Transfer function / diff. eq. question

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Discussion Overview

The discussion revolves around the formulation of a differential equation and transfer function for an RC circuit as part of a control systems assignment. Participants explore the mathematical derivation of the transfer function, its implications for system stability, and the process of obtaining the corresponding differential equation.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant describes their approach to forming the transfer function using Laplace transforms and voltage divider principles, leading to a specific expression for G(s).
  • Another participant points out a typo in the voltage divider expression, suggesting it should be corrected before proceeding.
  • A different participant asserts that having a transfer function with equal degree numerator and denominator is acceptable, citing examples of all-pass functions.
  • There is a suggestion to use standard circuit analysis methods, such as loop or node analysis, instead of relying solely on Laplace transforms for deriving the differential equation.
  • One participant explains the process of dividing the transfer function to separate the identity from the remainder, indicating that only the second term needs to be inverted for the differential equation.
  • The original poster later reports a breakthrough in deriving the differential equation independently and acknowledges the relevance of proper and strictly proper transfer functions in their understanding.
  • They mention successfully creating a Simulink model and simulating the system after receiving input from other participants.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to derive the differential equation, with some advocating for Laplace methods and others suggesting traditional circuit analysis techniques. There is no consensus on a single method being superior.

Contextual Notes

Some participants note the importance of correcting initial expressions and understanding the implications of transfer function properties, such as proper and strictly proper classifications, but these aspects remain unresolved in terms of their broader applicability to the problem at hand.

Who May Find This Useful

Students and practitioners in control systems, electrical engineering, and circuit analysis may find the discussion relevant, particularly those working on assignments involving transfer functions and differential equations in RC circuits.

mankku
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Hi everyone!

I'm stuck with an assignment in control systems and I need some opinions on what I've got so far.

Homework Statement



The problem relates to an RC circuit as displayed:

auto2060.png


The task is to form the differential equation and transfer function for the circuit, with V0(t) being the input and V2(t) being the output. The values for the components are given in the image. It is assumed that initial conditions (currents, voltages) are all zero.

Homework Equations



Well, basic electrical theory such as voltage, current, impedance as well as parallel and series combination of impedances, I guess.

The Attempt at a Solution



So I started out using the Laplace (frequency domain) impedances, attacking the problem from a voltage divider point of view. Thus, the resistor and capacitor on the left in the schematic were "combined" into a single impedance Z1, and the resistor and capacitor on the right were combined into Z2. The output voltage would thus be

[tex]V_{2}(s)=V_{0}(s) * \frac{Z_{2}(s)}{Z_{1}(s)+Z_{2}(s)}[/tex]

Thus the transfer function is obtained through dividing by V0(s) and is thus

[tex]G(s) = \frac{V_{2}(s)}{V_{0}(s)} = \frac{Z_{2}(s)}{Z_{1}(s)+Z_{2}(s)}[/tex]

The two impedances used are formed as:

[tex]Z_{1}(s) = R_1 || \frac{1}{sC} = \frac{\frac{R_1}{sC}}{\frac{sR_{1}C+1}{sC}} = \frac{R_1}{sR_{1}C+1}[/tex]

[tex]Z_{2}(s) = R_2 + \frac{1}{sC} = \frac{sR_{2}C+1}{sC}[/tex]

The sum of these two becomes:

[tex]Z_{1}(s)+Z_{2}(s) = \frac{sR_1C}{sC(sR_{1}C+1)}+\frac{(sR_{2}C+1)(sR_{1}C+1)}{sC(sR_{1}C+1)} = \frac{s^{2}R_{1}R_{2}C^{2}+s(2R_{1}C+R_{2}C)+1}{sC(sR_{1}C+1)}[/tex]

Forming the transfer function now then, we invert the above calculated sum and multiply by Z2(s) to obtain:

[tex]G(s) = \frac{Z_{2}(s)}{Z_{1}(s)+Z_{2}(s)} = \frac{sR_{2}C+1}{sC}*\frac{sC(sR_{1}C+1)}{s^{2}R_{1}R_{2}C^{2}+s(2R_{1}C+R_{2}C)+1}[/tex]

The sC eliminate each other and I am left with:

[tex]G(s) = \frac{(sR_{1}C+1)(sR_{2}C+1)}{s^{2}R_{1}R_{2}C^{2}+s(2R_{1}C+R_{2}C)+1}[/tex]

Which given a bit of cleaning turns into

[tex]G(s) = \frac{s^{2}R_{1}R_{2}C^{2}+s(R_{1}C+R_{2}C)+1}{s^{2}R_{1}R_{2}C^{2}+s(2R_{1}C+R_{2}C)+1}[/tex]

In case someone wants to plug in the values given for the components, the result becomes

[tex]G(s) = \frac{0.0000027s^2+0.0037s+1}{0.0000027s^2+0.0064s+1} = \frac{s^2+1370.37s+370370.37}{s^2+2370.37s+370370.37}[/tex]

4. The real question :-p

The reason why I'm asking is that this transfer function seems quite ok to start with. The zeroes and poles are all real, which means from a stability point of view that the system does not oscillate, as would be expected by an RC circuit. Also, the DC gain (i.e. s -> 0) is 1, which holds true seeing that if the input signal is DC, then the capacitor on the right is an open circuit and the resistor R_1 pulls the output to equal the input. So from this point of view, I'm satisfied.

On the other hand, is it possible to have a transfer function with numerator of equal degree as the denominator? I know that Simulink does not like that, and I don't know how to work around this. I have done a long division on the two polynoms and sure, it gave me 1+<something> where something had a numerator with degree less than that of the denominator, but IMO that does not help in this situation.

So to get the differential equation I would be tempted to use the expression for G(s) that I have, multiply with denominator and V0(s) and do an inverse Laplace transform. The problem is that I'm unsure whether I am allowed to do that and if that will yield the correct answer.

The diff.eq. I would get would thus be:

[tex]\ddot{v_2}(t)+2370.37\dot{v_2}(t)+370370.37v_{2}(t) = \ddot{v_0}(t)+1370.37\dot{v_0}(t)+370370.37v_{0}(t)[/tex]First of all, if anyone actually read through this in detail, thank you. :smile:

If you have any thoughts on the above, please reply because I am in desperate need of assistance. The assignment is due on Tuesday so there is still time, however I would like to get this sorted out of the way because it has been bothering me for several days now.

Cheers

// Mankku
 
Last edited:
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Right away, you have a problem with the voltage divider expression. The fraction should be:

Z2(s)
------------
Z1(s)+Z2(s)

I won't even bother to follow the rest of it until you fix this.
 
The Electrician said:
Right away, you have a problem with the voltage divider expression. The fraction should be:

Z2(s)
------------
Z1(s)+Z2(s)

I won't even bother to follow the rest of it until you fix this.

Which was indeed a typo on my behalf. The rest should be correct. Thanks for pointing the typo out, hopefully it hasn't caused others to stop reading through the post.

// Mankku
 
The transfer function is correct. There's nothing wrong with having the numerator and denominator of equal degree. All pass functions are an example of such a thing.

As far as "the differential equation" of the circuit, there are various ways of doing that. I would have thought you should solve the network with one of the standard methods such as the loop or node method, but using the constitutive relation (i(t) = C*de/dt) for the capacitor, instead of the Laplace formulation (using the variable s).
 
When you divide the numerator by the denominator of the transfer function you get
[tex]\frac{V_2(s)}{V_0(s)} = 1 + \frac{R(s)}{D(s)}[/tex], where D(s) is the denominator and R(s) the remainder of the division.
So, [tex]V_2(s) = V_0(s) + \frac{R(s)}{D(s)}V_0(s)[/tex].
You only have to worry with the inverse transform of the second term in the sum, since the first one is an identity.
 
Thanks for your replies!

In fact, the solution to the problem hit me this morning while I was having breakfast. It's kinda odd how the mind works, I wasn't even thinking about the assignment and suddenly *boing*, it hit me. And in five minutes I had derived the differential equation from scratch, and it corresponds to the transfer function I had worked out earlier.

When I got to work, I looked up the Wikipedia article on state space models, as the assignment next required me to form one for the system in question. There, I was introduced to the concept of proper and strictly proper transfer functions. In fact, I had performed the division which CEL suggested, but thought that it was of no use. Now I know otherwise.

I was able to make a Simulink model of the system, simulate with the required inputs and now all I have to do is write the report. Thanks to both of you guys for your input. :smile:

Cheers

// Mankku
 

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