Block diagram algebra

Example 3.1:

The textbook shows the solution as

\[z = ( P_{11} + P_{12} K ( I - P_{22} K )^{-1} P_{21} )w\]

We can do this calculation manually as follows:

\begin{align} b &= a + P_{22} c \\ c &= K b \\ c &= K a + K P_{22} c \\ c - K P_{22} c &= K a \\ (I - K P_{22}) c &= K a \\ (I - K P_{22})^{-1} (I - K P_{22}) c &= (I - K P_{22})^{-1} K a\\ c &= (I - K P_{22})^{-1} K a\\ &= K (I - P_{22} K )^{-1} a~\text{(push-through)} \\ z &= f + e\\ &= P_{21}w + P_{12}c \\ &= (P_{21} + P_{12}\underbrace{K(I - P_{22}K)^{-1}P_{12}}_c) w \end{align}

We should also be able to solve this problem symbolically using Sympy

import sympy
sympy.init_printing()

(w, a, b, c,
 d, e, f, z,
 P_21, P_22,
 K, P_12, P_11) = sympy.symbols("""w, a, b, c,
                                   d, e, f, z,
                                   P_21, P_22,
                                   K, P_12, P_11""", commutative=False
                               )
eqs = [a - P_21*w,
       b - (a + d),
       c - K*b,
       d - P_22*c,
       e - P_12*c,
       f - P_11*w,
       z - (e + f)]

Note this is the right call, unfortunately as of 2018-03-15, this causes an error.

# Uncomment to check if this solves now
# sympy.solve(eqs, (a, b, c, d, e, f, z))

If we are careful about the ordering, we can still get a sensible answer from SymPy without having to do tedious math ourselves.

# first we identify which variable can be eliminated by solving each equation
order = [a, b, c, d, e, f, z]
# This will store the solutions
solution = {}
for var, eq in zip(order, eqs):
    # We solve each equation for the new unknown given all the other solutions
    sol = sympy.solve(eq.subs(solution), var)[0]
    # and add that solution to the list of knowns
    solution[var] = sol
solution[z].collect(w)

$$w \left(P_{11} + P_{12} K P_{21} + P_{12} K \left(1 - P_{22} K\right)^{-1} P_{22} K P_{21}\right)$$