# Explanation of transfer function from electrical design

I am a bio-engineer student so not that profound with electric cirquits. But I understand how transfer function works. My problem is the following: I have found a source from literature that derives a transfer function from a given electrical circuit. The electrical circuit is the one I've attached as an image. It is actually a biological circuit based on a model of the mechanics in the lungs. Now the transfer function from the paper is given by: It is possible that the above transfer function is incorrect (Rst does appear in the transfer function and not in the diagram, which is weird). User rioaxe pointed out that the transfer function above is actually the impedance of the circuit, that is Pressure (P) / Flow (F) = Vin / Iin, which is very logical because the equation relating pressure to flow is: P = Raw * F, with Raw the airway resistance. Equivalent to Vin = R * Iin in electrical circuits.

The problem that I face now is how to derive the transfer function Vin / Iin from what rioaxe has given as answer. I never worked with electrical circuits like that before (in series + parallel).

If someone can help me derive the TF this will be very kind. And thanks for who already tried to help. Cheers

Here is a second picture of the (viscoelastic) model: • Have you tried Laplace Transform before? If no read this you will get a glimpse of guidance. On the other hand, if you answered yes, please do edit your question and share with us the steps you have done. As this site encourages new users to chip in their effort first before help and guidance were given. Dec 30, 2014 at 1:11
• Seems like there is something missing in the diagram- There's no load on the right side, and there is reference to Raw and Rst which do not appear in the diagram. As it stands there's no current path in the circuit. Dec 30, 2014 at 1:31
• @JohnD I think R == Rst == Raw. And it is actually a biological circuit of the lungs that is based on an electric circuit,so I don't know how to account for the current path. Dec 30, 2014 at 6:13
• @Sarenya I know how the Laplace transform works, if I use it for example in a feedback system. But I have no idea how to use it in such an electronical system as in the picture. That is getting the differential equation from such a system. Dec 30, 2014 at 6:18
• I've added another picture that is exactly the same circuit as the first one. If you assume Vin (= Flow) as input from the left side and Vout (= Pressure) as output at the right side, could you derive the differential equation from the circuit? Cause I really don't know how to do that... Dec 30, 2014 at 6:27

In the model, it is not Flow in on the left side and Pressure out on the right side. But rather: simulate this circuit – Schematic created using CircuitLab

$$Transfer\ Function = \frac{Pressure}{Flow} = \frac{Vin}{Iin} = impedance\ of\ the\ circuit$$

To analyze the circuit in Laplace s domain, substitute impedance of capacitor with 1/sC (impedance of resistor is simply R).

• I think you are correct. This seems like a better representation of the circuit (instead of Vout/Vin). I still don't know how to put this form into a transfer function, though. Can you maybe help me out here? I am really curious to see how the transfer function looks like. Dec 30, 2014 at 12:02
• Yes - the given function is NOT the transfer function but the INPUT impedance of the shown circuit. But you have to incorporate the following modifications: (1) The circuit is to be grounded (as shown in the above picture), (2) R=RAW and (3) Denominator: Replace RST with RVE.
– LvW
Dec 30, 2014 at 13:31
• @LvW With the modifications u mentioned, do you then obtain the same input impedance of the circuit (i.e. the transfer function Pressure/Flow that is given from the paper)? Dec 30, 2014 at 16:41
• YES! I have learned how to use impedances and got exactly the same transfer function as in the paper (Rst in denominator should be Rve). So satisfied. Thanks to all! Dec 30, 2014 at 18:56
• Rafi: Note that the terms "flow" and "pressure" as shown on the first figure make no sense in connection with the interpretation of the given function (input impedance).
– LvW
Dec 31, 2014 at 9:20