I was reading a paper about a Controller Design for Temperature Control of Heat Exchanger and I couldn't understand how to model the valve used in the system.

No need to read the paper as I will wrap up the whole issue:


  • capacity of control valve is 1.6kg/sec
  • time constant is 3 sec
  • valve input is pressure varying from 3 to 15 psi

The resulting transfer function is:

transfer function of the valve

It is obvious that they considered the valve as a first order transfer function.The gain Kp was calculated by: (1.6kg/sec)/(15-3)psi=1.6/12=0.133. Time constant=3 seconds and that's it.

But the problem is: shouldn't a step input of 15psi output a signal (1-exponential) with a final value of 1.6kg/sec. But that transfer function won't actually output that. It still needs some kind of a shift. So Am I missing something? Is the model wrong?

I also simulated the transfer function response with MATLAB SIMULINK along with a suggested alternative that probably solve the offset thing: SIMULINK MODEL

The resulting waveform:


zoomed in near time=0seconds: Waveform zoomed in

  • The yellow waveform is the input step function
  • The blue waveform is the output of the transfer function of the valve according to the paper.
  • The red waveform is the waveform of the transfer function with an offset.

  • problem with the blue waveform: wrong final value.
  • problem with the yellow waveform: starts from a negative value (-0.4 kg/sec)

So what is the correct model?


2 Answers 2


The paper is a bit sloppy about transfer functions. If \$G_V\$ is the transfer from function from control pressure to flow, then it needs an offset. Step response should be measured for a step from 3 to 15 psi.

Unless you also model the DAC and the "current to pressure converter" with correct gains and offsets, any PID gain will be meaningless anyway. The charts have a scale from 0 to 1 for "output". That does again not correspond with control voltage or pressure.

Use common sense to implement this in Simulink!


The 15 unit step is 15/s in the laplace domain. Multiply this by the valve function gives 2/3s^2+s. Since you are interested in the final value we can use the final value theorem. You may recall it says that the limit as t approaches infinity is equal to sF(s) as s goes to 0. sF(s)= 2s/3s^2+S. since the limit is undefined because of 0/0 we have to use L'Haspotal's rule. This says take the derivative of the numerator and denominator and take the limit of the result. After the derivatives we have 2/3s+1. The limit as s goes to zero is two which is what you got. I'll leave the units up to you.


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