ELEC4631 Chap.8 Optimal Control and LQR: Riccati Equations and Cost Weighting
Optimal Control and LQR: Riccati Equations and Cost Weighting
LQR chooses state feedback by minimising a quadratic cost rather than prescribing poles. Q weights state deviation and is normally symmetric positive semidefinite; R weights control effort and is symmetric positive definite. On a finite horizon the value matrix P(t) satisfies a Riccati differential equation integrated backward from a terminal cost. In the time-invariant infinite-horizon problem it settles into the continuous algebraic Riccati equation A^TP+PA-PBR^-1B^TP+Q=0. The relevant solution is the stabilising positive-semidefinite branch, not any symmetric algebraic root. The gain is K=R^-1B^TP and the optimal value is x0^TPx0. Substitution shows that P is also a closed-loop Lyapunov certificate. The numbered Lecture 9 deck is absent from the available set, so this chapter stays within the LQR/Riccati chain supported by the official schedule, Tutorial 6 and later lecture dependencies.
What this chapter covers
- 01Quadratic state and input costs
- 02Q and R interpretation
- 03Finite-horizon Riccati differential equation
- 04Terminal condition and time-varying gain
- 05Infinite-horizon CARE
- 06Stabilising solution and optimal value
- 07LQR versus exact pole placement
Select the stabilising scalar CARE root
- CAREHere a=b=r=1 and q=4. The scalar CARE is 2P-P^2+4=0.
- rootsRearrange to P^2-2P-4=0, giving P=1 plus or minus sqrt(5).
- selectionChoose P=1+sqrt(5); the other root is negative and does not give the stabilising value matrix.
- gainK=R^-1 B^T P=1+sqrt(5), so u*=-(1+sqrt(5))x.
- checkThe closed-loop pole is 1-K=-sqrt(5)<0, verifying the stabilising branch.
- valueThe optimal infinite-horizon value is J*=x0^2(1+sqrt(5)).
Key terms
- LQR
- Linear-quadratic regulator: linear dynamics, quadratic performance and state feedback chosen to minimise that cost.
- Q matrix
- Symmetric state-deviation weight. Larger penalties generally push the design to suppress the corresponding states more strongly.
- R matrix
- Symmetric positive-definite input-effort weight, ensuring a well-defined minimisation and inverse.
- Riccati differential equation
- Backward finite-horizon equation for the time-varying value matrix P(t), with terminal condition supplied at t_f.
- CARE
- The continuous algebraic Riccati equation for the infinite-horizon time-invariant problem.
- Stabilising solution
- The Riccati solution whose recovered feedback makes A-BK Hurwitz under the theorem’s assumptions.
Optimal Control and LQR: Riccati Equations and Cost Weighting FAQ
How is LQR different from pole placement?
Pole placement starts with exact desired eigenvalues. LQR starts with Q and R and chooses the gain minimising the resulting cost; closed-loop poles are consequences. LQR does not promise specified poles unless weights happen to produce them.
Do larger Q entries always mean larger gains?
They generally make state deviation more expensive and often lead to more aggressive feedback, while larger R makes effort more expensive. These are qualitative tendencies. Coupling, scaling and plant directions still require solving the Riccati equation and checking the result.
Why are there multiple CARE roots?
The CARE is nonlinear in P. Algebraic solutions need not all be positive or stabilising. The infinite-horizon control theorem selects the stabilising positive-semidefinite branch under appropriate stabilisability and detectability conditions.
May I use the CARE for a finite horizon?
Not automatically. A finite horizon has a time-varying P(t) driven backward from terminal data and normally a time-varying K(t). Use the CARE only when an infinite-horizon, time-invariant setting and its assumptions are established.
What should appear in a complete infinite-horizon LQR solution?
Write the plant, cost and assumptions first: Q symmetric positive semidefinite, R symmetric positive definite and the relevant stabilisability/detectability conditions. State the CARE and solve for a symmetric P. If several algebraic roots exist, test positivity and closed-loop stability rather than choosing by appearance. Form K=R^-1B^TP, state u*=-Kx and verify A-BK is Hurwitz. Substitute into the CARE or evaluate its residual. Finally state J*=x0^TPx0. Explain the weight trade-off in relation to the supplied Q and R, not through a universal slogan. This order turns P from an equation root into a value function and stability certificate and distinguishes the requested optimal problem from ordinary pole placement.
How can the CARE be remembered rather than memorised?
Assume a quadratic infinite-horizon value V=x^TPx. The Hamiltonian contains x^TQx+u^TRu+2x^TP(Ax+Bu). Differentiating with respect to u gives the minimiser u=-R^-1B^TPx. Substitute it into the stationary value equation: the resulting quadratic coefficient is A^TP+PA-PBR^-1B^TP+Q, which must vanish. This route recovers the minus sign and the two B/P factors. It also shows why P simultaneously measures remaining optimal cost and supplies a Lyapunov derivative equal to the negative running cost under optimal feedback. Re-deriving this scalar/matrix pattern is safer in a closed-book examination than remembering an isolated line whose transpose order may drift.
What numerical or symbolic checks should follow an LQR calculation?
Confirm P is symmetric and has the required semidefinite sign, evaluate the full Riccati residual, form K with the correct R inverse and B transpose, and check A-BK is Hurwitz. For a scalar or diagonal case, substitute the candidate roots directly and compare the rejected branch. Simulate two initial states only after these analytical checks and compare the accumulated cost with x0^TPx0. If an AI tutor proposes new Q/R weights, predict the qualitative change in effort and state penalty before resolving the CARE. Keep the numbered Lecture 9 source gap visible: use the official schedule and tutorial notation, and confirm any offering-specific solver convention on Moodle.
Exam move
Read the cost before the dynamics: write horizon, terminal penalty, Q and R dimensions and definiteness. For a finite horizon, place the terminal condition at t_f and remember the backward integration direction. For an infinite horizon, solve the CARE, list every symmetric candidate, select the stabilising positive-semidefinite solution and check A-BK. Reconstruct the optimal value only after selecting P. Use the identity produced by substituting K into the CARE to connect optimality with a Lyapunov derivative. Compare one LQR design with a pole-placement design on the same plant, stating what each promises. Because the Lecture 9 deck is unavailable, take notation and assessment emphasis from current tutorials and Moodle rather than assuming a missing slide convention.
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