Determine the Laplace Transform of the functionf(t)f(t+4)={4,−4,0≤t<11≤t<4=f(t)𝑓(𝑡)={4,0≤𝑡<1−4,1≤𝑡<4𝑓(𝑡+4)=𝑓(𝑡)
Question
Determine the Laplace Transform of the function
Solution
The Laplace Transform of a piecewise function can be determined by breaking the function into its separate parts and finding the Laplace Transform of each part.
Given the function f(t) as:
f(t) = {4, 0 ≤ t < 1 -4, 1 ≤ t < 4
We can write this as:
f(t) = 4 for 0 ≤ t < 1 f(t) = -4 for 1 ≤ t < 4
The Laplace Transform of f(t) is given by:
L{f(t)} = ∫ from 0 to ∞ e^(-st) f(t) dt
We can break this integral up into the two parts of the function:
L{f(t)} = ∫ from 0 to 1 e^(-st) * 4 dt + ∫ from 1 to 4 e^(-st) * -4 dt
Solving these integrals gives:
L{f(t)} = [ -4/s * (e^(-s) - 1) ] + [ 4/s * (e^(-4s) - e^(-s)) ]
Simplifying this gives:
L{f(t)} = 4/s * (1 - e^(-s) + e^(-4s) - e^(-s))
So, the Laplace Transform of the given function is:
L{f(t)} = 4/s * (1 - 2e^(-s) + e^(-4s))
For the function f(t+4), since the Laplace Transform has the property that L{f(t-a)} = e^(-as)F(s) where F(s) is the Laplace Transform of f(t), we can find the Laplace Transform of f(t+4) as:
L{f(t+4)} = e^(-4s) * L{f(t)} = e^(-4s) * 4/s * (1 - 2e^(-s) + e^(-4s))
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