Find the Laplace transform of tU(t -9).
Answers
Answer:e^-9s/s
Step-by-step explanation:
In this required question we will have a area under the curve t-9 from the given range 0 to infinity. Now we will find a discrete value from where the value of the area under the curve will change from 0 to 1 which will be 9 in this case.
As we see if they value is greater than 9 it will be 0 else it'll be 1.
Hence the range breakup will be
Laplace of u(t-9) =
integral of { e^-st*u(t-9)dt} from 0 to infinity.
Now this will be broken as integral ranging from 0 to 9 and then 9 to infinity.
Since integral ranging from 0 to 9 is zero we will just consider
Integral of { e^-st*u(t-9)dt} from 9 to infinity
This will be equal to [e^-st/-s]from 9 to infinity
This will be after putting values
e^-9s/s
Answer:
Step-by-step explanation:
