Laplace analysis of RLC and RLCC circuits

In summary: Hi BackEMF,Thanks for your response. I did guess that you might have been distracted by other things – as you say, it happens to all of us.To answer your question, yes, I did make some progress. I was able to find a solution to the differential equation, but only by utilising techniques which are really beyond the scope of what I wanted to cover. I suppose I was hoping for a more elegant solution which might have just involved a bit more algebraic manipulation. Anyway, I have attached a pdf file which shows my solution. I then used this solution to find the voltage across each component as a function of time, and plotted these on a common plot. It is this plot which has provoked my next question which
  • #1
Muddyrunner
4
0
(I hope I have posted this in the appropriate forum. I have consulted the posting guidelines and this seems to be OK. The subject of my post certainly isn’t homework or academic coursework, rather it is self-study for its own sake. If this is the wrong place – apologies, and maybe an admin could move it to a better place).

For some weeks now I have been studying what happens to RLC circuits when they are subject to a step change in applied voltage. The real-life applications for this include, for instance, DC-DC converters and full-bridge motor drive circuits, where a network composed of inductive, resistive and capacitive elements can experience a rapid change in voltage when one of the switching components changes state. My goal has been to find general equations describing current “i”, and from this equations describing the voltages appearing across the various circuit elements. It is specifically the transient behaviour of the circuits that I am interested in.

So far I have had success with the series-RLC configuration, and one where R and C are in parallel, and that combination in series with L. In both cases I was able to use Laplace transforms to analyse the circuits and find general equations describing the currents and voltages. Also in both cases, the solutions involved transforming a second-order DE from the s-domain to the time domain. The results were found to be identical to those produced by Spice simulations of the same circuits.

I have now moved on to a four-component configuration (RLCC), and have run straight into a brick wall. After several attempts I am stuck at the same point, and I do not know how to proceed. I have laid out my working in the attached pdf, as it is rather long and I have no idea where any errors may have slipped in. The general approach I have adopted is identical to that used for the simpler (three-component) cases, but I keep getting stuck at the point where I have to transform a third-order DE.

Any pointers which anyone could offer would be gratefully received. Finally, I should also apologise for any inadvertent misuse of mathematical terminology. These aren’t the kind of discussions in which I normally become involved!

Thanks and regards,

MR.
 

Attachments

  • RLCC problem v02.pdf
    152.6 KB · Views: 266
Physics news on Phys.org
  • #2
At first look - and I feel guilty for not looking at it in more detail given the effort you put into formulating the question - it seems as if you've calculated the overall impedance and called it X, and then have taken the current to be
[tex]I(s) = \frac{V(s)}{sX(s)}[/tex] which it seems could be in error. If X is an impedance (and it certaintly looks to be the way you have calculated it) the correct formula for current would be
[tex]I(s) = \frac{V(s)}{X(s)}.[/tex] Perhaps you have confused impedance and reactance (X is usually the symbol for reactance and Z the symbol for impedance - if X were a reactance, the first equation above would make more sense, but you'd be assuming zero resistance which is not true for this circuit).

Also at the end it is quite possible that your answer, in terms of exponentials, could yield an oscillatory response if some the exponents are strictly complex/imaginary (along with their conjugates):
[tex] \text{e}^{(-1-\text{j}5)t} + \text{e}^{(-1+\text{j}5)t} = 2\text{e}^{-t} \cos(5t)[/tex] by Euler's formula.
 
  • #3
Hello BackEMF,
Thanks for your reply, I have inserted my comments in the relevant places below:

BackEMF said:
At first look - and I feel guilty for not looking at it in more detail given the effort you put into formulating the question - it seems as if you've calculated the overall impedance and called it X, and then have taken the current to be
[tex]I(s) = \frac{V(s)}{sX(s)}[/tex] which it seems could be in error. If X is an impedance (and it certaintly looks to be the way you have calculated it) the correct formula for current would be
[tex]I(s) = \frac{V(s)}{X(s)}.[/tex] Perhaps you have confused impedance and reactance (X is usually the symbol for reactance and Z the symbol for impedance - if X were a reactance, the first equation above would make more sense, but you'd be assuming zero resistance which is not true for this circuit).

This is a perfect example of the sloppy terminology which I warned might be in my post! I did mean impedance in the s-domain (not reactance) and so I guess that should have been Z(s). So my formula for i(s) perhaps should have been i(s) = V/sZ(s). The additional “s” term in the denominator comes from the fact that my voltage V makes a step at t=0 i.e. V x 1/s. Is this a “forcing function”? Maybe I’ve got the terminology wrong again…

BackEMF said:
Also at the end it is quite possible that your answer, in terms of exponentials, could yield an oscillatory response if some the exponents are strictly complex/imaginary (along with their conjugates):
[tex] \text{e}^{(-1-\text{j}5)t} + \text{e}^{(-1+\text{j}5)t} = 2\text{e}^{-t} \cos(5t)[/tex] by Euler's formula.

That is an interesting proposition – I need to study this further. As I said in my original post, solving the simultaneous equations for a, b and c will be quite difficult. I already had a preliminary stab at it. “Messy” is the adjective I would use.
I must say I am astonished at how much more complicated the problem is made just by adding a fourth component. In comparison, the three-component configurations were quite straightforward. This is what has made me wonder whether I have made a very basic error somewhere.

Regards,

MR
 
  • #4
Hi Muddyrunner. I meant to reply to this far earlier, and actually had it partially written but as often happens, I got distracted (or someone distracted me to be more accurate!).

Anyway first of all, you are completely correct, the 's' in the denomintor was fine, due to the unit step input. Have you had any luck in getting futher?

You say that you're suprised how much hard it is to solve than the simpler RLC case - well this will happen when you keep everything as variables. If you pick specific values for the circuit elements it would not be much more difficult than the previous case. This happens a lot e.g. it is easy to write down the roots of a general quadratic equation, a bit more difficult for cubic polynomials, still even harder for fourth order polynomials - and impossible for fifth! So you can see how the complexity can increase...

Anyway, hopefully you have made some progress.
 
  • #5


Dear MR,

Thank you for sharing your self-study progress on Laplace analysis of RLC and RLCC circuits. It is certainly a complex and challenging topic, and I am glad to hear that you have already made some successful progress with the series-RLC and parallel-RLC configurations. It is also great to see that you have used simulation tools like Spice to verify your results.

Moving on to the four-component configuration (RLCC), it is not surprising that you have encountered some difficulties. The additional complexity of the circuit can make it more challenging to apply the same techniques used for the simpler cases. From your attached pdf, it seems like you have a good understanding of the general approach and have been able to transform the second-order DEs successfully. However, transforming a third-order DE can be more challenging and may require some additional techniques.

My suggestion would be to consult some additional resources, such as textbooks or online tutorials, that specifically cover the analysis of higher-order RLC circuits using Laplace transforms. You may also want to consider reaching out to other experts in the field, either through online forums or professional networks, to ask for their insights and advice on how to approach this problem.

I wish you the best of luck in your continued self-study and exploration of this complex topic. Keep up the good work!

Best regards,
 

FAQ: Laplace analysis of RLC and RLCC circuits

What is Laplace analysis of RLC and RLCC circuits?

Laplace analysis is a mathematical method for analyzing the behavior of RLC (resistor, inductor, and capacitor) and RLCC (resistor, inductor, capacitor, and current source) circuits. It involves converting the circuit into its equivalent mathematical model in the s-domain, where s is the complex frequency variable. This allows for easier analysis and solution of circuit equations.

How is Laplace analysis different from other circuit analysis methods?

Laplace analysis is different from other methods, such as Kirchhoff's laws or nodal analysis, because it is based on complex numbers and the s-domain. This allows for a more comprehensive and efficient analysis of circuits with complex components, such as inductors and capacitors.

What are the advantages of using Laplace analysis for RLC and RLCC circuits?

One of the main advantages of Laplace analysis is that it allows for the analysis of circuits with any number of components, including complex components like inductors and capacitors, without the need for complex algebraic equations. It also provides a more complete understanding of circuit behavior, including transient and steady-state responses.

What are the limitations of Laplace analysis for RLC and RLCC circuits?

Laplace analysis assumes linear and time-invariant circuits, which may not always accurately model real-world circuits. It also requires knowledge of complex algebra and the s-domain, which can be challenging for some individuals. Additionally, it may not provide a visual representation of the circuit, making it difficult to understand the physical layout of the circuit.

How is Laplace analysis used in practical applications?

Laplace analysis is used extensively in the design and analysis of electronic circuits, especially in the design of filters, amplifiers, and control systems. It is also commonly used in the field of signal processing and communication systems. In practical applications, Laplace analysis helps engineers and scientists to understand and predict the behavior of complex circuits, allowing for more efficient and accurate circuit designs.

Similar threads

Replies
1
Views
668
Replies
2
Views
1K
Replies
5
Views
1K
Replies
22
Views
963
Replies
8
Views
842
Replies
2
Views
2K
Replies
1
Views
1K
Back
Top