Chapter 5, Problem 5.22P

To determine

The state-variable model in standard form for the transfer function $\frac{Y\left(s\right)}{F\left(s\right)}=\frac{s+2}{s^2+4s+3}$ using two methods. Also, relate the initial conditions for the state variables with the given initial values $y\left(0\right)and\dot{y}\left(0\right)$.

Answer to Problem 5.22P

The state model for the transfer function $\frac{Y\left(s\right)}{F\left(s\right)}=\frac{s+2}{s^2+4s+3}$ using two methods are as shown:

By the first method,

$\begin{array}{l}{\dot{x}}_1=-4x_1+x_2+f\\ {\dot{x}}_2=-3x_1+2f\\ y=x_1\end{array}$

By the second method,

$\begin{array}{l}{\dot{x}}_1=-x_1+\frac{f}{2}\\ {\dot{x}}_2=-3x_2-\frac{f}{2}\\ y=x_1-x_2\end{array}$

And the relation between initial values of state variables and $y\left(0\right)and\dot{y}\left(0\right)$:

For the state model using the first method,

$\begin{array}{l}{x}_1\left(0\right)=y\left(0\right)\\ x_2\left(0\right)=\dot{y}\left(0\right)+4y\left(0\right)-f\left(0\right)\end{array}$

While for the state model using the second method,

$\begin{array}{l}{x}_1\left(0\right)=\frac{1}{2}\dot{y}\left(0\right)+\frac{3}{2}y\left(0\right)-\frac{1}{2}f\left(0\right)\\ x_2\left(0\right)=\frac{1}{2}\dot{y}\left(0\right)+y\left(0\right)-\frac{1}{2}f\left(0\right)\end{array}$.

Explanation of Solution

Given:

The transfer function of the system is as shown:

$\frac{Y\left(s\right)}{F\left(s\right)}=\frac{s+2}{s^2+4s+3}$

With given initial values $y\left(0\right)and\dot{y}\left(0\right)$.

Concept Used:

Firstly, the highest order of the denominator of a transfer function is equal to the number of state variable for the model.

Secondly, the methodology for assuming the state variables in case of given transfer function is as follows:

First method: For the transfer function of the form $\frac{Y\left(s\right)}{F\left(s\right)}=\frac{as+b}{s+c}$, try to make a term of ‘1’ in the denominator such that

$\begin{array}{l}\frac{Y\left(s\right)}{F\left(s\right)}=\frac{a+\frac{b}{s}}{1+\frac{c}{s}}\\ \Rightarrow Y\left(s\right)=aF\left(s\right)+\frac{b}{s}F\left(s\right)-\frac{c}{s}Y\left(s\right)\\ \Rightarrow Y\left(s\right)=aF\left(s\right)+\frac{1}{s}\left(bF\left(s\right)-cY\left(s\right)\right)\end{array}$

Here, the second term on the right side is the input to an integrator, $\frac{1}{s}$ thus take this as the state variable for the model and solve further to obtain the state model.

Second method: For a transfer function as $\frac{Y\left(s\right)}{F\left(s\right)}=\frac{s+a}{s+b}$, transfer the function as shown

$Y\left(s\right)=\left(s+a\right)\frac{F\left(s\right)}{s+b}$

And take $\frac{F\left(s\right)}{s+b}$ as the state variable for the model and solve further for the state model.

Calculation:

Given transfer function is as shown,

$\frac{Y\left(s\right)}{F\left(s\right)}=\frac{s+2}{s^2+4s+3}$

Using the first method,

Divide the numerator and denominator of transfer function by $s^2$ such that

$\begin{array}{l}\frac{Y\left(s\right)}{F\left(s\right)}=\frac{\frac{1}{s}+\frac{2}{s^2}}{1+\frac{4}{s}+\frac{3}{s^2}}\\ \Rightarrow Y\left(s\right)=-\frac{4}{s}Y\left(s\right)-\frac{3}{s^2}Y\left(s\right)+\frac{1}{s}F\left(s\right)+\frac{2}{s^2}F\left(s\right)\\ \Rightarrow Y\left(s\right)=\frac{1}{s}\left(\left(F\left(s\right)-4Y\left(s\right)\right)+\frac{1}{s}\left(2F\left(s\right)-3Y\left(s\right)\right)\right)\end{array}$

On taking,

$\begin{array}{l}{X}_1\left(s\right)=Y\left(s\right)=\frac{1}{s}\left(\left(F\left(s\right)-4Y\left(s\right)\right)+\frac{1}{s}\left(2F\left(s\right)-3Y\left(s\right)\right)\right)\\ and{X}_2\left(s\right)=\frac{1}{s}\left(2F\left(s\right)-3Y\left(s\right)\right)\end{array}$

Therefore,

$\begin{array}{l}Y\left(s\right)=X_1\left(s\right)\\ \Rightarrow y=x_1\left(1\right)\end{array}$

Similarly,

$\begin{array}{l}{X}_2\left(s\right)=\frac{1}{s}\left(2F\left(s\right)-3Y\left(s\right)\right)\\ \Rightarrow s{X}_2\left(s\right)=2F\left(s\right)-3X_1\left(s\right)\because From\left(1\right)\\ \Rightarrow {\dot{x}}_2=-3x_1+2f\left(2\right)\end{array}$

$\begin{array}{l}{X}_1\left(s\right)=Y\left(s\right)=\frac{1}{s}\left(\left(F\left(s\right)-4Y\left(s\right)\right)+\frac{1}{s}\left(2F\left(s\right)-3Y\left(s\right)\right)\right)\\ \Rightarrow X_1\left(s\right)=\frac{1}{s}\left(\left(F\left(s\right)-4X_1\left(s\right)\right)+X_2\left(s\right)\right)\because From\left(1\right)and\left(2\right)\\ \Rightarrow s{X}_1\left(s\right)=F\left(s\right)-4X_1\left(s\right)+X_2\left(s\right)\\ \Rightarrow {\dot{x}}_1=-4x_1+x_2+f\left(3\right)\end{array}$

Thus, from equations (1), (2) and (3), we have the state model in standard form as follows

$\begin{array}{l}{\dot{x}}_1=-4x_1+x_2+f\\ {\dot{x}}_2=-3x_1+2f\\ y=x_1\end{array}$

From equation (1) at t=0, we get

$x_1\left(0\right)=y\left(0\right)(4)$

Similarly, from equation (3) at t=0, we have

$\begin{array}{l}{\dot{x}}_1=-4x_1+x_2+f\\ \Rightarrow x_2\left(0\right)={\dot{x}}_1\left(0\right)+4x_1\left(0\right)-f\left(0\right)\\ \Rightarrow x_2\left(0\right)=\dot{y}\left(0\right)+4y\left(0\right)-f\left(0\right)(5)\because From(4)\end{array}$

Now, using the second method for the transfer function

$\frac{Y\left(s\right)}{F\left(s\right)}=\frac{s+2}{s^2+4s+3}$

This could be also written as

$\begin{array}{l}\frac{Y\left(s\right)}{F\left(s\right)}=\frac{s+2}{s^2+4s+3}\\ \Rightarrow Y\left(s\right)=\left(s+2\right)\frac{F\left(s\right)}{\left(s+3\right)\left(s+1\right)}\\ \Rightarrow Y\left(s\right)=\left(s+2\right)\frac{F\left(s\right)}{2}\left(\frac{1}{\left(s+1\right)}-\frac{1}{\left(s+3\right)}\right)\end{array}$

On taking the state variables as follows

$X_1\left(s\right)=\frac{F\left(s\right)}{2\left(s+1\right)}and{X}_2\left(s\right)=-\frac{F\left(s\right)}{2\left(s+3\right)}$

We have

$\begin{array}{l}{X}_1\left(s\right)=\frac{F\left(s\right)}{2\left(s+1\right)}\\ \Rightarrow s{X}_1\left(s\right)=\frac{F\left(s\right)}{2}-X_1\left(s\right)\\ \Rightarrow {\dot{x}}_1=-x_1+\frac{f}{2}\left(6\right)\end{array}$

Similarly,

$\begin{array}{l}{X}_2\left(s\right)=-\frac{F\left(s\right)}{2\left(s+3\right)}\\ \Rightarrow s{X}_2\left(s\right)=-\frac{F\left(s\right)}{2}-3X_2\left(s\right)\\ \Rightarrow {\dot{x}}_2=-3x_2-\frac{f}{2}\left(7\right)\end{array}$

Therefore,

$\begin{array}{l}Y\left(s\right)=\left(s+2\right)\frac{F\left(s\right)}{2}\left(\frac{1}{\left(s+1\right)}-\frac{1}{\left(s+3\right)}\right)\\ \Rightarrow Y\left(s\right)=\left(s+2\right)\left(X_1\left(s\right)+X_2\left(s\right)\right)\\ \Rightarrow Y\left(s\right)=s{X}_1\left(s\right)+s{X}_2\left(s\right)+2X_1\left(s\right)+2X_2\left(s\right)\\ \Rightarrow Y\left(s\right)=\frac{F\left(s\right)}{2}-X_1\left(s\right)+-\frac{F\left(s\right)}{2}-3X_2\left(s\right)+2X_1\left(s\right)+2X_2\left(s\right)\because From\left(6\right)and\left(7\right)\\ \Rightarrow Y\left(s\right)=X_1\left(s\right)-X_2\left(s\right)\\ \Rightarrow y=x_1-x_2\left(8\right)\end{array}$ Thus, from equations (6), (7) and (8) the state model for the transfer function is as

$\begin{array}{l}{\dot{x}}_1=-x_1+\frac{f}{2}\\ {\dot{x}}_2=-3x_2-\frac{f}{2}\\ y=x_1-x_2\end{array}$

On subtracting equations (6) and (7) at t=0, we get

$\begin{array}{l}{\dot{x}}_1\left(0\right)-{\dot{x}}_2\left(0\right)=-x_1\left(0\right)+\frac{f\left(0\right)}{2}+3x_2\left(0\right)+\frac{f\left(0\right)}{2}\\ \Rightarrow {\dot{x}}_1\left(0\right)-{\dot{x}}_2\left(0\right)=-x_1\left(0\right)+3x_2\left(0\right)+f\left(0\right)\\ \Rightarrow \dot{y}\left(0\right)=-x_1\left(0\right)+3x_2\left(0\right)+f\left(0\right)\because y=x_1-x_2\Rightarrow \dot{y}={\dot{x}}_1-{\dot{x}}_2\\ \Rightarrow \dot{y}\left(0\right)=-x_1\left(0\right)+3\left(x_1\left(0\right)-y\left(0\right)\right)+f\left(0\right)\because y=x_1-x_2\\ \Rightarrow \dot{y}\left(0\right)=2x_1\left(0\right)-3y\left(0\right)-f\left(0\right)\\ \Rightarrow x_1\left(0\right)=\frac{1}{2}\dot{y}\left(0\right)+\frac{3}{2}y\left(0\right)-\frac{1}{2}f\left(0\right)\left(9\right)\end{array}$ Therefore,

$\begin{array}{l}{x}_2\left(0\right)=x_1\left(0\right)-y\left(0\right)\because y=x_1-x_2\\ \Rightarrow x_2\left(0\right)=\frac{1}{2}\dot{y}\left(0\right)+\frac{3}{2}y\left(0\right)-\frac{1}{2}f\left(0\right)-y\left(0\right)\because From\left(9\right)\\ \Rightarrow x_2\left(0\right)=\frac{1}{2}\dot{y}\left(0\right)+y\left(0\right)-\frac{1}{2}f\left(0\right)\left(10\right)\end{array}$.

Conclusion:

The obtained standard form of state-model for the transfer function $\frac{Y\left(s\right)}{F\left(s\right)}=\frac{s+2}{s^2+4s+3}$ is as follows:

By the first method,

$\begin{array}{l}{\dot{x}}_1=-4x_1+x_2+f\\ {\dot{x}}_2=-3x_1+2f\\ y=x_1\end{array}$

By the second method,

$\begin{array}{l}{\dot{x}}_1=-x_1+\frac{f}{2}\\ {\dot{x}}_2=-3x_2-\frac{f}{2}\\ y=x_1-x_2\end{array}$

And the relation between initial values of state variables and $y\left(0\right)and\dot{y}\left(0\right)$:

For the state model using the first method,

$\begin{array}{l}{x}_1\left(0\right)=y\left(0\right)\\ x_2\left(0\right)=\dot{y}\left(0\right)+4y\left(0\right)-f\left(0\right)\end{array}$

While for the state model using the second method,

$\begin{array}{l}{x}_1\left(0\right)=\frac{1}{2}\dot{y}\left(0\right)+\frac{3}{2}y\left(0\right)-\frac{1}{2}f\left(0\right)\\ x_2\left(0\right)=\frac{1}{2}\dot{y}\left(0\right)+y\left(0\right)-\frac{1}{2}f\left(0\right)\end{array}$

And the output equation would be

$y=x_1$.