Laplace Transformation Help: Simplifying Denominators and Finding Solutions

AI Thread Summary
The discussion revolves around finding the current i0(t) using Laplace transformations and KVL equations. The user initially derived the expression i0 = (s + 4)/(3s^2 + 4s + 1) but struggled to match it to a desired form for inverse transformation. After revising their equations and confirming the results, they successfully simplified the expression and found the inverse Laplace transform. The final solution for i(t) was determined to be 1 - (2/3)e^(-4/3)t, confirming the correct application of the transformation principles. The conversation highlights the importance of careful algebraic manipulation and understanding the transition from the Laplace domain back to the time domain.
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1. Homework Statement [/b]
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I'm required to find the current i0(t), i wrote KVL equations for each loop, I0 can be expressed interms of i1 and i3, i3 being the current source on the right. i managed to get the solution down to the following :

i0 = (s + 4)/(3*s^2 + 4*s + 1)

I simplified the denominator by completing the square and got:

i0 = (s + 4)/((3*(s+4/6)^2 - 12/36))

Now I'm kind of stuck, i one of solutions is in the form (s+a)/((s+a)^2 + w^2) I'm not sure how i can get my solution to be similar to this, any tips would be awesome :)
thankyou!



Homework Equations





The Attempt at a Solution

 
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Can you show your initial loop equations? I'm not seeing the same result that you have for io.

(Try using the x2 and x2 icons in the edit panel header to produce exponents and subscripts :wink: )
 
hmm i think i made a few mistakes, i tried it again and this is what i get :

kvl 1:
-4/s + i1 + i1/s - i0/s
=> 4/s = i1 + i1/s - i0/s

kvl 2:
2io + io - 1/(s+1) + i0/s - i1/s = 0

Now i rearranged kvl 2 to find i1 and i got:
i1/s = 3i0 - 1/(s+1) + i0/s
= i0(3 + 1/s) - 1/(s+1)
i1 = s(i0(3+1/s) - 1(s+1)) = i0(3s + 1) - s/(s+1)

so how i got my original answer is a bit of a mystery haha. I tried it again and substituted i1 into kvl 1 and got the answer from wolphram alpha

(s+4)/(3s2 + 4s)

Is this what you got?
 
Yup, that's what I found.
 
Ok, now that i got that right :P i need to convert it back to the time domain, but no solution in my list seems to match. any ideas?
 
ahh, the answer was so simple, i used the inverse laplace transformation

the equation is in the form:

s+4 = A/s + B/(3s+4)
A = 1
B = -2

1/s = u(t)
B/(3(s+4/3) = -2e-(4/3)t*(3)

i'm not to sure do i just multiply the 3 to the equation?
 
I suspect that the 3 should divide the equation, not multiply. And the u(t) should be transformed, too.
 
well A = 1, 1/s = u(t) so what else could i do to that?
my final solution was:

1/u(t) - (2/3)*e-(4/3)t/(s+(4/3))
 
There should be no s's in the time domain result, only t's and constants. If time is assumed to begin at t=0 for the circuit, the u(t) becomes a constant 1.
 
  • #10
Oh ok i think i follow, i realized i had a small mistake (not sure why i had the extra s ahah) it should be:

u(t) - (2/3)*e-(4/3)t

i need to find I for t>0, so at t>0 u(t) = 1, is that what you're implying?

so my answer should be:

1 - (2/3)*e-(4/3)t
 
  • #11
That looks good to me :smile:
 

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