Question

Let \(f\) satisfy \(f(t+T)=f(t)\) for all \(t \geq 0\) and for some fixed positive number \(T ; f\) is said to be periodic with period \(T\) on \(0 \leq t<\infty .\) Show that $$ \mathcal{L}\\{f(t)\\}=\frac{\int_{0}^{T} e^{-s t} f(t) d t}{1-e^{-s T}} $$

Short answer

Question: Prove that for a periodic function \(f(t)\) with period \(T\), its Laplace Transform is given by \(\mathcal{L}\\{f(t)\\}=\frac{\int_{0}^{T} e^{-s t} f(t) d t}{1-e^{-s T}}\).

Step-by-step solution

1. Write Laplace Transform's definition for function f(t)

Recall the definition of the Laplace Transform for a function \(f(t)\): $$ \mathcal{L}\\{f(t)\\} = \int_{0}^{\infty} e^{-st} f(t) dt $$

2. Split the integral using the period T

We can write the integral from \(0\) to \(\infty\) as a sum of multiple integrals, each with limits \(0\) to \(T\), \(T\) to \(2T\), and so on. This can be written as: $$ \mathcal{L}\\{f(t)\\} = \sum_{n=0}^{\infty} \int_{nT}^{(n+1)T} e^{-st} f(t) dt $$

3. Make a substitution in the integral

Let's make a substitution \(u = t - nT\). Therefore, \(t = u + nT\) and \(dt = du\). The integral limits also change as follows: - When \(t=nT\), \(u=0\) - When \(t=(n+1)T\), \(u=T\) Now, rewrite the Laplace Transform integral using this substitution: $$ \mathcal{L}\\{f(t)\\} = \sum_{n=0}^{\infty} \int_{0}^{T} e^{-s(u + nT)} f(u + nT) du $$

4. Use the periodic property of function f(t)

The periodic property of the function \(f(t)\) states that \(f(t+T)=f(t)\). We can apply this property here because \(f(u+nT)=f(u)\) for all \(n\geq0\). So, we have: $$ \mathcal{L}\\{f(t)\\} = \sum_{n=0}^{\infty} \int_{0}^{T} e^{-s(u + nT)} f(u) du $$

5. Separate the exponential multiplication and sum

Now, separate the exponential multiplication as \(e^{-su}e^{-snT}\) and take the sum out: $$ \mathcal{L}\\{f(t)\\} = \int_{0}^{T} f(u)e^{-su}\left(\sum_{n=0}^{\infty} e^{-snT}\right) du $$

6. Evaluate the sum using geometric series

The sum inside the parentheses can be treated as a geometric series with first term \(1\) and common ratio \(e^{-sT}\). Thus, the sum of an infinite geometric series is given by: $$ \frac{1}{1-e^{-sT}} $$

7. Substitute the sum value and finalize the proof

Now, substitute the sum value back into the Laplace Transform expression: $$ \mathcal{L}\\{f(t)\\} = \int_{0}^{T} f(u)e^{-su} du \times \frac{1}{1-e^{-sT}} $$ This proof shows that for a periodic function \(f(t)\) with period \(T\): $$ \mathcal{L}\\{f(t)\\}=\frac{\int_{0}^{T} e^{-s t} f(t) d t}{1-e^{-s T}} $$

Key concepts

Periodic Function Properties

Periodic functions are foundational in various areas of mathematics and engineering, particularly in the study of signals and systems. A function is called periodic if it repeats its values at regular intervals, known as the period. For any periodic function with period T, the function satisfies the property that f(t + T) = f(t) for all t in its domain. This repetition means the behavior of the function over one period will dictate its behavior over any number of periods.

In the context of the Laplace Transform, periodic functions exhibit a pattern that can be exploited to simplify their transformation into the s-domain. By recognizing the cyclic nature of these functions, we can decompose a problem into an analysis of just one period rather than attempting to solve for an infinite range. This approach not only simplifies calculations but also harnesses repetition to represent a seemingly complex time-domain signal with a more manageable expression in the frequency domain.

Geometric Series in Laplace Transform

The relevance of a geometric series to the Laplace Transform comes to the forefront when dealing with periodic functions. A geometric series is a series where each term after the first is found by multiplying the previous one by a fixed, non-zero number known as the common ratio. In mathematical terms, a geometric series can be represented as S = a + ar + ar^2 + ar^3 + ..., where a is the first term and r is the common ratio.

When applying the Laplace Transform to periodic functions, the integral of the function over a single period can be multiplied by this geometric series, where the common ratio is typically in the form of an exponential function related to the Laplace variable s and the period T. This series converges if the absolute value of the ratio is less than one, which is a condition met by the exponential functions in the Laplace domain. As a result, we can sum an infinite series to yield a finite result that elegantly expresses the Laplace Transform of a periodic function.

Exponential Functions in Laplace Transform

Exponential functions play a critical role in the calculation of Laplace Transforms. In the domain of the transform, represented by the variable s, exponentials are in the form e^{-st}, with t being the time variable. This exponential factor is what links the time domain to the complex frequency domain.

When dealing with periodic functions, the period T comes into play by introducing an exponential term of the form e^{-snT} in the analysis. Since nT represents complete cycles of the period, these exponential terms form the common ratio of the geometric series previously discussed. The complex frequency variable s influences the decay rate of these exponential terms, and thus dictates the convergence of the geometric series. When these exponentials are included in the Laplace Transform of periodic functions, they allow us to capture the recurring nature of the function within a compact and elegant s-domain expression.

Understanding how exponential functions relate to the transform enables students to grasp deeper concepts in signal analysis, control systems, and physics, where time-domain representations of periodic phenomena are translated into the frequency domain for analysis and design purposes.