Question

Find the inverse Laplace transform of the given function. $$ F(s)=\frac{3 !}{(s-2)^{4}} $$

Short answer

Answer: The inverse Laplace transform of the given function is f(t) = e^(2t) (d^3/dt^3) δ(t).

Step-by-step solution

Identify the given function

The given function is: $$ F(s)=\frac{3 !}{(s-2)^{4}} $$

Shift property of Laplace transform

Recall that the shift property of the Laplace transform is given by: $$ \Laplace^{-1}\{f(s-a)\} = e^{at}f(t) $$ We will use this property to adjust the given function in the standard form.

Find the inverse Laplace transform of the standard form

Since we are given a factorial in the numerator, we will use the following formula for the inverse Laplace transform of derivatives: $$ \Laplace^{-1}\{\frac{n!}{s^{n+1}}\} = \frac{d^n}{dt^n} \delta (t) $$ In our case, n = 3. So, $$ \Laplace^{-1}\{\frac{3!}{s^{4}}\} = \frac{d^3}{dt^3} \delta (t) $$

Apply the shift property

Now, we apply the shift property to our standard form inverse Laplace transform: $$ \Laplace^{-1}\{\frac{3!}{(s-2)^{4}}\} = e^{2t} \frac{d^3}{dt^3} \delta (t) $$

Final answer

The inverse Laplace transform of the given function is: $$ f(t) = e^{2t} \frac{d^3}{dt^3} \delta (t) $$

Key concepts

Shift Property of Laplace Transform

When working with Laplace transforms, one important tool is the shift property. This property is very useful when you need to find the inverse Laplace transform of functions that involve a shifted parameter, like \((s-a)\). The shift property allows us to transform a shifted function \(F(s-a)\) back to the time domain by including an exponential term: \(e^{at}\).

• The key idea here is that if we have \(F(s-a)\), the inverse transform will be \(e^{at} f(t)\) where \(f(t)\) is the inverse transform of \(F(s)\).
• This makes it possible to handle shifts in the 's' variable effectively, allowing for simpler computation when dealing with problems like \(F(s)= \frac{3!}{(s-2)^4}\).

Understanding this property is crucial, especially in cases like the one illustrated in the exercise. After adjusting to the shift, we can find the inverse Laplace transform of the unshifted function and then apply the exponential adjustment to finalize our answer.

Standard Form in Laplace Transform

The standard form in Laplace transforms refers to expressions that are easy to work with, typically of the form \(\frac{n!}{s^{n+1}}\). When you see a function like \(\frac{3!}{s^4}\), this is a signal that you're ready to use standard inverse formulas.
Here's why the standard form is handy:

• The expression \(\frac{3!}{s^4}\) corresponds closely to a well-known inverse formula.
• Once in this form, it's straightforward to use the inverse Laplace formula which expresses these in terms of derivatives of the Dirac delta function: \(\Laplace^{-1}\{\frac{n!}{s^{n+1}}\} = \frac{d^n}{dt^n} \delta (t)\).

In exercises featuring factorials in the numerator, like this one, converting to standard form simplifies calculating the inverse transform. Remember, this concept is all about getting your given Laplace transform to match a known inverse transform that you can easily handle.

Derivatives in Laplace Transform

In Laplace transforms, derivatives play a significant role in linking the function in the complex frequency domain back to the time domain. Understanding how to deal with derivatives in Laplace transforms is key to solving many problems effectively.
Here’s how derivatives factor into what we learned:

• With expressions such as \(\frac{3!}{(s-2)^4}\), recognizing the derivative connection in the inverse transform is important: \(\Laplace^{-1}\{\frac{n!}{s^{n+1}}\} = \frac{d^n}{dt^n} \delta(t)\).
• In the exercise context, using the derivative means you're effectively calculating the \(n\)-th derivative of the Dirac delta function \(\delta(t)\), which is a special function that "picks out" values even when very complex transformations are involved.

Thus, knowing how derivatives are applied in Laplace transforms allows you to convert complex algebraic forms back into time functions that provide useful real-world insights. This is critical for sound analysis, particularly in engineering and physics applications.