Unit step functions to find Laplace Transform

[izen]

Homework Statement

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

using unit step functions to find Laplace Transform

Homework Equations

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

please help thank you

 

[jbunniii]

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 $\cos(x + 4\pi)$?

 

[Ray Vickson]

Homework Statement

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

using unit step functions to find Laplace Transform

Homework Equations

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

please help thank you

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.

 

[izen]

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