*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.*

It is an exponential, but in this case, we’ll need to factor a 3 out of the denominator before taking the inverse transform. Whenever a numerator is off by a multiplicative constant, as in this case, all we need to do is put the constant that we need in the numerator.

The denominator of the third term appears to be #3 in the table with \(n = 4\). There is currently a 7 in the numerator and we need a \(4! We will just need to remember to take it back out by dividing by the same constant. \[\begin H\left( s \right) & = \frac - \frac \frac\\ & = 19\frac - \frac\frac \frac\frac\end\] So, what did we do here? We factored the 3 out of the denominator of the second term since it can’t be there for the inverse transform and in the third term we factored everything out of the numerator except the 4!

\[\begin G\left( s \right) & = \frac\frac \frac\\ & = \frac\frac \frac\end\] Notice that in the first term we took advantage of the fact that we could get the 2 in the numerator that we needed by factoring the 8.

The inverse transform is then, \[g\left( t \right) = \frac\sin \left( \right) \frac\sinh \left( \right)\] So, probably the best way to identify the transform is by looking at the denominator.

Be careful with negative signs in these problems, it’s very easy to lose track of them.

The second term almost looks like an exponential, except that it’s got a \(3s\) instead of just an \(s\) in the denominator.

The transform becomes, \[\begin F\left( s \right) & = 6\frac \frac\ & = 6\frac \frac\frac\end\] Taking the inverse transform gives, \[f\left( t \right) = 6\cos \left( \right) \frac\sin \left( \right)\] In this case the first term will be a sine once we factor a 3 out of the denominator, while the second term appears to be a hyperbolic sine (#17).

Again, be careful with the difference between these two.

So, with this advice in mind let’s see if we can take some inverse transforms.

From the denominator of the first term it looks like the first term is just a constant.

## Comments Laplace Transform Solved Problems Pdf

## The Laplace Transform - Illinois Institute of Technology

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