Content is from the course ECE327 at the University of Waterloo. Eventually I will clean it up with proper citations.
Summary
The course focuses on control systems for linear dynamic systems. The course also focuses on systems with few inputs, few outputs; clear first-principles models; clear sources of noise and disturbance.
- State-space models.
- Linearization technique.
- Stability analysis
- Frequency response.
- Poles and zeroes, dominant, and effects.
- Root-locus method
- PID controllers
- Frequency domain stability analysis
Modelling
This portion of the course finished May 14, 2024.
Linearization of dynamic systems
Approximate around a controlled equilibrium point \((x^*, u^*)\), where \(x\) is the state vector and \(u\) is the input vector.
Now define small perturbations around the equilibrium point:
\[\delta x(t) := x(t) - x^*\] \[\delta u(t) := u(t) - u^*\]
We had before \(\dot x = f(x^*, u^*) + A\delta x(t) + B\delta u(t) + \text{higher order terms}\).
Where \(A\) and \(B\) are the Jacobians of \(f\) with respect to \(x\) and \(u\).
When we are at the controlled equilibrium point, the linearized equation becomes \(\delta x(t) = A\delta x(t) + B\delta u(t)\).
- Select a desired controlled equilibrium \((x^*, u^*)\).
- Where it is depends on if it is a stabilization or tracking problem.
- These are two specific subsets of general problems in control eng.
- \(f(x^*, u^*) = 0\), and desired output is given \(y^* = h(x^*, u^*)\).
- Compute Jacobians \(A, B\) of \(f\) and \(C, D\) of \(h\).
- The linearized system is:
\[\dot{\delta x}(t) = A\delta x(t) + B\delta u(t)\] \[\delta y(t) = C\delta x(t) + D\delta u(t)\]
Under “mild conditions” (continuous differentiability of \(f, h\)), this linearized system is a valid approximation of the nonlinear one in a sufficieintly small neighborhood of the equilibrium point.
The final control system tracks deviation from the equilibrium and uses reference deviation instead of tracking position with reference position.
Transfer functions of LTI systems
Only LTIs have transfer functions. The transfer function is defined as a ratio of Laplace transforms, both taken with initial conditions (ICs) of zero.
\[G(s) = \frac{Y(s)}{U(s)}\]
In linear systems, this ratio doesn’t depend on the input you choose.
Given a nonlinear mass-spring-damper system:
\[M\ddot y = u - Ky - C(\dot y)\]
We cannot write a transfer function because the system is nonlinear. If it is linear, we can.
\[My^{(2)} = u - Ky - by^{(1)}\]
Taking Laplace transforms on both sides with ICs of zero:
\[Ms^2Y(s) = U(s) - KY(s) - b s Y(s)\] \[\frac{Y(s)}{U(s)} = \frac{1}{Ms^2 + bs + K}\]
The result is “rational.” Rational functions are ratios of polynomial functions.
To get a transfer function from a state-space model, we can use the following:
\[sX(s) = AX(s) + BU(s)\] \[Y(s) = CX(s) + DU(s)\]
Take first equation, isolate for \(X(s)\):
\[X(s) = (sI - A)^{-1}BU(s)\]
Plug into second equation:
\[Y(s) = C(sI - A)^{-1}BU(s) + DU(s)\]
Then transfer function is:
\[G(s) = \frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D\]
Linear systems
Given a transfer function \(G(s)\) and complex number \(z\in\mathbb{C}\):
- \(z\) is a pole if \(\lim{s\to z} \lvert G(s)\rvert = \infty\) (is not finite)
- \(z\) is a zero if \(\lim{s\to z} G(s) = 0\).
If \(G(s)\) is rational and its numerator and denominator are coprime (no common factors), then the poles are the roots of the denominator and the zeros are the roots of the numerator.
How can we tell if a system is stable?
State-space models: asymptotic stability
(Lyapunov stability)
Systems are asymptotically stable if \(x(t)\to 0\) as \(t\to\infty\) for every initial condition (every possible combination of parameters given at \(t=0\)).
Why is this important? Well, such a system is always going to converge to an equilibrium point given enough time, no matter how it is disturbed or what its initial conditions are. Think of a pendulumn: it will always eventually come to rest.
Log
Day 3
- Linearization of dynamic systems
- Block diagram controller implementation