Laplace Transformation Help: Simplifying Denominators and Finding Solutions

In summary, the student attempted to solve a KVL equation for io but had difficulty doing so. They then used a WolframAlpha-generated solution.
  • #1
stef6987
32
0
1. Homework Statement [/b]
NruJm.png

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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  • #2
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: )
 
  • #3
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?
 
  • #4
Yup, that's what I found.
 
  • #5
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?
 
  • #6
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?
 
  • #7
I suspect that the 3 should divide the equation, not multiply. And the u(t) should be transformed, too.
 
  • #8
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))
 
  • #9
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:
 

Related to Laplace Transformation Help: Simplifying Denominators and Finding Solutions

1. What is a Laplace transformation?

A Laplace transformation is a mathematical technique used to convert a function of time into a function of complex frequency. It is commonly used in engineering and physics to solve differential equations and analyze systems in the frequency domain.

2. How is a Laplace transformation performed?

A Laplace transformation is performed by integrating the function of time multiplied by a decaying exponential. The integration is performed from 0 to infinity and the resulting function is known as the Laplace transform.

3. What is the purpose of using Laplace transformation?

The purpose of using Laplace transformation is to simplify the analysis of systems by converting them from the time domain to the frequency domain. This allows for easier solving of differential equations and understanding of system behavior.

4. What are the benefits of using Laplace transformation?

Using Laplace transformation can help to solve complex differential equations, analyze system stability and response, and determine the frequency characteristics of a system. It also allows for easy manipulation of equations using algebraic techniques.

5. Are there any limitations to using Laplace transformation?

While Laplace transformation is a powerful tool, it is not always applicable. It can only be used for linear, time-invariant systems and cannot handle systems with discontinuous inputs. It also requires a good understanding of the underlying mathematical concepts and techniques.

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