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At the moment I can find the limit of my transfer function as follows: 

(5 4001.6`)/(5 4001.6` + (4 + s) (200 + s)) /. s -> 0

This works as I would expect and provides a final value of 0.961553 In this example my value of gain, k, is 4001.6` so re-writing the expression for arbitrary k would give:

   (5 k)/(5k + (4 + s) (200 + s)) /. s -> 0

I want to find a value of k that makes the result 0.95. To do this I was thinking about using a while loop.

I thought the while loop could start with k=1, compute the limit, test to see if less than 0.95, and if not increment k by one, and perform the test again, until the correct value of k is found:

This is what I have tried:

    k = 1; While[(5 k)/(5 k + (4 + s) (200 + s)) /. s < 0.95, k++]

However I get the error: "ReplaceAll::reps: {s<=0.95} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing. >>"

Any help would be most appreciated.

m_goldberg
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AugustCrawl
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    Shouldn't it be (... /. s -> 0) < 0.95? The error message is telling you that you forgot a ->. – Michael E2 Aug 14 '13 at 20:18
  • Ah, I think Michael has a better handle on your particular intent. Nevertheless consider reading: (1835) for some examples. – Mr.Wizard Aug 14 '13 at 20:19
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    By the way: FindRoot[0.95 == (5 k)/(5 k + (4 + s) (200 + s)) /. s -> 0, {k, 4000}] Out: {k -> 3040.} – Mr.Wizard Aug 14 '13 at 20:34
  • FWIW, Control`StaticGains[TransferFunctionModel[(5 k)/(5 k + (4 + s) (200 + s)), s]] will compute the steady-state gains. – Suba Thomas Aug 14 '13 at 21:52
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    There are easier ways to do this, but I am now confused about your TransferFunction. This can't be an actual transfer function. closed loop basic TF is $\frac{k H(s)}{1+ k H(s)}$ and what you do next is use RootLocusPlot to see where it crosses the $s$ axis for specific $k$. It is easy now to make a manipulate which varies $k$ and see which values causes it to become unstable. – Nasser Aug 14 '13 at 23:11
  • @Nasser this is just kH(s) [OL response] and by varying k and bode plotting the impulse response (response to a dirac delta function) we can see where the poles & zeros lie. It is an approximation of the CL system -> see "http://homepages.engineering.auckland.ac.nz/~covic/sctrl_da.PDF" for details :) – AugustCrawl Aug 15 '13 at 07:39

1 Answers1

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Rather than try to code the numerical solver loop yourself, Mathematica has great built in functions. In the comments, @Mr.Wizard suggested FindRoot, but Solve may be easier for you:

Solve[((5 k)/(5 k + (4 + s) (200 + s)) /. s -> 0) == .95, k]

{{k -> 3040.}}
0xFE
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