Lecture Slides Corrections

p26

First line should be

C(s) = \frac{5}{s(s+5)} = \frac 1s \frac{1}{1+0.2s}

System and Block Diagrams

Laplace Transform

F(s) = \mathscr L {F(t)} = \int^\infty_0 f(t)e^{-st} \mathrm{d}t

where s = \alpha + j\omega

The function F(s) is often much easier to manipulate than periodic function f(t).

Final Value Theorem

As f(t) tends to infinity, sF(s) tends to 0.

Example

\dot x_o = ax_o = ax_i

where x_o is the output and x_i is the input

Take the Laplace transform:

sX_o(s) + aX_o(s) = aX_i(s)

Rearrange to get equation for the transfer function:

G(s) = \frac{X_o}{X_i} = \frac{a}{s+a}

 X_o = GX_i

If X_i is a unit step, then:

X_i = \frac1s

and

X_o = \frac{a}{s(s+a)}

Taking the inverse gives:

X_0 = 1 - e^{-at}

Non-Linearity

Sometimes, components of a system will not reduce to a simple linear relationship. When this is the case superposition and Laplace transforms do not apply/are not valid.

Reasons for this include:

saturation
backlash
clearance
coulomb friction
material non-linearity
flow through an orifice (choked flow)

Linearisation

System behaviour is approximated to a linear relationship near the "nominal" operating point:

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