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(Infinite gain margin of LQ feedback.) As above, suppose that \(\Pi\) is a positive definite solution of the ARE and that \(Q\) is also positive definite. Pick any \(\rho \in[1 / 2, \infty)\) and let \(F:=-\rho R^{-1} B^{\prime} \Pi\). Show that the closed-loop matrix \(A_{c l}=A+B F\) is Hurwitz. The result in Lemma 5.7.18 is needed in the next proof. This states that the operator $$ \mathcal{L}: \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^{n \times n}, \quad \mathcal{L}(X):=M X+X N $$ is invertible if both \(M\) and \(N\) are Hurwitz.

Short Answer

Expert verified
Define the matrices and operator involved: \(A\), \(B\), \(R\), \(\Pi\), \(Q\), \(F:=-\rho R^{-1} B^{\prime} \Pi\), \(A_{cl}=A+BF\), and \(\mathcal{L}(X) = MX + XN\). Rewrite \(\mathcal{L}(X)\) in terms of \(A_{cl}\) and \(F\). Check the conditions for \(\mathcal{L}(X)\) to be invertible using Lemma 5.7.18, and connect the Lyapunov Equation with the Algebraic Riccati Equation (ARE). Conclude that \(M\) and \(N\) are Hurwitz, and therefore \(A_{cl}\) is also Hurwitz as required.
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Step 1: Define the given matrices and operator

We are given the matrices \(A\), \(B\), and \(R\), which are part of a linear quadratic regulator problem, as well as a positive definite solution \(\Pi\) of the Algebraic Riccati Equation. The matrix \(Q\) is also positive definite. We define \(F:=-\rho R^{-1} B^{\prime} \Pi\), where \(\rho \in [1 / 2, \infty)\), and the closed-loop matrix \(A_{cl} = A + BF\). Additionally, Lemma 5.7.18 gives us an operator \(\mathcal{L}(X) = MX + XN\) that will be invertible if both \(M\) and \(N\) are Hurwitz.

Step 2: Rewrite operator \(\mathcal{L}(X)\) in terms of the closed-loop matrix and \(F\)

Let's rewrite the operator \(\mathcal{L}(X)\) in terms of the closed-loop matrix and the matrix \(F\). Let \(M = A + \rho(R^{-1}B^{\prime}\Pi - BF)\) and \(N = A + \frac{1}{\rho}(R^{-1}B^{\prime}\Pi - BF)\) . Thus, consider the operator $$\mathcal{L}(X) = (A+\rho(R^{-1}B^{\prime}\Pi - BF))X+X(A+\frac{1}{\rho}(R^{-1}B^{\prime}\Pi - BF)).$$ From now on, we want to show that both \(M\) and \(N\) are Hurwitz.

Step 3: Check the conditions for \(\mathcal{L}(X)\) to be invertible

Using Lemma 5.7.18, we know \(\mathcal{L}(X)\) is invertible if both \(M\) and \(N\) are Hurwitz. Recall that a matrix is Hurwitz if all of its eigenvalues have negative real parts. Moreover, if either of these matrices is Hurwitz, then \(A_{cl}\) will be Hurwitz since it is a linear combination of them. To check this condition, consider the Lyapunov equation for \(M\) and \(N\): $$MX + XM^{\prime} + Q = 0$$ $$NX + XN^{\prime} + Q = 0.$$

Step 4: Connect the Lyapunov Equation with the Algebraic Riccati Equation (ARE)

Note that the given \(\Pi\) is a positive definite solution of the ARE. The given equation, \(A\Pi+\Pi A^{\prime}-\Pi B R^{-1} B^{\prime} \Pi+Q=0\), is also a Lyapunov equation. Therefore, we need to connect the given Lyapunov equations for \(M\) and \(N\) with the given ARE. By substituting the defined \(M\) and \(N\) from Step 2 into their respective Lyapunov equations, we can obtain the following equations: $$-(A\Pi+\Pi A^{\prime}-\rho\Pi B R^{-1} B^{\prime} \Pi+Q)=0$$ $$-(A\Pi+\Pi A^{\prime}-\frac{1}{\rho}\Pi B R^{-1} B^{\prime} \Pi+Q)=0.$$ Since \(\rho \geq 0.5\), we can rearrange the equations above to match the given ARE: $$A\Pi+\Pi A^{\prime}-\Pi B R^{-1} B^{\prime} \Pi+Q=0.$$

Step 5: Conclude that \(M\) and \(N\) are Hurwitz

Since the Lyapunov equations for \(M\) and \(N\) have the same form as the given ARE and \(\Pi\) is positive definite, we can conclude that both \(M\) and \(N\) are Hurwitz. This is because a positive definite solution to a Lyapunov equation implies that the associated matrix is Hurwitz.

Step 6: Conclude that \(A_{cl}\) is Hurwitz

Finally, since both \(M\) and \(N\) are Hurwitz, the operator \(\mathcal{L}(X)\) defined in Step 2 is invertible according to Lemma 5.7.18. Thus, \(A_{cl} = A + BF\) must also be Hurwitz, as required.

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Most popular questions from this chapter

Chapter 8

If \(\Sigma\) is a controllable time-invariant linear continuous-time system over \(\mathbb{K}=\mathbb{R}\), then there exists an \(m \times n\) real matrix \(F\) such that \(A+B F\) is a Hurwitz matrix. The above Corollary is, of course, also a consequence of the Pole-Shifting Theorem, which establishes a far stronger result. Next we consider a discretetime analogue; its proof follows the same steps as in the continuous- time case.

Chapter 8

Consider the case when \(B=0, Q=I\), and \(S\) approaches zero. Show that the formulas for least-squares observation in Section \(6.3\) can be recovered from the results in this section. (Hint: The equation for \(\widetilde{P}\) can be solved with final condition zero, and its solution at the initial time can be expressed directly in terms of the Wronskian \(W\).)

Chapter 8

Prove, without using the Pole-Shifting Theorem: If \(\Sigma\) is a controllable time-invariant linear discrete-time system over \(\mathbb{K}=\mathbb{R}\), then there exists an \(m \times n\) real matrix \(F\) such that \(A+B F\) is a convergent matrix.

Chapter 8

Suppose that \(L_{f} V(x) \leq 0\) for all \(x\) and that \(\dot{x}=f(x)+G(x) u\) is globally stabilized by \(u=-(\nabla V(x) \cdot G(x))^{\prime}\), as in Proposition 5.9.1. Show that \(u=k(x)\) is an optimal feedback, and \(V\) is the value function, for some suitably chosen cost. (Hint: Let $Q(x):=-L_{f} V(x)+\frac{1}{2} L_{G} V(x)\left(L_{G} V(x)\right)^{\prime}$, which gives (8.64) for which \(R\) ? Use Exercise 8.5.5.)

Chapter 8

Show that, if \((A, C)\) is observable, then $\Pi:=\lim _{t \rightarrow \infty} \Pi(t)$ exists and that it satisfies the dual algebraic Riccati equation $$ A \Pi+\Pi A^{\prime}-\Pi C^{\prime} Q C \Pi+B R^{-1} B^{\prime}=0 . $$

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