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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

f(t) f(t)

f(t)={4,0t<14,1t<4 f(t) = \begin{cases} 4, & 0 \leq t < 1 \\ -4, & 1 \leq t < 4 \end{cases}

f(t+4)=f(t) f(t+4)=f(t)

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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))

This problem has been solved

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