Model is
dx/dt = Ax + Bu
y = Cx + Du
where A = 0, B = 0, C = I, and D = 0.
The optimal cost-to-go is the P that satisfies
AᵀP + PA − PBR⁻¹BᵀP + Q = 0
Let A = Aᵀ and B = Cᵀ for state observers.
AP + PAᵀ − PCᵀR⁻¹CP + Q = 0
Let A = 0, C = I.
−PR⁻¹P + Q = 0
Solve for P. P, Q, and R are all diagonal, so this can be solved element-wise.
−pr⁻¹p + q = 0
−pr⁻¹p = −q
pr⁻¹p = q
p²r⁻¹ = q
p² = qr
p = sqrt(qr)
Now solve for the Kalman gain.
K = PCᵀ(CPCᵀ + R)⁻¹
K = P(P + R)⁻¹
k = p(p + r)⁻¹
k = sqrt(qr)(sqrt(qr) + r)⁻¹
k = sqrt(qr)/(sqrt(qr) + r)
Multiply by sqrt(q/r)/sqrt(q/r).
k = q/(q + r sqrt(q/r))
k = q/(q + sqrt(qr²/r))
k = q/(q + sqrt(qr))
For q = 0 and r ≠ 0,
k = 0/(0 + sqrt(0))
k = 0/0
Apply L'Hôpital's rule to k with respect to q.
k = 1/(1 + r/(2sqrt(qr)))
k = 2sqrt(qr)/(2sqrt(qr) + r)
k = 2sqrt(0)/(2sqrt(0) + r)
k = 0/r
k = 0
For q ≠ 0 and r = 0,
k = q / (q + sqrt(0))
k = q / q
k = 1