ME3120 · Dynamic Modelling
Theme 1 · Foundations
The state of a system
Why knowing where something is isn't enough to predict its future — you also need to know how fast it's moving.
Source: course notes, Week 1 foundations.
Before you start
What you need first
- Generalized coordinates \(q\) — one position variable per degree of freedom (previous topic).
What you'll be able to do
- Explain why position alone can't predict motion.
- Write the state of a system as \([q,\ \dot q]\).
- Count how many states a system has.
Position alone cannot predict motion
A ball is at height 10 m. Where is it in 1 second?
You cannot answer until you also know its speed right now:
Dropped (speed 0) → it just falls.
Thrown up at 5 m/s → it first rises, then falls.
Same position, different futures. Position + velocity
together can predict the motion; position alone cannot.
States: position and speed
Each degree of freedom needs two numbers, called states: the position \(q\) and the velocity \(\dot q\).
$$\text{state}=\big[\,q,\ \dot q\,\big]$$
| Symbol | Meaning | SI unit |
|---|---|---|
| q | position (a generalized coordinate) | m or rad |
| \(\dot q\) | velocity — how fast \(q\) changes | m/s or rad/s |
Why two? The laws of motion give the
acceleration \(\ddot q\) — they are second order. To get
the future motion you add the acceleration up (integrate) twice: first to
get the velocity — which needs a starting speed — then to get the
position — which needs a starting place. Two integrations → two starting
facts: where it is now, and how fast it moves now.
🔭 Looking ahead: a 2-link arm → 2 DOF → 4 states.
A 6-joint robot → 12 states. The bookkeeping grows; the idea does not.
✏️ Try it yourself
For each system, give the number of DOF and the number of states, and write the state vector:
- A mass on a spring.
- A single pendulum.
- A flat 3-link robot arm.
1. 1 DOF → 2 states:
\([x,\ \dot x]\).
2. 1 DOF → 2 states:
\([\theta,\ \dot\theta]\).
3. 3 DOF → 6 states:
\([\theta_1,\theta_2,\theta_3,\ \dot\theta_1,\dot\theta_2,\dot\theta_3]\). Rule: states \(=2\times\) DOF.
Recap — the whole topic on one screen
| Idea | What you own now |
|---|---|
| Position isn't enough | Same place, different speed → different future |
| State | \([q,\ \dot q]\) — position and velocity together |
| Why two per DOF | 2nd-order law → integrate twice → two starting facts |
| Counting | states \(= 2 \times\) DOF |