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I have a block diagram that I am trying to get the transfer function for but can't seem to figure it out, I am sure that I am making it more difficult than it needs to be but still can't get it. The diagram is below and any help would be great!

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Edit
So far I have got myself to this point but I am unsure if it is correct: $$\frac{\theta }{M_p}=\frac{G(s)(I(s)-K_t)}{1+D(s)G(s)(I(s)+K_t)}$$ I feel I should try solving the inner loop and then the outer but I can't work out how to put the inner loop in one function. Any ideas?

EDIT INNER LOOP SIMPLIFICATION
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  • \$\begingroup\$ I see two loops - one inner and one outer loop. Do you know how to write the transfer function for a single feedback loop? \$\endgroup\$
    – LvW
    Commented Apr 30, 2016 at 15:54
  • \$\begingroup\$ I usually set the inner loop as some function (A simple variable that lives within the overall expression) and then get the transfer function of the overall; Let \$Loop_{inner}\$ be \$F_{li}\$; now get the transfer function of the overall loop and plug in the inner loop afterwards. \$\endgroup\$
    – Peter Smith
    Commented Apr 30, 2016 at 16:58
  • \$\begingroup\$ @PeterSmith That's the thing, I can't work out what the Inner loop is when there are two inputs... I've got the outer no problemo as I've already had to get that - just can't decipher the inner. Outer Loop = \$\frac{\theta}{M_P}=\frac{U(s)I(s)}{1+D(s)U(s)I(s)}\$ where \$U(s)\$ is the inner loop \$\endgroup\$
    – MrPhooky
    Commented Apr 30, 2016 at 17:00
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    \$\begingroup\$ I would separate the inputs at the summing node; in the inner loop use the feedback from \$K_t\$ and \$M_p\$ and then use this as a summing node from the output of \$D_s\$. A summing node with 3 inputs cann be viewed as two summing nodes. \$\endgroup\$
    – Peter Smith
    Commented Apr 30, 2016 at 17:04
  • \$\begingroup\$ @PeterSmith Hi, I think I get what you've said, I've tried to do, please see the edit in the question - is this what you meant? \$\endgroup\$
    – MrPhooky
    Commented Apr 30, 2016 at 17:13

1 Answer 1

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I find it useful to work backwards from the system output signal to a convenient summing junction output, and then equate this output to the summing junction's inputs.

Thus, the output from the 2nd SJ is \$\large\frac{\theta}{I(s)G(s)}\$

The inputs to that SJ sum to: \$ M_P-\theta \:D(s)-\large\frac{\theta}{I(s)}K_T\$

Equating these two expressions gives: $$\frac{\theta}{M_P}=\frac{I(s)G(s)}{1+K_T G(s)+D(s)I(s)G(s)}$$

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  • \$\begingroup\$ That is what I was getting in the end! Great, thanks - however, if it would be possible, how would I get this in a form to use in a root locus? \$\endgroup\$
    – MrPhooky
    Commented Apr 30, 2016 at 20:04
  • \$\begingroup\$ I have worked it out now thanks, spent hours being a fool! \$\endgroup\$
    – MrPhooky
    Commented Apr 30, 2016 at 20:57
  • \$\begingroup\$ For root locus, the characteristic equation is: \$1+K\frac{GID}{1+K_TG}=0\$ , where \$0\le K\le \infty\$, is the forward gain \$\endgroup\$
    – Chu
    Commented Apr 30, 2016 at 21:02

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