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Problem 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$.

Expert verified
We studied a linearized pendulum system with a PD control and analyzed its behavior under different damping factors, $$\zeta$$. We found that for $$\zeta<1$$ (underdamped case), all solutions are decaying oscillations. For $$\zeta = 1$$ (critically damped) and $$\zeta>1$$ (overdamped case), all solutions with $$\varphi(0) \neq 0$$ have $$\varphi(t) = 0$$ for at most one $$t>0$$. If $$\zeta \geq 1$$, and the pendulum starts from rest at a displaced position, every solution approaches zero monotonically without overshooting. The plots of the solutions show unique behaviors for each case: decaying oscillations for underdamped, non-oscillatory decay without overshoot for critically damped and overdamped cases, with overdamped cases having slower decay compared to critically damped cases.
See the step by step solution

## Step 1: 1. Find the general solution for the given differential equation

This is a second-order linear ordinary differential equation (ODE) with constant coefficients. To solve the given ODE, we will try the exponential solution, $$\varphi(t) = e^{rt}$$, where $$r$$ is a constant. Substituting this into the given ODE, we get: $e^{rt}(r^2 + 2 \zeta \omega r + \omega^2) = 0$ This is a quadratic equation in $$r$$, we will find its roots and then use them to get the general solution.

## Step 2: 2. Find the roots of the quadratic equation

We have a quadratic equation $$r^2 + 2 \zeta \omega r + \omega^2 = 0$$. Using the quadratic formula, we can find the roots as: $r = \frac{-2 \zeta \omega \pm \sqrt{(2 \zeta \omega)^2 - 4 \omega^2}}{2}$ Simplify the expression to get: $r = -\zeta \omega \pm \omega \sqrt{\zeta^2 - 1}$

## Step 3: 3. Study the three cases of damping ratio $$\zeta$$

Using the roots obtained in the previous step, we will analyze the three cases of damping ratio, $$\zeta$$. Case 1: $$\zeta<1$$ (underdamped): In this case, $$\zeta^2 - 1 < 0$$, so the roots are complex conjugates: $$r = -\zeta \omega \pm j\omega \sqrt{1-\zeta^2}$$. The general solution is: $\varphi(t) = e^{-\zeta \omega t}(C_1 \cos{\omega_d t} + C_2 \sin{\omega_d t})$ where $$\omega_d = \omega \sqrt{1-\zeta^2}$$ and $$C_1$$ and $$C_2$$ are constants determined by the initial conditions. All solutions are decaying oscillations as the exponential term decays over time. Case 2: $$\zeta=1$$ (critically damped): In this case, $$\zeta^2 - 1 = 0$$, so we have repeated real roots: $$r = -\omega$$. The general solution is: $\varphi(t) = (C_1 + C_2 t)e^{-\omega t}$ All solutions with $$\varphi(0) \neq 0$$ are such that $$\varphi(t)=0$$ for at most one $$t>0$$. Case 3: $$\zeta>1$$ (overdamped): In this case, $$\zeta^2 - 1 > 0$$, so the roots are real and distinct: $$r_1 = -\zeta \omega - \omega \sqrt{\zeta^2 - 1}$$ and $$r_2 = -\zeta \omega + \omega \sqrt{\zeta^2 - 1}$$. The general solution is: $\varphi(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}$ Similar to the critically damped case, all solutions with $$\varphi(0) \neq 0$$ are such that $$\varphi(t) = 0$$ for at most one $$t>0$$. If $$\zeta \geq 1$$ and $$\dot{\varphi}(0) = 0$$, every solution approaches zero monotonically, as both exponential terms decay over time without oscillations.

## Step 4: 4. Plot typical solutions for each case

For the underdamped case ($$\zeta < 1$$), the plot will show decaying oscillations, starting from a displaced position. The oscillations' amplitude will decay exponentially over time. For the critically damped case ($$\zeta = 1$$), the plot will show an initial displacement and a smooth, non-oscillatory decay to zero without overshooting. For the overdamped case ($$\zeta > 1$$), the plot will show an initial displacement and a smooth, non-oscillatory decay to zero without overshooting, but it will be slower compared to the critically damped case.

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