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Problem 2

Show that if \(e \neq 0\) and \(\alpha>1, \beta>0\), then no solution of $$ \ddot{\varphi}(t)+\beta \dot{\varphi}(t)+(\alpha-1) \varphi(t)=e $$

Expert verified

Under the given conditions \(e \neq 0\), \(\alpha > 1\), and \(\beta > 0\), there is no particular solution to the second-order linear inhomogeneous differential equation \(\ddot{\varphi}(t) + \beta \dot{\varphi}(t) + (\alpha - 1) \varphi(t) = e\). Thus, this differential equation has no solution under these conditions.

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Chapter 1

Find numbers \(\alpha, \beta, \mu\) with the following property: For each $e \in \mathbb{R}$, all of the solutions of the system of equations $$ \begin{aligned} \dot{x}_{0}(t) &=x_{1}(t) \\ \dot{x}_{1}(t) &=x_{2}(t) \\ \dot{x}_{2}(t) &=-\mu x_{0}(t)+(1-\alpha) x_{1}(t)-\beta x_{2}(t)+e \end{aligned} $$ converge to \((e / \mu, 0,0)\). (Hint: First solve this problem for the homogeneous system that results if \(e=0\) by finding parameters that make the matrix associated to the equations have all eigenvalues with negative real parts. Then show that the same parameters work in the general case.)

Chapter 1

Consider again the pendulum linearized about its unstable upper position, given by the equation \(\ddot{\varphi}(t)-\varphi(t)=u(t)\), and assume that we use the PD control law \(u(t)=-\alpha \varphi(t)-\beta \dot{\varphi}(t)\) to obtain an asymptotically stable closed-loop system $\ddot{\varphi}(t)+b \dot{\varphi}(t)+a \varphi(t)=0\( (with \)a=\alpha-1>0\( and \)b=\beta>0$ ). Introduce the natural frequency \(\omega:=\sqrt{a}\) and the damping factor \(\zeta:=b /(2 \sqrt{a})\), so that the equation now reads $$ \ddot{\varphi}(t)+2 \zeta \omega \dot{\varphi}(t)+\omega^{2} \varphi(t)=0 . $$ (A) Prove the following facts: 1\. If \(\zeta<1\) (the "underdamped" case), all solutions are decaying oscillations. 2\. If \(\zeta=1\) (the "critically damped" case) or if \(\zeta>1\) ("overdamped"), then all solutions for which \(\varphi(0) \neq 0\) are such that \(\varphi(t)=0\) for at most one \(t>0\). 3\. If \(\zeta \geq 1\), then every solution that starts from rest at a displaced position, that is, \(\varphi(0) \neq 0\) and \(\dot{\varphi}(0)=0\), approaches zero monotonically. (In this case, we say that there is no "overshoot.") 4\. Show rough plots of typical solutions under the three cases $\zeta<1, \zeta=1\(, and \)\zeta>1$.

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