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## Q. 7.EX.7

Analog-Computer Implementation
Find a state-variable description and the transfer function of the third-order system shown in Fig. 7.5 whose differential equation is

$\overset{...}{y} + 6\ddot{y} +11 \dot{y} +6 y =6u.$ ## Verified Solution

We solve for the highest derivative term in the ODE to obtain

$\overset{…}{y} = −6\ddot{y} − 11\dot{y} − 6y + 6u.$             (7.8)

Now we assume that we have this highest derivative and note that the lower order terms can be obtained by integration as shown in Fig. 7.6(a). Finally, we apply Eq. (7.8) to complete the realization shown in Fig. 7.6(b). To obtain the state description, we simply define the state-variables as the output of the integrators $x_1 = \ddot{y}, x_2 = \dot{y}, x_3 = y,$ to obtain

$\dot{x}_1 = −6x_1 − 11x_2 − 6×3 + 6u, \\ \dot{x}_2 = x_1, \\ \dot{x}_3 = x_2,$

which provides the state-variable description

$\pmb F= \left[\begin{matrix} -6 & -11 & -6 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right] , \ \ \ \pmb G= \begin{bmatrix} 6 \\ 0 \\ 0 \end{bmatrix} , \ \ \ \pmb H = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} , \ \ \ J = 0.$

The MATLAB statement ## Script Files

[num,den] = ss2tf(F,G,H,J);

will yield the transfer function

$\frac{Y(s)}{U(s)} = \frac{6}{s^3 + 6s^2 + 11s + 6}.$

If the transfer function were desired in factored form, it could be obtained by transforming either the $ss \text{ or } tf$ description. Therefore, either of the MATLAB statements

% convert state-variable realization to pole–zero form
[z,p,k] = ss2zp(F,G,H,J)

and

% convert numerator-denominator to pole–zero form
[z,p,k] = tf2zp(num,den)

would result in

$z=[], \ \ \ p= \begin{bmatrix} -3 & -2 & -1 \end{bmatrix}^{\prime} , \ \ \ k=6.$

This means that the transfer function could also be written in factored form as

$\frac{Y(s)}{U(s)} = \frac{6}{(s + 1)(s + 2)(s + 3)}.$