.The signal flow graph for a system is given in the figure. Find resultant gain.

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    \$\begingroup\$ Apply Mason's formula. There are two forward paths and 4 loops and two non touching loops \$\endgroup\$ – aparna Jul 7 '19 at 3:47
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    \$\begingroup\$ While Sam Mason's paper, Feedback Theory -- Further Properties of Signal Flow Graphs, is an excellent read, I still find it so much easier to just label each signal and write down the set of simultaneous equations. That's simple, it's hard to make mistakes, it works, and it relies upon prosaic theory, widely taught and used (no need to drill and remember yet another set of tools.) \$\endgroup\$ – jonk Jul 7 '19 at 5:37
  • \$\begingroup\$ Can you please elaborate ? \$\endgroup\$ – Modulus Jul 7 '19 at 5:50
  • \$\begingroup\$ Is the answer 44/9. \$\endgroup\$ – Modulus Jul 7 '19 at 5:51
  • \$\begingroup\$ @Modulus I can't read the "find gain" fraction. The numerator is obvious (\$c\$.) But what's the denominator there? Is it the input node at the far left? Or something else? \$\endgroup\$ – jonk Jul 7 '19 at 6:00

Let's take a far more complex case, just to show how easy this all is using simultaneous equations:

enter image description here

(I picked the above problem from this site.)

Using sympy as my solver, enter the following lines:

var('ga gb gc gd ge gf gg gh gi gj n2 n3 n4 n5 n6')
s2 = Eq( n2, ga*n1 + gj*n3 )
s3 = Eq( n3, gb*n2 + gh*n5 )
s4 = Eq( n4, gc*n3 + gi*n5 )
s5 = Eq( n5, gd*n4 + gg*n3 + gf*n5 )
s6 = Eq( n6, ge*n5 )
ans = solve( [s2,s3,s4,s5,s6], [n2,n3,n4,n5,n6] )

pprint( ans[n6]/n1 )

                      ga⋅gb⋅ge⋅(gc⋅gd + gg)
gb⋅gd⋅gi⋅gj + gb⋅gf⋅gj - gb⋅gj - gc⋅gd⋅gh - gd⋅gi - gf - gg⋅gh + 1

Please take a moment and go above to read each of the equations I set up (\$s_2\$, etc.) You should easily be able to see how it is that I wrote those out. It's very easy. For example, node 2 (\$n_2\$) only has two terms added together: \$N_1\cdot A\$ and \$N_3\cdot J\$. You should easily see why I wrote out the equation for \$s_2\$ in the way I did.

The solver does the rest. And the answer is correct. (You can verify it by simply looking at the site I mentioned earlier. Please note that they did not use simultaneous equations to arrive at their solution. They used Mason's gain formula. But their solution is exactly the same one that I arrived at using a very simple and very well-known approach.)

Now, given the above process I've laid out, do you think you can write the appropriate equations for your case? (It's fewer equations and simpler to do.) The result, if you get it correctly handled, will have the numerator you mentioned (44) but not the same denominator value you mentioned.

(If you still need help, I'll add more details directly targeting your solution.)

In your circumstance, you need to label your nodes (you haven't done that, yet.) I believe your gains are the finite values shown on your diagram. So you can just use those values, directly, in your equation set up.

For example, I've labeled two of your nodes below:

enter image description here

The equation for it is: \$X_1=1\cdot R_s - 1\cdot X_2=R_s-X_2\$. You should be able to develop the equations for all of \$X_1\$ through \$X_4\$ (\$C_s=X_4\$, so that's trivial.)

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    \$\begingroup\$ More help required. I don't know sympy. I don't know where are the equation kindly explain for a layman. \$\endgroup\$ – Modulus Jul 7 '19 at 7:03
  • \$\begingroup\$ @Modulus You don't need to know sympy. The text is relatively readable without knowing the precise details of the language. If you know anything about simultaneous equation solutions, it should be relatively understandable how each equation was set up. And if you have ever performed a simultaneous equation solution of this kind, you either have the software tools available or else you know how to use Cramer's rule, by hand. Is this completely obscure to you? \$\endgroup\$ – jonk Jul 7 '19 at 7:10
  • \$\begingroup\$ Quite obscure, however Cramer's rule I have read long time back. Lemme revise. \$\endgroup\$ – Modulus Jul 7 '19 at 7:14
  • \$\begingroup\$ @Modulus I hadn't realized you were doing self-study, earlier. Your other question clarifies this. If you write back with more questions about how to proceed, I'll add more to the answer I've written here. \$\endgroup\$ – jonk Jul 7 '19 at 20:12
  • \$\begingroup\$ In fact it's kind of urgent.......However all the qns asked or to be asked in future will figure here only after sincere efforts have not yielded an answer and my solution will be right there. \$\endgroup\$ – Modulus Jul 7 '19 at 23:56

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