Math 213: Introduction to Control Systems and Differential Equations

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Separable ODE

Goal is to separate functions and derivatives of x and y to either side of the equation, dy/dx(expression of y) = f(x), then take the integral of both sides

Exact ODE

Uncommon, but we want to find a substitution u(x,y) = C

First-Order Linear ODE (variable coefficients)

Nth-Order Linear ODE (Homogeneous)

Nth-Order Linear ODE (Homogeneous with constant coefficients)


Ordinary Differential equations are ones where the unknown function depends on a single independent variable t


Studying linear, time-invariant dynamic systems using differential equations

Lecture 2

A general solution for an ODE is an expression for y(t) satisfying the equation and containing n "arbitrary constants/constants of integration"

Laplace Transforms

Helps us solve differential equations. Transform takes you from one set of functions to another set of functions. f(t) -> F(s), different domains.

Some known Laplace transforms:

Properties of Laplace using notation L{ f(t) } = F(t):

  1. L{ c1f(t) + c2g(t) } = L{c1f(t)} + L{c2g(t)} by properties of integrals, the Laplace transform is a linear operator
  2. L{ f(ct) } = 1/c F(s/c) time-scaling
  3. L{ e^(at)f(t) } = F(s-a) translates the transform
  4. L{ μc(t) f(t-c) } = e^(-cs) F(s)
  5. L{ t f(t) } = - d/ds F(s)
  6. L{f'(t)} = s • L{f(t)} - f(0) by integration by parts. This turns derivatives into multiplication.

The transform can be inverted by means of the inversion integral: f(t) = 1/(2π • j) integral of F(s) e^st from σ - j∞ to σ + j∞ so somehow, by taking the integral over the imaginary range, we get the inverse laplace transform operator

Part 3: Laplace for differential equations

So the reason that Laplace transforms help us solve differential equations is because we apply a one-to-one transform on both sides of an equation

y'' + 5y' + 6y = 0, y(0) = 2, y'(0) = 3
L{y''} + 5L{y'} + 6L{y} = L{0}
s^2 L{y} - s(2) - 3 + 5(s L{y} - 2) + 6L{y} = 0
s^2L{y} + 5s L{y} + 6L{y} - 2s - 13 = 0
L{y}(s^2+5s+6) -2s -13 = 0
L{y}(s+3)(s+2) = 2s + 13
L{y} = (2s+13)/((s+3)(s+2))
y = L^-1{(2s+13)/((s+3)(s+2))}  partial fraction expansion now
  = L^-1{ A/(s+3) + B/(s+2)

Part 4-1: Signals and Systems

Essential some function or transformation. Properties of systems:

  1. CT, DT, hybrid describe types of in/out functions
  2. Memoryless (pure function) vs. Dynamic (depends on state)
  3. Causality, dependent only on prior values of the input with respect to time
  4. Multivariable vs. Scalar
  5. Linearity, if the input is a linear combination of signals, then the output is a linear combination of their individual responses
  6. Time Invariance, doesn't change with time (response can be horizontally shifted without change shape)
  7. Lumped vs. distributed-parameter systems

Part 4-2: Zero-state and Zero-input

We'll look at linear, time-invariant, lumped-parameter systems

The big conclusion from this section is that we're studying zero-input response. It tells us all we need to know about the system's dynamics

Part 4-3: Convolution

Convolution of (f * g)(t) is the integral from 0 to t of f(t-τ)g(τ)dτ

The big conclusion of this section is that we get a lot of information about a system's transient (think oscillation part) by looking at its poles on the complex plane

Part 5-1: Frequency Responses of LTI systems

Previously, we transform problems in analysis of LTI systems into Laplace to be solved algebraically

It seems the frequency response contains just as much information as the impluse or step responses.

Part 5-2: Bode Plots

We know that putting sinusoids as input to a system (the frequency response) restricts what algebraic transformations can occur. This happens to mean that for sinusoids, this is only two things that can change: amplitude and phase. So for a fixed frequency input, the output has the exact same frequency. So a Bode Plot is just some graphical representation of what our system response looks like for all possible inputs (changing the frequency)

We can use transfer functions again let's us find the bode plot of a system

Graphing Bode Plots by hand from the transfer function, with straight line approximation

So transfer functions in LTI systems can be decomposed into a product of four types of components:

  1. Constants (where the gain is constant, log K and phase is 0 or 180 for ±K)
  2. Poles/Zeroes at origin (where gain is a linear function (dB vs. decades of ω) slope -20 dB/dec, phase is ±90 deg)
  3. Real Poles/Zeroes
  4. Complex Poles, we have to check a third case for magnitude, where ω = 1/τ, in which case we get a bump in the graph

Part 6: Fourier Series

Sinusoids are easier to analyze because of the frequency response and the fact that it is only ever changed in amplitude or horizontal translations