Question

Express the given function \(f\) in terms of unit step functions and use Theorem 8.4 .1 to find \(\mathcal{L}(f) .\) Where indicated by, graph \(f\). $$ f(t)=\left\\{\begin{array}{cl} 2, & 0 \leq t<1, \\ -2 t+2, & 1 \leq t<3, \\ 3 t, & t \geq 3. \end{array}\right. $$

Short answer

Based on the step-by-step solution provided, the short answer for the Laplace transform of the given function \(f(t)\) is: \(\mathcal{L}(f) = \frac{2}{s} + e^{-s}\left(\frac{2}{s^2} - \frac{4}{s}\right) + e^{-3s}\left(\frac{5}{s^2} - \frac{4}{s}\right)\)

Step-by-step solution

(Step 1: Express function in terms of unit step functions)

To express the function in terms of unit step functions, we need to break the function into three separate equations, one for each of the defined intervals: 1. \(2, 0 \leq t < 1\): This expression can be represented as \(2u(t)\) where \(u(t)\) is the unit step function. 2. \(-2t + 2, 1 \leq t < 3\): To represent this expression, we can write it as \((-2t + 2 - 2)u(t - 1) + 2u(t - 1) = (-2t + 4)u(t - 1)\). 3. \(3t, t \geq 3\): To represent this expression in terms of unit step functions, we can write it as \((3t -(-2t + 4))u(t - 3) = (5t - 4)u(t - 3)\). Now, we can write \(f(t)\) as: $$ f(t) = 2u(t) + (-2t + 4)u(t - 1) + (5t - 4)u(t - 3) $$

(Step 2: Apply Theorem 8.4.1 to find Laplace transform)

According to Theorem 8.4.1, for each term: 1. \(\mathcal{L}(2u(t)) = \frac{2}{s}\) 2. \(\mathcal{L}((-2t + 4)u(t - 1)) = e^{-1s}\mathcal{L}((-2(t + 1) + 4)) = e^{-s}\left(\frac{2}{s^2} - \frac{4}{s}\right)\) 3. \(\mathcal{L}((5t - 4)u(t - 3)) = e^{-3s}\mathcal{L}((5(t + 3) - 4)) = e^{-3s}\left(\frac{5}{s^2} - \frac{4}{s}\right)\) Now, we can find the Laplace transform \(\mathcal{L}(f)\) by combining the above results: $$ \mathcal{L}(f) = \frac{2}{s} + e^{-s}\left(\frac{2}{s^2} - \frac{4}{s}\right) + e^{-3s}\left(\frac{5}{s^2} - \frac{4}{s}\right) $$

Key concepts

Unit Step Functions

Unit step functions, often denoted as u(t), are crucial in representing piecewise functions, particularly when dealing with time-dependent processes in engineering and differential equations. A unit step function u(t-a) essentially 'turns on' at time t=a, jumping from a value of 0 to 1.

Imagine flipping a switch to illuminate a bulb; before flipping, the light is off (0), and after flipping, it's on (1). The unit step function embodies this binary nature, remaining at 0 before t=a and switching to 1 for t ≥ a. By shifting the function along the time axis, you can dissect complex functions into simpler segments that activate at specified moments.

For instance, a change in a system's behavior at t = 3 seconds can be modeled by multiplying the change's magnitude by u(t-3). This translates a segment of the function to start influencing the system at t = 3. This approach simplifies analyzing systems with multiple phases or operational stages.

Laplace Transform Theorem

The Laplace transform is instrumental in converting a time-domain function f(t) into a frequency-domain representation F(s), which simplifies differential equations into algebraic ones. This powerful conversion allows for easier manipulation and solution, particularly for systems described by linear time-invariant differential equations.

One of the fundamental Laplace transform theorems is the linearity theorem. It states that if F(s) = L[f(t)] and G(s) = L[g(t)], then for any constants a and b, the Laplace transform of af(t) + bg(t) is aF(s) + bG(s). The linearity property ensures the transform of a sum is the sum of the transforms, a feature that greatly aids in solving piecewise and discontinuous functions.

The theorem permits the transformation of intricate time-domain behaviors into a more manageable format for frequency-domain analysis, thereby streamlining the process of finding solutions to various system behaviors and responses.

Piecewise Function Representation

Representing a function that has different expressions over different intervals is often a necessity in mathematics and engineering, hence the importance of piecewise functions. These functions are defined by multiple sub-functions, each corresponding to a specific piece of the domain.

When modeling real-world scenarios, it is common to encounter situations where a response or behavior changes abruptly at certain points. Using piecewise functions, you can accurately reflect these changes by defining rules for every segment. Each rule indicates the function's form on a particular interval. For example, a thermostat controlled heating system may operate at one intensity in a specific temperature range and switch to another intensity in a different range - a scenario that naturally maps onto a piecewise function.

Describing such functions in terms of unit step functions simplifies their handling, especially when applying mathematical operations like Laplace transforms, as it allows us to break the function down into manageable pieces activated at the needed points in time.

Exponential Shift Theorem

The Exponential Shift Theorem is a vital tool in the arsenal of Laplace transforms, particularly when dealing with delayed or shifted functions. This theorem states that if L[f(t)] = F(s), then L[e^{at}f(t)] = F(s-a). In essence, multiplying a time-domain function by an exponential results in a shift in the s-domain.

This theorem is especially useful when working with unit step functions, which inherently include delays or 'shifts' as part of their formulation. By recognizing that a function f(t) starting at time t = b can be represented as f(t)u(t-b), the Exponential Shift theorem provides us with a straightforward method to find its Laplace transform by shifting the corresponding F(s) as per the theorem.

When solving differential equations or analyzing system responses, this theorem simplifies the process of taking Laplace transforms of delayed input signals or time-shifted functions, providing a clear path to incorporating time delays into the frequency domain.