# Maple issue - inverse laplace transform equation from a basic series RLC circuit

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maple issue -- inverse laplace transform equation from a basic series RLC circuit

Pretty simple for some of you I know, but I have a laplace transform equation from a basic series RLC circuit

((sV-v(0))/L+si(0))/(s^2+sR/L+1/LC) = I(s)

I want to take the inverse laplace of it, I am given all initial conditions, and constants for R,L, and C.

I've done laplace but cannot figure out how to take the inverse of this one, and cannot figure how to check it with maple..

Any help appreciated.

EDIT: initial conditions are L = 10H, v = 10v C = 1000uF i(0) = 1A R = 100ohms

Can I just fill in values at this point and attain 2s-.5/s^2+10s+100 and simplify this and bring it back to the time domain?

Sorry, im asked for the current response in the time domain.. So I think this is the right approach... maybe I have made a mistake.

Any help with the maple would still be appreciated greatly. I am new to the program.

I get 2e^-5tcos(sqrt75)t - (21/2(sqrt75))e^-5tsin(sqrt75)t

I was expecting a cleaner result....

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Pretty simple for some of you I know, but I have a laplace transform equation from a basic series RLC circuit

((sV-v(0))/L+si(0))/(s^2+sR/L+1/LC) = I(s)

... Can I just fill in values at this point and attain 2s-.5/s^2+10s+100 and simplify this and bring it back to the time domain?

$$I(s) = \frac{2s-0.5}{s^2+10s+100}$$
I haven't checked your substitutions, but assuming the above is correct, then you should complete the square on denominator before attempting to take the inverse Laplace transform.

In general you would either factorize or complete the square. In this case it's easy to show that the denominator has no real roots, so completing the square is more appropriate.

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Yes, and thankyou for the response.

I did complete the square, and seperate the expression taking two inverse laplace transforms, I did not show the work.

I actually also got it figured out in maple, and received the correct answer by hand. The one up there is not correct, but it is close.

The correct solution is -4.5/sqrt75sinsqrt75t + cossqrt75t

I am having a larger issue now, when the capacitor is charged before t=0, and wether or not I can use the same laplace transforms for a capacitor that is now supplying power.

It was my initial response to simply put it on the other side of my kirchoffs voltage equation, as a rise. However, I am not receiving correct results.

My professor adds a source in series to his capacitors and inductances, rather than simply laplace them.. he laplaces them seperately.

I am at wits end !

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Ratch

FOIWATER,

What is V? What is v(o)? In sV-v(0))/L, does L divide the numerator, or is it sV-v(0))/(L+si(0)))? The solution to the LaPlace expression by uart is in the attachment. Did you read the help file for MAPLE?

Ratch

#### Attachments

• FOIWATER.JPG
10.6 KB · Views: 584
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Actually I simplified it wrong friend

but I have it figured out now in terms of manual calculation. Thankyou for the attachment, I now understand how to put it into maple.

the correct solution was

L = 10H v = 10v C = 1000uF i(0) = 1A R = 100ohm

v(t) = Ri(t) + Ldi/dt + 1/cintegrali(t)dt

taking the laplace and simplifying I got I(s) = V(s)/L + Si(s) - v(0)/L all over SR/L + S^2 + 1/LC) which when I inverse laplaced, I got i(t) = -4.5/sqrt75sinsqrt75t + cossqrt75t which is correct

Thanks

Anyone know how to add equations here so it looks better than im doing?

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Here's what I put into maple I did not know about with(inttrans) Thanks, helped

#### Attachments

• maple, inverse laplace.png
5.7 KB · Views: 641