a m and delta k a is taken out from the plant model and then we can do analysis of the systems. So, I will stop it now here.