Using Mason rule to obtain the transfer function from state space description

Nasser M. Abbasi

September 21, 2015 page compiled on September 21, 2015 at 10:23pm

Given the observable form


Show that the transfer function is              3   2
YU-= γ +  s4β+3αs3+s3β+2sα2+sβ2+1sα+1sβ+0α0-   using Mason rule.

The transfer function can ofcourse be found using Y- = C (sI − A )− 1B +  D
S  which gives


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 x1, x2,x3,x4   . From the matrix equations we obtain


Solving for x  now, and in the process we change  ′
x to sx  by taking Laplace transforms of all variables, and we add the output equation as well


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


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.


We now need to calculate the associated Δk  for each of the above forward loops. Δk  is found by removing Fk  from the graph and then calculating the main mason Δ  of what is left in the graph. When remove F1   no loops remain, hence Δ1 =  1  . When removing F2   all the loops remain, hence Δ2  = 1 − (L1 + L2 + L3 )  , and when removing F3, F4,F5   no loops remain, hence Δ3 = Δ4  = Δ5  = 1  . Therefore

Y-    F1Δ1-+-F2-Δ2-+-F3-Δ3-+-F4-Δ4-+-F5Δ5--
U  =           1 − (L  +  L  + L )
                     1    2    3

There is no other combinations of loops. The above becomes



|G (s) = Y- = γ + ---β3s--+-β2s-+--β1s +-β0---|