Problem 6 - Differential Equations and Linear Algebra

Question

Determine the Laplace transform of the given function. $$f(t)=\left\\{\begin{array}{ll}2 t / \pi, & 0 \leq t<\pi / 2 \\\\\sin t, & \pi / 2 \leq t<\pi\end{array}\right.$$ where \(f(t+\pi)=f(t)\).

Step-by-Step Solution

Answer

The Laplace transform of the given function is: \[ L\{f(t)\} = \frac{1}{1-e^{-s\pi}}\left(\frac{2}{s^2\pi} - \frac{2}{s \pi}e^{-s\pi/2} + \frac{s}{s^2+1}\left(e^{-s\pi/2}-e^{-s\pi}\right)\right)\]

Step 1: Find the Laplace transform of \(2t/\pi\)

To find the Laplace transform of \(2t/\pi\) for \(0 \leq t < \pi / 2\), we will use the definition of the Laplace transform: \[ L\{f(t)\} = \int_0^{\infty}f(t)e^{-st} \, dt\] However, we have limits for this piece of the function, so we will modify the integral as follows: \[ L\{2t/\pi\} = \int_0^{\pi/2} \frac{2t}{\pi}e^{-st} \, dt\] Now, let's calculate the integral: \[ L\{2t/\pi\} = \frac{2}{\pi}\int_0^{\pi/2} te^{-st} \, dt\] #2. Calculate Laplace transform of the second piece

Step 2: Find the Laplace transform of \(\sin t\)

To find the Laplace transform of \(\sin t\) for \(\pi / 2 \leq t < \pi\), we will use the definition of the Laplace transform: \[ L\{f(t)\} = \int_0^{\infty}f(t)e^{-st} \, dt\] Again, we have limits for this piece of the function, so we will modify the integral as follows: \[ L\{\sin t\} = \int_{\pi/2}^\pi \sin t \cdot e^{-st} \, dt\] Now, let's calculate the integral: \[ L\{\sin t\} = \int_{\pi/2}^\pi \sin t \cdot e^{-st} \, dt\] #3. Combine the two pieces

Step 3: Combine the Laplace transforms of pieces

Now we combine the two Laplace transforms of the two pieces and use the formula for the Laplace transform of a periodic function: \[ L\{f(t)\} = \frac{1}{1-e^{-s\pi}}\left(\int_0^{\pi/2} \frac{2t}{\pi}e^{-st} \, dt + \int_{\pi/2}^\pi \sin t \cdot e^{-st} \, dt\right)\] #4. Final Solution

Step 4: Find the integrals

Now, we just need to integrate and simplify the expression: \[ L\{f(t)\} = \frac{1}{1-e^{-s\pi}}\left(\frac{2}{s^2\pi} - \frac{2}{s \pi}e^{-s\pi/2} + \frac{s}{s^2+1}\left(e^{-s\pi/2}-e^{-s\pi}\right)\right)\] The Laplace transform of the given function is: \[ L\{f(t)\} = \frac{1}{1-e^{-s\pi}}\left(\frac{2}{s^2\pi} - \frac{2}{s \pi}e^{-s\pi/2} + \frac{s}{s^2+1}\left(e^{-s\pi/2}-e^{-s\pi}\right)\right)\]

Key Concepts

Periodic Functions

Periodic functions are mathematical functions that repeat their values in regular intervals. A function \( f(t) \) is periodic with period \( T \) if \( f(t+T) = f(t) \) for all \( t \). In simpler terms, after every interval \( T \), the function exhibits the same behavior and outputs the same values.
Understanding periodicity is crucial in mathematics, especially in fields like signal processing and control systems, as it allows predicting the function's behavior over time.
For example, the function described in the exercise is periodic with a period of \( \pi \), denoted by the equation \( f(t+\pi) = f(t) \). This means every \( \pi \) units of time, the pattern of \( f(t) \) repeats itself. Recognizing this pattern helps in simplifying computations involving such functions, especially when using transformations like the Laplace transform, which can leverage symmetry and repeated patterns to simplify complex integrations.

Piecewise Functions

Piecewise functions are defined by different expressions depending on the interval of the input. They allow us to describe functions that behave differently over various portions of their domain.
This versatility is particularly useful in modeling real-world problems, where conditions and interactions often change at certain thresholds or boundaries.
In the exercise, the function \( f(t) \) is piecewise, defined by two distinct expressions:

• \( \frac{2t}{\pi} \) for \( 0 \leq t < \frac{\pi}{2} \)
• \( \sin t \) for \( \frac{\pi}{2} \leq t < \pi \)

By splitting into these distinctive pieces, it becomes easier to analyze and compute operations like the Laplace transform separately for each part. Combining them afterward ensures the whole function's transformation reflects all its varying behaviors across the entire input range.

Integration

Integration is a fundamental mathematical operation used to calculate the area under a curve. It is crucial when dealing with piecewise functions in Laplace transforms.
Integrating a function involves finding an antiderivative and is a central concept in calculus used for measuring cumulative quantities, such as area, volume, or total accumulation of a quantity over time.
In the context of the exercise, integration is employed separately for each piece of the piecewise function over its respective interval. For instance, to find the Laplace transform, you integrate \( \frac{2t}{\pi} \) over \( [0, \frac{\pi}{2}] \) and \( \sin t \) over \( [\frac{\pi}{2}, \pi] \).
This process requires setting up the correct integral bounds and solving each part independently. Later, you combine the results to find the complete Laplace transform of the function, ensuring the calculated expressions accurately portray the different behaviors exhibited by the original piecewise function over its domain.