Question 2.1.6: Find the complete response of the following equation. ẍ + 4...
Find the complete response of the following equation.
\ddot{x}+4 \dot{x}+53 x=15 u_s(t) \quad x(0)=8 \quad \dot{x}(0)=-19
We cannot use Table 2.1.1 for this part because the intial conditions are not zero. Comparing this equation with the form in Table 2.1.2, we see that a = 4, b = 53, and c = 15. The characteristic equation is s² + 4s + 53 = 0 and its roots are s = −2 ± 7j. This corresponds to entry 4 in the table, where 𝜎 = −2 and 𝜔 = 7. Thus, the form of the solution is
x(t)=e^{-2 t}\left(C_1 \sin 7 t+C_2 \cos 7 t\right)+\frac{15}{53}
and
\dot{x}(t)=-2 e^{-2 t}\left(C_1 \sin 7 t+C_2 \cos 7 t\right)+e^{-2 t}\left(7 C_1 \cos 7 t-7 C_2 \sin 7 t\right)
Now evaluate these using the initial conditions
x(0)=8=C_2+\frac{15}{53}
which gives C_2 = 409 ∕ 53. Also,
\dot{x}(0)=-19=-2\left(\frac{409}{53}\right)+7 C_1
Thus, C_1 = −27∕53, and the solution is
x(t)=e^{-2 t}\left(\frac{409}{53}\cos 7 t-\frac{27}{53}\sin 7 t\right)+\frac{15}{53}
The solution is plotted in Figure 2.1.10. The radian frequency of the oscillations is 7, which corresponds to a period of 2𝜋 ∕ 7 = 0.867. The oscillations are difficult to see for t > 2 because e^{−2t} < 0.02 for t > 2. So, for most practical purposes, we may say that the response is essentially constant at the steady-state response 15 ∕ 53 for t > 2.
| Table 2.1.1 Step response for zero initial conditions. | ||
| ODE | Roots | x(t) |
| \text{1.}\quad\dot{x}+a x=M u_s(t) | s = −a | x(t)=\frac{M}{a}\left(1-e^{-a t}\right) |
| \text{2.}\quad\ddot{x}+(a+b) \dot{x}+a b x=M u_s(t) | s = −a, −b a ≠ b | x(t)=M\left[\frac{e^{-a t}}{a(a-b)}+\frac{e^{-b t}}{b(b-a)}+\frac{1}{a b}\right] |
| \text{3.}\quad\ddot{x}+2 a \dot{x}+a^2 x=M u_s(t) | s = −a, −a | x(t)=\frac{M}{a^2}\left[1-(a t+1) e^{-a t}\right] |
| \text{4.}\quad\ddot{x}+b^2 x=M u_s(t) | s = ±bj b > 0 | x(t)=\frac{M}{b^2}(1-\cos b t) |
| \text{5.}\quad\ddot{x}+2 a \dot{x}+\left(a^2+b^2\right) x=M u_s(t) | s = −a ± bj b > 0 | x(t)=\frac{M}{a^2+b^2}\left[1-\left(\frac{a}{b}\sin b t+\cos b t\right) e^{-a t}\right] |
| Equation | Solution form |
| First order: \dot{x}+a x=b~~~a\neq0
Second order: \ddot{x}+a\dot{x}+b x=c\quad b\neq0 |
x(t)={\frac{b}{a}}+C e^{-a t} |
| 1. (a^{2}\gt 4b)\;\mathrm{distinct,real\;roots:}s_{1},s_{2} | x(t)=C_{1}e^{s_{1}t}+C_{2}e^{s_{2}t}+\frac{c}{b} |
| 2. (a^{2}=4b)\,\mathrm{repeated,\,real\,roots:}s_{1},s_{1} | x(t)=(C_{1}+t C_{2})e^{s_{1}t}+{\frac{c}{b}} |
| 3. \begin{array}{l c r}{{(a=0,b\gt 0)\operatorname*{imaginary}\operatorname{roots}s=\pm\,j\omega,}}\\ {{\omega={\sqrt{b}}}}\end{array} | x(t)=C_{1}\sin\omega t+C_{2}\cos\omega t+{\frac{c}{b}} |
| 4. \begin{array}{l c r}{{(a\neq0,\,a^{2}\lt 4b)\,\text{complex roots}:\,s=\sigma\pm j\omega,}}\\ {{\sigma=-a/2,\,\omega={\sqrt{4b-a^{2}}}/2}}\end{array} | x(t)=e^{\sigma t}(C_{1}\sin\omega t+C_{2}\cos\omega t)+{\frac{c}{b}} |
