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
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\$\begingroup\$ everything seems right except \$1/7\delta(t)\$ will be there instead of only 1/7 \$\endgroup\$– RohitNov 6, 2017 at 16:29
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1 Answer
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\$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.
answered Nov 6, 2017 at 16:41
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\$\begingroup\$ But the LT of \$-\frac {6}{49}e^{-\frac{6t}{7}}\$ is not \$\frac{s}{7s+6}\$ \$\endgroup\$– ChuNov 7, 2017 at 0:06
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\$\begingroup\$ @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. \$\endgroup\$ Nov 7, 2017 at 0:12
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\$\begingroup\$ It depends on whether \$i(t)\$ has a step at \$t=0\$; there are two possible solutions if this condition is not specified. \$\endgroup\$– ChuNov 7, 2017 at 0:39
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\$\begingroup\$ @Chu Then supply your answer and take all the glory. \$\endgroup\$ Nov 7, 2017 at 0:56