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I have a scenario where a valve q_in(t) is pumping water into a tank of height h(t) and area A, which is then pumped out by a second valve q_out(t) that has a resistance to water flow R.

I got this transfer function:

$ \frac{H(s)}{Q_i(s)} = \frac{R}{A R s + 1} $

Where R= 1.5m(m^3/min) and A= 2m^2

I would be grateful if someone could tell me how to get the value of Rs so I can plot the transfer function in Labview!

MelanieW403
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  • Do you mean $ \frac{H(s)}{Q_i(s)} = \frac{R}{A R s + 1} $? If so, please edit your question with the appropriate parentheses, or replace your relevant text-math with `$ \frac{H(s)}{Q_i(s)} = \frac{R}{A R s + 1} $`. And you ask for the value of "Rs" -- do you mean "R times s", where s is the Heavyside operator? – TimWescott Oct 14 '19 at 16:43
  • Yes sorry, R times s is what I want! I will change the equation now! – MelanieW403 Oct 14 '19 at 16:49
  • Thanks for your help about the equation! – MelanieW403 Oct 14 '19 at 18:22
  • If you could tell me how to get R times s so I could plot the values in a graph I would be very grateful. – MelanieW403 Oct 14 '19 at 18:24
  • This seems too trivial -- multiply the value of R by the value of s at whatever frequency you're interested in. Are you trying to make a Bode plot? Where are you stuck? – TimWescott Oct 14 '19 at 18:41
  • I'm trying to plot the response of an open-loop system with a step input on labview. – MelanieW403 Oct 14 '19 at 18:44
  • But when I go to put values in for my transfer function i need values for Rs – MelanieW403 Oct 14 '19 at 18:44
  • As well as R and A to get results on the graph. – MelanieW403 Oct 14 '19 at 18:44
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    $s$ is an **operator**, not a variable per se. A transfer function is a shorthand for a differential equation that describes the behavior of a linear system. You can use a transfer function almost directly to get a *frequency response* of a system by substituting values into $s$, but to get a response to a signal in the time domain, you need to simulate the system, or solve the inverse Laplace transform of the output signal. This should be a matter of pushing the right buttons in Labview, but I'm not familiar with that program. – TimWescott Oct 14 '19 at 18:47
  • $R$ and $A$ are given, surely. – TimWescott Oct 14 '19 at 18:48
  • Yes, R= 1.5m(m^3/min) and A= 2m^2 – MelanieW403 Oct 14 '19 at 18:49
  • By the output signal do you mean the transfer function? I'm a little confused! – MelanieW403 Oct 14 '19 at 19:19
  • You're missing some big chunks of fairly basic control theory. A transfer function is a *description* of the *behavior* of a *system*. An output is a *signal*. **Signals are not systems**, and **systems are not signals**, so a transfer function cannot be an output signal, or visa versa. I think you need to Google on "transfer function" and study up. – TimWescott Oct 14 '19 at 19:29
  • Sorry, I get that a transfer function descibe the behaviour of a system in that it shows the ratio of output to the input of a control system. It's just I'm a bit clueless with laplace transforms! I've gone over it before but don't have a set method of how to work it out that makes it easy to understand and remember. – MelanieW403 Oct 14 '19 at 19:34
  • Do you still need an answer? – useless-machine Nov 17 '19 at 21:18

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