MATLAB: Reducing transfer functions using Append & Connect technique vs feedback, cloop, series & parallel command technique

Question

The problem here was when I tried to counter check my answer using algebra and the two techniques above I got three different outputs all together.

The code:

G1=tf([0 1],[1 10]);    
G2=tf([0 1],[1 1]);
G3=tf([1 0 1],[1 4 4]);
G4=tf([1 1],[1 6]);
H1=tf([1 1],[1 2]);
H2=tf([0 2],[0 1]);
H3=tf([0 1],[0 1]);
k=tf([2 12],[1 1]);      % H2/G4 - I moved G4 before H2
feedback((series((feedback((series((feedback((series(G3,G4)),H1,+1)),G2)),k,-1)),G1)),H3,-1)
%me checking for errors
syms s;
g1=1/(s+10); 
g2=1/(s+1);
g3=((s^2)+1)/((s^2)+4*s+4);
g4=(s+1)/(s+6);
h1=(s+1)/(s+2);
h2=2;
h3=1;
simplify(g3g4)
series(G3,G4)
simplify((g3g4)/(1-(g3g4h1)))   % so far so good
feedback((series(G3,G4)),H1,+1)
simplify((g2g3g4)/(1-(g3g4h1)))

series((feedback((series(G3,G4)),H1,+1)),G2) %and here's where it fell apart
%me trying another approach
G1=tf([0 1],[1 10]);
G2=tf([0 1],[1 1]);
G3=tf([1 0 1],[1 4 4]);
G4=tf([1 1],[1 6]);
H1=tf([1 1],[1 2]);
H2=tf([0 2],[0 1]);
H3=tf([0 1],[0 1]);
R=tf([0 1],[0 1]);
t1=append(G1,G2,G3,G4,H1,H2,H3,R);
q=[1 8 -7;2 1 -6;3 2 5;4 3 0 ;5 4 0 ;6 3 0 ;7 4 0;];
input= 8;
output=4;
ts=connect(t1,q,input,output);
tf(ts)  % It gave a whole new different answer

Can anyone point out where I went wrong?

Answer 1

Assuming the two outputs from G4 are the same (= Y), add a block, \$\frac{1}{G4}\$, in series with H2 and move the input to this path to Y. Now the task is much easier; just work outwards from the innermost loop.

Answer 2

I'd try and sound smart but...I think I found why using the append&connect commands and the feedback,series,parallel,cloop commands have different outputs...

Using feedback,series,parallel,cloop commands:

Using append&connect commands: (gets rid of the '10' from the denominator)

the later simplifies the reduced form even more.

Answer 3

I can`t work with Matlab - nevertheless, it is clear that

• there is only ONE SINGLE solution for the transfer function, but there are

• at least two DIFFERENT SOLUTIONS for arranging a corresponding block diagram (which means: There are alternatives for the distribution between forward and feedback blocks).

• As a consequence, we also have more than one single solution for the loop gain.