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Problem 5

# (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.

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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