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49329 Control Of Mechatronic Systems

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49329 Control Of Mechatronic Systems

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49329 Control Of Mechatronic Systems

0 Download4 Pages / 755 Words

Course Code: 49329
University: University Of Technology Sydney

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Country: Australia

Question:
Consider the spring-mass-dashpot system mounted on a massless cart that the cart is moved at a constant speed u(t)=constant at t>0 and is the input to system. The displacement y(t) of the mass (Relative to the ground) is the output. In the system, m denotes the mass, b denotes the viscous-friction coefficient, and k is the spring constant. We assume that the friction force of the dashpot is proportional to ????? − ????? and the spring is a linear spring; that is, the spring force is proportional.
Answer: 
Using Laplace transform to convert from time domain to s-domain, 
The transfer function is given as, 
To obtain the state vector to get the state-space illustration of the system. The standard state-space illustration is given as,  
The second order representation of the mass spring dashpot system is reduced to a set of two first order differential equations represented by the position and velocity variables as,  
Mechanical system with . 
Choose the state vector to be, 
The equation of the motion of the spring mass system with, m=1, 
Acknowledging the frictional forces,  
State space representation 
To obtain the eigen values, 
Expanding the cofactors, we obtain,   
The eigenvalues of the system equation are 0,0, 0.4677, -0.4677.
Only one eigenvalue is strictly positive and another is strictly negative alongside two zero values. This is an open-loop and the system is unstable. Inspecting the matrix further to assess for controllability, 
It expands to form, 
The square matrix is fully row ranked hence the system is controllable. The Rank, n=4. The determinant of the matrix R is given as a non-zero number for any real valued inputs. 
A system is considered to be stable in the event that the eigenvalues of  all have negative real values. The MATLAB place function is used to obtain the solution. The poles are given as -5, -15, -20, -25. Its feedback control law is u=-Kx. The desired K values to get the desired poles are given as,  
Analyze these attributes on the system based on diverse values of .

Asymptotic stability
Controllability
Observability

A system is controllable if it performs whatever it is set to do with the given dynamic system with respect to control input. The system is considered observable if one can perceive what is going on inside the system. The range of , using MATLAB, the three conditions are tested.  
The poles of H(s) are uncancelled eigenvalues of A. system is stable if all the eigen values of A are stable.  
For
PART I   
a=0;
A=[-2 a;3 -5]
B=[0;1]
C=[1,-1]
D=0 
sys=ss2tf(A,B,C,D) 
All the eigenvalues are (0, -1, -2) Hence the system is asymptotically stable.
PART II
%Test for controllability
Co=ctrb(A,B)
% Determining the uncontrollable state
unco=length(A)-rank(Co) 
The system is NOT controllable as the rank is equal to the row value of the system state matrix A.
Rank =2
The controllability matrix is given as, 
PART III
% Test for observability
Ob=obsv(A,C)
% Number of unobservable states
unob=length(A)-rank(Ob) 
The system is NOT observable. The observability matrix is given as, 
For
PART I
a=3;
A=[-2 a;3 -5]
B=[0;1]
C=[1,-1]
D=0 
sys=ss2tf(A,B,C,D) 
All the eigenvalues are (0, -1, 1) Hence the system is NOT asymptotically stable.
PART II
%Test for controllability
Co=ctrb(A,B)
% Determining the uncontrollable state
unco=length(A)-rank(Co)
The system is controllable as the rank is equal to the row value of the system state matrix A.
Rank =2
The controllability matrix is given as, 
PART III
% Test for observability
Ob=obsv(A,C)
% Number of unobservable states
unob=length(A)-rank(Ob) 
The system is observable. The observability matrix is given as, 
For
PART I
a=5;
A=[-2 a;3 -5]
B=[0;1]
C=[1,-1]
D=0 
sys=ss2tf(A,B,C,D) 
All the eigenvalues are (0, -1, 3) Hence the system is NOT asymptotically stable. PART II
%Test for controllability
Co=ctrb(A,B)
% Determining the uncontrollable state
unco=length(A)-rank(Co) 
The system is controllable as the rank is equal to the row value of the system state matrix A.
Rank =2
The controllability matrix is given as, 
PART III
% Test for observability
Ob=obsv(A,C)
% Number of unobservable states
unob=length(A)-rank(Ob) 
The system is observable. The observability matrix is given as,  
Mobile robot trajectory. 
The states are given as, 
The inputs are given as, 
The non-linear state space model is given as, 
When the trajectory produces a reference, assuming the full state feedback, it is denoted as, 
The desired trajectory is referenced as, 
The linearized dynamics,  
To obtain the state feedback control law is given using the feedforward using the path curvature and speed such that,

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