. Laplace Transform of Periodic Functions If satisfies for all , where is some fixed positive number, then is called a periodic function with period .
The Correct Answer and Explanation is:
The problem requires us to prove the formula for the Laplace Transform of a periodic function. A function f(t) is periodic with a period T if f(t + T) = f(t) for all t ≥ 0. The formula to be proven is:
L{f(t)} = [∫₀ᵀ e⁻ˢᵗ f(t) dt] / [1 – e⁻ˢᵀ]
Here is the step by step proof and explanation.
The proof begins with the fundamental definition of the Laplace transform, which is an improper integral from zero to infinity:
L{f(t)} = ∫₀^∞ e⁻ˢᵗ f(t) dt
The key insight for a periodic function is to decompose this single infinite integral into a sum of integrals over each period. We can break the interval [0, ∞) into an infinite sequence of intervals: [0, T], [T, 2T], [2T, 3T], and so on. This allows us to rewrite the transform as an infinite series:
L{f(t)} = Σₙ₌₀^∞ ∫ₙₜ^⁽ⁿ⁺¹⁾ᵀ e⁻ˢᵗ f(t) dt
To evaluate this series, we can simplify each integral term. We use a change of variables to shift the integration interval for each term back to [0, T]. Let us define a new variable u such that t = u + nT. From this, we get dt = du. We also update the limits of integration: when t = nT, u = 0, and when t = (n+1)T, u = T.
Substituting these into the integral gives:
∫₀ᵀ e⁻ˢ(ᵘ⁺ⁿᵀ) f(u + nT) du
Now we use the given property that the function is periodic. Since f(u + nT) = f(u), we can simplify the expression. We also use the property of exponents to separate e⁻ˢ(ᵘ⁺ⁿᵀ) into e⁻ˢᵘ * e⁻ˢⁿᵀ. The integral becomes:
∫₀ᵀ e⁻ˢᵘ * e⁻ˢⁿᵀ f(u) du
The term e⁻ˢⁿᵀ is a constant with respect to the integration variable u, so we can factor it outside the integral:
e⁻ˢⁿᵀ ∫₀ᵀ e⁻ˢᵘ f(u) du
Substituting this simplified form back into our infinite series, we get:
L{f(t)} = Σₙ₌₀^∞ [e⁻ˢⁿᵀ ∫₀ᵀ e⁻ˢᵘ f(u) du]
The integral ∫₀ᵀ e⁻ˢᵘ f(u) du is a constant value with respect to the summation index n, so it can be factored out of the series:
L{f(t)} = (∫₀ᵀ e⁻ˢᵘ f(u) du) * (Σₙ₌₀^∞ e⁻ˢⁿᵀ)
The remaining summation, Σₙ₌₀^∞ (e⁻ˢᵀ)ⁿ, is a standard geometric series with a common ratio r = e⁻ˢᵀ. For the Laplace transform to converge, s must be sufficiently large and positive, which ensures that |r| < 1. The sum of an infinite geometric series is 1 / (1 – r). Therefore:
Σₙ₌₀^∞ (e⁻ˢᵀ)ⁿ = 1 / (1 – e⁻ˢᵀ)
Finally, we substitute this result back into our equation and replace the dummy integration variable u with t to yield the final formula:
L{f(t)} = [∫₀ᵀ e⁻ˢᵗ f(t) dt] / [1 – e⁻ˢᵀ]
