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I'm trying to solve this using Laplace transform:

zadani

(U - voltage, I - current). I am to get Uc. My attempt was to calculate I and then get Uc using ohm's law, but I wasn't able to find the I yet.

Here it's for t < 0, to get initial conditions

enter image description here

And then I try to solve it for t > 0:

t gt 0

(I with hat is current in laplace domain)

There I got kinda stuck, I don't know how to proceed. Also, it's possible I've made some mistake.

How do I get current in the time domain? It should be some kind of damped wave, looking at the poles, but I'm not sure how to do the inverse transform of this.

Thanks for help!

(ps sorry if I use incorrect terminology, I'm translating it from Czech and am not sure about some things.. I hope you can understand it like this.)

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  • \$\begingroup\$ I don't understand what you mean, what is kvl ckt? I can't get any further, if I stuff this Laplace image in wolfram, it gives a hideous mess which is quite clearly wrong, so I must have some mistake in it already. \$\endgroup\$ – MightyPork Feb 4 '15 at 12:53
  • \$\begingroup\$ Wolfram shows an answer for the input (800000*s + 0.4)/(.004*(s^2) + 32*s + 100000) . See the widget here - >"wolframalpha.com/input/?i=inverse+Laplace+transform+%28800000*s+%2B+0.4%29%2F%28.004*%28s^2%29+%2B+32*s+%2B+100000%29" ....... (Kvl = Kirchoffs voltage law ; that working image for t > 0 didnt load, so i wasnt able to figure out what the problem is) \$\endgroup\$ – Plutonium smuggler Feb 4 '15 at 13:08
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I don't think I can give you all the details, but this may help you get started. Fortunately, there are lots of information on the Web for this. A typical way of reverse transforming an expression like yours is to look up a table of Laplace transforms, rewrite the expression such that the components match one or more of the forms in the table. With your expression, it is going to fit into these two forms (copied from Wikipedia): enter image description here

Now, rewrite your expression such that \$I(s) = AP(s) + BQ(s), A,B\$ are constants.
The first step of rewriting would probably be rewriting the denominator by "completing the square". And you would be factoring out constants during the rewrite.
After you get P(s) and Q(s) to match the forms in the table exactly, then, \$i(t) = Ap(t) + Bq(t)\$ by using the original functions given by the table.

New edit: I was curious, so finished the reverse transform as below

$$ I=\frac{8 \times 10^{-5}s + 0.4}{4 \times 10^{-3} s^2 + 32s +10^5} =\frac{0.02s + 100}{s^2 + 8000s + 25000000} $$ $$ =0.02 (\frac{s + 5000}{s^2 + 8000s + 25000000}) =0.02 (\frac{s + 5000}{(s+4000)^2 + 3000^2}) $$ \$\alpha=4000\$, \$\omega=3000\$ These match the roots you calculated for the poles. $$ I = 0.02 (\frac{s + 4000 - 4000 + 5000}{(s+4000)^2 + 3000^2}) = 0.02 (\frac{s + 4000}{(s+4000)^2 + 3000^2}+\frac{1000}{(s+4000)^2 + 3000^2}) = 0.02 (\frac{s + 4000}{(s+4000)^2 + 3000^2}+\frac13\frac{3000}{(s+4000)^2 + 3000^2})$$ $$ i(t) = 0.02(e^{-4000t}cos(3000t) + \frac13e^{-4000t}sin(3000t))$$

What got me curious was I tried to get the reverse transform from WolframAlpha also, and got some really complicated answer in real form. My guess is that it uses a mechanical method that produces an answer too complicated for it to reduce. So if one just plug numbers into a computer to get answers, sometimes simpler relationships may stay hidden.

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I have tabulated and plotted my solution given below in Excel (fully general/universal solution, i.e. all 3 cases, see below; I can provide an interested person with it via e-mail; I'm Czech as the inquirer MightyPork most probably is) and have also simulated the circuit in PSpice (receiving identical results with Excel graph at the bottom). Instead of 'I with hat' I have used I(s) notation during my derivation. enter image description here

http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=7c762190486dfb47dca59a9a1f8cb1a8&title=Inverse%20Laplace%20Transform%20Calculator&theme=orange&i0=(s-z0)/(s(s-p1)(s-p2))&i1=s&i2=t&podSelect=&includepodid=Input&includepodid=Result enter image description here

enter image description here

(plot added 2015-02-13) enter image description here

Appendix: (added 2015-02-11)

MightyPork wrote:

…There I got kinda stuck, I don't know how to proceed. Also, it's possible I've made some mistake. How do I get current in the time domain? It should be some kind of damped wave, looking at the poles, but I'm not sure how to do the inverse transform of this…

  1. Yes, you've made a little mistake missing a minus sign in exponent (rather a sort of a "typing" error :) :

enter image description here

  1. To do the inverse transform of that, then, if you want to use the "WolframAlpha Calculator" mentioned above (or some other similar tool, tabulated expressions etc.), you should find roots of the numerator (zeroes) and denominator (poles; you've already done it) and rewrite the right side as follows:

enter image description here

For our expression you can find the inverse Laplace transform as:

enter image description here

So, if we are interested just in our particular case (with complex poles), then we can write:

enter image description here

so, the final result (as rioraxe has already stated) is:

enter image description here


Looking for the voltage \$ u_C(t) \$, we have to integrate the calculated current as follows (we know that \$ u_C(0) = 0 \$):

(I have used a variable x instead of t within the function i(t) so as not to confuse the independent variable with the definite integral limits which finally will become the independent variable in the result (after integration and substituting them)

enter image description here

If we wanted to do it manually, it'd be better to use the above "exponential expression" because enter image description here

and the integration itself becomes quite easy, but let's do it using the WolframAlpha Calculator:

Requesting integration of a generic expression 'e^ax (c sin(bx) + d cos(bx))'

by entering the "integrate e^ax (c sin(bx) + d cos(bx))" command at

http://www.wolframalpha.com/calculators/integral-calculator/

we receive:

enter image description here

and for the following given values \$ \> a \> = \> –4000; \> b = \> 3000; \> c \> = \> \frac{1}{3}; \> d \> = \> 1, \> C \> = \> 0,4 \cdot 10^{–6} \> \$ :

enter image description here

enter image description here

… astonishingly, the same result as earlier :) (having used the Laplace transform till the end of calculation previously)

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