# laplace inverse transformation

So I have have an equation and I need to do laplace inverse transformation $$I(s)=\frac{s}{7s+6}$$ and then for partial fractions I got $$I(s)=1/7-\frac{6}{49s+42}$$ and now for inverse transform I have $$I(t)=\frac{-6}{49}*e^{-(6t/7)}+\frac{1}{7}$$ I can't get the right answers with this so can somebody tell me is the transformation done wrong
or is just my starting values wrong

• everything seems right except $1/7\delta(t)$ will be there instead of only 1/7 Nov 6, 2017 at 16:29

$I(s)=\frac{s}{7s+6}$

Multiply by $\frac{1}{7}\frac{1}{\frac{1}{7}}$

$I(s)=\frac{1}{7}\frac{s}{s+\frac{6}{7}}$

Move out s from the numerator so we got a 1 there so we can use the laplace transform of $\frac{1}{s+a}\mathcal{L}e^{-at}$

$I(s)=s\frac{1}{7}\frac{1}{s+\frac{6}{7}}$

So a factor s in laplace domain is a derivation in time domain, a factor $\frac{1}{7}$ is just a scaling, the last factor is the $e^{-at}$

The answer is the derivative of this expression: $\frac{1}{7}e^{-\frac{6t}{7}}$

I will let you do that step so I don't take all the glory.

Here's a link if you need more help.

• But the LT of $-\frac {6}{49}e^{-\frac{6t}{7}}$ is not $\frac{s}{7s+6}$
– Chu
Nov 7, 2017 at 0:06
• @Chu I assume you didn't click the link for more help. I will copy paste the important part. "Evaluate $e^{-at}$ at point 0 = 1 => $-ae^{-at}+1$". Lo and behold. I got all the glory. Nov 7, 2017 at 0:12
• It depends on whether $i(t)$ has a step at $t=0$; there are two possible solutions if this condition is not specified.
– Chu
Nov 7, 2017 at 0:39
• @Chu Then supply your answer and take all the glory. Nov 7, 2017 at 0:56