Problem 10 - Differential Equations and Linear Algebra

Question

Make a sketch of the given function on \([0, \infty)\) and express it in terms of the unit step function. $$f(t)=\left\\{\begin{aligned}2, & 0 \leq t<2 \\\1, & 2 \leq t<4 \\\\-1, & t \geq 4 \end{aligned}\right.$$

Step-by-Step Solution

Answer

First, we sketch the function according to the given conditions: 1. A horizontal line at \(f(t) = 2\) for \(0 \leq t < 2\) 2. A horizontal line at \(f(t) = 1\) for \(2 \leq t < 4\) 3. A horizontal line at \(f(t) = -1\) for \(t \geq 4\) Then we express \(f(t)\) using the unit step function \(u(t)\) with appropriate time shifts and weights: $$f(t)= 2u(t) + (u(t-2) - u(t-4)) - u(t-4)$$

Step 1: Sketching the function

First, let's sketch the function according to the definition. The line segments will be: 1. A horizontal line at \(f(t) = 2\) for \(0 \leq t < 2\) 2. A horizontal line at \(f(t) = 1\) for \(2 \leq t < 4\) 3. A horizontal line at \(f(t) = -1\) for \(t \geq 4\)

Step 2: Express the function in terms of the unit step function

Now we'll express \(f(t)\) using the unit step function \(u(t)\). Recall that \(u(t) = 1\) for \(t \geq 0\) and \(0\) for \(t < 0\). To do this, we will use step functions with appropriate time shifts and weights. 1. For the first segment, we need a constant 2 for \(0 \leq t < 2\). We can achieve this by having the function \(2u(t), 0 \leq t < 2\). This will give us 2 for the interval \(0 \leq t < 2\) and will keep the value of the function 0 elsewhere. 2. For the second segment, we need a constant 1 for \(2 \leq t < 4\). We can achieve this by having the function \(u(t-2)-u(t-4)\). In this case, the function will take the value of 1 for \(2 \leq t < 4\) while its value will be 0 elsewhere. 3. For the third segment, we need a constant -1 for \(t \geq 4\). We can achieve this by having the function \(-u(t-4)\) which will give us -1 for \(t \geq 4\) and will keep the value of the function 0 elsewhere. Now, to find the final function, we add the results from steps 1, 2, and 3. So the representation of \(f(t)\) in terms of the unit step function is: $$f(t)= 2u(t) + (u(t-2) - u(t-4)) - u(t-4)$$

Key Concepts

Piecewise Functions

Piecewise functions are functions defined by different expressions over different intervals of the domain. They allow us to model situations where a function has distinct behavior at different parts of its domain. In the given exercise, the function \(f(t)\) is defined with three different constants over specified intervals. These intervals are:

• From \(0\) to \(2\), where \(f(t) = 2\)
• From \(2\) to \(4\), where \(f(t) = 1\)
• For \(t \geq 4\), where \(f(t) = -1\)

Each segment represents a constant value, illustrating the piecewise nature. Understanding piecewise functions is crucial for analyzing scenarios with different conditions at varying intervals, particularly when translating into expressions like the unit step function. This translation helps simplify complex functions by breaking them into manageable segments.

Laplace Transform

The Laplace transform is a powerful integral transform used to convert functions in the time domain into the s-domain (frequency domain). It is widely used in engineering and physics for solving differential equations and analyzing linear time-invariant systems. By transforming functions into the s-domain, calculus operations such as differentiation and integration become simpler arithmetic operations.
In the context of the given problem, the Laplace transform plays a role in handling the unit step function representation of \(f(t)\). When functions are expressed in terms of unit step functions, their Laplace transforms can be systematically derived using properties of linearity and known transforms of step functions like \(u(t-a)\). This enables easier manipulation and solving of differential equations involving piecewise functions.

Discontinuity Handling

Discontinuity handling is an essential concept when dealing with piecewise functions, where abrupt changes or jumps occur at specific points. For the given problem, the function \(f(t)\) experiences discontinuities at \(t = 2\) and \(t = 4\) where the function value shifts sharply.
To manage these discontinuities, the unit step function is used. It acts like a switch, turning a function on or off at given points. For instance, \(u(t-2)\) turns the value on at \(t = 2\). Hence, the expression \(u(t-2) - u(t-4)\) captures the switch from 1 to 0 in the interval \([2, 4)\). Similarly, \(-u(t-4)\) reflects the function's behavior past \(t = 4\).

• The use of unit step functions allows capturing of sharp transitions without losing information.
• This method is particularly useful when you wish to use calculus techniques on functions that change values sharply.

By incorporating the unit step function to manage discontinuities, analyses involving sudden changes become algebraically manageable.