# Laplace transforms of heaviside step functions

• danj303
In summary, we have an initial value problem with a given function fk(t) and a condition for its value. We can convert fk(t) to a Heaviside function and use the Laplace transform to solve the problem. However, when taking the inverse Laplace transform, we must use the linearity property and partial fraction decomposition to get the correct solution.
danj303

## Homework Statement

Consider the initial value problem
y'' + 1/3y' + 4y = fk(t)

with y(0) = y'(0) = 0,

fk(t) = 1/2k for 4 - k < t < 4 + k
0 otherwise

and 0 < k < 4.

(a) Write fk(t) in terms of Heaviside step functions and then solve the initial value problem.

## The Attempt at a Solution

I can convert it to a heaviside function and do the laplace transform and get

fk(t)= 1/2k H(t-(4-k)) - 1/2k H(t+(4-k))

and then taking the laplace transform of the entire equation get

Y(s) = 1/2k(s2+1/3 s + 4) * (e-(4-k)s/s - e-(4+k)s/s)

Im assuming this is corrent. But then how do I take the inverse laplace transform of this?

danj303 said:
I can convert it to a heaviside function and do the laplace transform and get

fk(t)= 1/2k H(t-(4-k)) - 1/2k H(t+(4-k))

I think if you graph this out, you'll find it is not the function you want

$$H(t-4+k)-H(t+4-k)=\left\{\begin{array}{lr}-1 &, t< 4-k \\ 1 &, t>4-k\end{array}\right.$$

and then taking the laplace transform of the entire equation get

Y(s) = 1/2k(s2+1/3 s + 4) * (e-(4-k)s/s - e-(4+k)s/s)

Im assuming this is corrent. But then how do I take the inverse laplace transform of this?

Much like the other problem you posted, you'll want to use the linearity of the inverse LT:

$$\mathcal{L}^{-1}\left[\frac{e^{-(4-k)s}-e^{-(4+k)s}}{2ks\left(s^2+\frac{1}{3}s+4\right)}\right]=\mathcal{L}^{-1}\left[\frac{e^{-(4-k)s}}{2ks\left(s^2+\frac{1}{3}s+4\right)}\right]-\mathcal{L}^{-1}\left[\frac{e^{-(4+k)s}}{2ks\left(s^2+\frac{1}{3}s+4\right)}\right]$$

For each of the two individual inverse LTs, you might want to use the method of partial fraction decomposition

## 1. What is a Heaviside step function?

A Heaviside step function, also known as the unit step function, is a mathematical function that has a value of 0 for all negative inputs and a value of 1 for all positive inputs.

## 2. What is a Laplace transform?

A Laplace transform is a mathematical tool used to convert a function from the time domain to the frequency domain. It is commonly used in engineering and physics to solve differential equations.

## 3. How do you take the Laplace transform of a Heaviside step function?

To take the Laplace transform of a Heaviside step function, you can use the formula: L{u(t-a)} = e^-as/s, where a is the time shift and s is the complex frequency variable. This formula can also be extended to include other constants and variables in the function.

## 4. What are the applications of Laplace transforms of Heaviside step functions?

The applications of Laplace transforms of Heaviside step functions include solving differential equations, analyzing control systems, and studying the behavior of electrical circuits. They are also used in signal processing and image processing.

## 5. Are there any limitations to using Laplace transforms of Heaviside step functions?

One limitation of using Laplace transforms of Heaviside step functions is that they can only be applied to functions with finite and well-defined Laplace transforms. Additionally, they may not be suitable for functions with discontinuities or singularities.

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