Question

Find the Laplace transform of the given function. In Problem 27 assume that term-by-term integration of the infinite series is permissible. $$ f(t)=\left\\{\begin{array}{ll}{1,} & {0 \leq t<1} \\ {0,} & {1 \leq t<2} \\\ {1,} & {2 \leq t<3} \\ {0,} & {t \geq 3}\end{array}\right. $$

Short answer

In order to find the Laplace transform of the given piecewise function, we first split the function into different intervals and find the Laplace transform for each part separately. Then, we sum the Laplace transforms of each part to find the total Laplace transform. The Laplace transform of the given function is: $$ \mathcal{L}\{f(t)\} = F(s) = \frac{1}{s} - \frac{1}{s}e^{-s} + \frac{1}{s}e^{-2s} - \frac{1}{s}e^{-3s}. $$

Step-by-step solution

Analyze the given function and split into intervals

The given function is a piecewise function defined over different intervals: $$ f(t)=\left\\{\begin{array}{ll}{1,} & {0 \leq t<1} \\\ {0,} & {1 \leq t<2} \\\ {1,} & {2 \leq t<3} \\\ {0,} & {t \geq 3}\end{array}\right. $$ We will break it into four parts and analyze them in each interval separately.

Define and find Laplace Transform for interval 0 ≤ t < 1

In this interval, the function is defined as f(t) = 1. We can find its Laplace transform using the formula: $$ \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t) dt $$ Let \(F_1(s)\) be the Laplace transform of f(t) in this interval. We integrate from 0 to 1. $$ F_1(s) = \int_0^1 e^{-st} \cdot 1 dt $$ We evaluate this integral by applying the antiderivative of \(e^{-st}\) w.r.t t and then evaluating at the limits. $$ F_1(s) = \left[-\frac{1}{s}e^{-st}\right]_0^1 = -\frac{1}{s}e^{-s} + \frac{1}{s} $$

Define and find Laplace Transform for interval 1 ≤ t < 2

For this interval, the function is defined as \(f(t) = 0\). Since the function is zero, its Laplace transform should also be zero. Let \(F_2(s)\) be the Laplace transform of f(t) in this interval. $$ F_2(s) = \int_1^2 e^{-st} \cdot 0 dt = 0 $$

Define and find Laplace Transform for interval 2 ≤ t < 3

For this interval, the function is again defined as \(f(t) = 1\). However, since this interval starts from t=2, we need to account for the time delay in our Laplace transform. Let \(F_3(s)\) be the Laplace transform of f(t) in this interval. We integrate from 2 to 3. $$ F_3(s) = \int_2^3 e^{-st} \cdot 1 dt $$ We evaluate this integral by applying the antiderivative of \(e^{-st}\) w.r.t t and then evaluating at the limits. $$ F_3(s) = \left[-\frac{1}{s}e^{-st}\right]_2^3 = -\frac{1}{s}e^{-3s} + \frac{1}{s}e^{-2s} $$

Define and find Laplace Transform for interval t ≥ 3

For this interval, the function is again defined as \(f(t) = 0\). We can easily find its Laplace transform as zero. Let \(F_4(s)\) be the Laplace transform of f(t) in this interval. $$ F_4(s) = \int_3^\infty e^{-st} \cdot 0 dt = 0 $$

Combine the solutions in all intervals

Now, we have found the Laplace transforms for all the different intervals. We just need to add them up to get the final Laplace transform of the given function. $$ \mathcal{L}\{f(t)\} = F(s) = F_1(s) + F_2(s) + F_3(s) + F_4(s) $$ Plugging in the values of \(F_1(s)\), \(F_2(s)\), \(F_3(s)\), and \(F_4(s)\), we get $$ F(s) = -\frac{1}{s}e^{-s} + \frac{1}{s} + 0 -\frac{1}{s}e^{-3s} + \frac{1}{s}e^{-2s} + 0 $$ Simplifying and combining terms, we have $$ F(s) = \frac{1}{s} - \frac{1}{s}e^{-s} + \frac{1}{s}e^{-2s} - \frac{1}{s}e^{-3s} $$ Thus, the Laplace transform of the given function is: $$ \mathcal{L}\{f(t)\} = F(s) = \frac{1}{s} - \frac{1}{s}e^{-s} + \frac{1}{s}e^{-2s} - \frac{1}{s}e^{-3s}. $$

Key concepts

Piecewise Functions

When working with Laplace transforms, understanding piecewise functions is crucial. Piecewise functions are defined by different expressions based on the interval of the input variable, in this case, time \(t\). They allow us to describe situations where a function behaves differently across certain ranges. In this exercise, the function \(f(t)\) showcases a typical piecewise structure:

• \(f(t) = 1\) for \(0 \leq t < 1\)
• \(f(t) = 0\) for \(1 \leq t < 2\)
• \(f(t) = 1\) for \(2 \leq t < 3\)
• \(f(t) = 0\) for \(t \geq 3\)

This is a simple yet powerful way to describe periods of activity and inactivity. By splitting intervals and defining the function separately over each interval, we can easily compute the Laplace transform of each portion, which can then be summed up to find the overall transform. Understanding the piecewise nature of functions helps in breaking down complex problems into manageable parts.

Integral Calculus

Integral calculus plays a vital role in finding the Laplace transform. The Laplace transform of a function \(f(t)\) is defined by the integral:\[\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t) \, dt\]Here, the integral allows us to analyze how the function \(f(t)\) behaves over time and to transform it into a form that is easier to work with for solving differential equations, among other applications.

Each piecewise component of \(f(t)\) requires evaluating this integral over its specific interval. For example:

• For \(0 \leq t < 1\), we compute
  \[\int_0^1 e^{-st} \, dt\]
• For \(1 \leq t < 2\), since \(f(t) = 0\), the integral is simply 0.

The power of integral calculus in this context is in its ability to handle functions defined piecewise, allowing each segment to be processed in isolation before piecing everything back together.

Step Functions

Step functions, sometimes known as unit step functions or Heaviside functions, significantly simplify working with piecewise functions in Laplace transforms. Step functions are perfect for representing changes in a function at distinct points in time.

For the given function \(f(t)\), you can think of representing it using step functions that "turn on" and "turn off" at specific times. This concept can simplify the definition of \(f(t)\) by just combining step functions:

• A step function for "1" from \(0 \leq t < 1\) can be written as \(u(t) - u(t-1)\)
• Another step function to represent "1" from \(2 \leq t < 3\) becomes \(u(t-2) - u(t-3)\)

The step function helps visualize time delays and shifts, making it easier to apply the properties of Laplace transforms. Overall, using step functions can streamline calculations and better predict how the piecewise function behaves over time.