 So, last class we are discussing about the solution of H N U V D control problem state space formulation framework. So, our planned description is x dot is equal to A x plus B 1 W B 2 U, W is the exogenous input whose dimension m cross m 1 cross 1 and this is the control input whose dimension is m 2 cross 1 and z is the regulated output whose dimension you can write it P 1 cross 1 and W U R as I mentioned earlier and y is your measured output that will be used for control purpose to generate the control U. So, that dimension you can write P cross P 2 cross 1. So, this is the description of the plan and if you see we have a 2 inputs U control input and W is the exogenous input in all equations. So, as we know our let us call we if we have a system x dot is equal to A x plus B U y is equal to C x plus D U which we have discussed earlier then what is the trans function of this one nothing, but if from the knowledge of the A B C D matrix we can write is C S i minus A whole inverse B plus D when D is equal to 0 the transformation will be strictly power trans functions and this thing the transformation if C with the knowledge of A B C D we can find out the trans function of the system. So, pack notation means in short hand we can write G S if you know the information A B C D I can easily generate trans function it does not mean G S is equal to this one mind it G S if you know the A B C D matrix I can write this trans function into this form. So, keeping this thing in mind then we can write our given set of equation in state phase form we can write in packed form on short hand form our A is this one as I told you our input are 2 inputs are there if you see earlier equation that these are the 2 inputs one is input control input and another is the exogenous input in all equations. So, this thing if you want to find out the trans function of this one we can write in compact form A B is combined B 1 and B 2 C is combined C 1 and C 2 and D is all inputs with this input. So, similarly I can write it trans function of P of S C this is our C you can compare D this is our S i minus A this is our B and this is our D. So, if you expand and write it into a matrix trans function form we will get 4 block like this way and now you see this 4 blocks where the 4 blocks are the what are the inputs w and u w is the exogenous input and u is the control input and z is the regulated output y is the measured output. So, as we mentioned that this trans function P of S is nothing but a C S i minus A B plus D and if you expand this expression and write it into this form you will get it this C i S i minus A whole inverse then B i plus this this one D 1 1 that is our P 1 1 S similarly you can write C 2 S i minus A whole inverse this is B 1 then B 1 plus D 2 1 and what will be this one again now this column C 1 S i minus A whole inverse B 2 plus D 1 2 now this block C 2 S i minus A whole inverse B 2 plus D 2 2. So, if you just this column this matrix multiplied by this then multiplied by this then add with this one you will get it this one in other way this is nothing but if you see this one this is nothing but a P 1 1 S this is nothing but our P 1 1 S this is nothing but a P 1 2 S and this is nothing but a our P 2 1 S and this is nothing but our P 2 2 S. So, in other words you can say when input is 0 find out the trans function between w to z w to z when input is and that transform this from is not there and that will be this one what we got it here now similarly the transformation between w to y when u is 0 that transformation P 2 1 is nothing but a P 2 1 is nothing but a you see C 2 S i minus 1 inverse B 1 plus D 2 and that I can write it straight away from the description of that state base equation if you see the state base equation that you can write it when input is not there agree then one input is now transformation between w to w to z what is the transformation u is not there this is not there. So, it is a b c d I can write it simply when transformation between w to z when u is not there these two term will not be there this matrix is a b c d that is what I am writing for all this cases then P 1 2 P 1 2 you see when w is 0 when w is 0 then what is the transformation between z to 2 y that transformation is C 1 S i minus a inverse B 2 plus this and that you can write from basic state base description of our plant from there you can write it similarly I can write it P 2 2 S if you see I can write it the P 2 2 S transformation between y to u when w is 0 that transformation is that one how can write it as I mentioned from the plant description of this one from this one I can write transformation between u and y when w is 0 this. So, or a b c d w is 0 means this term will not be there. So, each block 4 block representation P 1 1 P 1 2 P 2 1 P 2 2 we can find out is transformation just this way now if somebody interested because our ultimate aim to reduce the effect of w on measure what is called regulated output that. So, what is the transformation between these two things as you recall you can recall that our 4 block representation is that two inputs w is the exogenous input u is the control input z is the regulated output y is the measured output which will be used for control purpose. So, this P 1 as a if you see P of S we have like this way P 1 1 P 1 2 P 2 1 P 2 2. So, if you want to find out the transformation between the w to z the lower loop this is the lower loop of P this is the lower loop of P is closed with the block P. If you want to find out the transformation between w to z transformation and the k of s in the lower block of P that is closed with the block k s. So, how to do it when you are lower block. So, this is the lower block matrices again. So, it will be starting with the upper block matrices you see P 1 1 and it suffixes 1 1. So, I will start with first suffix. So, what will be P 1 2 s into k of s this is we have to merge into P block break it I now P 2 2 s P 1 1 already P 2 2 s into k s whole inverse. So, I have already taken the what is called that first suffixes already taken over the second suffix I have to take last first suffix I have to taken first and second suffix I have to take last. So, it will be P 2 1. So, second suffix will be last this one P 2 1. So, this is the transformation between this to this of this one. So, our if you look our statement of the problem is solution of h infinity control problem standard h infinity control problem means that design a suboptimal control k of s that this you design a suboptimal control k of s such that our closed loop this closed loop system should be internally stable not only that and effect of disturbance on regulated output should be less than some pre specified positive quantity. So, our goal is this design of suboptimal what is called design of suboptimal h infinity controller means or standard h infinity controller means design a suboptimal controller k of s such that closed loop system is internally closed loop system is internally stable agree and not only stable our motivation to design k such that h infinity norm of transformation between w to z will be less than some gamma. So, there are two folds in our motivation they have to design k of s in such a way that it should stabilize the closed loop system internally as well as the h infinity norm between the signal w, exogenous input w and regulated output z h v norm will be less than gamma and gamma is some p specified value positive p is greater than 0. So, before I as I mention as I am now telling you detail derivation of standard h infinity control problem we will not discuss derived only we will give you the algorithmic steps for this one what the before that we have to make some assumptions the h infinity suboptimal solution is obtained by the following assumptions the h infinity suboptimal solution is obtained by the following assumption what is this following assumption first thing to controller just to obtain is stabilizing controller k s to obtain stabilizing controller k s the necessary and sufficient condition is these two a b 2 this pair b 2 associate with u in the state equation and c 2 associated c 2 associated in the output equation in the output equation c 2. So, our a b 2 must be stabilizable and a c 2 must be detectable. So, these two ensures these two condition is the necessary and sufficient condition for existence of for existence of stabilizing controller. So, this is the first before we apply the h infinity suboptimal control problem solution of suboptimal we have to check whether the system is stabilizable or detectable a b 2 n s n next assumption is because when you will design the h infinity controller we have to solve two Riccati equation and the existence of two Riccati equation agree guarantees that this rank must be satisfied this rank must be satisfied what is the first a j omega i b 2 associate with a u 2 u and c 1 in measurement equation matrix c 1 and d 1 2 and this matrix should have full column for all frequency whatever the frequency you train you let us call frequency is 0.1 because I know a b a b 2 c 1 d 2 find out the for particular frequency find out the rank of this one and that rank must be full column rank for all frequency when you will sweep the frequency from 0 to infinity the rank should not differentiate from full rank that means n plus m 2 this is the a 2 1 condition and a 3 condition is the rank of a b 2 c 1 d 2 find out the j omega i b 1 associate with the exonation put c 2 is associate with the what is called measured output and d 2 d 2 1 associate with the exonation that rank should be a full row rank full row rank for all frequency means some frequency omega is equal to omega 1 find out the rank that rank must be n plus m 2 another frequency omega to find out the rank of this one that also must be n plus m 2 for all frequencies this rank should be full row rank. So, this dimension column is n this column will be how much m 1 and this row is p 2 and naturally this one will be n. So, this should be a full row rank. So, a 2 and a 3 condition ensure that the two Riccati equation which will be necessary to design the h infinity controller one Riccati equation for designing the controller another Riccati equation will be necessary for designing a estimator. So, these two existence of solution of these two Riccati equation ensures the rank row rank full row rank and full column must satisfied and not only this the h infinity controller when you will design it it will ensure there is no pole zero cancellation on the imaginary axis on the imaginary axis. So, this ensure that the system will be internally stable. So, this is the condition three and condition four is the rank of d 1 2 d 1 2 associate if is the state page description our original problem d 1 2 associate with the input in measurement equation and d 2 1 associate with the output equation in the what is called exogenous input. So, this should be full column means rank should be m 2 and this should be full rank that means rank of this one will be p. So, this ensures this two condition ensure that whatever the controller you will get it after designing h infinity controller that controller will be proper that controller will be proper not only this you will be able to realize this controller it will not be improper trans function of that one that controller. So, this condition also necessary to be satisfied. So, condition 1 to 4 is essential to design a h infinity controller standard h infinity controller or suboptimal controller when you will go to design based on h infinity controller approach then this a to 4 that condition must be satisfied. Other condition is the simplify the our control algorithms. So, a 5 conditions that means fifth condition this condition is simplifies the h infinity control problem only. So, if b 1 d 2 1 d 2 1 dimension is p 2 m 1. If b 1 d 2 1 transpose is equal to 0 physically it implies b 1 associate with the noise in the state equation means noise means exogenous input d 2 1 if you see the d 2 1 is associate with the measurement noise d 2 1 associate with the measurement noise. Though we have considered both are same that means they are correlated if this will be 0 when they are uncorrelated this that mean process noise and the measurement noise are uncorrelated then this product of this one will be 0 because this product will come in the h infinity control algorithms. There will if it uncorrelated there itself we can write it b 1 d 2 1 transpose whenever we will get it will put it 0. So, from this one I can write it d 2 1 always we can make into this structure 0 i by using that what is called some transformation I can any d 2 1 I can convert into this structure agree. So, if you make it this structure then d 2 1 d 1 2 transpose is identity from this you see identity further implies that b 1 d 2 1 if you multiply by d 2 1 transpose this into this 0 this into this is i identity matrix this. So, if the structure of d 2 1 is this one I can write b 1 d 2 1 and d 2 1 into d 2 transpose is identity and b 1 d 2 1 is 0 when process noise and measurement noise are uncorrelated. This is the fifth condition, but that is not the essential to design a h infinity controller. So, sixth condition is again d 2 1 transpose d c 1 if it is 0 and d 1 2 d 1 2 associate with the regulated output noise it is d 1 2 associate with the regulated output equations and it is associate with the existence input. So, if this structure d 1 2 structure if it is 0 i I can always make even it is not that structure I can always make it 0 and i structure. If it is this implies d 1 2 transpose d 1 2 is identity matrix and further I can write it d 1 2 c 1 d 1 2 that this multiplied by c 1 d 1 multiplied by c 1 this is transpose agree d 1 2 transpose c 1 is 0 that is we have seen and d 1 2 transpose d 1 is identity degree. So, this simplification is if you can simplify the our h infinity standard control problem this is not the essential what is called assumption that 5 and 6 similarly 7 also. What is 7 d 1 2 is 0 if you see the d 1 2 is 0 state way we can see from this equation I will show you this one if d 1 2 is 0 if this is 0 that means our p 1 1 s is equal to c 1 s i minus a whole inverse v 1. So, this p 1 1 matrix which has transformation between w 2 z will be strictly proper transformations if d 2 2 is 0 see if d 2 2 is 0 just here that is if d 2 2 is 0 that is then c 2 s i minus inverse v 2. So, p 2 2 the transformation between u 2 y the transformation p 2 s is your will be strictly proper transformation. So, these two assumptions d 1 2 0 d 2 2 is 0 is not essential, but if it is 0 p 1 1 s will be strictly proper transformations and this is will be strictly proper transformation p 2 2 if d 1 1 is 0 p 1 1 will be strictly proper transformation and this will be strictly this assumption as I told you not needed, but it is simplifies the what is called the solution of Riccati equation other things and also the transformation between w 2 z. So, our main basic assumption a 1 to a 4 and other assumption 5 6 7 is not a essential assumption to design the standard h infinity control problem. So, let us as I told you I will not go for details derivation of the h infinity standard h infinity control problem we will just write the algorithm steps how to design sub optimal h infinity controller k of s. So, our algorithm step for sub optimal h infinity control problem solution. So, first step you write in state page description of the plan into this form this is a state equation and is x is the state vector whose dimension n cross n w is the exogenous input whose dimension m 1 cross 1 u is the control input m 2 cross 1 and z is the regulated output and w is exogenous input u is exogenous u is the control input and hence I can find out the dimension of a b 1 b 2 c 1 d 1 1 d 1 2 and y t is the what is called measured output since I know the dimension of w d u and x I can find out the dimension of c 2 d 2 1 d 2 2. This is the first you have to make it into this structure if you want to design h infinity controller based on state page model approach then you have to make it in this structure it is called four block representation is something like this form two inputs w and another input is u and this input is your output is z regulated output this output is y and we have a controller u we have designed based on this our controller k of s. So, this is our p of s the four block representation you have to write like this way. Next step is that as I told you the check the assumption one check the assumption one a 1 a 2 a 3 are satisfied now if they are not a 2 and a 3 this is the corresponding a 2 and a 3 if it is not satisfied then what you have to do it then you have to modify that our system by introducing the weights. So, this as I mentioned what is the useful assumption for a 1 is what for a 2 a 3 for what you are doing a 1 is assumption is that a b 2 a c 2 must be stabilizable and detectable in order to get the controller stabilizing controller. So, this is the second step check the all assumption a 2 a 1 a 2 a 3 the third is check the assumption a 4 a 7 though I told you it is not necessary in a 4 and a 7 you see the a 4 and a 7 assumption is what it means the a 4 the assumption full rank column rank full row rank is used for power getting the proper controller even if it is not proper it is a by proper then hence we can this will give you the proper what is called transfer function of the controller. Hence, you can realize for a 4 and a 5 a 6 a 7 agree this assumption a 4 assumption check the assumption a 4 and a 7 a 7 assumption is what a 7 assumption d 1 2 is equal to d 1 1 0 d 2 2 is 0 means to get the p 1 1 p 2 2 as strictly proper transform and that does simplify the algorithm only. So, check, but as I mentioned earlier in our assumption this assumption this is our assumption if you see this our assumption is this is the assumption is 5 we have considered 0, but this assumption you can relaxed this if you see b 1 d 2 1 we assigned in assumption 5 0. So, that assumption you can relaxed and this is the assumption 6 if you see that assumption also you can relaxed it this assumption. So, I am relaxing that assumption step 4 select a large positive value of gamma positive value of gamma where the h and v norm will be less than gamma. Gamma is positive quantity large positive initially you have considered large positive number of gamma where gamma interprets is what the h v norm between the exogenous input to the regulated output solve two Riccati equations. Now, next step is after checking all these things you have to solve two Riccati equation. What is this two Riccati equation before I write the two Riccati equation in details that our recall our LQG problem or LQR problem what is this or LQG or when we are going to design a controller we have to solve a Riccati equation this is called algebraic Riccati equation. And this algebraic A transpose P P A minus P B R inverse B transpose P plus Q you have to solve it where you know A B R is the designatures Q is designatures you have to solve P once you solve P you can easily find out the controller gain R inverse B transpose P the controller gain in R inverse B transpose P. If you want to if you once you solve this one and you see I need the information of A B Q R agree if that Hamiltonian matrix this is called Hamiltonian matrix A minus B R inverse B transpose minus Q is the Q is the weight is associated with the state vector and R is the weight is in the control vector and minus 8. So, this is the Hamiltonian matrix and from the Hamiltonian matrix I can create what is called the Riccati equation once you know the Hamiltonian matrix of this one I can create the what is called algebraic Riccati now how would it see A transpose A A transpose P then transpose of the P A minus P this quantity minus P this quantity B R inverse B P minus P this quantity B R inverse B P this quantity that minus will come and then P plus Q what is the sign is there plus Q. So, from the in short I am writing Riccati of this one means from the knowledge of Hamiltonian matrix this is nothing but a Hamiltonian matrix this is nothing but a Hamiltonian matrix from the knowledge of Hamiltonian matrix one can create the algebraic Riccati equation. So, keeping this thing in mind what I am doing it now I am writing the Riccati equation that we have to solve for the controller. So, now you see the x infinity is a solution of Riccati equation for designing controller. So, this our Riccati equation this is the Hamiltonian matrix this is the Hamiltonian matrix and Riccati equation I can form the knowledge of Hamiltonian matrix what is this I can consider in equivalently you can consider this is A this is this whole thing equivalently and this is whole thing equivalently I can consider if you consider L Q R then this is nothing but a B R inverse B R inverse this is equivalent if you consider A this is whole thing equivalently B R inverse B transpose this whole thing equivalently you can consider minus Q and if you this is A then it will be A transpose this whole thing you can minus A transpose. So, I can easily write the Riccati equation A transpose P sorry A transpose P means our x infinity here I have considered the matrix the solution of Riccati equation A is x infinity A transpose P then P A then minus because minus then what is I am writing P that quantity that quantity minus is there here then P. Then this quantity this quantity is minus Q plus U we have to write it this. So, this is the Riccati equation and what is the Riccati equation for designing the controller H N V D controller designing therefore, controller part and what is here we introduce the new thing that D 1 to bar D 1 to bar is D 1 transpose D 1 2 that is I am considering D 1 to bar inverse again the whole thing is D 1 to bar inverse. Now, if you consider the special structure of D 1 2 as I mentioned earlier D 1 2 as I mentioned earlier if you see that one anyway the 0 I 5 just minute. If you consider D 1 2 structure you see D 1 to transpose D 1 2 is identity. So, if you that structure this will be identity again it does not matter even if you not that structure it will be you will calculate and put the value is here, but it is known to you. So, you have to solve the Riccati equation we have already discussed how to solve the Riccati equation different methods that you recall. Once you know the solution of the Riccati equation x infinity then controller gain k consequently again I can use by using that one and this term will be 0 if this term is 0 if the if this is 0 then this will not come into the picture in our things. So, our controller gain is that one this is 0 if this quantity is 0 then this term will not be there agree or you forget it is this will not be 0 if this is 0 that what we have consider in this assumption. If D 1 2 transpose C 1 is 0 then naturally this will not come into the picture if this is if. So, similarly that what is called that what is called we can design this is the controller this is the your estimator this is the Riccati is the estimator Riccati equation. What is the Riccati and this is the knowledge of A you again you can consider this is A this is our minus B R inverse R inverse B transpose then this is our Q this is our if it is A this will be A transpose. So, you can write it and our D 2 1 is what that one and B 1 bar expressed by this one. So, D 2 1 bar if D 2 1 has a having a special structure then this quantity will be identity even though it is not a special structure one can compute by knowing the knowledge of D 2 1. So, from this Hamiltonian matrix this is nothing but a Hamiltonian matrix I can write the Riccati equation. What is this A I have written A transpose means that matrix only this whole thing is A A transpose means this will go A transpose P then P A P A transpose A what is this A will be that will transpose I missed it here P A then P this matrix is written this matrix into P because it is minus is there minus plus Q that one. So, this is the algebraic Riccati equation this is you can say algebraic Riccati equation you have to solve the algebraic Riccati equation why infinity of this one this you know all matrices then you can solve, but this is a matrix and non-linear algebraic equation. So, this is a step 8 is the solution of what is called Riccati equation for estimator agree that you have to solve it. Now, step 9 is estimator gain for estimator or observer gain you have to find out is nothing but a that z infinity into K e what is z infinity is related with this expression. I know x infinity I know y infinity gamma I have assigned initially very large. So, that the z is no z infinity is known and K e is what y infinity you know all these matrices you know you can find out K e not in many K e. So, I can write this is nothing but a estimated gain K e l. As you know estimated equation now estimated equation x hat dot from basic equation I can write it from basic equation I can write the estimated equation you see the our basic equation see that this is where the basic equation estimated equation I can write it x hat dot is equal to a hat b 1 w hat plus u 1 w hat plus correction term. So, I am write x hat dot is equal to a hat plus b 1 w hat plus w 2 b 2 u of t plus correction term and this is the output measurement error is the actual measurement and this is the estimated measurement that multiplied by what is called estimated gain this is called estimated gain. So, this is the how is calculated w hat that this expression in this right hand side everything is known to us from this one and the y hat that output estimated value of the output is that one. And if you see this one our standard what is called LQG problem where the estimator is there this term and this term are coming extra in H infinity H infinity controller in comparison with LQG. This term and this term agree and this term and this term is coming extra for this one. So, and w hat you know how to calculate w hat this one estimated for this. So, I know the estimated equation well defined now. So, I have given the equation number 1 is our if you see the our equation number 1 is the our controller gain control law this is the equation where u is equal to k c x hat of t equation this is the estimated equation is 2 and w hat y hat is 3 and 4. So, now one can easily find out the if you put this if you put the value of y hat here w hat here y hat here and after that u you put u is equal to minus k c x hat in this expression then I can simplify this equation in final form. So, that is why we are writing the using equation number 1 then 2, 4, 5, 6, 7, 8, in equation 4 we get this expression is nothing but a if you see you can get this expression by putting the value of y hat by this expression sorry w hat and y hat by this expression in this and u u u of t you can put minus k c x hat of t in this expression then finally, estimated equation this is the estimated equation you will get this one. Now, if somebody wants to find out the transformation of the controller as you see from the very beginning I have written a transformation of this controller here the transformation of the controller between y to w y to u. So, one can easily find out the transformation of this one step 11 transformation from y, y is the input u is the output of the controller input to the controller y output to the controller u how you will find out using equation 5 and our equation 1 and our equation 1 is what is nothing but a u is equal to minus k c x hat of t this is our equation 1. So, from this one I can easily find out the this is find out the transformation k of s what is it is something like that x dot is equal to a x plus b u y is equal to c x plus d find out the transformations. So, in our case this whole thing will be our a y of t is the input of the controller. So, this is our b our output is u of t. So, this will be our c. So, it is a c here b here and d there is no d term here d is 0 d bar and whole thing is you can think of is a bar b bar all this thing. So, our controller transfer function this is the transfer function of the controller generally this controller transfer function will be very large order the even it is more than the our original system order. So, one can if you one one can find out the transfer function of the controller in this one. Then finally, there exist a stabilizing controller k s if and only if necessary and sufficient condition two Riccati equation one Riccati equation for controller another Riccati equation for the estimator are solvable and they are solvable provided the assumption which we made it in the beginning of the solution of the problem assumption that must be satisfied. And following condition is satisfied that means rho x infinity y infinity of less than gamma. That means spectral radius of the of this matrix spectral radius of this matrix spectral radius means maximum Eigen value of the matrix maximum Eigen value of the matrix. If it is less than gamma square again that condition also must satisfy this condition following condition the following condition must also satisfied. That means there exist a controller, but that does not mean that controller what we have designed the controller that does not mean the h infinity norm of h infinity norm of T z w will be minimum will be minimum. So, what have to do it next is whatever the initial value of gamma we have selected decrease the gamma and proceed this whole process that following repeat the you see repeat the steps six and eight six is then you when it is you got you have got the gamma initially you have the large value of gamma and we got this solution. Let us call this condition we satisfied then what we mean the gamma value we can reduce it reduce the gamma value. And then again you solve what you have solved it the two Riccati equation find out the controller gain observer gain or the estimated gain agree and then again you check this two condition. If it is less than this one further you reduce the gamma in this way process you go on until unless you are able to reduce the value of gamma. So, if this condition is satisfied that means and repeat the steps six and eight what is the steps six and eight see repeat six and eight that this is a six and eight that solution of controller Riccati equation and eight is the solution of what is called observer Riccati equation observer Riccati equation. Once you solve x and y infinity you can easily find out this if this is less than that one then further you reduce and solve x six and eight agree step six and eight. So, this process will continue until and unless you will be able to reduce the value of gamma in the sense that when you have solved the two Riccati equation spectrum norm of this two Riccati product of two Riccati equation spectrum norm should be less than gamma square if you get it then further you reduce it. So, finally I can say compute this you see this is important compute the spectral radius of product of this two Riccati equation until satisfactory solution is obtained then compute finally you compute k c, k l, k s s etcetera to design our controller. So, this is the algorithmic steps of that one now let us see basically how you will realize this h n v d controller in block diagram representation. This is the block diagram form basically I am doing if you see this one basically I am representing the plant first then I am representing the estimator that is all. So, I am just showing you the plant equation that plant this is the plant equation that I am just doing first thing and I made it assumptions that d 1 2 that assumptions anyway. So, this is the equation the plant descriptions plant description is this one I made an assumption in order to draw the what is called that block diagram form I made assumption d 1 1 0 d 2 2 0 this indicates that our the transformation p 1 1 s and p 2 2 s will be strictly proper transformation. In other words we made an assumption that our what is assumption number I think 6 assumption number 6 possibly yes 7 sorry this assumption I made it in order to do this one. So, I made an assumption assumption 7 is please say assumption a 7 means assumption 7 is assumed a 7 is assumed. So, if a 7 is assumed that means d 1 1 d 1 2 d 1 2 d 1 1 d 2 2 is 0. So, first you see plant I have represented with a block dotted box. So, what is x dot look at that plant description x dot is equal to A x. So, A into x plus b 1 into w. So, b 1 this is b 1 w it is coming signal then b 2 u. So, b 2 u that is x dot. So, x dot and our regulated output z, z is what c 1 x d 1 1 I have considered 0. So, there is no input from w. So, d 1 u down part is 0 then what will come d 1 2 u. So, it is coming d 1 2 u this signal. So, that is the regulated output and that dimension is if you consider p 1 cross 1. So, our regulated output this a measured output is what y c 2 into x c 2 into x plus d 2 d 2 1 into what w d 2 1 into w you see d 2 1 into w. So, this is the regulated output. So, this dotted box is the plant description this is the plant descriptions made it and in other words I can 4 block representation I just can make it p s form. Now, it is a estimated equation if you see the estimated equation here that is we just written it estimated equation this is the estimated that is I am now representing in block diagram form. Now, you see the estimated equation is what A this is the estimated y hat what is y hat y hat is a c c 2 x hat c 2 x hat and also you see this one c 2 x hat and also that that one x hat into gamma square you see c 2 x hat and x c 2 x hat where is c 2 x hat and the gain see this one estimated equation. I just written all this equation from the basic estimated equation see from this equation putting the w and y hat and u print I got this block diagram again this with the dotted things and controller is what k c into x hat and k c into x hat with the this is a x hat k c into x hat and our estimator is a A into x hat plus B 1 into y hat B 1 into w hat B 1 into w hat a w hat value you put it this is B 1 into w hat again then B 2 into u then B 2 into u you see B 2 into u this block is coming then will be correction term what is correction term k l into y y hat y minus y hat this is the correction term is coming. So, this and what is y hat y hat is nothing but a your c 2 into x hat plus d 2 1 plus d 2 1 into d 2 1 w hat. So, you write it w hat this one so note that this block and this block gamma square d 2 1 gamma square d 2 1 gamma square B 1 transpose x and gamma square B 1 B 2 transpose these two blocks are extra term in comparison with L Q you see this block and this block where you have written here and here are the extra terms in comparison with L Q R. This is the HMVD standard control problem in structure wise this one estimated is there and this block is called estimated this dotted board is estimated and that is obtained using equation 2 to 3 this whole block is obtained this one. So, this is the all about the course then I will give you the assignment sheet and the one problem I will solve give you the in the assignment sheet how to solve HMVD control problem using MATLAB program because hand calculation as I told may manual calculation is just impossible for higher order system. So, I will work out one example before I give the what is called the assignment sheet you will find everything in then assignment sheet how to solve the HMVD control problem for higher dimensional problems. So, this is all about the course and you will get your assignment sheets from static optimization and dynamic optimization and HMVD control problems all assignment sheets you will get it in the website. Thank you very much.