Problem 55 - Advanced Engineering Mathematics

Question

Write each function in terms of unit step functions. Find the Laplace transform of the given function. $$ f(t)=\left\\{\begin{array}{lr} 2, & 0 \leq t<3 \\ -2, & t \geq 3 \end{array}\right. $$

Step-by-Step Solution

Answer

The Laplace transform of \( f(t) \) is \( \frac{2}{s} - \frac{4e^{-3s}}{s} \).

Step 1: Express the Function Using Unit Step Functions

To express the function in terms of unit step functions, we use the unit step function, which is defined as \( u(t-a) \) where it switches from 0 to 1 at \( t = a \). Our function is given by: - \( f(t) = 2 \) for \( 0 \leq t < 3 \).- \( f(t) = -2 \) for \( t \geq 3 \).This can be written as the combination of two step functions:\[f(t) = 2u(t) - 4u(t-3) + 2\]The term \( 2u(t) \) represents the step at \( t = 0 \), and \( -4u(t-3) \) accounts for the switch at \( t = 3 \) to \( -2 \). The constant term 2 is added to adjust the equation correctly at \( t = 3 \).

Step 2: Find the Laplace Transform

The Laplace transform of a function \( f(t) \) using unit step functions can be calculated using the linearity property and known transforms.The Laplace transform of a constant, like \( 2 \, \), is:\[\mathcal{L}\{2u(t)\} = \frac{2}{s}\]And for \( -4u(t-3) \), we use the property for shifted unit steps \( \mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) \), where \( F(s) \) is the Laplace transform of \( f(t) \):\[\mathcal{L}\{-4u(t-3)\} = -4 \cdot e^{-3s} \cdot \frac{1}{s} = -\frac{4e^{-3s}}{s}\]So, the Laplace transform of the entire function \( f(t) \) is:\[\mathcal{L}\{f(t)\} = \frac{2}{s} - \frac{4e^{-3s}}{s}\]

Key Concepts

Unit Step Function

The Unit Step Function, often denoted as \( u(t-a) \), is a fundamental tool in signal processing and control systems. It acts as a switch that "turns on" at a specific moment in time, \( t = a \). Prior to that point, the function equals zero, and thereafter, it equals one. This makes it extremely useful for representing piecewise functions, where different segments of the function vary over different intervals.
In the exercise example, the function \( f(t) \) was piecewise defined as 2 for \( 0 \leq t < 3 \) and -2 for \( t \geq 3 \). By using unit step functions, we can express this function as a single, cohesive formula: \( f(t) = 2u(t) - 4u(t-3) + 2 \).

• The term \( 2u(t) \) indicates that the function starts at a value of 2 from \( t = 0 \).
• The term \( -4u(t-3) \) adjusts the function at \( t = 3 \), dropping its value by 4 to ensure it reaches -2.
• And finally, the constant \(+2\) balances the overall equation.

This elegant approach through unit step functions allows us to handle complex temporal changes smoothly.

Function Shift

The concept of Function Shift involves adjusting the starting point of a function based on time, essentially moving it along the time axis. This is crucial when dealing with functions like the unit step function to map out when certain conditions of a piecewise function apply.
In our context, the function shift can be seen in the term \( u(t-3) \). Here, \( u(t-3) \) represents a shifted unit step function that "activates" at \( t = 3 \). This shifting technique helps us refine our piecewise function expression.

• The original value of the function is unaffected until the shift point is reached.
• Beyond this point, additional conditions can be applied, such as changing the function's value.

Only after the shift point, the function undergoes the desired change, allowing more precise control over the function's behavior without altering its entire domain.

Laplace Transform Properties

Laplace Transform Properties allow us to solve complex differential equations and analyze systems in the s-domain by converting functions from the time domain. The linearity and shifting properties of Laplace transforms help deal with complex functions involving units steps and shifts effectively.
To solve the given function \( f(t) = 2u(t) - 4u(t-3) + 2 \) using Laplace transforms, we must apply these properties:

• Linearity: The Laplace Transform is linear, which means we can handle each term of the function independently. For example, \( \mathcal{L}\{2u(t)\} \) and \( \mathcal{L}\{-4u(t-3)\} \) can be transformed separately.
• Shifting Property: When handling the shifted unit step functions, we use the property \( \mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) \). It helps in managing terms like \( -4u(t-3) \).
• Transformation of Constants: The transform of a constant like 2 is simply \( \frac{2}{s} \).

The resultant Laplace transform for \( f(t) \) is \( \frac{2}{s} - \frac{4e^{-3s}}{s} \), effectively combining these properties to provide a solution in the frequency domain. This way, the Laplace transform becomes an essential tool for engineers and scientists for system analysis and design.