The mathematics of change

Derivatives, integrals, initial-value problems and the connection between poles and responses.

A derivative is a rate of change

If rotor speed increases by 4 rad/s in 0.2 s, average acceleration is 20 rad/s². The derivative is the limiting rate over a short interval. A horizontal speed trace has nearly zero acceleration; a steep upward trace has large positive acceleration. Zero speed does not imply zero acceleration.

An integral accumulates change

At a constant acceleration of 10 rad/s² for 0.3 s, speed increases by 3 rad/s. If ω₀ = 5 rad/s, the result is 8 rad/s. An integral is a signed area: negative acceleration subtracts speed. Integrating current gives charge, and integrating power gives energy. Units provide a useful check: W·s = J.

From a physical law to an equation

With constant applied torque and viscous friction, the current speed determines its own slope. Increasing speed raises friction and reduces acceleration until M = bω. A numerical solver follows these local slopes without knowing ω(t) beforehand. An initial condition such as ω(0) = 0 selects starting from rest; the equation alone describes many possible motions.

The state vector

For a motor, collect current and speed as y = [i, ω]. The right-hand side receives both states and returns their derivatives in the same order. Never change that order between equations, initial conditions and output interpretation.

def rhs(t, y):
    current, speed = y
    di = (12 - 2*current - 0.1*speed) / 0.02
    dw = (0.1*current - 0.002*speed) / 0.002
    return [di, dw]

These returned numbers are rates, not the next current and speed. Their units are A/s and rad/s². The solver, not this function, integrates them.

Initial conditions and the Laplace transform

A transfer function G(s) = K/(τs + 1) describes the input–output relation with zero initial state. Nonzero initial energy contributes a separate term. For x′ = Ax + Bu, the operator equation is (sI − A)X = BU + x₀. With constant inputs and an existing equilibrium, x(t) = x∞ + exp(At)(x₀ − x∞).

Second order: roots and motion

For ζ > 1, the poles are distinct negative real numbers; at ζ = 1 they coincide; for 0 < ζ < 1 they are complex conjugates. The oscillatory natural response has an exp(−ζωₙt) envelope and damped frequency ωd = ωₙ√(1 − ζ²). Negative real parts produce decay; zero real parts do not give asymptotic stability.

These peak formulas apply to the standard second-order system without zeros, for a unit step and zero initial conditions. They do not automatically apply to every output of a multivariable model.

Further reading: Swarthmore College: second-order poles and responses. Notation may differ from this course.

Independent exercises

1. Solve 0.2x′ + x = 5 with x₀ = 1. Find X(s), x(0.2) and x′(0⁺).

2. Find the poles, damping ratio, damped frequency and overshoot of x″ + 6x′ + 25x = 25u.

Practical work

Check an analytical first-order solution by substitution and by its initial value. List the units of every derivative in the motor right-hand side. Compare real and complex pole cases. Explain why a second-order equation needs two independent initial conditions.

Calculated illustrations

Key takeaways

Derivatives describe local change and integrals accumulate it. State the variable order, derivative units and initial conditions before using a solver.