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Examples

Example 2.1: derive and analyze a state-space model

Consider a mass–spring–damper system with m=2m=2 kg, c=3c=3 N s/m, and k=8k=8 N/m. The control input is force uu and the measured output is displacement y=xy=x.

Step 1: Write the dynamic equation.

2x¨+3x˙+8x=u.2\ddot{x}+3\dot{x}+8x=u.

Step 2: Define the states.

x1=x,x2=x˙.x_1=x, \qquad x_2=\dot{x}.

Step 3: Write the first-order model.

x˙1=x2,x˙2=4x11.5x2+0.5u,\dot{x}_1=x_2, \qquad \dot{x}_2=-4x_1-1.5x_2+0.5u,

so

x˙=[0141.5]x+[00.5]u,y=[10]x.\dot{\mathbf{x}}= \begin{bmatrix}0&1\\-4&-1.5\end{bmatrix}\mathbf{x} +\begin{bmatrix}0\\0.5\end{bmatrix}u, \qquad y=\begin{bmatrix}1&0\end{bmatrix}\mathbf{x}.

Step 4: Analyze stability. The eigenvalues are the roots of

λ2+1.5λ+4=0,\lambda^2+1.5\lambda+4=0,

which are

λ=0.75±j1.854.\lambda=-0.75\pm j1.854.

Their real parts are negative, so the open-loop equilibrium is asymptotically stable.

Step 5: Interpret the result. The system oscillates, but the oscillations decay. If the designer raises stiffness to k=18k=18 without changing damping, the oscillation frequency increases and the best feedback tuning may change.

Example 2.2: effect of PD feedback

Suppose Example 2.1 uses

u=6x4x˙.u=-6x-4\dot{x}.

The closed-loop equation is

2x¨+(3+4)x˙+(8+6)x=0,2\ddot{x}+(3+4)\dot{x}+(8+6)x=0,

with characteristic equation

2λ2+7λ+14=0.2\lambda^2+7\lambda+14=0.

The closed-loop poles have a more negative real part than the open-loop poles, indicating increased damping and faster decay. Feedback improves the response by altering the effective dynamics.

Example 2.3: identify signals in cruise control

Consider a simple vehicle moving along a straight road:

mv˙=Ftcrvmgsinθ,m\dot{v}=F_t-c_rv-mg\sin\theta,

where vv is vehicle speed, FtF_t is traction force, crc_r is a drag or rolling-resistance coefficient, and θ\theta is road slope.

A sensible description is:

This first-order model reminds us that a useful state set need not contain many variables.

Common modeling mistakes

  1. Treating outputs as states without justification. Outputs may equal states, but they do not have to.

  2. Ignoring disturbances. This usually creates unrealistic conclusions about controller performance.

  3. Using too many states. More states are not always better; the state set should be minimal and meaningful.

  4. Forgetting units. Dynamic equations must be dimensionally consistent.

  5. Assuming plant design changes only constraints or cost. In CCD, plant design often changes the differential equations themselves.

  6. Confusing stability with performance. A stable system is not necessarily fast, efficient, comfortable, or optimal.