# Simple circuit analysis

Discussion in 'Electronics Homework Help' started by YoeriB_deleted, Aug 11, 2014.

1. ### YoeriB_deleted

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Aug 11, 2014
Hello all,

I am trying to solve a simple circuit, but I think I am missing out on one equation. Can somebody please comment on what I am missing? The goal is to calculate the voltage at U_3 and the AC source is a perfect voltage source. I eventually want to simulate this circuit using these equations, in order to quickly vary parameters (frequency, voltage and element values).

I am using a quite straightforward method for solving this, starting at U3, working my way back.

Code:
```U_3 = i_{Rf} * R_f   (Ohms law)

i_{Rf} = i_{Ro} + i_C (KCL)

i_{Ro}  = U_{Ro} / Ro (Ohms law)

U_{Ro} = U_2 - U_3 (KVL)

U_2 = U1 - i_{Rs}*Rs (Ohms law)

i_{C} = C d/dt U_{Co} (Capacitor-elemental)

U_{Co} = U_{Ro}
```
However now the value of i_{Rs} is still unknown, but I cannot get it from any equation used previously, as I will destroy the system of equations. I cannot find which step I am missing.

Thanks a million.

2. ### Harald KappModeratorModerator

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Nov 17, 2011
Welcome to our forum.

I moved your thread to the homework section where I think it belongs.

I'm sure you'll find the answer to your question yourself by looking at the difference between I_{Rf} and I_{Rs}.

3. ### YoeriB_deleted

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Aug 11, 2014
Thank you for moving the topic.

You are right, i_{Rf} is equal to i_{Rs}, but then I will have to introduce i_{Rf} in the KVL equation for U_{Ro}, resulting in a differential equation which has to be solved to continue. Substituting and rewriting to the 2nd equation (KCL) yields:

d/dt(i_{Rf}) - i_{Rf}/Ro = C/Rs * d/dt (U1 - U3) + (U1 - U3)/(Ro*Rs)

The end result does not have to be solved analytically by the way, as the simulation will be done numerically.

4. ### Laplace

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Apr 4, 2010
Have you learned anything about steady-state AC analysis yet? Or the complex frequency domain? It's the smarter way.

No one solves ordinary circuit equations using differential equations except beginning students who must solve their obligatory first problem using basic component physics with differential equations. I assume that teaching is done in this manner so that students will appreciate the smarter way that will be taught next.

So just suck it up, have fun fiddling with the math, and realize this may be the last time you see a differential equation, at least until you get to field theory.

5. ### YoeriB_deleted

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Aug 11, 2014
Just to be clear: I am not a student (anymore). This example is related to personal projects. I've had some electronics in my mechanical engineering bachelor, but its been a while.

As I understand, you propose to derive the system transfer, by H(w) = U3/U1, where the RC-couple is then described by a complex function of something like Ro/(jwCRo + Ro), correct?

6. ### Laplace

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Apr 4, 2010
It would depend upon whether you want the steady-state solution at t=∞ or the transient solution from when the AC source is switched on at t=0. Using the term jω for characterizing the impedance will give the steady state solution.

7. ### Ratch

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Mar 10, 2013
Use the voltage divider method, and you will only have to write one equation.

Ratch

8. ### YoeriB_deleted

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Aug 11, 2014
I am looking for a steady-state solution, although it must be time-dependent in the steady state. ie: I want to simulate the individual periods. The startup behaviour is not of interest.

9. ### Laplace

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Apr 4, 2010
In that case then use the voltage divider with steady state AC analysis. Not sure if there is any mechanical analog to the electrical voltage divider, i.e., that the voltage across a resistor/impedance in a series string is proportional to the value of the resistance/impedance. So here the AC voltage is divided between Rs, Rf, and Zp. Zp is the complex impedance of the parallel combination of Ro & C.

The general formula for parallel resistance/impedance is 1/Zp = 1/Z1+1/Z2; Zp = Z1∙Z2/(Z1+Z2). In this case Zp=Ro(1/jωC)/(Ro+1/jωC).

V3 = Vac∙Rf/(Rs+Rf+Zp) And apply some complex algebra to get the result in a form you can use.

Last edited: Aug 12, 2014
10. ### YoeriB_deleted

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Aug 11, 2014
Thank you for the input. I have already looked at this approach but decided it was not suitable for some reason... Probably my lack of knowledge. I'll have another go at it!

11. ### YoeriB_deleted

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Aug 11, 2014
As expected, this was the solution to my problem. Thank you Laplace (oh the irony).

The result is now some complex-valued rectangular number for each timestep (ie: 5 + 2i or so). To transform this numerical data back into a time-based signal, I can take the real part and be done with it. However the imaginary part says something about the phase, correct?

12. ### Laplace

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Apr 4, 2010
By the phrase "time-based signal" I assume you mean what your voltmeter would measure. But the voltmeter measures the absolute magnitude of the vector sum of the real and imaginary parts of the voltage. For steady-state AC analysis the voltages are often plotted on a phasor diagram (with real and imaginary axes) in order to show the phase relationships.

13. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
We find his posts are often transformative!

14. ### YoeriB_deleted

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Aug 11, 2014
By time-based I mean time domain. To give you some more insight: the model will be used to match some parameters to measurements that I am doing. To do that, I need to transform the output of the model back to the time domain. If i take the real part of the model output, I do get the result as I expect, but no phase information.

By the way: this circuit is a simplification of reality (obviously), so Rf will be replaced with some more complicated circuitry and especially Ro and C are dependent on some parameters on my setup.

15. ### Laplace

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Apr 4, 2010
I don't quite understand what it means to take the real part with no phase information. Any voltage in the model will have the form V=A∙sin(ωt+φ) where 'A' is the vector magnitude and 'φ' is the phase angle with respect to some reference, usually the voltage source 'Vac'. The values for A & φ are calculated using steady-state AC analysis at a given frequency for any voltage node in the circuit.

It is impossible to take the real part without the phase information, whereas it is possible to take the magnitude 'A' without the phase information simply by measuring with an RMS voltmeter and adjusting for the peak value, A∙sin(ωt).

16. ### YoeriB_deleted

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Aug 11, 2014
Then probably I did not fully understand your previous comments on working in the frequency domain.

I now basically have a single equation V3 =Vac∙Rf/(Rs+Rf+Zp), with Zp = Ro/(jwCRo + Ro). This then has a real and imaginary component.
Then I use a sinusoidal voltage for Vac in my model, which is a time-domain signal. Maybe this is illegal, as I am now using time and freqency domain within one equation?

17. ### Laplace

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Apr 4, 2010
The stated problem is one that can be solved by steady-state AC analysis. This is a rather simple method because it provides a restricted view of both the time and frequency domain. In the time domain the signal source is a constant frequency sine wave with a constant amplitude, and any start-up transients have died out so the circuit has achieved a steady state. All voltage nodes in the circuit are also sine waves of the same constant frequency, each with their own magnitude and phase. It is the same in the frequency domain; all voltages are at a single frequency, each with their own magnitude and phase.

It is because of this steady-state simplification that AC analysis can be performed using the same basic forms as DC analysis, albeit with complex algebra (real & imaginary parts) to account for the AC magnitude and phase rather than just the DC magnitude. So it does not matter which domain you think you are in, all that you see is just magnitude and phase. Unless, of course, in the time domain you are interested in instantaneous voltages -- then you would put a time value in the sine function with its magnitude and phase and calculate the exact voltage for that instant of time. But I can't think of any reason why that would be an interesting thing to do. For problems in steady-state AC analysis it is the magnitude and phase that are most interesting.

PS. It occurred to me that there may be a perceptual problem with the terminology, i.e. that a 'real' voltage has a physical presence whereas an 'imaginary' voltage does not really exist. But the terms actually refer to real numbers and imaginary numbers. An 'imaginary' AC voltage has as much physical presence as a 'real' AC voltage, but it has a 90 degree phase shift.

Last edited: Aug 17, 2014