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Math  /  Calculus

QuestionFind the inverse Laplace transform of F(s)=8e−9ss2+49F(s)=\frac{8 e^{-9 s}}{s^{2}+49} f(t)=1f(t)=1 ( Use step (t−c)(t-c) for uc(t)u_{c}(t).)

Studdy Solution

STEP 1

What is this asking? We're trying to find the *inverse Laplace transform* of a given function F(s)F(s), which involves an exponential term and a quadratic in the denominator.
Essentially, we're going from the "s-domain" back to the "t-domain"! Watch out! Don't forget about that exponential term – it introduces a time shift, so we need to be careful with our step function!

STEP 2

1. Rewrite the Function
2. Inverse Laplace Transform

STEP 3

Alright, let's **rewrite** our function F(s)F(s) to make it look a little friendlier.
We can **factor out** the constant in the numerator:
F(s)=8e−9s⋅1s2+49F(s) = 8e^{-9s} \cdot \frac{1}{s^2 + 49}

STEP 4

Now, notice that the second part looks almost like the Laplace transform of sin⁡(at)\sin(at), which is as2+a2\frac{a}{s^2 + a^2}.
In our case, a2=49a^2 = 49, so a=7a = 7.
We're missing the aa in the numerator, so let's **multiply and divide** by **7**:
F(s)=87e−9s⋅7s2+49F(s) = \frac{8}{7}e^{-9s} \cdot \frac{7}{s^2 + 49}Now we're ready to rock and roll with the inverse transform!

STEP 5

Remember that exponential term e−9se^{-9s}?
That's going to introduce a **time shift** of **9** units.
The inverse Laplace transform of e−csF(s)e^{-cs}F(s) is uc(t)f(t−c)u_c(t)f(t-c), where uc(t)u_c(t) is the unit step function (sometimes written as u(t−c)u(t-c) or step(t−c)(t-c)).
In our case, c=9c = 9.

STEP 6

The inverse Laplace transform of 7s2+49\frac{7}{s^2 + 49} is simply sin⁡(7t)\sin(7t).
So, putting it all together, we have:
f(t)=87u9(t)sin⁡(7(t−9))f(t) = \frac{8}{7}u_9(t)\sin(7(t-9))

STEP 7

Let's **clarify** this one last time.
The 87\frac{8}{7} is our **constant multiplier**.
The u9(t)u_9(t) is our **unit step function**, which is **0** for t<9t < 9 and **1** for t≥9t \ge 9.
Finally, sin⁡(7(t−9))\sin(7(t-9)) is our **sine function**, shifted by **9** units to the right.

STEP 8

f(t)=87u9(t)sin⁡(7(t−9))f(t) = \frac{8}{7}u_9(t)\sin(7(t-9))

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