
#1
Feb1512, 03:43 AM

P: 21

I've been looking at nonlinear control system and fuzzy control in particular.
If I simply want to do trajectory control (have the system to follow a certain trajectory in phase space), can I use much simpler local linear control? I mean, for each point in phase space, do a local linearization and find the local linear control transfer function, as long as the system stay close to the intended trajectory this would work right? 



#2
Feb2412, 12:04 PM

P: 67

What to you refer to by "phase space"?
And yes linear controllers usually can be used successfully within certain limits on nonlinear systems. Most real systems are nonlinear, some much more so than others. 



#3
Mar1912, 09:49 PM

P: 4

First the trajectory setting. It is easier to follow a trajectory if the path has the following conditions: 1. position follows a continiuos function through the path. lets say x=f(x). 2. speed function is continous in the segment, then df(x) continous 3. aceleration ddf(x) is continous. this will prevent from trying to follow a difficult or impossible trajectory. Choose the input variable to use as control input. Usually we use time in order to decide the position. This would mean that time will be use in order to find the error, and you will have some errors in position. perhaps you want to input position and try to accomplish the positioning in the time required, giving you errors in time and no in position. I usually use position rather than time because I prefer to have erros in time and no in position. If you can create a plant or function on what "should" be the inputs in order to follow the trajectory (Output), use it as the main entrance, then add a PI controll and you will have a good result. 



#4
Mar2012, 11:03 AM

P: 1

Nonlinear Control problem
Hello ,Guys!
I have been trying to solve an optimal control problem that needs a code written for it.I have used the forward RK 4th order method (for the state variable) and backward RK 4th order method (for the adjoint equation) but the results I obtained are not right.I am saying this because the article I found the problem from includes the graph of the solution which is nothing like mine .Now I want you to help me find the bug if any or suggest a better algorithm. Here is the Fortran code: !=========================== PROGRAM opct IMPLICIT NONE !Declaration real,DIMENSION(1010)::t,z,eta,w,u,x_1,x_2,A,B,C,D INTEGER::k,i,j real :: h,n,k1,k2,k3,k4,l1,l2,l3,l4,dz,deta,tol real:: gama,g_1,g_2,phi_1,phi_2,z_0,eta_T,z_1,z_2,u_1,u_2 !Initial conditions x_1(1)=427.22;x_2(1)=157.10 z(1)= x_2(1)/x_1(1) !z(1)=0.37 phi_1=0.14;phi_2=0.1;gama=0.5;g_1=0.67;g_2=1.0;u_1=0.01;u_2=1.0 u(1)=0.05;n=1000.00;t(1)= 0.0;z_1=0.1033; z_2=31.5502 t(1001) = 5.0;eta(1001)=0.0;h=(t(1001)t(1))/n;tol=0.001 w(1)= (phi_1 + phi_2 * (z(1)**gama))*z(1)/(g_1 * z(1) + g_2) !print*,w(1) !The header of the output write(19,100) 100 format(5X,"t",10X,"z",9X,"eta",8X,"u") !201 format (2f10.6,2f10.6,2f16.6,2f10.6) !do i = 1,1001 !Do loop of RungeKutta fourth order forward method to get state variable do i = 1,1001 do j = 1,1001 k1 = h*dz(t(j),z(j)) k2 = h*dz(t(j) + h/2.0,z(j) + k1/2.0) k3 = h*dz(t(j) + h/2.0,z(j) + k2/2.0) k4 = h*dz(t(j) + h,z(j) + k3) !The values of z will be calculated z(j+1) = z(j) + (k1 + 2.0*k2 + 2.0*k3 + k4)/6.0 t(j+1) = t(j)+ h !Getting set the time for the next iteration w(j+1) = (phi_1 + phi_2 * (z(j+1)**gama))*z(j+1)/(g_1 * z(j+1) + g_2) !print*, k1,k2,k3,k4 A(j)=k1;B(j)=k2;C(j)=k3;D(j)=k4 end do ! for z !print*,w !end do !write(19,101) !101 format(5X,"A",10X,"B",9X,"C",8X,"D") !202 format (2f10.6,2f10.6,2f10.6,2f10.6) !Do loop of RungeKutta fourth order backward method to get adjoint variable !do i=1001,2,1 do k = 1001,1,1 l1 = h*deta(t(k),z(k),eta(k)) l2 = h*deta(t(k)  h/2.0,z(k)+A(k)/2.0,eta(k) +l1/2.0) l3 = h*deta(t(k)  h/2.0,z(k)+B(k)/2.0,eta(k) + l2/2.0) l4 = h*deta(t(k)  h,z(k)+ C(k),eta(k) + l3) !=========================== !The values of lambda will be calculated eta(k1) = eta(k)  (l1 + 2.0*l2 + 2.0*l3 + l4)/6.0 t(k1)=t(k)h !Getting set the time for the next iteration !write(8,*) l1,l2,l3,l4 !write(9,*) eta(k) end do !for eta !end do !print*,eta !Do i=1,1001 if (eta(i) .lt. (g_1/g_2)) then u(i) = u_1 else if (eta(i).eq.(g_1/g_2)) then if (z(i) .le. z_1) then u(i)= u_1 else if (z_1 .lt. z(i) .and. z(i) .lt. z_2) then u(i)= w(i) else u(i)= u_2 end if else u(i)=u_2 end if end do !for u !======================================= saved slopes do j=1,100 write(*,*)j,A(j),B(j),C(j),D(j) end do !======================================= !if (abs(x(i)x(i+1)) + abs(u(i)u(i+1))+ abs(lamd(i)lamd(i+1)) .lt. tol )then ! stop !else 201 format (2f10.6,2f10.6,2f16.6,2f10.6) do i=1,1001 write(19,201) t(i),z(i),eta(i),u(i) end do !end if !print*,u end program opct !====================================================================== ================= function dz(t,z) integer::i real, intent(in)::t,z real,DIMENSION(1010)::u dz = (phi_1 + phi_2*(z**gama))*z + (g_1*z + g_2)*u(1) end function dz !================================ function deta(t,z,eta) integer::i real, intent(in)::t,z,eta real,DIMENSION(1010)::u deta = ( phi_1 + (1gama)*phi_2*(z**gama))*eta + g_2*u(1)*(etag_1/g_2)*eta(gama*phi_2/(z**(1gama))) end function deta !================================================= Thanks a lot! 


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