Calculating an in-loop signal as part of a hierarchical control loop I've got a control system with two feedback paths, with each path going to a different actuator that corrects the error in the system. One feedback path provides feedback at low frequencies, and the other at high frequencies. Here is the loop:



I want to calculate the signal on one of the actuators (the node at the output of the "slow" block) while the loop is closed, given a disturbance that I inject before the summing junction. The signal I want to calculate is "A". The input is normally zero (I just want to suppress disturbances to zero, not some set-point).

I initially started by calculating the open loop gain provided by the inner plant-slow-sum loop, i.e. A = slow * plant * disturbance, and then divided the disturbance by this open loop gain to give me something that approximates the disturbance suppressed by the closed loop, but this is not correct because in reality some of the disturbance is also suppressed by the outer plant-fast-sum-sum loop. In effect, the disturbance that the "slow" loop will see at high frequencies will be much lower, since the "fast" loop suppresses it there.

How do I calculate the closed-loop signal at A due to the disturbance, including the effect of the outer loop?
UPDATE: My original answer incorrectly assumed the outer loop would not impact the result, but as pointed by Ben, this transfer function is dependent on the fast feedback. 

The resulting equations are found by solving for $A/X$ from the block diagram below, with X representing the disturbance input:



Solving this:

$$Y = P(SY+FY+X)$$
$$A/S = P(A + FA/S + X)$$
$$A = PAS + PFA + PXS$$
$$A(1- PS - PF) = PXS$$

$$\frac{A}{X} = \frac{PS}{1-PS-PF}$$

Note that this follows the general equation for the closed loop gain using:

$$G_{CL} = \frac{G_F}{1+ G_{OL}}$$

Where:

$G_{CL}$: Closed Loop Gain

$G_{F}$: Forward Gain (In this case $G_{F}  = PS$)

$G_{OL}$: Open Loop Gain (In this case $G_{OL} = -P(S+F)$

Note how negative feedback is assumed in the generic expression, but since the block diagram does not show negation at the summations, the open loop gain is then negative as given above. Using the general equation and recognizing that the parallel result of S and F as a gain given as S+F allows us to solve this from inspection.