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I am given the following block diagram of a control system (I think it also applies to electrical circuits as well with linear components. Correct me if I'm wrong). My textbook says that C(s) is connected in series with the system characterized by input v(t) and output y(t). Why is it that?

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  • \$\begingroup\$ It kind of depends on what them abstract boxes represent... \$\endgroup\$
    – Lundin
    Mar 26, 2020 at 11:59
  • \$\begingroup\$ Could you give a more precise question. What you really want to know? "My textbook says that C(s) is connected in series with the system characterized by input v(t) and output y(t)" is not really a good question. \$\endgroup\$ Mar 26, 2020 at 12:12
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    \$\begingroup\$ @Lundin these are standard labels in control engineering. \$\endgroup\$
    – jonathanjo
    Mar 26, 2020 at 12:29

2 Answers 2

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I have no explanation for the quoted text...it makes no sense.

Block diagram manipulation:

When the output of block C(s) is added AFTER the block G(s) the top path contains the product C(s)*G(s). In this case, we could say that C(s)*G(s) is in parallel (not in series) to the system characterized by Y(s)/V(s) (when C(s) would be removed from the system).

The resulting transfer function is:

Y/V=H(s)=GR/(1+GR) + GC/(1+GR)

H(s)=(GR+GC)/(1+GR)

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  • \$\begingroup\$ Well, I don't think that "\$C(s)G(s)\$ is in parallel [to \$Y(s)/V(s)\$]", since \$Y(s)/V(s)\$ is the transfer function of the whole system, and by having \$Y(s)/V(s)\$ in parallel to \$C(s)G(s)\$ you would have that the transfer function of the whole system is \$Y(s)/V(s)+C(s)G(s)\$, but it is just \$Y(s)/V(s)\$. So that path including \$C(s)\$ is already included in the response of \$Y(s)/V(s)\$. \$\endgroup\$
    – jDAQ
    Mar 26, 2020 at 15:36
  • \$\begingroup\$ jDAQ...I know what you mean - but if you read my answer carefully, you will realize that i wrote "...when C(s) would be removed from the system". That means: The (former) transfer function without the upper forward path. This can be seen from the final result in my answer: The first part is the transfer function without the forward path .....now, when C is added, the second part is added in parallel (with the same feedback path.). \$\endgroup\$
    – LvW
    Mar 26, 2020 at 16:53
  • \$\begingroup\$ Sorry, but I disagree, it might be the lack of an actual diagram (as wordy descriptions can get quite confusing even for careful readers). But, "the former TF [removing C]" is the feedback block with TF $$\frac{R(s)G(s)}{1+R(s)G(s)}$$ and if that were to be in parallel with \$C(s)G(s)\$ the whole system would have a TF $$\tilde{H}(s)=\frac{R(s)G(s)}{1+R(s)G(s)}+C(s)G(s) $$. Your TF for \$H(s)\$ is correct, and it is not equal to \$\tilde{H}\$ \$\endgroup\$
    – jDAQ
    Mar 26, 2020 at 17:11
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Series connections are two blocks "in-line" the output of one gets feed to the next one, that would be R and G, or C and G in your diagram.

A parallel connection is when the same signal feeds two different blocks and their outputs get added (you might have different operations, but in the simple case you are summing them).

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  • \$\begingroup\$ Wrong !! No, R and C are NOT in parallel - both have not the same input signal! \$\endgroup\$
    – LvW
    Mar 26, 2020 at 14:37

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