 So, last class we have discussed if you have a actual plant trans function is G of s and the plant model G m of s and there is a discrepancy between the actual plant and the plant model. And the controller is designed based on this model plant model. Now, our question is whether the design controller based on the plant model whether it will be able to stabilize the part of model that whatever the discrepancy between this and this that are compensated with a part of model. And if it is able to withstand the stability of this system then we have to find out how big that perturbation is there just before becoming the system is unstable find out that big how big that perturbation is there. So, if you recollect this one our we did it that this is the plant model this is the controller controller is designed based on the plant model and the discrepancy between the plant actual plant and plant model we have compensated with a delta a which is connected across the plant. So, this is nothing but a additive in the sense what is the plant model is there with the plant model the perturbation model is added we have made an assumption that delta s is a is stable one. One can find out the maximum upper or upper bound of delta a with the knowledge of plant model and actual system frequency response and plant model frequency response from that you can find out the upper bound of delta a anyway. So, how to study the stability of this system in presence of that controller and controller is designed in presence of that in considering the plant model. So, you take that input to the output of the part of model is v and input to the part of model is u. So, you take out this block and find out the trans function between the input v and the output u and that we have found out that trans function is m of s that trans function is m of s and by using the small gain theorem we have to see whether the system is stable if it is stable what is the how big the perturbation is there just before it is become unstable. So, if you look at this one that small gain theorem we have applied and we found out that upper bound of delta a is less than that mod of that what c j this is the controller trans function and s j is sensitive function this delta a must be less than this one. So, long this condition is satisfied the design controller is will be able to stabilize the part of plant and that design controller is based on the plant model, but I already mentioned this provides you the what is called conservative results. It means that if there is a perturbation, but this condition is not satisfied it does not mean the system will not be stable it may be a stable one. So, that is that is why it is given next we consider the multiplicative perturbations. So, the perturbation before the plant model and this is this and delta m is the multiplicates m stands for multiplicative a stands for additive. So, this function is if you equivalent block is that 1 plus delta m. So, if you see the total block of that one this this transform the g m into 1 plus delta m s. So, it is a multiplicative that 2 trans function. So, actual plant is a is model plant per in multiplied by part of model with 1. So, again in order to study the stability of the system in presence of multiplicative uncertainties then what you have to do mind it that C F O S design based on the plant model then there is a perturbation I want to check it whether the system will be stable of not under that perturbation and that perturbation is multiplicative perturbation. And in order to find out this one you take out the block of that one part of block that means output of the part of block is v input to the part of block is u. So, you take it out find out the trans function between the input v and the output u after taking out. So, again once again we got the that m of s is nothing but a that what is called this is the output this is u of s sorry it is not a u u of s by v s this one the trans function we denoted by m of s. So, this nothing but a this again now look at this one this is the sensitive function sensitive function is that one not this is this one this is the sensitive function that t 0 of s the complemented sensitive function sensitive function is a 0 and complemented sensitive t of t 0 of s this is the complement. So, u s by v s is equal to that t 0 t 0 this c 0 is not that t 0 this one, but g m you see this one from here to here the trans function v s to y s g m 1 plus g f is I am sorry this is this is a y s this is y s sorry and when where this is the y s when you are finding m s is nothing but a input signal that output signal this is t s. So, now in order to find out the trans function of part of plan you have to again push it the part of model delta of m now find using the small gain theorem this is the condition we have arrived this c s is not there. So, again this yields the conservative results that is what we have studied it now next is we have consider a general control configuration with uncertainty that mean let us see we have a mass then you have a damper and spring and the force is applied f of t now this m 0 k 0 b 0 is the nominal parameters the model of the mass spring and this and actual plant actual mass this is something else. So, that we are compensated with a delta m delta a b delta a k. So, we have considered additive perturbation and k stands for spring b stands for damping and m stands for mass. So, this is the dynamic equation of this one and f is the forcing function or input to the systems. So, naturally that when you give an input the before it comes to steady state there will be dynamics is involved this one and dynamic equation is described by this one. So, this thing I can easily convert into our main aim to take out delta m delta m delta a b delta a k from the plant model separate this from this model. So, we define the state variable x is the displacement x dot is the velocity x double dot is the acceleration. So, displacement of the object is x 1 new state variable x dot is equal to x 2 dot x 2 which is nothing but x 1 dot. Now, from equation 1 from this equation 1 I can easily write that expression if you see this one. So, this I am now representing into a block diagram form if you see this block diagram x 2 dot x 2 dot is nothing but a x 1 multiplied by this x 2 multiplied by this and forcing function divided by some quantity 1 by m 0 by delta m. So, x 2 dot so, we have this is a second order systems if you see the second order differential equation we need to realize in state phase model for two first order differential equations. So, we consider this is our x 2 dot which output is x 1 x 2 which is nothing but a x 1 dot and again you can put it integrated this is x. So, if you see x 1 multiplied by delta k k 0 plus delta a k x 1 multiplied delta a k this one. So, this block is nothing but a this block is nothing but a that one then x 2 multiplied by b 0 plus delta b k. So, this combination is nothing but a that one. So, this two terms already we got it this and this term and f of t. So, this two signal is coming and going here. So, this is not a that one that means I have told you that x 1 dot is equal to x 2 that is we do not consider that this is your x 2 dot. So, this signal is nothing but this signal plus this signal and then f t plus f t that is the f t. So, that whole thing will be divided by that 1 by m delta m 0 1 by delta m 0 is this block if you see they are from here to here trans function it will be 1 by m 0 delta m which I have shown it here. So, this is the block representation of that one in state based model next is our job is how to write the state based model of this one. So, we have seen this is a second order mass damper and this is a second order system there will be a two state variable one is x 1 x 2 physical meaning of x 1 is a displacement x and x 2 is the velocity and x 2 dot is the accelerations. So, now we can write it this block diagram representation if you see that one I am not drawing once again this one, but we are telling it that let us call this signal is I am denoted by z 1 of t then output of part of model what is this one that is d 1 similarly this signal what you are getting it here that is z 2 t and output of that signal that output of this signal is d 2 similarly this signal I am considering is z 3 of t this output of that signal is your is d 3. So, now I am just writing the state based equation from this block diagram nothing else. So, if you look this look that one you see if you look that one I can write it what is our x 1 dot this x 1 dot x 1 dot is what x 1 dot is nothing, but your x 2 you see x 1 dot is nothing, but x 2. So, if I write this expression x 1 dot of t is nothing, but our x 2. So, x 1 of t x 2 of t and 0 and 1 there is no other terms involved in this x 1 into 0 x 2 0 plus there are we will write this thing there is no other terms is there in here for the time being. Now, x 2 dot see this x 2 dot x 2 dot is what see x 2 dot this signal this signal add together then this signal passing through this sound is x 2. So, it is nothing, but a if you see x 1 into x 1 into k 0 1 signal going x 1 into k 0 this only this signal I am considering x 1 in k 0 going this way divided by this one passing through this one that 1 by m 0. So, this is going here coming here going here then passing through let us call with this block. So, I can write it x 2 dot of t is nothing, but a so x 1 1 by m 0. So, it is a minus y minus is considering that minus sign is that signal that signal is going with minus sign agree. So, this is a minus k 0 by m 0 now this is a x 2 you see this is the x 2 x 2 b 0 going this signal minus sign and 1 by m 0. So, this is a k b 0 m 0 x 2 this is not there are some other signals also there in x 2 now coming to the perturbation side x 1 multiplied by delta a k what is this output that output is d 3. So, d 3 is coming this way minus sign again m by 1 by m 0. So, I can write it d 1 d 2 d 3 agree I can write d 1 d 2 d 3. So, if you consider d 3 this one d 3 this x 2 x 1 multiplied by delta x 1 is equal to d 3 d 3 going this way d 3 going this way agree multiplied by 1 by m 0. So, it will be minus 1 by m 0 now d 2 is x 2 multiplied by delta a k d 2 then going like this way in this way minus multiplied by 1 by m. So, minus sign is there. So, it is a d 2. So, it will be a minus 1 by m 0 similarly d 1 is going like this way of m 0 this this minus 1 0. So, it is a minus 1 by m 0. So, and other terms is not there. So, this is 0 0 0 agree. So, what is left only left is signal is f f is going and this is multiplied by f 0 you see f is going multiplied by 1 by m 0. So, this term is coming where it is associate with the x 2 dot. So, it will come plus 0 1 by m 0 into f of t see this one f of t multiplied by 1 by m 0 and that signal is coming here. So, x 2 dot is combination of second that is second row is combination of all this signals is x 2 dot. So, we can consider this is our forcing function means u of t I can consider this is a matrix at and this is the state x of t and this is our denote this is our b of 1 matrix and this is our w of t that there is our disturbance or part of part of model output and this is our we consider this vector is our b 2 of t and this is nothing, but a x dot of t. So, this is we have written it for this one. So, let us call this is equation number one then further we can write it you see the expression of z 1 z 1 is how related to with d 1 how related to z 1 is nothing, but a if you see this one z 1 multiplied by delta a m agree and z 1 is what is nothing, but a x 2 dot z 1 is nothing, but a x 2 dot. So, I can write it z 1 expression you same as the x 2 dot expressions this one this is nothing, but our z 1 of t if you see. So, if I write the that signals z 1 of t z 2 of t z 3 of t we can write it this into state form in express in terms of the state in terms of part of model output and the input will filled up this thing and which is nothing, but a u of t. Now, z 2 is nothing, but our if you see x 2 dot and x 2 dot we got it this one you just filled up like this way. So, what is that one minus x 0 m 0 minus b 0 m 0 and this is minus 1 by m 0 minus 1 by m 0 minus 1 by m 0. So, z 1 is nothing, but a x 1 into this plus x 2 is this d 1 into this d 2 is this and d 3 into this what we got it that one plus one term is there z 1 which is equal to 1 by you see this term is there 1 by m 0 f of t 1 by m 0 f of t. So, this is why I have written z 1 what is z 2 see this expression z 2 is what is nothing, but a z 2 is nothing, but a x 2. So, what we will write it z 2 is nothing, but a x 2. So, you will write it simply 0 1 other thing is 0 0 0 and this is 0. So, this is our z 2 what is z 3 sorry z 3 is you see z 3 is x 1 z 3 is nothing, but a z 3 is nothing, but a your x 1 z 3 nothing, but a x 1. So, z 3 is x 1 means 1 0 0 0 0 0 0. So, this is we have express the part of model output relation between the what is called z 1 z 2 z 3 expression like this way what is the main purpose of writing this one I will tell you later, but this is we can write it this one. So, what is this one I will write it now this is nothing, but our if you see this is nothing, but our x 2 and this is nothing, but our x 1. So, we consider this is our c 1 and this is our x of t and this is our d 1 1 and this is our w of t or in general that is d 1 2 and this is u of t that this is a u of t then another equation we can write output equation. What is our output if you consider the output our system is x 1 this is our output let us call this multiplied by 1 is our output. If this is the output we will write the output equation y t is the output equal to x 1 which you can write it is equal to c 2 x of t in general expression d 2 1 w of t d 2 2 u of t in our case this term is 0 d 2 1 0 and d 2 2 is 0, but in general structure I can write it like this way in our case this is equal to 0 this equal to 0 and c 2 c 2 in your case is 1 0 this is our c 2 in your case. So, in general structure is like this way now I can write the general structure of any part of model all this thing we can if it is additive perturbations is there we can easily take out the perturbation model perturbation term from the model agree. So, our state equation in general we can write it in general x dot of t is equal to a x of t plus b 1 of w of t plus b 2 of u of t and z of t is equal to c 1 of x of t plus d 1 1 w of t plus d 1 2 u of t plus y of t is equal to c 2 x of t plus d 2 1 w of t plus d 2 2 u of t. So, I am now I am telling this is x is a state vector x is this equation is called state equation and x is a state vector whose dimension is I am writing this dimension is let us call this dimension is n cross 1 n cross 1 and this is our exogenous exogenous or you can write it this is the our input what input part of system input. So, our you or you can write it this exogenous exogenous input or signal this is the input signal and this is the regulated output this is the regulated output and this is the measured output and this is a structure of as I mentioned a general structure of control problem if there is a perturbation all these things there plant perturbation all these I can write it into this structure of this one. Now, see this one what so this dimension let us call the exogenous input dimension is p cross 1 and input dimension is m cross 1 once I define the exogenous input agree and that this the input is m cross 1 that m cross 1 and this is pre cross 1 of this one let us call then immediately I can find out this dimension immediately, but let us call this dimensions is p 1 cross 1 and p 2 cross 1 this dimension or you better you just put it is a m 1 cross m 2 cross input this is input this is also input, but this is the exogenous input with this input we do not have any control you see in our case this is d 1 d 2 d 3. So, this is the output of the part of plant we do not have any control with this input agree to the plant that is d 1 d 2 d 3. So, that is why it is exogenous input and this is the input signal which we can generate this control action and that regulated output is p 1 cross 1 and measured output p 2 cross 1 and once you give this dimension every all matrices dimension is fixed now. So, it is the general structure of that one. So, now we are going to what is called the structure p delta form structured p delta m. So, now if we conclude this one you see if you recollect this one our actual system was this is our exogenous being mass damper systems and nominal plant is m 0 k 0 and b 0 and since there is a discrepancy between the actual plant and the plant model. So, this discrepancy is compensated by part of model and that system we can represent into a this structure form where w is d 1 d 2 d 3 and what is d 2 d 1 d 2 d 3 is nothing but a the output of the part of model output of the part of model d 1 d 2 d 3 that is we have seen it. So, in general any physical system is given we can express into this form one is state equation one is regulated output will more tell about the regulated output later and is a measured output where this output will be used for control to generate the control action that means you and this is the regulated output that means we want to regulate the state the error vector again then we have to control the control signal we have to controlled signal we have to I mean control then this that is why it is called regulated signal controlled signal some signal we have to control it, but all these things z is direct effect of this exogenous input and exogenous input are such input which we do not have any control such as the sensor noise and your that external disturbance and your what is called the command input command signal that is our exogenous input sensor signal section noise disturbances and command input generally we considered is a exogenous input and regulated output is what as I mentioned state error state vector then your control signal controlled signal mind it the signal which are going to control the signal this. So, we are now talking about the structure signal that what is the model we got it this one how we can describe this in this in trans function model form. So, this model you see look what I am doing it here. So, I am representing the state phase model of that system again into a 4 block representation this is you can get 4 block representation. So, we have a P m this one P 11 P 12 P 21 P 22 and we have a input now look says how many inputs are there there are 2 inputs one is exogenous input another is your input signal how many outputs are there 2 outputs regulated output and the measured output which will be used for our control purpose. So, naturally I can show it into 2 inputs 2 output systems this is our z this is our y you can say y and this is our exogenous input and this is our control input u of t. And this q of t I told you y of t is used for control purpose designing a to get a control signal that let us call k of s is the controller the measured output is used for measured output is used for generating the control signal u of t in terms of this you can get time this then you see there is a perturbation is associated what is w in our example if you see the w is what if you see this is d 1 d 2 d 3 is equal to our w how they are related if you see how they are related if you see this expression how d 1 is related z into delta a m z 1 into delta m is d 1 z 2 into delta a b is d 2 z 3 into delta a k is d 3. So, we can if you see this one for our example we can write it that expression from the figure itself z 1 z 1 is equal to delta a m delta a m into sorry d 1 is equal to d 1 is equal to z m into delta 1 d 2 is equal to delta a b into z 2 d 3 is equal to delta a k into z 3. Now, this thing you see how we can write it d 1 vector from your d 2 d 3 I can write is delta a m delta a b and delta a k multiplied by z 1 z 2 z 3 which can be written as that is nothing but a d which is nothing but a w is equal to delta a this is I denoted this is all elements are 0 this is 3 by 3 dimension is 3 by 3 is this that I denoted by a this matrix I denoted by a diagonal matrix which denoted by delta a delta a into that z that is I denoted z into z and that is I denoted with a d which is nothing but a is equal to w I denoted. So, now, see this one so what is you can think of a delta a multiplied by z. So, it is acting as a input then output is the part of model output is the. So, now, ultimately you see the state base equation what we got it from there delta a is taken out. So, our plant model is now in transformation domain it is a that one this whole thing equivalently I can represent by this one then how to find out that p 1 1 p 1 2 p 1 2 1 p 2 2 that is next question the how to find out this one. So, easily you can find out from the knowledge of that things how will you find out p 1 1 see this one p m if you consider the p m just I am writing p m p m is what p m is p 1 1 p 1 2 sorry p 2 1 p 1 2 p 2 2 and what is this input input is that is in transformation form input is w r u z this. So, your transformation you can say then I can write it now z of s y of s is nothing but your p 1 1 of s p 1 2 of s p 2 1 of s p 2 2 of s into the your input w of s and u of s how what is p 1 1 of s experimentally I can find out p 1 1 s I do not excite with a input agree I gave it the excitation with this one then what will be the z z 1 if you see this has a two parts p 1 1 s into w of s plus p 1 2 s u of s it is a z of s y of s is equal to what is this p 2 1 of s w of s p 2 2 s u of s. So, how to find out p 1 1 1 s you excite with only u of s assume u of 0 is 0 then you will get it p 1 1 1 of s agree u 1 get p 1 1 of s then how will you get it p 2 1 s because I excite with a w of s only w u of s is 0. So, it will p 2 s is when you excite this one ratio of this two things or y of y 1 of s is a p 2 1 s this I can get it this one similarly p 1 2 s you excite only with u of s then you will get it. So, from this one now I can write it what is p 1 of s can you tell me what will be the p 1 of s p 1 of s u 1 of s is u of s u of t is 0 agree. So, if you recollect this one I am just telling you this you see suppose we have a system x dot is equal to A x plus B u say I am just recall you please recall this expression. How to find out the transformation from the state pace equation how to find out the transformation from the state pace equation is the transformation is g of s from state pace equation to transformation model g of s is equal to c into s i minus a whole inverse B plus D agree that is the transformation same thing I will apply here I will apply here to find out the transformation of this one. You know already that this is a state pace I want to find out the transformation of p 1 of s then what will be the u of s is 0 means u of t is 0 u of t is 0 then what is this one you will get u of t is 0 this term will not be there this term will not be there. So, between what between w and z is p of s you see between w and z will give p 1 1 of s. So, can you tell me what is the p 1 of s now from this one let us see from this one p 1 of s p 1 of s p 1 1 s will be equal to if you see this one this is 0 this is 0 now between the state equation is there output equation is this one regular data. So, what is our transformation when this is not there between this two thing transformation c 1. So, c 1 then s i minus a whole inverse then b b 1 then d d 1 1 then how will you find out the again the what is called p 2 1 s p 2 1 s the input signal w s and y s is the output when u s 0. So, between state equation and this when u is 0. So, again I will write p 2 s is equal to c 2 s i minus a whole inverse then your b 1 then your d 2 1 that is p 1 when this is means when u of t is equal to 0 this two quantities I can get it when u 2 is 0 then I can find out p 1 2 s p 2 2 s when w of t is equal to 0. Now, when w of t is 0 you see here I want to try p 2 1. So, this quantity w of t 0 when w s is 0 then I can find out the input is u s output is z. So, what is p 2 what between the input is y output is z when w is 0. So, how will you find out that that one again p 1 2 is c 1 s i minus a whole inverse b 2 plus d 1 2 d 1 2 similarly that between input and the output when w is 0 w of t is 0 then again you can write it c 2 s i minus a whole inverse then your b 2 then d 2 2. So, I can easily find out what is our that matrices what is that one this matrices we can easily find out from the knowledge of the state page equation of this one. Now, this is this one. So, now I am writing it what is z z is the z is represents z represents those output of the system whose dependence dance on w of t. For example, I told you states for examples states error signal error signal and controlled mind it is not a controlled signal controlled signal the signal where which you are going to signals control and that are dependent on w effect of w on z will consider this and this is the measured output measured output which will be used for designing a controller or generating the u of t. And this we have written this is the control environment w is in general w is is the exogenous input as I told you x that we do not have any control that noise disturbances noise noises disturbances and command input inputs that we do not have any control. So, this is the any system now short any system we can represent into a four block diagram representation of this is the block four signals of their input output two input two outputs are there in general I can represent in this one if there is a perturbation is there at the perturbation we can take it out from the plant model and the controller is this one. So, this is the general structure of then I we know how to find out that one. Now, next question is if you want to merge that you see the if you want to merge the perturbation block this is called perturbation block perturbation block. So, this is can call upper loop of p this is the p m upper loop of p m agree. So, this if now if p m of s and plant uncertainty if you want to merge together p m and delta you can merge together then how you can do it if the plant p m s and plant uncertainty delta a are combined to give p delta of s. Now, how to do it this one we know this expression z y is equal to p 1 1 p 1 2 p 2 1 p 2 2 w and u and also we know that w is equal to delta a z and delta a perturbation block which we can place it in diagonal form all perturbation in the elements we can put it in diagonal form. So, now I can write it z is equal to p 1 1 w plus p 1 2 w u 1 expression y is equal to p 2 1 w p 2 2 u. Now, from this eliminating w from this expression eliminating the how will we eliminate w w expression I can put it in terms of z then I can express z in terms of u agree. So, you that if you express z interval let us see how to eliminate the eliminating w form y expression that y expression first thing we can write it y is equal to p 2 1 w what is w z a I am sorry delta a into z plus p 2 2 u agree. So, what is z now what we can write it this y delta a this. So, y is equal to now and another expression I know z is equal to what z is equal to I know p 1 1 delta a z p 1 1 w the for w I am writing delta a z plus p 1 2 u. So, z is equal to bring it that side agree. So, I can write it i minus p 1 delta a whole z is equal to p 1 2 u. So, z is equal to i minus p 1 1 delta a whole z. Now, one can easily find out the dimension of p 1 1 from the basic equations immediately you can find out that what is the it depends on the dimension of this one depends on dimension of z and w similarly. So, number of inputs is in this case number of inputs is w and number not number of inputs is w and w dimension if you specify and z is the measured output whose dimension is known. So, I will note the dimension of p 1 1 s. So, now I can write it in place of z put this one p 2 1 agree p 2 1 then what is the delta a delta a then z is what I minus p 1 1 delta a into delta a into what is our z sorry it is not a z it is a u. So, this is our p 2 1 delta a agree then this just to meet this is your because you see I am taking the inverse both side this one. So, it will be a p 1 2. So, there will be a p 1 2. So, p 1 2 this delta a as it is z I have expressed by that one into that is u is already there. So, you have a u plus p 2 2 u. So, if you now take it common. So, p 2 2 plus p 2 1 delta a I minus p 1 1 delta a inverse p 1 2 then u. So, now you see how to remember that that one when you want to make the outer loop outer upper loop this is the lower loop outer upper loop is delta a outer lower loop is delta a when you want to merge the outer upper loop block into a p then input output relationship input output relation y and u relationship trans function you are getting this now how to remember when you are putting the upper block agree to merge into this block then you consider the lower part of this one p 2 2. So, p 2 2 the first index second is p 2 2. So, first index will be there p index is 2 p 2 1 delta a the perturbation we are merging that delta a is merging delta a then multiplied by I minus p 2 2 only we consider what is left is p 1 1 is left multiplied by delta a whole inverse then first index we have taken it p 2 1 then this second index we have to take second means second index. So, p 1 2 is a second index it is coming multiplied by u. So, this you can write it now block diagram form if you see it it is like this way p delta of s this is w this is z this is our u and this is our y you see the upper block of 4 block upper loop is merged and that is denoted by p delta a. So, this is nothing but a p delta s u of this and now this is the controller. So, this thing this whole expression we can write it is a function of p and delta a u and that is called this is called LFT this is called linear fractional transfer function linear fractional transfer function transfer LFT. And since the upper block is merged to this block 4 block it is called U LFT U LFT since the upper block is merged to this one it is called U upper linear fractional transfer function transfer transformation or not transformation transformation. So, this is called linear fractional linear fractional LFT linear fractional transformation since the upper block is merged it is called upper linear fractional transformation again. Similarly, one can do if you like it if you like it now you see the outer loop of the block is merged to the outer loop of the p block is merged to the p block. Now one can do the outer upper loop the outer lower loop can be merged to the p block and that will looks like what that will be looks like this. So, we will define this is a p k delta of s. So, tomorrow we will discuss this one how one can do this one and this is w and this is z. So, this is a lower block is merged to the p block. So, it is called lower LFT lower fractional transformation the lower block lower loop is merged to this one. So, tomorrow we will discuss the rest of the things. So, our main job is how to any system given is description perturbation noise everything we want to represent into a flow 4 block representation and it is analysis in trans function domain.