Balancing by pole placement

Choose the eigenvalues

The lean rule failed because two eigenvalues of $A - BK$ stayed on the unit circle. So turn it around: decide where all four closed-loop eigenvalues, the poles, should be, and compute the $K$ that puts them there. That's pole placement.

A pole $p$ makes its mode shrink by $p$ each period:

• $p$ close to 1 (0.95): a slow mode. The cart wanders a long way before it comes back, and much closer to 1 it reaches the stops.
• $p$ closer to 0 (0.7): a fast mode, but the motor has to push hard. From 6°, poles at 0.7 ask for 25 N; this motor gives 10.
• Negative or outside the unit circle: oscillating or unstable.

Somewhere between 0.8 and 0.95 works well here.

Ackermann's formula

The eigenvalues of $A - BK$ are the roots of its characteristic polynomial. You want that to be

For one input, Ackermann's formula gives the gain directly:

• $p(A) = A^4 + a_1 A^3 + a_2 A^2 + a_3 A + a_4 I$ is the polynomial evaluated at the matrix $A$.
• $\mathcal{C}$ is the controllability matrix. It's invertible exactly when the force can steer every state, and then any poles you like can be placed.

np.poly(poles) returns $[1, a_1, \dots, a_n]$, and np.linalg.solve(C, pA) computes $\mathcal{C}^{-1} p(A)$ without forming the inverse.

Your task

1. Implement place(A, B, poles) for any number of states $n$. Return $K$ as a $1 \times n$ array.
2. Pick four POLES.

The program plots your poles and the eigenvalues of $A - BK$ together, then balances for 4 s from 4–6°. To pass, the pole must be within 1° of upright by 1.5 s, the force must stay within 10 N, the cart must stay off the stops, and it must end up back at the centre.