Calculating Inverse Laplace Using Convolutions

• ColdFusion85
In summary, the task is to use the convolution theorem to compute the inverse Laplace transform of the given function. Using a table, it is found that the inverse Laplace transform of 1/(s^2+16) is 1/4*sin(4t) and the inverse Laplace transform of e^(-2s) is the Dirac delta function, delta(t-2). The resulting integral is evaluated using the properties of the delta function, where the function f(a) is substituted for the delta function when the limits of integration include x=a.
ColdFusion85

Homework Statement

We are to use the convolution theorem to compute the inverse Laplace transform of the function

$$L=\frac{1}{s^2 + 16}e^{-2s}$$

Homework Equations

Using a table, I find that $$L^{-1}[\frac{1}{s^2 + 16}] = \frac{1}{4}sin(4t)$$
and
$$L^{-1}[e^{-2s}] = \delta(t-2)$$

The Attempt at a Solution

Using the convolution theorem, $$L^{-1}[\frac{1}{s^2 + 16}e^{-2s}] = \int_{0}^{t} \frac{1}{4}sin(4\tau)\delta(\tau-2)d\tau$$

My question is, how do I evaluate that integral? I know that when you have the delta function, e.g. $$\delta(t-2)$$ times a function f(x), and the limits are negative infinity to infinity, it is just f(2). But what if the limits are 0 to t?

Can anyone help?

$$\int f(x)\delta(x-a) dx= f(a)$$
as long as the interval of integration includes x= a and is 0 if it doesn't. It doesn't have to be from $-\infty$ to $\infty$

In particular, with a variable upper limit (and assuming a> x0),
$$\int_{x_0}^t f(x)\delta(x-a) dx$$
is the function that is 0 if t< a, f(a) if t$\ge$ a.

1. What is the purpose of calculating inverse Laplace using convolutions?

The purpose of calculating inverse Laplace using convolutions is to find the original function in the time domain that corresponds to a given Laplace transform in the frequency domain. This is useful in solving differential equations and analyzing systems in engineering and physics.

2. How does convolution help in calculating inverse Laplace?

Convolution is a mathematical operation that combines two functions to produce a third function. In the context of calculating inverse Laplace, convolution allows us to break down a complex Laplace transform into simpler components that can be more easily inverted. This makes the process of finding the inverse Laplace transform more manageable.

3. What is the general formula for calculating inverse Laplace using convolutions?

The general formula for calculating inverse Laplace using convolutions is:

f(t) = L-1{F(s)} = ∫0t g(t-u)h(u)du

Where f(t) is the original function, g(t) is the Laplace transform of f(t), and h(t) is the inverse Laplace transform of g(t).

4. Can any Laplace transform be inverted using convolutions?

No, not all Laplace transforms can be inverted using convolutions. The function must meet certain conditions, such as being continuous and having a finite number of discontinuities, for the inverse Laplace transform to exist. Additionally, some Laplace transforms may require more advanced methods for inversion.

5. Are there any limitations to using convolutions for calculating inverse Laplace?

One limitation of using convolutions for calculating inverse Laplace is that it may not always be the most efficient method. In some cases, using partial fractions or other techniques may be faster and more straightforward. Additionally, convolutions may become more complex and time-consuming for more complicated Laplace transforms.

Similar threads

• Calculus and Beyond Homework Help
Replies
1
Views
423
• Calculus and Beyond Homework Help
Replies
10
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
616
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
495
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
407
• Calculus and Beyond Homework Help
Replies
1
Views
857
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
2K