# Unit step functions to find Laplace Transform

• izen
In summary, the problem involves finding the Laplace Transform of a function using unit step functions. The function is f(t) = cos(pi*t) if 1 <= t < 4 and 0 elsewhere. The unit step function f(t) = cos(pi*t) u(t-1) - cos(pi*t) u(t-4) is used to apply the second shifting theorem, but a mistake is made in the process. Instead, the original form of the function can be used to find the Laplace Transform. The final step involves using the trig identity cos(x+y) = cos(x)cos(y) - sin(x)sin(y).
izen

## Homework Statement

f(t) = cos (pi*t) if 1$\leq$ t <4 and 0 elsewhere

using unit step functions to find Laplace Transform

## The Attempt at a Solution

I came up with the unit step function f(t) = cos(pi*t) u(t-1) - cos(pi*t) u(t-4)
in order to use the second shifting theorem f(t) must in the format of f(t-a) in this case a = 1 and 4.

I add 1 and subtract 1 , add 4 and subtract 4 and I get

f(t) = cos(pi(t-1+1) u(t-1) - cos(pi(t-4+4) u(t-4)
= cos (pi(t-1)) u(t-1) +u(t-1) - cos(pi*(t-4) u(t-4)+u(t-4) << i think this step is not correct
It should get -cos (pi(t-1)) u(t-1) - cos(pi*(t-4) u(t-4) but i don't know how to get there

izen said:
I came up with the unit step function f(t) = cos(pi*t) u(t-1) - cos(pi*t) u(t-4)
in order to use the second shifting theorem f(t) must in the format of f(t-a) in this case a = 1 and 4.

I add 1 and subtract 1 , add 4 and subtract 4 and I get

f(t) = cos(pi(t-1+1) u(t-1) - cos(pi(t-4+4) u(t-4)
= cos (pi(t-1)) u(t-1) +u(t-1) - cos(pi*(t-4) u(t-4)+u(t-4) << i think this step is not correct
It is not correct. Instead, let's go back to the previous step:
$$f(t) = \cos(\pi(t-1+1)) u(t-1) - \cos(\pi(t-4+4)) u(t-4)$$
I can write this as follows:
$$f(t) = \cos(\pi(t-1)+\pi) u(t-1) - \cos(\pi(t-4)+4\pi) u(t-4)$$
Now what can you say in general about ##\cos(x + \pi)##? What about ##\cos(x + 4\pi)##?

izen said:

## Homework Statement

f(t) = cos (pi*t) if 1$\leq$ t <4 and 0 elsewhere

using unit step functions to find Laplace Transform

## The Attempt at a Solution

I came up with the unit step function f(t) = cos(pi*t) u(t-1) - cos(pi*t) u(t-4)
in order to use the second shifting theorem f(t) must in the format of f(t-a) in this case a = 1 and 4.

I add 1 and subtract 1 , add 4 and subtract 4 and I get

f(t) = cos(pi(t-1+1) u(t-1) - cos(pi(t-4+4) u(t-4)
= cos (pi(t-1)) u(t-1) +u(t-1) - cos(pi*(t-4) u(t-4)+u(t-4) << i think this step is not correct
It should get -cos (pi(t-1)) u(t-1) - cos(pi*(t-4) u(t-4) but i don't know how to get there

Does the problem statement say you MUST use the unit step function? If not, just using the original form is as easy a way as any:
$$\cal{L}(f)(s) = \int_0^4 e^{-st} \cos(\pi t) \, dt.$$

Thank you jbunniii I got it After that we use the trig identity of cos(x+y)=cos(x)-cos(y)-sin(x)sin(y)
Thank you Ray Vickson for the tip

## 1. What is a unit step function?

A 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. It is commonly denoted as u(t) or θ(t).

## 2. How is a unit step function used to find Laplace Transform?

The unit step function is used in the definition of Laplace Transform to indicate the starting point of the function. It allows us to calculate the integral from the starting point to infinity, making it easier to find the Laplace Transform.

## 3. Can a unit step function have different starting points?

Yes, a unit step function can have different starting points, denoted as u(t-a). This means that the function will have a value of 0 for all inputs less than a, and a value of 1 for all inputs greater than or equal to a.

## 4. How do you convert a unit step function to a piecewise function?

To convert a unit step function to a piecewise function, you can use the definition of the unit step function and write it as a piecewise function. For example, u(t) = 0 for t < 0 and u(t) = 1 for t ≥ 0.

## 5. What is the inverse Laplace Transform of a unit step function?

The inverse Laplace Transform of a unit step function u(t-a) is given by e^(-as), where s is the variable in the Laplace domain. This means that the function will have a jump of magnitude 1 at the point a in the time domain.

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