Question

For each \(c>0\) let \(f_{c}\) be the \(2 c\)-periodic function on the interval \([0, \infty)\) for which $$ f_{c}(t)= \begin{cases}t, & 0 \leq t \leq c, \\ 2 c-t, & c \leq t \leq 2 c .\end{cases} $$ Calculate the Laplace transform of \(f_{c}\).

Short answer

The Laplace transform of the function is given by \[ L\{f_c(t)\} = \frac{1}{1 - e^{-2cs}} \left( \frac{-c e^{-sc}}{s} + \frac{1}{s^2} + \frac{2c}{s} \right). \]

Step-by-step solution

Definition of Laplace Transform

The Laplace transform of a function \( f(t) \) is given by the formula \[ L\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt \] where \( s \) is a complex number. In this problem, we need to consider that \( f_{c}(t) \) is a piecewise function.

Periodicity Consideration

Since \( f_c(t) \) is a \( 2c \)-periodic function, we can use the property that the Laplace transform of a periodic function \( f(t) \) with period \( T \) is given by \[ L\{ f(t) \} = \frac{1}{1 - e^{-sT}} \int_{0}^{T} e^{-st} f(t) \, dt. \] Here, \( T = 2c \). Thus, we need to compute the integral over one period \([0, 2c]\).

Integrate from 0 to c

In this interval, \( f_{c}(t) = t \). So the integral becomes \[ \int_{0}^{c} e^{-st} t \, dt. \] This can be solved using integration by parts:Let \( u = t \) and \( dv = e^{-st} \, dt \). Then \( du = dt \) and \( v = \frac{-e^{-st}}{s} \). The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Applying this:\[ \int_{0}^{c} e^{-st} t \, dt = \left[ -\frac{t e^{-st}}{s} \right]_{0}^{c} - \int_{0}^{c} \left( -\frac{e^{-st}}{s} \right) \, dt. \] This simplifies to:\[ \left[-\frac{c e^{-sc}}{s} + \frac{0}{s} \right] + \frac{1}{s}\left[ \frac{e^{-st}}{s} \right]_{0}^{c} = \frac{-c e^{-sc}}{s} + \frac{1}{s^2}\left[ 1 - e^{-sc} \right]. \]

Integrate from c to 2c

In this interval, \( f_{c}(t) = 2c - t \). So the integral becomes \[ \int_{c}^{2c} e^{-st} (2c - t) \, dt. \] This can also be solved using integration by parts. Let \( u = 2c - t \) and \( dv = e^{-st} \, dt \). Then \( du = -dt \) and \( v = \frac{-e^{-st}}{s} \). Applying integration by parts:\[ \int_{c}^{2c} e^{-st} (2c - t) \, dt = \left[ -\frac{(2c - t) e^{-st}}{s} \right]_{c}^{2c} - \int_{c}^{2c} \frac{e^{-st}}{s} \, dt. \] Simplifying:\[ \left[\frac{2c e^{-2sc}}{s} - 0\right] - \frac{1}{s} \left[\frac{e^{-st}}{s}\right]_{c}^{2c} = \frac{2c(1 - e^{-2sc})}{s} - \frac{1}{s^2}(e^{-sc} - e^{-2sc}). \]

Combine Integrals

Now combine the results from both parts:\[ L\{f_c(t)\} = \frac{1}{1 - e^{-2cs}} \left( \frac{-c e^{-sc}}{s} + \frac{1 - e^{-sc}}{s^2} + \frac{2c(1 - e^{-2sc})}{s} - \frac{1}{s^2}(e^{-sc} - e^{-2sc}) \right). \] Combining and simplifying these terms leads us to the final form. This may require algebraic manipulations to factor and combine like terms.

Key concepts

Fourier Series

A Fourier series aims to express a periodic function as a sum of simple oscillating functions, like sine and cosine functions. The idea is that even complex or discontinuous periodic functions can be represented by these simpler trigonometric components because of their periodic nature. Fourier series are extensively used in signal processing, heat transfer, and vibration analysis.

Here are some key points about Fourier series:

• A function must be periodic to have a Fourier series representation.
• Each term in a Fourier series is associated with a specific frequency, known as the harmonics of the original function.
• There are two main forms in Fourier series representation: sine and cosine forms, or complex exponential form using Euler’s identity.

Understanding Fourier series is crucial for converting a piecewise function into a form that is easier to work with mathematically when dealing with periodic functions.

Piecewise Function

A piecewise function is defined by different expressions over different intervals within its domain. These functions are practical when a single expression does not suffice to describe the function's behavior across all its domain, or when the function exhibits distinct characteristics in different regions.

Considerations when working with piecewise functions:

• Each piece is defined over a specific interval.
• Conditions or inequalities specify where each piece applies.
• Jumps or discontinuities might occur at the boundaries of the intervals.
• When integrating or differentiating piecewise functions, address each piece separately, then combine the results.

In the exercise, the function \( f_{c}(t) \) is a piecewise function with distinct expressions in the intervals \([0, c]\) and \([c, 2c]\). Understanding how to handle piecewise functions is essential for calculating their Laplace transforms.

Periodic Function

A periodic function is a function that repeats its values at regular intervals, described by its period. In mathematical terms, a function \( f(t) \) is periodic with period \( T \) if \( f(t + T) = f(t) \) for all values in its domain. These functions frequently occur in physical phenomena like pendulum swings and sound waves.

Characteristics of periodic functions include:

• The smallest positive \( T \) for which the function repeats itself is called the "fundamental period."
• Periodic functions can be constructed by extending non-periodic functions with periodic extensions using their Fourier series.
• In the context of Laplace transforms, a transformation of a periodic function simplifies due to known repetitive behavior over intervals equivalent to its period.

In the exercise, \( f_{c}(t) \) is a \( 2c \)-periodic function, meaning it repeats its form every \( 2c \) units. This property is key to simplifying its Laplace transform computation.

Integration by Parts

Integration by parts is a mathematical technique used to integrate products of functions. It is derived from the product rule of differentiation and is particularly useful when the integral involves a product of a polynomial function and an exponential, logarithmic, or trigonometric function.

The integration by parts formula is:\[ \int u \, dv = uv - \int v \, du, \]where:

• \( u \) is a function to differentiate.
• \( dv \) is a function to integrate.
• To apply the method, choose \( u \) and \( dv \) wisely to simplify the integral \( \int v \, du \).

In the given exercise, integration by parts helps evaluate integrals over the intervals \([0, c]\) and \([c, 2c]\) for the piecewise function \( f_c(t) \). This technique makes the otherwise complex integration more manageable.