Laplace transform and partial fractions

[Mechaman]

I am trying to set up this Laplace transform question for the next step, partial fractions. What I'm left with doesn't look ready to apply partial fractions to yet?

 

[ksdhart2]

Assuming that your notation means \( Y = \mathcal{L} \left\{ y(t) \right\} \) then what you've done so far is good, and you're on the right track to think of partial fraction decomposition for the first fraction. You'll want to find an appropriate \(A\) and \(B\) so that:

\(\displaystyle \frac{1}{s(s+7)} = \frac{A}{s} + \frac{B}{s + 7}\)

\(\displaystyle \frac{1}{s(s+7)} = \frac{A(s + 7) + Bs}{s(s+7)} \implies A(s + 7) + Bs = 1\)

This must be true for every value of \(s\) so we can make it easier on ourselves by testing for simple value, say \(s = 0\):

\(\displaystyle A(0 + 7) + B(0) = 1 \implies 7A = 1 \implies A = \frac{1}{7}\)

Then back substitute and discover:

\(\displaystyle 1 = \frac{1}{7}(s + 7) + Bs \implies Bs + \frac{s}{7} = 0 \implies B = -\frac{1}{7}\)

Therefore:

\(\displaystyle \frac{1}{s(s+7)} = \frac{1}{7s} - \frac{1}{7(s + 7)}\)

And so:

\(\displaystyle Y = \frac{1}{7s} - \frac{1}{7(s + 7)} + \frac{3}{s+7}\)

\(\displaystyle y(t) = \mathcal{L}^{-1} \left( \frac{1}{7s} - \frac{1}{7(s + 7)} + \frac{3}{s+7} \right)\)

I trust you can finish up from here?

 

[Deleted member 4993]

Assuming that your notation means \( Y = \mathcal{L} \left\{ y(t) \right\} \) then what you've done so far is good, and you're on the right track to think of partial fraction decomposition for the first fraction. You'll want to find an appropriate \(A\) and \(B\) so that:

\(\displaystyle \frac{1}{s(s+7)} = \frac{A}{s} + \frac{B}{s + 7}\)

\(\displaystyle \frac{1}{s(s+7)} = \frac{A(s + 7) + Bs}{s(s+7)} \implies A(s + 7) + Bs = 1\)

This must be true for every value of \(s\) so we can make it easier on ourselves by testing for simple value, say \(s = 0\):

\(\displaystyle A(0 + 7) + B(0) = 1 \implies 7A = 1 \implies A = \frac{1}{7}\)

Then back substitute and discover:

\(\displaystyle 1 = \frac{1}{7}(s + 7) + Bs \implies Bs + \frac{s}{7} = 0 \implies B = -\frac{1}{7}\)

Therefore:

\(\displaystyle \frac{1}{s(s+7)} = \frac{1}{7s} - \frac{1}{7(s + 7)}\)

And so:

\(\displaystyle Y = \frac{1}{7s} - \frac{1}{7(s + 7)} + \frac{3}{s+7}\)

\(\displaystyle y(t) = \mathcal{L}^{-1} \left( \frac{1}{7s} - \frac{1}{7(s + 7)} + \frac{3}{s+7} \right)\)

I trust you can finish up from here?

A quicker way to calculate the "partial fraction" coefficients:

\(\displaystyle \frac{1}{s(s+7)} = \frac{A}{s} + \frac{B}{s + 7}\)

Multiply both sides by 's':

\(\displaystyle \frac{1}{(s+7)} = A + \frac{B * s}{s + 7}\)

Calculate \(\displaystyle \lim_{s \to 0}\) of both sides:

1/7 = A

Now - again

\(\displaystyle \frac{1}{s(s+7)} = \frac{A}{s} + \frac{B}{s + 7}\)

Multiply both sides by '(s+7)':

\(\displaystyle \frac{1}{s} = B + \frac{A * (s+7)}{s}\)

Calculate \(\displaystyle \lim_{s \to {-7}}\) of both sides:

-1/7 = B

Same answer as Ksd - just a bit quicker!

 

[tkhunny]

This must be true for every value of \(s\) so we can make it easier on ourselves by testing for simple value, say \(s = 0\):

Never been a fan of this. s = 0 is not in the Domain. The method works fine. I just find it fundamentally unsatisfying.

 

[Mechaman]

Thanks for the responses. I understand a bit better now!