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### Course: Differential equations>Unit 3

Lesson 2: Properties of the Laplace transform

# Inverse Laplace examples

Using our toolkit to take some inverse Laplace Transforms. Created by Sal Khan.

## Want to join the conversation?

• Mr. Sal,
I tried to solve the same problem using MATLAB and I got the following answer:
2*cos(t - 2)*heaviside(t - 2)*exp(t - 2)
where the heaviside is the unit step function. Would you explain the reason how MATLAB did it with (t-2) in the unit step function?
• Also notice how Sal has the answer as u₂(t) with a sub-scripted 2.
This is the same as "the step function starting from two".
Which is also the same as "shifting the step function two to the right" (the step function starts at 0), or u(t-2). Notice no subscripts with this answer.
• How can I do the inverse laplace transform of 1/(s-a)?
(1 vote)
• That's 1/s shifted by a. And you can shift by a by multiplying your function f(t) with e^-at. Since L{1} = 1/s and therefore f(t) = 1, L{ e^-at*f(t) } = 1/(s-a). Of course if s is equal or greater than a.
• At why not cancel out (s-1) from numerator & denominator before we continue? (which would leave ( 2 e^-2s ) / s )
• You can only cancel factors if they are actually factors both in the numerator and in the denominator.

In this case `(s-1)` is a factor of the numerator (it's multiplying the whole numerator), but in the denominator you have `(s-1)² + 1`; since you have 2 terms on the denominator, you don't have any factors that you could cancel out.
• Was the fact that there was a 2 constant, and also an e^(-2s) just a coincidence that they were both 2's?
• Yes it is a coincidence. The 2 constant in the question is the 2 constant in the answer. The 2s from u2(t)*e^(t-2)*cos(t-2) is a result of the 2 in e^(-2s).
• what is the inverse laplace of 1 ?
L^-1{1}
• The inverse laplace of 1 is the dirac delta function d(t). The inverse laplace transform of any number (K) is K*d(t).
• How did it become e^(t-2) ?
(1 vote)
• Okay, so in the video, Sal establishes that f(t) = e^t * cos(t).

Whenever you change or shift the variable in the "f" portion of the equation, you equivalently change the other side. Do you remember, back in algebra/algebra II that you might be given something like "f(x) = 2x^2 + 3x, find f(2x+3)"? Those kinds of things show what Sal is doing here. We know f(t), but we want to find f(t-2): therefore, we substitute in t-2 for every t in the f(t) function.
• If I understand correctly,
ℒ{ μ_c(t) f(t-c) } = e^(cs) F(s)
Shifting by f(t) to the right by c in the time domain shrinks F in the frequency domain by e^(-cs)
ℒ{ e^(at) f(t)} = F(s-a)
Multiplying f(t) by e^(at) in the time domain shifts F in the frequency domain to the right by a, resulting in F(s-a)
(1 vote)
• I think you have a typo in second line. Should be e^(-cs) F(s), but other than that I'd say you summed it up well.
(1 vote)
• Hi to the khan academy team, would you please practice section in this lesson?
(1 vote)
• Isn't there a 1/s factor missing in the original equiation for the time domain answer to contain a step function? I believe that a lone e^-ts factor is an indication of the impulse function not the step function.
(1 vote)
• Why do these inverses work so well with s^(-n) but not with s^(n)?
(1 vote)
• Because for functions that are polynomials, the Laplace transform function, F(s), has the variable ("s") part in the denominator, which yields s^(-n). However, there's no restriction on whether we have/use "+n" or "-n" so just make sure you pay attention to your (-) signs!
(1 vote)