Laplace analysis of RLC and RLCC circuits

[Muddyrunner]

(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.

 

[BackEMF]

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
$$I(s) = \frac{V(s)}{sX(s)}$$ 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
$$I(s) = \frac{V(s)}{X(s)}.$$ 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):
$$\text{e}^{(-1-\text{j}5)t} + \text{e}^{(-1+\text{j}5)t} = 2\text{e}^{-t} \cos(5t)$$ by Euler's formula.

 

[Muddyrunner]

Hello BackEMF,
Thanks for your reply, I have inserted my comments in the relevant places below:

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
$$I(s) = \frac{V(s)}{sX(s)}$$ 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
$$I(s) = \frac{V(s)}{X(s)}.$$ 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…

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):
$$\text{e}^{(-1-\text{j}5)t} + \text{e}^{(-1+\text{j}5)t} = 2\text{e}^{-t} \cos(5t)$$ 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

 

[BackEMF]

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.