Problems

The ODE is already in standard form; therefore, identify \[\begin{aligned} \omega_n &= \sqrt{25} \\ &= 5. \\ \zeta &= \frac{5} {2 \omega_n} \\ &= 1/2. \end{aligned}\]

From table 6.2 for damping \(\zeta \in (0,1)\) and forcing \(f(t) = u_s(t)\), \[\begin{aligned} y_\text{ch}(t) &= \dfrac{1} {\omega_n^2} \left ( 1 - \dfrac{e^{-\zeta \omega_n t}}{\sqrt{1-\zeta^2}} \cos (\omega_d t + \psi) \right ) \end{aligned}\] where \(\omega_d = \omega_n \sqrt{1-\zeta^2}\) and \(\psi = -\arctan(\zeta/\sqrt{1-\zeta^2})\).

From superposition, the forced response is \[\begin{aligned} y_\text{fo}(t) &= 2 \dot{y}_\text{ch} + 3 y_\text{ch}, \end{aligned}\] where \[\begin{aligned} \dot{y}_\text{ch} &= \frac{e^{- \omega_n \zeta t}}{\omega_n^2 \sqrt{1 - \zeta^2}} \left(\omega_d \sin\left(\omega_d t + \psi\right) + \omega_n \zeta \cos\left(\omega_d t + \psi\right)\right)\\ &= \dfrac{e^{-\zeta \omega_n t}}{\omega_n \sqrt{1-\zeta^2}} \sin (\omega_d t). \end{aligned}\] This last equality can be derived by applying a two-to-one formula from or recognizing that \(\dot{u}_s = \delta\) and using the characteristic response formula corresponding to \(\delta\) in table 6.2.

The ODE is not given in the standard form of a first-order ODE. Dividing both sides by \(3\), \[\begin{aligned} \frac{1} {3} \dot{y} + y &= \frac{2} {3} \dot{u} + \frac{1} {3} u. \end{aligned}\] Therefore, by inspection, the time constant \(\tau = 1/3\).

From , for a unit step input, the characteristic response is \[\begin{aligned} y_s &= 1 - e^{-t/\tau}. \end{aligned}\]

Beginning with \(y_s\) and applying superposition and the derivative property, the forced response is \[\begin{aligned} y_\text{fo} &= 3\left(\frac{2} {3} \dot{y_s} + \frac{1} {3} y_s\right) \\ &= 2 \dot{y_s} + y_s \\ &= 2\cdot 3 e^{-3 t} + 1 - e^{-3 t} \\ &= 1 + 5 e^{-3 t}. \end{aligned}\]

The free response \(y_\text{fr}\) to initial condition \(y(0) = -4\) is given by to be \[\begin{aligned} y_\text{fr}(t) &= y(0)\, e^{-t/\tau}. \end{aligned}\]

The total response \(y_t\) when both the input \(u\) and initial condition \(y(0)\) are applied simultaneously is, by superposition, merely the sum \[\begin{aligned} y_t &= y_\text{fr} + y_\text{fo} \\ &= y(0)\, e^{-t/\tau} + 1 + 5 e^{-3 t} \\ &= -4\, e^{-3 t} + 1 + 5 e^{-3 t} \\ &= 1 + e^{-3 t}. \\ \end{aligned}\]

From the standard form of the I/O ODE, from the \(y\) coefficient, \(\omega_n^2 = 25\), which implies \(\omega_n = 5\) (\(\omega_n\) is positive by definition). From the \(\dot{y}\) coefficient, \[\begin{aligned} 2 \zeta \omega_n &= 5 \\ \zeta &= \frac{5} {2\omega_n} = 1/2. \end{aligned}\]

From table 6.2, with \(0 < \zeta < 1\), the characteristic response is \[\begin{aligned} y_\delta(t) = \frac{e^{-\zeta \omega_n t}}{\omega_n \sqrt{1-\zeta^2}} \sin (\omega_d t), \end{aligned}\] where \(\omega_d = \omega_n \sqrt{1 - \zeta^2}\) is the damped natural frequency.

For the forcing function \(\dot{u} + 7 u\), we can combine the superposition and derivative properties of LTI stystems to conclude that \[\begin{aligned} y_\text{fo} &= \dot{y}_\delta + 7 y_\delta \\ &= \frac{1} {\omega_n \sqrt{1-\zeta^2}} \left( -\zeta \omega_n e^{-\zeta \omega_n t} \sin (\omega_d t) + \omega_d e^{-\zeta \omega_n t} \cos (\omega_d t) \right) \\ &= \frac{e^{-\zeta \omega_n t}}{\omega_n \sqrt{1-\zeta^2}} \left( -\zeta \omega_n \sin (\omega_d t) + \omega_d \cos (\omega_d t) \right)\\ &= e^{-\zeta \omega_n t} \left( \frac{\zeta} {\zeta^2 - 1} \sin (\omega_d t) + \cos (\omega_d t) \right). \end{aligned}\] It would perhaps be good to combine the \(\sin\) and \(\cos\) terms into a single sinusoid, but this will suffice for our purposes.

The free response for initial conditions \(y(0) = 11\) and \(\dot{y}(0) = 0\) is given by of to be \[\begin{aligned} y_\text{fr}(t) &= y(0) \dfrac{e^{-\zeta \omega_n t}}{\sqrt{1-\zeta^2}} \cos (\omega_d t + \psi) \\ &= 11 \dfrac{e^{-\zeta \omega_n t}}{\sqrt{1-\zeta^2}} \cos (\omega_d t + \psi) \end{aligned}\] where the phase \(\psi\) is \[\psi = -\arctan\dfrac{\zeta} {\sqrt{1-\zeta^2}}.\]

The total response \(y_t\) is, from superposition, \[\begin{aligned} y_t(t) = y_\text{fr}(t) + y_\text{fo}. \end{aligned}\]

The equilibrium output \(y(t) = \overline{y}\) obtains when \(\ddot{y} = \dot{y} = \dot{u} = 0\). Therefore, from the original ODE, \[\begin{aligned} 25 \overline{y} &= 7 \overline{u} \\ \overline{y} &= \frac{7} {25} \overline {u}. \end{aligned}\]

The roots of the characteristic equation are given in of to be \[\begin{aligned} \lambda_1 \text{, } \lambda_2 = -\zeta \omega_n \pm j \omega_d. \end{aligned}\] Therefore \(\Re(\lambda_1) = \Re(\lambda_2) = -\zeta \omega_n\), which is strictly negative. Therefore, the system is asymptotically stable.