I want to find the closed loop gain of this block diagram:

I remember a rule that since these 2 loops collide with each other we ignore them so is the closed loop gain just equal to $\frac{H_{1}(s)}{1+H_{1}(s)H_{2}(s)}$?

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    $\begingroup$ Homework? What have you done to try to solve this so far? $\endgroup$
    – TimWescott
    Jun 13, 2023 at 23:58
  • $\begingroup$ @TimWescott I am preparing for a exam.And I have already posted a guess the answer should be Y/N. $\endgroup$
    – Volpina
    Jun 14, 2023 at 4:31
  • $\begingroup$ Is there a source for this block diagram ? A signal line where signal flows in two directions is not usually seen in text books. This diagram may simply be wrong. Please provide a source. $\endgroup$
    – AJN
    Jun 14, 2023 at 16:14
  • $\begingroup$ I didn't consider that making a wild guess (and a wrong one, at that) shows enough effort. Something like the answer below, with a math error in the middle -- that counts as effort, and worthy of effort on my part. $\endgroup$
    – TimWescott
    Jun 14, 2023 at 23:22

1 Answer 1


I've redrawn your diagram below and added some of my own labels:

enter image description here

We can then find an expression for $Y(s)/U(s)$ as follows: \begin{align} Y(s) &= X_1(s) - X_2(s) \\ &= H_1(s)X_3(s) - H_2(s)Y(s) \\ &= H_1(s)(U(s) - X_2(s)) - H_2(s)Y(s) \\ &= H_1(s)U(s) - H_1(s)X_2(s) - H_2(s)Y(s) \\ &= H_1(s)U(s) - H_1(s)(H_2(s)Y(s)) - H_2(s)Y(s) \\ &= H_1(s)U(s) - H_1(s)H_2(s)Y(s) - H_2(s)Y(s) \\ Y(s) + H_1(s)H_2(s)Y(s) + H_2(s)Y(s) &= H_1(s)U(s) \\ (1 + H_1(s)H_2(s) + H_2(s))Y(s) &= H_1(s)U(s) \\ \frac{Y(s)}{U(s)} &= \frac{H_1(s)}{1 + H_1(s)H_2(s) + H_2(s)} \\ \end{align} The general strategy for dealing with block diagrams is to make sure that all inputs and outputs of summing junctions are labelled, and then we can express all labelled signals as a function of either $Y(s)$ or $U(s)$. We then solve for $Y(s)/U(s)$.

  • $\begingroup$ Note that this is the "brute force" method for reducing a block diagram. It always works, so you should know it in the event that block diagram reduction fails. If you're doing a lot of block diagram reduction, then memorizing all the rules for it is helpful -- but those rules don't cover all the cases, and sometimes you'll end up doing more work than if you just write out the math. $\endgroup$
    – TimWescott
    Jun 14, 2023 at 23:25
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    $\begingroup$ @TimWescott I think the “brute-force” method is not emphasized enough when first learning about block diagram reduction for the first time. In my experience I learned the basic block diagram reduction techniques and had to memorize them, but knowing the brute-force method after a long time was illuminating. $\endgroup$
    – mhdadk
    Jun 14, 2023 at 23:29
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    $\begingroup$ I'm not sure if it's not emphasized enough, or if people just forget it because they've memorized a table of rules to get through an exam. But -- the value can't be emphasized more, for sure. If I've been away from control systems for a while and need to reduce block diagrams, I check my reduction skills using the brute-force method. $\endgroup$
    – TimWescott
    Jun 15, 2023 at 0:04

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