# What is the reduced form of this block diagram?

What is the reduced form of this block diagram? I can't see any solution way :(

• If you for example want to find C/R then you could introduce a temporary variable for example between G2 and G3, solve for it in terms of R and C and use that to solve for C/R. – fibonatic Mar 30 '15 at 9:31

Starting with your system and some guidelines for block diagram reduction (PDF):

1. Move the pickoff point between G2 and G3 to before G2
2. Combine G2 and G3 into one block (G2G3)
3. Combine G4 with the G2G3 block (G4 + G2G3)
4. Combine G2 and H1 (G2H1)
5. Redraw the rats nest after G1
6. Duplicate the H2 line to give a feedback loop around (G4 + G2G3) and a feedback line towards (G2H1)
7. Collapse the (G4 + G2G3) and H2 loop
8. Make room for more work!
9. Move the summing junction located just before G1 to just after G1
10. Move the pickoff point just before the big transfer function to just after it (what was the G4 + G2G3 stuff)
11. At this point everything should be easy
12. Collapse summing junctions into summed blocks
13. Collapse feedback loops
14. Collapse feedback loops
15. Collapse feedback loops

Done! (see the pic below for details) Please check my work; I did everything in paint, but this is the route you should take. Moving summing junctions and pickoff points is key to solving this, as it is in nearly every instance.

• Thanks lol. Still easier than Visio... I hate that program. So much. – Chuck Jun 4 '15 at 18:34

From my experience, it is often easier to stop thinking in terms of "block diagrams" and start to think in terms of equations that these block diagrams represent.

I introduce a few new nodes E, A and B:

Now let's write the equations for each of them \begin{align} E &= R-C\\ A &= H_2C+G_1(E+H_1B)\\ B &= G_2A\\ C &= G_3B + G_4A\\ \end{align}

Insert B in A and C gives \begin{align} E &= R-C\\ A &= H_2C+G_1(E+H_1G_2A)\\ C &= (G_3G_2 + G_4)A\\ \end{align}

Insert E in A gives \begin{align} A &= H_2C+G_1(R-C+H_1G_2A) = H_2C+G_1R-G_1C+H_1G_1G_2A \\ &= (H_2-G_1)C +G_1R +H_1G_1G_2A \\ &=\frac{(H_2-G_1)C +G_1R}{1-H_1G_1G_2}\\ C &= (G_3G_2 + G_4)A\\ \end{align}

Insert A in C gives \begin{align} C &= (G_3G_2 + G_4)\frac{(H_2-G_1)C +G_1R}{1-H_1G_1G_2}\\ (1-H_1G_1G_2)C &= (G_3G_2 + G_4)(H_2-G_1)C +(G_3G_2 + G_4)G_1R\\ [(1-H_1G_1G_2)-(G_3G_2 + G_4)(H_2-G_1)]C &= (G_3G_2 + G_4)G_1R\\ \end{align}

And finally

\begin{align} \frac{C}{R} &=\frac{(G_3G_2 + G_4)G_1}{(1-H_1G_1G_2)-(G_3G_2 + G_4)(H_2-G_1)}\\ &\\ &= \frac{G_1G_2G_3 + G_1G_4}{1-G_1G_2H_1-G_2G_3H_2 - G_4H_2 + G_1G_2G_3 + G_1G_4}\\ \end{align}