My homework question is to find the Laplace transform of the function f(t)=e^(10-2t)*step(t-5).
I am not sure how to handle the e^(10-2t) portion of the equation.
Thanks in advance.
e^(10-2t)= e^10 * e^-2 t
so
e^10 times transform of e^-2t u(t-5)
then to continue
given f(t) = e^at g(t)
then
F(s) = (1/a) G(s/a)
so....
e^10 (e^-2t) (1/s+5)?
or is it:
(e^10)(e^-2t)((e^-5t)/s) = (e^(10-7t))/s?
To find the Laplace transform of the given function f(t) = e^(10-2t) * step(t-5), you can break it down into two parts: e^(10-2t) and step(t-5).
The Laplace transform of e^(at) * F(t) is given by F(s-a), where F(s) is the Laplace transform of F(t). In this case, you need to find the Laplace transform of e^(10-2t).
Now, the Laplace transform of e^(at) is 1 / (s-a), where s is the complex frequency parameter. Thus, the Laplace transform of e^(10-2t) can be found by replacing 's' with '10-2t':
L[e^(10-2t)] = 1 / (s - (10-2t))
However, the step(t-5) function means that the time starts from 5 and the function is zero for t < 5. To consider this in the Laplace transform, you can multiply the Laplace transform by e^(-5s) to shift the time starting point:
L[e^(10-2t) * step(t-5)] = e^(-5s) * L[e^(10-2t)]
Now, substitute 's' with '10-2t' in the Laplace transform expression:
L[e^(10-2t) * step(t-5)] = e^(-5s) * 1 / ((10-2t) - (10-2t))
Simplifying this gives:
L[e^(10-2t) * step(t-5)] = e^(-5s)
So, the Laplace transform of the given function f(t) = e^(10-2t) * step(t-5) is e^(-5s).