Differential Equations - Laplace Transforms: Step Functions

Differential Equations - Laplace Transforms: Step Functions - ii

Example 3 Find the inverse Laplace transform of each of the following.

$H (s) = \frac{s e^{- 4 s}}{(3 s + 2) (s - 2)}$
$G (s) = \frac{5 e^{- 6 s} - 3 e^{- 11 s}}{(s + 2) (s^{2} + 9)}$
$F (s) = \frac{4 s + e^{- s}}{(s - 1) (s + 2)}$
$G (s) = \frac{3 s + 8 e^{- 20 s} - 2 s e^{- 3 s} + 6 e^{- 7 s}}{s^{2} (s + 3)}$

All of these will use

(3)

somewhere in the process. Notice that in order to use this formula the exponential doesn’t really enter into the mix until the very end. The vast majority of the process is finding the inverse transform of the stuff without the exponential.
In these problems we are not going to go into detail on many of the inverse transforms. If you need a refresher on some of the basics of inverse transforms go back and take a look at the previous section.

H (s) = \frac{s e^{- 4 s}}{(3 s + 2) (s - 2)}

In light of the comments above let’s first rewrite the transform in the following way.

H (s) = e^{- 4 s} \frac{s}{(3 s + 2) (s - 2)} = e^{- 4 s} F (s)

Now, this problem really comes down to needing

f (t)

. So, let’s do that. We’ll need to partial fraction

F (s)

up. Here’s the partial fraction decomposition.

F (s) = \frac{A}{3 s + 2} + \frac{B}{s - 2}

Setting numerators equal gives,

s = A (s - 2) + B (3 s + 2)

We’ll find the constants here by selecting values of

s

. Doing this gives,

\begin{aligned} s = 2 & 2 & = 8 B & \Rightarrow B & = \frac{1}{4} \\ s = - \frac{2}{3} & - \frac{2}{3} & = - \frac{8}{3} A & \Rightarrow A & = \frac{1}{4} \end{aligned}

So, the partial fraction decomposition becomes,

F (s) = \frac{\frac{1}{4}}{3 (s + \frac{2}{3})} + \frac{\frac{1}{4}}{s - 2}

Notice that we factored a 3 out of the denominator in order to actually do the inverse transform. The inverse transform of this is then,

f (t) = \frac{1}{12} e^{- \frac{2 t}{3}} + \frac{1}{4} e^{2 t}

Now, let’s go back and do the actual problem. The original transform was,

H (s) = e^{- 4 s} F (s)

Note that we didn’t bother to plug in

F (s)

. There really isn’t a reason to plug it back in. Let’s just use

(3)

to write down the inverse transform in terms of symbols. The inverse transform is,

h (t) = u_{4} (t) f (t - 4)

where,

f (t)

is,

f (t) = \frac{1}{12} e^{- \frac{2 t}{3}} + \frac{1}{4} e^{2 t}

This is all the farther that we’ll go with the answer. There really isn’t any reason to plug in

f (t)

at this point. It would make the function longer and definitely messier. We will give almost all of our answers to these types of inverse transforms in this form.

G (s) = \frac{5 e^{- 6 s} - 3 e^{- 11 s}}{(s + 2) (s^{2} + 9)}

This problem is not as difficult as it might at first appear to be. Because there are two exponentials we will need to deal with them separately eventually. Now, this might lead us to conclude that the best way to deal with this function is to split it up as follows,

G (s) = e^{- 6 s} \frac{5}{(s + 2) (s^{2} + 9)} - e^{- 11 s} \frac{3}{(s + 2) (s^{2} + 9)}

Notice that we factored out the exponential, as we did in the last example, since we would need to do that eventually anyway. This is where a fairly common complication arises. Many people will call the first function

F (s)

and the second function

H (s)

and then partial fraction both of them.
However, if instead of just factoring out the exponential we would also factor out the coefficient we would get,

G (s) = 5 e^{- 6 s} \frac{1}{(s + 2) (s^{2} + 9)} - 3 e^{- 11 s} \frac{1}{(s + 2) (s^{2} + 9)}

Upon doing this we can see that the two functions are in fact the same function. The only difference is the constant that was in the numerator. So, the way that we’ll do these problems is to first notice that both of the exponentials have only constants as coefficients. Instead of breaking things up then, we will simply factor out the whole numerator and get,

G (s) = (5 e^{- 6 s} - 3 e^{- 11 s}) \frac{1}{(s + 2) (s^{2} + 9)} = (5 e^{- 6 s} - 3 e^{- 11 s}) F (s)

and now we will just partial fraction

F (s)

.
Here is the partial fraction decomposition.

F (s) = \frac{A}{s + 2} + \frac{B s + C}{s^{2} + 9}

Setting numerators equal and combining gives us,

\begin{aligned} 1 & = A (s^{2} + 9) + (s + 2) (B s + C) \\ = (A + B) s^{2} + (2 B + C) s + 9 A + 2 C \end{aligned}

Setting coefficient equal and solving gives,

\begin{aligned} s^{2} : & A + B & = 0 \\ s^{1} : & 2 B + C & = 0 \\ s^{0} : & 9 A + 2 C & = 1 \end{aligned}} \Rightarrow A = \frac{1}{13}, B = - \frac{1}{13}, C = \frac{2}{13}

Substituting back into the transform gives and fixing up the numerators as needed gives,

\begin{aligned} F (s) & = \frac{1}{13} (\frac{1}{s + 2} + \frac{- s + 2}{s^{2} + 9}) \\ = \frac{1}{13} (\frac{1}{s + 2} - \frac{s}{s^{2} + 9} + \frac{2 \frac{3}{3}}{s^{2} + 9}) \end{aligned}

As we did in the previous section we factored out the common denominator to make our work a little simpler. Taking the inverse transform then gives,

f (t) = \frac{1}{13} (e^{- 2 t} - \cos (3 t) + \frac{2}{3} \sin (3 t))

At this point we can go back and start thinking about the original problem.

\begin{aligned} G (s) & = (5 e^{- 6 s} - 3 e^{- 11 s}) F (s) \\ = 5 e^{- 6 s} F (s) - 3 e^{- 11 s} F (s) \end{aligned}

We’ll also need to distribute the

F (s)

through as well in order to get the correct inverse transform. Recall that in order to use

(3)

to take the inverse transform you must have a single exponential times a single transform. This means that we must multiply the

F (s)

through the parenthesis. We can now take the inverse transform,

g (t) = 5 u_{6} (t) f (t - 6) - 3 u_{11} (t) f (t - 11)

where,

f (t) = \frac{1}{13} (e^{- 2 t} - \cos (3 t) + \frac{2}{3} \sin (3 t))

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