% Solutions to flowgraph problems using the Symbolic Toolbox
% Flowgraphs are a staple of s-parameter based network analysis.
% Although there are "simple" rules (Sam Mason, MIT) for their analysis,
% even simple rules can be forgotten as time passes. The symbolic toolbox
% is a great tool for analyzing flowgraphs, and no "rules" need be remembered. It
% is simply a matter of writing the equations for the flowgraph in an
% orderly manner, and using the "solve" function of the symbolic toolbox to
% do the tedious and error prone algebra.
%
% See http://en.wikipedia.org/wiki/Mason's_rule
%
% The notation used in this example is as follows:
% summing node: O
% signal flow : > < V ^ (right, left, down, up)
% signal node : *
% branches without a coefficient: default "gain" = 1
%
%
% As a first example, consider a simple feedback topology ....
%
% x1
% u*--K->---O---G->----*----*y
% | |
% ---<-(-H)---
%
% Goal: solve for output y in terms of input u and parameters K,G,H
% Dick Benson, The Mathworks
% Copywrite 2002-2010 The MathWorks, Inc.
%% start by defining symbols ..
clc
clear
syms u y K G H x1
% then write the two equations
eq1 = x1*G - y; % y = G*x1
eq2 = -H*y + K*u - x1; % x1 = K*u - H*y
solution_1 = solve(eq1,eq2,x1,y); % use the solve function on the equations
disp('answer to problem 1: ')
y = solution_1.y % solution for y in terms of input x
% y = K*u/(H*G+1)*G
%% That was a simple example that could have easily be worked by hand.
% For a more challenging case, consider the error factors (Exy) involved with S parameter measurements (Sxy)
% a1 = incident, b1 = reflected, b2 = transmitted (power)
%
% ------------------------- Exf->------------------------------
% | |
% | x1 x3 |
% a1*---->--*--->-----O--->------*-----S21->-----O---->----*-----Etf->----O------*b2
% | | | | |
% V Edf ^ Esf V S11 ^ S22 V Elf
% | | | | |
% b1*--<----O--<-Erf--*--<-------O-----<-S12-----*----<----O
% x4 x2
%
% Goal: solve for outputs b1 and b2 in terms of input a1
%
syms a1 b1 b2 Edf Erf Esf S11 S21 S22 S12 Elf Etf Exf x1 x2 x3 x4
eq1 = Esf*x4 + a1 - x1; % x1 = Esf*x4 + a1
eq2 = Elf*x3 - x2; % x2 = Elf*x3
eq3 = S21*x1 + S22*x2 - x3; % x3 = S11*x1 + S22*x2
eq4 = S11*x1 + S12*x2 - x4; % x4 = S11*x1 + S12*x2
eq5 = Erf*x4 + Edf*a1 - b1; % b1 = Erf*x4 + Edf*a1
eq6 = Etf*x3 + Exf*a1 - b2; % b2 = Etf*x3 + Exf*a1
solution = solve(eq1,eq2,eq3,eq4,eq5,eq6,x1,x2,x3,x4,b1,b2);
disp('answer to problem 2: ')
b1=solution.b1
b2=solution.b2
% In case you do not have the symbolic Toolbox, the solution is :
% b1 = a1*(Erf*S21*S12*Elf-Erf*S22*Elf*S11+Erf*S11-Edf*S21*Esf*S12*Elf+Edf*S22*Elf*S11*Esf-Edf*S22*Elf-Edf*S11*Esf+Edf)/(-S21*Esf*S12*Elf+S22*Elf*S11*Esf-S22*Elf-S11*Esf+1)
% b2 = a1*(Etf*S21-Exf*S21*Esf*S12*Elf+Exf*S22*Elf*S11*Esf-Exf*S22*Elf-Exf*S11*Esf+Exf)/(-S21*Esf*S12*Elf+S22*Elf*S11*Esf-S22*Elf-S11*Esf+1)
% The chances of getting this answer correct is slim to none for most of us mortals !
