Chapter 16, Problem 95P

To determine

Find the outputs $y_1\left(t\right)$ and $y_2\left(t\right)$ for given state equation.

Answer to Problem 95P

The outputs $y_1\left(t\right)$ and $y_2\left(t\right)$ for given state equation are,

$\underset{\_}{\left(-2.4+4.4e^{-3t}\cos t-0.8e^{-3t}\sin t\right)u\left(t\right)}$ and $\underset{\_}{\left(-1.2-0.8e^{-3t}\cos t+0.6e^{-3t}\sin t\right)u\left(t\right)}$ respectively.

Explanation of Solution

Given data:

The state equation of input and output are,

$\dot{\text{x}}\text{=}\left[\begin{array}{cc}-2& -1\\ 2& -4\end{array}\right]x+\left[\begin{array}{cc}1& 1\\ 4& 0\end{array}\right]\left[\begin{array}{c}u\left(t\right)\\ 2u\left(t\right)\end{array}\right]$        (1)

$\text{y=}\left[\begin{array}{cc}-2& -2\\ 1& 0\end{array}\right]x+\left[\begin{array}{cc}2& 0\\ 0& -1\end{array}\right]\left[\begin{array}{c}u\left(t\right)\\ 2u\left(t\right)\end{array}\right]$        (2)

Formula used:

Write the standard form of the state variable equation and for the output equation as follows.

$\dot{\text{x}}\text{=}Ax+Bz$        (3)

$\text{y=}Cx+Dz$        (4)

Here,

A, B, C and D are state matrices.

x and z are variable matrices.

Write the formula to find the variable solution $x$ as follows.

$x=\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]$

Apply the Laplace transform for x,

$X\left(s\right)=\left[\begin{array}{c}{X}_1\left(s\right)\\ X_2\left(s\right)\end{array}\right]$

Then the solution for the state variable equation can be written as follows.

$\left[\begin{array}{c}{X}_1\left(s\right)\\ X_2\left(s\right)\end{array}\right]={\left(sI-A\right)}^{-1}BZ\left(s\right)$        (5)

Here,

I is the identity matrix.

Calculation:

Compare equation (1) with equation (3) and equation (2) with equation (4).

$A\text{=}\left[\begin{array}{cc}-2& -1\\ 2& -4\end{array}\right],B=\left[\begin{array}{cc}1& 1\\ 4& 0\end{array}\right]$

$C=\left[\begin{array}{cc}-2& -2\\ 1& 0\end{array}\right],D=\left[\begin{array}{cc}2& 0\\ 0& -1\end{array}\right]$

$z=\left[\begin{array}{c}u\left(t\right)\\ 2u\left(t\right)\end{array}\right]$

Apply the Laplace transform to above the z matrix.

$Z\left(s\right)=\left[\begin{array}{c}\frac{1}{s}\\ \frac{2}{s}\end{array}\right]\left\{\because L\left[u\left(t\right)\right]=\frac{1}{s}\right\}$        (6)

From equation (2) expand the output equations as follows.

$\begin{array}{l}\left[\begin{array}{c}{y}_1\left(t\right)\\ y_2\left(t\right)\end{array}\right]\text{=}\left[\begin{array}{cc}-2& -2\\ 1& 0\end{array}\right]\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& -1\end{array}\right]\left[\begin{array}{c}u\left(t\right)\\ 2u\left(t\right)\end{array}\right]\left\{\because y=\left[\begin{array}{c}{y}_1\left(t\right)\\ y_2\left(t\right)\end{array}\right],x=\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]\right\}\\ \left[\begin{array}{c}{y}_1\left(t\right)\\ y_2\left(t\right)\end{array}\right]\text{=}\left[\begin{array}{c}-2x_1\left(t\right)-2x_2\left(t\right)\\ x_1\left(t\right)\end{array}\right]+\left[\begin{array}{c}2u\left(t\right)\\ -2u\left(t\right)\end{array}\right]\end{array}$

Then the output equations are,

$y_1\left(t\right)=-2x_1\left(t\right)-2x_2\left(t\right)+2u\left(t\right)$        (7)

$y_2\left(t\right)=x_1\left(t\right)-2u\left(t\right)$        (8)

Substitute $\left[\begin{array}{cc}-2& -1\\ 2& -4\end{array}\right]$ for A, $\left[\begin{array}{cc}1& 1\\ 4& 0\end{array}\right]$ for B, and $\left[\begin{array}{c}\frac{1}{s}\\ \frac{2}{s}\end{array}\right]$ for $Z\left(s\right)$ in equation (5).

$\begin{array}{l}\left[\begin{array}{c}{X}_1\left(s\right)\\ X_2\left(s\right)\end{array}\right]={\left(s\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}-2& -1\\ 2& -4\end{array}\right]\right)}^{-1}\left[\begin{array}{cc}1& 1\\ 4& 0\end{array}\right]\left[\begin{array}{c}\frac{1}{s}\\ \frac{2}{s}\end{array}\right]\left\{\because I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right\}\\ \left[\begin{array}{c}{X}_1\left(s\right)\\ X_2\left(s\right)\end{array}\right]={\left(\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}-2& -1\\ 2& -4\end{array}\right]\right)}^{-1}\left[\begin{array}{c}\frac{1}{s}+\frac{2}{s}\\ \frac{4}{s}+0\end{array}\right]\\ \left[\begin{array}{c}{X}_1\left(s\right)\\ X_2\left(s\right)\end{array}\right]={\left[\begin{array}{cc}s+2& 1\\ -2& s+4\end{array}\right]}^{-1}\left[\begin{array}{c}\frac{3}{s}\\ \frac{4}{s}\end{array}\right]\left\{\because {\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}^{-1}=\frac{1}{ad-bc}{\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]}^{-1}\right\}\\ \left[\begin{array}{c}{X}_1\left(s\right)\\ X_2\left(s\right)\end{array}\right]=\frac{1}{\left(s+2\right)\left(s+4\right)-\left(1\right)\left(-2\right)}\left[\begin{array}{cc}s+4& -1\\ 2& s+2\end{array}\right]\left[\begin{array}{c}\frac{3}{s}\\ \frac{4}{s}\end{array}\right]\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}\left[\begin{array}{c}{X}_1\left(s\right)\\ X_2\left(s\right)\end{array}\right]=\frac{1}{s^2+4s+2s+10}\left[\begin{array}{c}\left(s+4\right)\frac{3}{s}-\frac{4}{s}\\ \frac{6}{s}+\left(s+2\right)\frac{4}{s}\end{array}\right]\\ \left[\begin{array}{c}{X}_1\left(s\right)\\ X_2\left(s\right)\end{array}\right]=\frac{1}{s^2+6s+10}\left[\begin{array}{c}\frac{3s+12-4}{s}\\ \frac{4s+8+6}{s}\end{array}\right]\\ \left[\begin{array}{c}{X}_1\left(s\right)\\ X_2\left(s\right)\end{array}\right]=\frac{1}{{\left(s+3\right)}^2+1^2}\left[\begin{array}{c}\frac{3s+8}{s}\\ \frac{4s+14}{s}\end{array}\right]\end{array}$

From the above equation,

$X_1\left(s\right)=\frac{3s+8}{s\left({\left(s+3\right)}^2+1^2\right)}$        (9)

$X_2\left(s\right)=\frac{4s+14}{s\left({\left(s+3\right)}^2+1^2\right)}$        (10)

Apply the partial fraction expansion for the function in equation (9).

$\frac{3s+8}{s\left({\left(s+3\right)}^2+1^2\right)}=\frac{P}{s}+\frac{Qs+R}{{\left(s+3\right)}^2+1^2}$        (11)

$\begin{array}{l}\frac{3s+8}{s\left({\left(s+3\right)}^2+1^2\right)}=\frac{P\left({\left(s+3\right)}^2+1^2\right)+s\left(Qs+R\right)}{s\left({\left(s+3\right)}^2+1^2\right)}\\ 3s+8=P\left(s^2+6s+10\right)+s\left(Qs+R\right)\\ 3s+8=\left(P+Q\right)s^2+\left(6P+R\right)s+10P\end{array}$

Compare the coefficients of $s^2$, s, and constants on the both sides.

$\begin{array}{l}P+Q=0\Rightarrow P=-Q\\ 6P+R=3\\ 10P=8\end{array}$

Then, the value of P is,

$\begin{array}{l}P=\frac{8}{10}\\ P=0.8\end{array}$

Therefore, from P, Q relation,

$Q=-0.8$

And

$\begin{array}{l}6\left(0.8\right)+R=3\\ R=3-4.8\\ R=-1.8\end{array}$

Substitute 0.8 for P, –0.8 for Q and –1.8 for R in equation (11).

$\begin{array}{l}\frac{3s+8}{s\left({\left(s+3\right)}^2+1^2\right)}=\frac{0.8}{s}+\frac{-0.8s-1.8}{{\left(s+3\right)}^2+1^2}\\ X_1\left(s\right)=\frac{0.8}{s}+\frac{-0.8\left(s+3\right)+0.6}{{\left(s+3\right)}^2+1^2}\\ X_1\left(s\right)=\frac{0.8}{s}-0.8\frac{\left(s+3\right)}{{\left(s+3\right)}^2+1^2}+0.6\frac{1}{{\left(s+3\right)}^2+1^2}\end{array}$

Apply the inverse Laplace transform to the above equation using the formulas

${ℒ}^{-1}\left[\frac{1}{s}\right]=u\left(t\right),{ℒ}^{-1}\left[\frac{s+a}{{\left(s+a\right)}^2+b^2}\right]=\left(e^{-at}\cos bt\right)u\left(t\right)$

and ${ℒ}^{-1}\left[\frac{b}{{\left(s+a\right)}^2+b^2}\right]=\left(e^{-at}\sin bt\right)u\left(t\right)$, then the time domain function $x_1\left(t\right)$ is,

$x_1\left(t\right)=\left(0.8-0.8e^{-3t}\cos t+0.6e^{-3t}\sin t\right)u\left(t\right)$        (12)

Apply the partial fraction expansion for the function in equation (10).

$\frac{4s+14}{s\left({\left(s+3\right)}^2+1^2\right)}=\frac{E}{s}+\frac{Fs+G}{{\left(s+3\right)}^2+1^2}$        (13)

$\begin{array}{l}\frac{4s+14}{s\left({\left(s+3\right)}^2+1^2\right)}=\frac{E\left({\left(s+3\right)}^2+1^2\right)+s\left(Fs+G\right)}{s\left({\left(s+3\right)}^2+1^2\right)}\\ 4s+14=E\left(s^2+6s+10\right)+s\left(Fs+G\right)\\ 4s+14=\left(E+F\right)s^2+\left(6E+G\right)s+10E\end{array}$

Compare the coefficients of $s^2$, s and constants on the both sides.

$\begin{array}{l}E+F=0\Rightarrow E=-F\\ 6E+G=4\\ 10E=14\end{array}$

Then, the value of E is,

$\begin{array}{l}E=\frac{14}{10}\\ E=1.4\end{array}$

Therefore, from E, F relation,

$F=-1.4$

And

$\begin{array}{l}6\left(1.4\right)+G=4\\ G=4-8.4\\ G=-4.4\end{array}$

Substitute 1.4 for E, –1.4 for F and –4.4 for G in equation (13).

$\frac{4s+14}{s\left({\left(s+3\right)}^2+1^2\right)}=\frac{1.4}{s}+\frac{-1.4s-4.4}{{\left(s+3\right)}^2+1^2}$

$\begin{array}{l}{X}_2\left(s\right)=\frac{1.4}{s}+\frac{-1.4\left(s+3\right)-0.2}{{\left(s+3\right)}^2+1^2}\\ X_2\left(s\right)=\frac{1.4}{s}-1.4\frac{s+3}{{\left(s+3\right)}^2+1^2}-0.2\frac{1}{{\left(s+3\right)}^2+1^2}\end{array}$

Apply inverse Laplace transform for the above equation.

$x_2\left(t\right)=\left(1.4-1.4e^{-3t}\cos t-0.2e^{-3t}\sin t\right)u\left(t\right)$        (14)

Substitute equation (12) and equation (14) in equation (7) to find $y_1\left(t\right)$.

$\begin{array}{l}{y}_1\left(t\right)=\left\{\begin{array}{l}-2\left[\left(0.8-0.8e^{-3t}\cos t+0.6e^{-3t}\sin t\right)u\left(t\right)\right]\\ -2\left[\left(1.4-1.4e^{-3t}\cos t-0.2e^{-3t}\sin t\right)u\left(t\right)\right]+2u\left(t\right)\end{array}\right\}\\ y_1\left(t\right)=\left\{-1.6+1.6e^{-3t}\cos t-1.2e^{-3t}\sin t-2.8+2.8e^{-3t}\cos t+0.4e^{-3t}\sin t+2\right\}u\left(t\right)\\ y_1\left(t\right)=\left(-2.4+4.4e^{-3t}\cos t-0.8e^{-3t}\sin t\right)u\left(t\right)\end{array}$

Substitute equation (12) in equation (8) to find $y_2\left(t\right)$.

$\begin{array}{l}{y}_2\left(t\right)=\left(0.8-0.8e^{-3t}\cos t+0.6e^{-3t}\sin t\right)u\left(t\right)-2u\left(t\right)\\ y_2\left(t\right)=\left(0.8-0.8e^{-3t}\cos t+0.6e^{-3t}\sin t-2\right)u\left(t\right)\\ y_2\left(t\right)=\left(-1.2-0.8e^{-3t}\cos t+0.6e^{-3t}\sin t\right)u\left(t\right)\end{array}$

Conclusion:

Thus, the outputs $y_1\left(t\right)$ and $y_2\left(t\right)$ for given state equation are $\underset{\_}{\left(-2.4+4.4e^{-3t}\cos t-0.8e^{-3t}\sin t\right)u\left(t\right)}$ and $\underset{\_}{\left(-1.2-0.8e^{-3t}\cos t+0.6e^{-3t}\sin t\right)u\left(t\right)}$ respectively.