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Both of the terms will also need to have their numerators fixed up.Here is the transform once we’re done rewriting it.If there is only one entry in the table that has that particular denominator, the next step is to make sure the numerator is correctly set up for the inverse transform process.
If there is more than one entry in the table that has a particular denominator, then the numerators of each will be different, so go up to the numerator and see which one you’ve got.
If you need to correct the numerator to get it into the correct form and then take the inverse transform.
The third term also appears to be an exponential, only this time \(a = 3\) and we’ll need to factor the 4 out before taking the inverse transforms.
So, with a little more detail than we’ll usually put into these, \[\begin F\left( s \right) & = 6\,\,\frac - \frac 4\frac\\ & f\left( t \right) = 6\left( 1 \right) - 4\left( \right)\\ & = 6 - 4\end\] The first term in this case looks like an exponential with \(a = - 2\) and we’ll need to factor out the 19.
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The first one has an \(s\) in the numerator and so this means that the first term must be #8 and we’ll need to factor the 6 out of the numerator in this case.
The second term has only a constant in the numerator and so this term must be #7, however, in order for this to be exactly #7 we’ll need to multiply/divide a 5 in the numerator to get it correct for the table.
If there is more than one possibility use the numerator to identify the correct one. These are a little more involved than the first set.
Fix up the numerator if needed to get it into the form needed for the inverse transform process. From the denominator of this one it appears that it is either a sine or a cosine.