Chapter 8, Problem 8.19P

To determine

(a)

The response $x\left(t\right)$ for the model equation $\ddot{x}+4\dot{x}+8x=2u_s\left(t\right)$.

Answer to Problem 8.19P

The response $x\left(t\right)$ is $x(t)=\frac{1}{4}-\frac{1}{4}{e}^{-2t}\cos 2t-\frac{1}{4}{e}^{-2t}\sin 2t$.

Explanation of Solution

Given:

The given model equation is as:

$\ddot{x}+4\dot{x}+8x=2u_s\left(t\right)$

With initial conditions as follows:

$x(0)=0$ and $\dot{x}(0)=0$

Concept Used:

Laplace transform is used for obtaining the response of the model equation.

Calculation:

Equation to be solved is as:

$\ddot{x}+4\dot{x}+8x=2u_s\left(t\right)$

By taking the Laplace of this equation that is,

$\begin{array}{l}\ddot{x}+4\dot{x}+8x=2u_s\left(t\right)\\ \Rightarrow \left(s^{2}X(s)-sx(0)-\dot{x}\left(0\right)\right)+4\left(sX(s)-x(0)\right)+8X(s)=\frac{2}{s}\\ \Rightarrow \left(s^{2}X(s)-0-0\right)+4\left(sX(s)-0\right)+8X(s)=\frac{2}{s}\because x(0)=0,\dot{x}\left(0\right)=0\\ \Rightarrow X(s).(s^2+4s+8)=\frac{2}{s}\\ \Rightarrow X(s)=\frac{2}{s(s^2+4s+8)}\end{array}$

On using partial fraction expansion, this expression could be expressed as follows:

$\begin{array}{l}X(s)=\frac{2}{s(s^2+4s+8)}=\frac{A}{s}+\frac{Bs+C}{s^2+4s+8}\\ \Rightarrow \frac{2}{s(s^2+4s+8)}=\frac{A(s^2+4s+8)+s\left(Bs+C\right)}{s(s^2+4s+8)}\\ \Rightarrow 2=s^2(A+B)+s\left(4A+C\right)+8A\\ \Rightarrow A=\frac{1}{4},B=-\frac{1}{4},C=-\text{1}\end{array}$

Therefore, $X(s)=\frac{2}{s(s^2+4s+8)}=\frac{1}{4s}-\frac{1}{4}\cdot \frac{\left(s+4\right)}{{\left(s+2\right)}^2+{\left(2\right)}^2}$

Now, on taking the inverse Laplace of the above transfer function, we get

$\begin{array}{l}X(s)=\frac{1}{4s}-\frac{1}{4}\cdot \frac{\left(s+4\right)}{{\left(s+2\right)}^2+{\left(2\right)}^2}=\frac{1}{4s}-\frac{1}{4}\cdot \frac{\left(s+2\right)}{{\left(s+2\right)}^2+{\left(2\right)}^2}-\frac{1}{4}\cdot \frac{2}{{\left(s+2\right)}^2+{\left(2\right)}^2}\\ \Rightarrow x(t)=\frac{1}{4}-\frac{1}{4}{e}^{-2t}\cos 2t-\frac{1}{4}{e}^{-2t}\sin 2t\end{array}$

Thus, this is the response for the given model equation.

Conclusion:

The total response $x\left(t\right)$ for the model equation $x(t)=\frac{1}{4}-\frac{1}{4}{e}^{-2t}\cos 2t-\frac{1}{4}{e}^{-2t}\sin 2t$.

To determine

(b)

The response $x\left(t\right)$ for the model equation $\ddot{x}+8\dot{x}+12x=2u_s\left(t\right)$.

Answer to Problem 8.19P

The response $x\left(t\right)$ is $x(t)=\frac{1}{6}-\frac{1}{6}{e}^{-4t}\cosh 2t-\frac{1}{3}{e}^{-4t}\sinh 2t$.

Explanation of Solution

Given:

The given model equation is as:

$\ddot{x}+8\dot{x}+12x=2u_s\left(t\right)$

With initial conditions as follows:

$x(0)=0$ and $\dot{x}(0)=0$

Concept Used:

Laplace transform is used for obtaining the response of the model equation.

Calculation:

Equation to be solved is as:

$\ddot{x}+8\dot{x}+12x=2u_s\left(t\right)$

By taking the Laplace of this equation that is,

$\begin{array}{l}\ddot{x}+8\dot{x}+12x=2u_s\left(t\right)\\ \Rightarrow \left(s^{2}X(s)-sx(0)-\dot{x}\left(0\right)\right)+8\left(sX(s)-x(0)\right)+12X(s)=\frac{2}{s}\\ \Rightarrow \left(s^{2}X(s)-0-0\right)+8\left(sX(s)-0\right)+12X(s)=\frac{2}{s}\because x(0)=0,\dot{x}\left(0\right)=0\end{array}$

$\begin{array}{l}\Rightarrow X(s).(s^2+8s+12)=\frac{2}{s}\\ \Rightarrow X(s)=\frac{2}{s(s^2+8s+12)}\end{array}$

On using partial fraction expansion, this expression could be expressed as follows:

$\begin{array}{l}X(s)=\frac{2}{s(s^2+8s+12)}=\frac{A}{s}+\frac{Bs+C}{s^2+8s+12}\\ \Rightarrow \frac{2}{s(s^2+8s+12)}=\frac{A(s^2+8s+12)+s\left(Bs+C\right)}{s(s^2+8s+12)}\\ \Rightarrow 2=s^2(A+B)+s\left(8A+C\right)+12A\\ \Rightarrow A=\frac{1}{6},B=-\frac{1}{6},C=-\frac{4}{3}\end{array}$

Therefore, $X(s)=\frac{2}{s(s^2+4s+8)}=\frac{1}{6s}-\frac{1}{6}\cdot \frac{\left(s+8\right)}{{\left(s+4\right)}^2-{\left(2\right)}^2}$

Now, on taking the inverse Laplace of the above transfer function, we get

$\begin{array}{l}X(s)=\frac{1}{6s}-\frac{1}{6}\cdot \frac{\left(s+8\right)}{{\left(s+4\right)}^2-{\left(2\right)}^2}=\frac{1}{6s}-\frac{1}{6}\cdot \frac{\left(s+4\right)}{{\left(s+4\right)}^2-{\left(2\right)}^2}-\frac{1}{3}\cdot \frac{2}{{\left(s+4\right)}^2-{\left(2\right)}^2}\\ \Rightarrow x(t)=\frac{1}{6}-\frac{1}{6}{e}^{-4t}\cosh 2t-\frac{1}{3}{e}^{-4t}\sinh 2t\end{array}$.

Conclusion:

The total response $x\left(t\right)$ for the model equation $x(t)=\frac{1}{6}-\frac{1}{6}{e}^{-4t}\cosh 2t-\frac{1}{3}{e}^{-4t}\sinh 2t$.

To determine

(c)

The response $x\left(t\right)$ for the model equation $\ddot{x}+4\dot{x}+4x=2u_s\left(t\right)$.

Answer to Problem 8.19P

The response $x\left(t\right)$ is $x(t)=\frac{1}{2}-\frac{1}{2}{e}^{-2t}-t{e}^{-2t}$.

Explanation of Solution

Given:

The given model equation is as:

$\ddot{x}+4\dot{x}+4x=2u_s\left(t\right)$

With initial conditions as follows:

$x(0)=0$ and $\dot{x}(0)=0$

Concept Used:

Laplace transform is used for obtaining the response of the model equation.

Calculation:

Equation to be solved is as:

$\ddot{x}+4\dot{x}+4x=2u_s\left(t\right)$

By taking the Laplace of this equation that is,

$\begin{array}{l}\ddot{x}+4\dot{x}+4x=2u_s\left(t\right)\\ \Rightarrow \left(s^{2}X(s)-sx(0)-\dot{x}\left(0\right)\right)+4\left(sX(s)-x(0)\right)+4X(s)=\frac{2}{s}\\ \Rightarrow \left(s^{2}X(s)-0-0\right)+4\left(sX(s)-0\right)+4X(s)=\frac{2}{s}\because x(0)=0,\dot{x}\left(0\right)=0\\ \Rightarrow X(s).(s^2+4s+4)=\frac{2}{s}\\ \Rightarrow X(s)=\frac{2}{s{(s+2)}^2}\end{array}$

On using partial fraction expansion, this expression could be expressed as follows:

$\begin{array}{l}X(s)=\frac{2}{s(s^2+4s+4)}=\frac{A}{s}+\frac{B}{s+2}+\frac{C}{{\left(s+2\right)}^2}\\ \Rightarrow \frac{2}{s(s^2+4s+4)}=\frac{A{\left(s+2\right)}^2+Bs\left(s+2\right)+Cs}{s{\left(s+2\right)}^2}\\ \Rightarrow 2=s^2(A+B)+s\left(4A+2B+C\right)+4A\\ \Rightarrow A=\frac{1}{2},B=-\frac{1}{2},C=-\text{1}\end{array}$

Therefore, $X(s)=\frac{2}{s(s^2+4s+4)}=\frac{1}{2s}-\frac{1}{2}\cdot \frac{1}{\left(s+2\right)}-\frac{1}{{\left(s+2\right)}^2}$

Now, on taking the inverse Laplace of the above transfer function, we get

$\begin{array}{l}X(s)=\frac{1}{2s}-\frac{1}{2}\cdot \frac{1}{\left(s+2\right)}-\frac{1}{{\left(s+2\right)}^2}\\ \Rightarrow x(t)=\frac{1}{2}-\frac{1}{2}{e}^{-2t}-t{e}^{-2t}\end{array}$

Thus, this is the response for the given model equation.

Conclusion:

The total response $x\left(t\right)$ for the model equation $x(t)=\frac{1}{2}-\frac{1}{2}{e}^{-2t}-t{e}^{-2t}$.