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Given the observable form $$\begin{array}{rl}{x}^{\prime }& =\left(\begin{array}{cccc}0& 0& 0& -{\alpha }_0\\ 1& 0& 0& -{\alpha }_1\\ 0& 1& 0& -{\alpha }_2\\ 0& 0& 1& -{\alpha }_3\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right)+\left(\begin{array}{c}{\beta }_0\\ {\beta }_1\\ {\beta }_2\\ {\beta }_3\end{array}\right)u\\ y& =\left(\begin{array}{cccc}0& 0& 0& 1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right)+\left[\gamma \right]u\end{array}$$

Show that the transfer function is $\frac{Y}{U}=\gamma +\frac{{\beta }_3{s}^3+{\beta }_2{s}^2+{\beta }_{1}s+{\beta }_0}{s^4+{\alpha }_3{s}^3+{\alpha }_2{s}^2+{\alpha }_{1}s+{\alpha }_0}$ using Mason rule.

The transfer function can ofcourse be found using $\frac{Y}{S}=C{(sI-A)}^{-1}B+D$ which gives$$\begin{array}{rl}\frac{Y}{U}& =\left(\begin{array}{cccc}0& 0& 0& 1\end{array}\right){(\left(\begin{array}{cccc}s& 0& 0& 0\\ 0& s& 0& 0\\ 0& 1& s& 0\\ 0& 0& 1& s\end{array}\right)-\left(\begin{array}{cccc}0& 0& 0& -{\alpha }_0\\ 1& 0& 0& -{\alpha }_1\\ 0& 1& 0& -{\alpha }_2\\ 0& 0& 1& -{\alpha }_3\end{array}\right))}^{-1}\left(\begin{array}{c}{\beta }_0\\ {\beta }_1\\ {\beta }_2\\ {\beta }_3\end{array}\right)+\gamma \\ & =\gamma +\frac{{\beta }_3{s}^3+{\beta }_2{s}^2+{\beta }_{1}s+{\beta }_0}{s^4+{\alpha }_3{s}^3+{\alpha }_2{s}^2+{\alpha }_{1}s+{\alpha }_0}\end{array}$$

But we want to use Mason rule here. The first step is to write the equations so that the nodes variable are on the left side of the equation. The node variables are the states $x_1,x_2,x_3,x_4$. From the matrix equations we obtain$$\begin{array}{rl}{x}_1^{\prime }& =-{\alpha }_0{x}_4+{\beta }_{0}u\\ x_2^{\prime }& =x_1-{\alpha }_1{x}_4+{\beta }_{1}u\\ x_3^{\prime }& =x_2-{\alpha }_2{x}_4+{\beta }_{2}u\\ x_4^{\prime }& =x_3-{\alpha }_3{x}_4+{\beta }_{3}u\end{array}$$

Solving for $x$ now, and in the process we change $x^{\prime }$ to $sx$ by taking Laplace transforms of all variables, and we add the output equation as well$$\begin{array}{rl}{X}_4& =-\frac{s}{{\alpha }_0}{X}_1+\frac{{\beta }_0}{{\alpha }_0}U\\ X_1& =s{X}_2+{\alpha }_1{X}_4-{\beta }_{1}U\\ X_2& =s{X}_3+{\alpha }_2{X}_4-{\beta }_{2}U\\ X_3& =(s+{\alpha }_3)X_4-{\beta }_{3}U\\ Y& =X_4+\gamma U\end{array}$$

We now draw the Mason diagram, putting $U$ on the left most node and $Y$ on the right most node (the input and output). Here is the result

The forward paths from $U$ to $Y$ are$$\begin{array}{rl}{F}_1& =\frac{{\beta }_0}{{\alpha }_0}\\ F_2& =\gamma \\ F_3& =(-{\beta }_2)\left(s\right)(-\frac{s}{{\alpha }_0})\\ F_4& =(-{\beta }_1)(-\frac{s}{{\alpha }_0})\\ F_5& =(-{\beta }_3)\left(s\right)\left(s\right)(-\frac{s}{{\alpha }_0})\end{array}$$

Now we find all the loops. There are only three loops. A loop is one that starts from a node and returns back to it without visiting a node more than once.$$\begin{array}{rl}{L}_1& =(-\frac{s}{{\alpha }_0})\left({\alpha }_2\right)\left(s\right)\\ L_2& =(-\frac{s}{{\alpha }_0})(s+{\alpha }_3)\left(s\right)\left(s\right)\\ L_3& =(-\frac{s}{{\alpha }_0})\left({\alpha }_1\right)\end{array}$$

We now need to calculate the associated ${\mathrm{\Delta }}_k$ for each of the above forward loops. ${\mathrm{\Delta }}_k$ is found by removing $F_k$ from the graph and then calculating the main mason $\mathrm{\Delta }$ of what is left in the graph. When remove $F_1$ no loops remain, hence ${\mathrm{\Delta }}_1=1$. When removing $F_2$ all the loops remain, hence ${\mathrm{\Delta }}_2=1-(L_1+L_2+L_3)$, and when removing $F_3,F_4,F_5$ no loops remain, hence ${\mathrm{\Delta }}_3={\mathrm{\Delta }}_4={\mathrm{\Delta }}_5=1$. Therefore$$\frac{Y}{U}=\frac{F_1{\mathrm{\Delta }}_1+F_2{\mathrm{\Delta }}_2+F_3{\mathrm{\Delta }}_3+F_4{\mathrm{\Delta }}_4+F_5{\mathrm{\Delta }}_5}{1-(L_1+L_2+L_3)}$$ There is no other combinations of loops. The above becomes$$\begin{array}{rl}\frac{Y}{U}& =\frac{F_1+\gamma (1-(L_1+L_2+L_3))+F_3+F_4+F_5}{1-(L_1+L_2+L_3)}\\ & =\gamma +\frac{F_1+F_3+F_4+F_5}{1-(L_1+L_2+L_3)}\\ & =\gamma +\frac{\frac{{\beta }_0}{{\alpha }_0}+(-{\beta }_2)\left(s\right)(-\frac{s}{{\alpha }_0})+(-{\beta }_1)(-\frac{s}{{\alpha }_0})+(-{\beta }_3)\left(s\right)\left(s\right)(-\frac{s}{{\alpha }_0})}{1-((-\frac{s}{{\alpha }_0})\left({\alpha }_2\right)\left(s\right)+(-\frac{s}{{\alpha }_0})(s+{\alpha }_3)\left(s\right)\left(s\right)+(-\frac{s}{{\alpha }_0})\left({\alpha }_1\right))}\\ & =\gamma +\frac{\frac{{\beta }_0}{{\alpha }_0}+{\beta }_2\frac{s^2}{{\alpha }_0}+{\beta }_1\frac{s}{{\alpha }_0}+{\beta }_3\frac{s^3}{{\alpha }_0}}{1-(-\frac{s^2}{{\alpha }_0}{\alpha }_2-\frac{s^2}{{\alpha }_0}(s+{\alpha }_3)-\frac{s}{{\alpha }_0}{\alpha }_1)}\\ & =\gamma +\frac{\frac{{\beta }_3{s}^3+{\beta }_2{s}^2+{\beta }_{1}s+{\beta }_0}{{\alpha }_0}}{1+\frac{s^3+s^2{\alpha }_3+s^2{\alpha }_2+s{\alpha }_1}{{\alpha }_0}}\end{array}$$

Hence$$G\left(s\right)=\frac{Y}{U}=\gamma +\frac{{\beta }_3{s}^3+{\beta }_2{s}^2+{\beta }_{1}s+{\beta }_0}{s^3+s^2{\alpha }_3+s^2{\alpha }_2+s{\alpha }_1+{\alpha }_0}$$