Chapter 6: Problem 10
Find the Laplace transform of the given function. $$ f(t)=u_{1}(t)+2 u_{3}(t)-6 u_{4}(t) $$
Short Answer
Expert verified
Answer: The Laplace transform of the given function is: $$L(f(t)) = \frac{e^{-s} + 2e^{-3s} - 6e^{-4s}}{s}$$
Step by step solution
01
Find the Laplace transform of the first term
The first term is $$u_1(t)$$. The Laplace transform of a unit step function with amplitude A and time shift c is given by $$\frac{Ae^{-cs}}{s}$$. In this case, A=1 and c=1. Therefore, the Laplace transform of the first term is:
$$
L(u_1(t)) = \frac{e^{-s}}{s}
$$
02
Find the Laplace transform of the second term
The second term is $$2u_3(t)$$. Here, A=2 and c=3. So, the laplace transform of the second term is:
$$
L(2u_3(t)) = \frac{2e^{-3s}}{s}
$$
03
Find the Laplace transform of the third term
The third term is $$-6u_4(t)$$. This time, A=-6 and c=4. Thus, the Laplace transform of the third term is:
$$
L(-6u_4(t)) = \frac{-6e^{-4s}}{s}
$$
04
Find the Laplace transform of the entire function
Now, we have the Laplace transforms of all three terms. To find Laplace transform for the given function, we simply add the transforms of all three terms:
$$
L(f(t)) = L(u_1(t)) + L(2u_3(t)) + L(-6u_4(t))
$$
$$
L(f(t)) = \frac{e^{-s}}{s} + \frac{2e^{-3s}}{s} + \frac{-6e^{-4s}}{s}
$$
So, the Laplace transform of the given function is:
$$
L(f(t)) = \frac{e^{-s} + 2e^{-3s} - 6e^{-4s}}{s}
$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Unit Step Function
The unit step function, often denoted as u(t - c), is a mathematical function that jumps from 0 to 1 at a given point in time, c. Think of it like flipping a light switch on at time c: before time c, the light is off (the value is 0), and at time c and after, the light is on (the value is 1).
In the context of Laplace transforms, the unit step function is particularly interesting because it allows us to consider functions that start 'being' at a certain moment. For a unit step function initiated at time c, the Laplace transform is given by L(u_c(t)) = \( \frac{e^{-cs}}{s} \). This relation is very useful because it introduces the time-shift property, which will be expanded upon in the next section.
In the context of Laplace transforms, the unit step function is particularly interesting because it allows us to consider functions that start 'being' at a certain moment. For a unit step function initiated at time c, the Laplace transform is given by L(u_c(t)) = \( \frac{e^{-cs}}{s} \). This relation is very useful because it introduces the time-shift property, which will be expanded upon in the next section.
Time Shift
Time shifting is a property that deals with functions that start at a time other than t = 0. In the realm of Laplace transforms, this means adjusting our function to accommodate a 'delay' in when the function becomes active. A unit step function is a perfect example to illustrate this.
The time shift property of the Laplace transform says that if you have a function f(t) delayed by c units in time, represented as f(t - c)u_c(t), its Laplace transform includes a multiplicative factor of e^{-cs}. So, if you have a delayed version of a function with its own Laplace transform, L{f(t)}, the transform of the delayed function is L{f(t-c)u_c(t)} = e^{-cs}L{f(t)}. This principle was applied in the textbook solution to handle the given unit step functions with varying delays.
The time shift property of the Laplace transform says that if you have a function f(t) delayed by c units in time, represented as f(t - c)u_c(t), its Laplace transform includes a multiplicative factor of e^{-cs}. So, if you have a delayed version of a function with its own Laplace transform, L{f(t)}, the transform of the delayed function is L{f(t-c)u_c(t)} = e^{-cs}L{f(t)}. This principle was applied in the textbook solution to handle the given unit step functions with varying delays.
Laplace Transform Properties
The Laplace transform is a powerful tool in solving differential equations because it simplifies many operations. One of its strongest aspects is the number of properties it obeys that can simplify complex equations. Some key properties include linearity, where the transform of a sum is the sum of the transforms; time shifting, as detailed above; and differentiation and integration in the time domain, which correspond to multiplication and division by s in the Laplace domain.
Other properties of significance include scaling in the time domain, which scales the transform inversely in the Laplace domain and the initial and final value theorems, which provide shortcuts to finding these values without doing inverse transforms. These properties enable us to decompose more complex functions into simpler parts, take their Laplace transforms individually, and then reconstruct the transform of the complex function, as was showcased in the solution of the problem provided.
Other properties of significance include scaling in the time domain, which scales the transform inversely in the Laplace domain and the initial and final value theorems, which provide shortcuts to finding these values without doing inverse transforms. These properties enable us to decompose more complex functions into simpler parts, take their Laplace transforms individually, and then reconstruct the transform of the complex function, as was showcased in the solution of the problem provided.
Differential Equations
Differential equations are equations that relate a function with its derivatives. They appear frequently in science and engineering to describe various phenomena. Solving differential equations can sometimes be straightforward, but more often, they require sophisticated methods.
The Laplace transform is a strategic tool to solve differential equations because it converts the potentially intricate differential equations in the time domain into more manageable algebraic equations in the Laplace domain. After solving for the transformed function, we find the original function by applying the inverse Laplace transform. The use of the Laplace transform is especially prevalent in problems involving linear time-invariant systems, such as in control theory and signal processing. It is worth noting that the execution of a step by step Laplace transform, especially when dealing with piecewise functions and unit step functions as in the provided exercise, demands a keen understanding of both the transform itself and the nature of the differential equation at hand.
The Laplace transform is a strategic tool to solve differential equations because it converts the potentially intricate differential equations in the time domain into more manageable algebraic equations in the Laplace domain. After solving for the transformed function, we find the original function by applying the inverse Laplace transform. The use of the Laplace transform is especially prevalent in problems involving linear time-invariant systems, such as in control theory and signal processing. It is worth noting that the execution of a step by step Laplace transform, especially when dealing with piecewise functions and unit step functions as in the provided exercise, demands a keen understanding of both the transform itself and the nature of the differential equation at hand.
