The Dark Art of Inverting the Transform

The Dark Art of Inverting the Transform

The Dark Art of inverting the transform.

Invert the transform:

(1) (2) (3) (4) (5)

(6) (7) (8)

(9) Solve using laplace transforms:

(10) Invert

(11) Invert


(1) . This one's fairly easy – in most of these problems, the denominator is the key thing to look at first – does said denominator conform to one of the formulas on p. 319? Clearly, it does - #5 and #6 look good. Now we go to the numerator, and we see that we have a constant in our problem, which tells me that #5 should be my choice, as #6 has the variable s in the numerator. Now, note that the denominator should be 're-conceived':


and that in the formula the constant being squared (2) in the denominator has to 'match' the constant in the numerator – thus we have to rewrite using some algebra:

, and

(2) . Again, we begin looking for something resembling the denominator in the list. The only transforms that involve powers other than squares seem to be #4 and #11, and #11 seems like the only one that will work. As the power of the parenthetical expression is a cube, we would take our n to be n = 2, i.e.:

. But note that in the formula, we have to have n! = 2! = 2 on top.

So, rig it up that way: and

(note that the a in the formula was a = 1).

(3) First, when you see a trinomial like this on the bottom, you're in for some partial fraction action:

If there's no variables in the numerator, just pull these newly-found A and B out front of the terms, then invert:

(4) Again, we look to partial fractions:

Now, bring the constants out front and invert the transform:

(5) . The denominator won't factor, and in general we try completing the square to try and use #9 and #10 in such situations:

, Now note again that we have to make things 'match up' with the formulas. Note that #10 requires that the numerator of the expression be the same as the terms in parentheses on the bottom (s – a). That means we need an

s + 1 on top to use the formula (note that #9 doesn't have a variable on top, so we need to look at #10 first). This can be accomplished by simply factoring out 2 (a common trick, though usually requiring more steps afterward as well).

(6) The bottom will factor – let it rip with the partial fraction action:

You know the routine – bust it up with constants out front and invert:

(7) Bottom won't factor – complete the square: and again we're aiming to use #9 and #10, but things are a little more complicated this time – remember that in order to use #10 (which is indicated by the presence of the variable in the numerator) – the numerator has to 'match' the parenthetical expression (s – 1) on the bottom. The only way that it will work is to factor out the lead coefficient 2 and figure out k in this equation:

2s + 1 = 2(s – 1) + k,

2s + 1 = 2s – 2 + k,

k = 3, and:

Now, the first term is good to go with transform #10, but we're going to have to use #9 for the second term – the numerator has to match the loose constant 1 in the denominator – so just pull the three in front and we're good:

(8) We are going to use partial fractions, but one of the factors is an irreducible quadratic, which requires special measures:

Recall that here, you need to just pick a couple of s's and generate a system of equations for B and C:

Alright, now we need to make this 'fit' some transforms from the list – the first term is easy w/ #1 (just pull the 3 in front). The denominator of the second term indicates that we should try using #5 and #6. All told, she goes:

note how I 'rigged' the last term, since the numerator has to be the square root of the constant in the denominator to use #5. Now, invert it:

Ok, let's solve a problem from scratch now:

Laplace both sides:

Factor out the Y(s), move the trash over to the right, then solve for Y(s) :

Now, use partial fractions on the right:

Finally, invert the transform to obtain the solution to the initial-value DE:

Factoring, solving, partial fractions – all that old math come back to haunt you.


(14) Ok – I'm going to temporarily ignore the exponential term, and do partial fractions on the rest:

Now, I need to invert the transform using Theorem 6.3.1. Note that the 'c' in the theorem is the 'translation constant' (yes, I made that up), the coefficient on s in the coefficient, and here c = 2. 1/(s + 2) and 1/(s – 1) are normally the translates of and , respectively, but here we have to translate them by c = 2, incorporating the step function as per the Theorem:

Note how the translation took place – we subtracted the c = 2 from both the exponents (instances of t generally), and inserted the pulsing function u2(t).

(10). Alright, when we break off the exponential, the denominator won't factor, so we complete the square:

Note how the parenthetical expression in the numerator matches the one in the denominator – good news, as we're ready to rip with #10 from our list – it would normally invert to , but we have to translate it by c = 3. No problema:

Note that I translated both the instances of t – in the exponential and cos as well.

(11) Invert . I'm going to wing it a little: factor out the exponential, use partial fractions, and try to figure out what's going on.

At this point, you have to be a little careful, I recognize that and are the translates of e-2t and et, respectively, but I have to remember that when I invert back to a step function (again, the tipoff that I'm going to do that is the presence of the e-cs, where

c = 2 here), that I have to 'translate' by c:

y(t) =


translate the inverse transforms by c= 2