Question 9.5.2: Using Integrator Outputs Determine the model for the output ...

The input to an integrator block 1/s is the time derivative of the output. Thus, by examining the inputs to the two integrators shown in the diagram we can immediately write the time-domain equations as follows.
\dot{x} = g(t) + y       \dot{y} = 7w  −  3x           w = f (t)  −  4x
We can eliminate the variable w from the last two equations to obtain \dot{y} = 7 f (t)  −  31x . Thus, the model in differential equation form is
\dot{x} = g(t) + y                  \dot{y} = 7 f (t)  −  31x
To obtain the model in transfer function form we first transform the equations.
s X(s) = G(s) + Y (s)                 sY (s) = 7F(s) − 31X(s)
Then we eliminate Y (s) algebraically to obtain
X(s) = \frac{7}{s^{2}  +  31} F(s) + \frac{s}{s^{2}  +  31} G(s)
There are two transfer functions, one for each input-output pair. They are
\frac{X(s)}{F(s)} = \frac{7}{s^{2}  +  31}             \frac{X(s)}{G(s)} = \frac{s}{s^{2}  +  31}