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EEE8013 Revision lecture 1 Ordinary Differential Equations. Modeling. Model : Ordinary Differential Equations (ODE):. Dynamics : Properties of the system, we have to solve/study the ODE. First order systems: Study approaches. First order ODEs :. Analytic :
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EEE8013 Revision lecture 1 Ordinary Differential Equations Autumn 2008
Modeling Model: Ordinary Differential Equations (ODE): Dynamics: Properties of the system, we have to solve/study the ODE. Autumn 2008
First order systems: Study approaches First order ODEs: Analytic: Explicit formula for x(t) (a solution – separate variables, integrating factor) which satisfies INFINITE curves (for all Initial Conditions (ICs)). Autumn 2008
First order linear equations First order linear equations - (linear in x and x’) General form: Numerical Solution: k=5, u=0.5 Autumn 2008
Analytic solution: Step input Autumn 2008
Response to a sinusoidal input Autumn 2008
Response to a sinusoidal input Autumn 2008
Second order ODEs Second order ODEs: So I am expecting 2 arbitrary constants u=0 => Homogeneous ODE Let’s try a Autumn 2008
1.5 Overall solution 1 x 2 0.5 0 x 1 -0.5 0 1 2 3 4 5 6 Overdamped system Roots are real and unequal Autumn 2008
Critically damped system Roots are real and equal A=2, B=1, x(0)=1, x’(0)=0 => c1=c2=1 Autumn 2008
Underdamped system Roots are complex Underdamped system r=a+bj Theorem: If x is a complex solution to a real ODE then Re(x) and Im(x) are the real solutions of the ODE: Autumn 2008
Underdamped system, example A=1, B=1, x(0)=1, x’(0)=0 => c1=1, c2=1/sqrt(3) Autumn 2008
Undamped Undamped system A=0, B=1, x(0)=1, x’(0)=0 =>c1=1, c2=0: Autumn 2008
Stability In all previous cases if the real part is positive then the solution will diverge to infinity and the ODE (and hence the system) is called unstable. Autumn 2008
is the damping factor and is the natural frequency of the system. Natural frequency, damping frequency, damping factor Case 1: => Overdamped system implies that Autumn 2008
Natural frequency, damping frequency, damping factor Case 2: => Critically damped system implies that Case 3: => Underdamped systems implies called damped frequency or pseudo-frequency Case 4: No damping the frequency of the oscillations = natural frequency Autumn 2008
Summary Autumn 2008
If then cases 1-3 are the same but with Stability revised Autumn 2008
NonHomogeneous (NH) differential equations • u=0 => Homogeneous => x1 & x2. • Assume a particular solution of the nonhomogeneous ODE: xp If u(t)=R=cosnt => Then all the solutions of the NHODE are So we have all the previous cases for under/over/un/critically damped systems plus a constant R/B. If complementary solution is stable then the particular solution is called steady state. Autumn 2008
Example x(0)=1, x’(0)=0 => c1=-1, c2=-1/sqrt(3) Autumn 2008
State Space Very difficult to be studied => so we use computers Computers are better with 1st order ODE 1 nth => n 1st Powerful tools from the linear algebra Use sensors: Output = x => Autumn 2008
State Space Autumn 2008
Block Diagram • X is an n x 1 state vector • U is an q x 1 input vector • Y is an p x 1 output vector • A is an n x n state matrix • B is an n x q input matrix • C is an p x n output matrix • D is an p x m feed forward matrix (usually zero) Autumn 2008
State space rules The state vector describes the system => Gives its state => The state of a system is a complete summary of the system at a particular point in time. If the current state of the system and the future input signals are known then it is possible to define the future states and outputs of the system. • The choice of the state space variables is free as long as some rules are followed: • They must be linearly independent. • They must specify completely the dynamic behaviour of the system. • Finally they must not be input of the system. Autumn 2008
State space The system’s states can be written in a vector form as: • A standard orthogonal basis (since they are linear independent) • for an n-dimensional vector space called state space. Matlab definition Autumn 2008
Solution Autumn 2008
Assume is a parameter => A homogeneous linear system Solution II How can we solve that ??? (This last equation is the characteristic equation of the system, why???). Autumn 2008
Solution III I assume that a2=1 so a1=2 I assume that a1=2 so a2=-2 Matlab example Autumn 2008
General Solution The roots of this equation are called eigenvalues negative eigenvalues => stable positive eigenvalues => unstable repeated eigenvalues => eigenvectors are not linearly independent. Complex eigenvalues => conjugate and the eigenvector will be complex =>solution will consists of sines, cosines and exponential terms Autumn 2008
Properties of general solution If we start exactly on one eigenvector then the solution will remain on that forever. Hence if I have some stable and some unstable eigenvalues it is still possible (in theory) for the solution to converge to zero if we start exactly on a stable eigenvector. Determines the nature of the time response (stable, fast..) Determines the extend in which each state contributes to Determines the extend in which the IC excites the To find the eigenvalues and eigenvectors use the command eig() Autumn 2008
Example Autumn 2008
Example Autumn 2008
Example Autumn 2008
State Transition Matrix Until now the use of vector ODEs was not very helpful. We still have special cases => No special cases are needed then 4 ways to calculate it!!! Use the command expm (not exp)! X(0)=[1 2] X(5) =? Autumn 2008
State Transition Matrix Autumn 2008
State Transition Matrix Autumn 2008
Solution Autumn 2008
SS => TF??? TF Response to ICs Autumn 2008
SS => TF??? CE of the TF The TF is a matrix Autumn 2008
SS => TF??? Example: Find the TF of Autumn 2008
Basic properties of state space State space transformations State space representations are not unique Same input/output properties, => same eigenvalues T is an invertible matrix Z is the new state vector Autumn 2008
Basic properties of state space Do these two systems have the TF? Matlab example Autumn 2008
Observability - Controllability Observability - Controllability Notice the structure of A and C Autumn 2008
Observability - Controllability There is a pole zero cancellation The cancellation is due to C=[3 0]. pzmap(ss_model) Matlab verification We can influence x2 through U but we cannot observe how it behaves and hence there is no way to feedback that signal to a controller!!! Autumn 2008
Observability - Controllability In this case we can see how both states behave but we can not change U in any way so that we can influence x2 due to the form of B. Unobservable & uncontrollable Minimal realisation. Difficult task if the system is nonlinear!!!! Autumn 2008
Observability - Controllability Check the rank >> rank(obsv(A,C)) >> rank(ctrb(A,B)) Autumn 2008