Chapter 7: Problem 20
Find the Laplace transforms of the functions given in Problems. $$ f(t)=0 \text { if } t<1 ; f \overline{(t)} \equiv t \text { if } t \geq 1 $$
Short Answer
Expert verified
The Laplace transform of the function is \( \frac{e^{-s}}{s^2} \).
Step by step solution
01
Understand the Function
The function \( f(t) \) is defined piecewise: \( f(t) = 0 \) for \( t < 1 \), and \( f(t) = t \) for \( t \geq 1 \). This implies that \( f(t) \) starts at \( t = 1 \) as a ramp function (straight line) with a slope of 1.
02
Apply Laplace Transform Definition
The Laplace transform \( \mathscr{L}\{f(t)\} \) for a function \( f(t) \) is defined as \( \int_{0}^{\infty} e^{-st}f(t) \, dt \). Since \( f(t) = 0 \) for \( t < 1 \), the integral effectively starts at \( t = 1 \).
03
Shift Theorem in Laplace Transforms
Use the Heaviside step function to write \( f(t) \) as \( f(t) = (t-1)u(t-1) \). Here, \( u(t-1) \) is the unit step function that shifts the function to start at \( t = 1 \).
04
Laplace Transform with Heaviside Function
The Laplace transform of \( (t-a)u(t-a) \) is \( e^{-as} \mathscr{L}\{t\}(s) \). In our case, \( a = 1 \). Thus, the Laplace transform of the function is \( e^{-s} \times \frac{1}{s^2} \), since \( \mathscr{L}\{t\}(s) = \frac{1}{s^2} \).
05
Conclusion
The final Laplace transform of the given function \( f(t) \) is \( \mathscr{L}\{f(t)\} = \frac{e^{-s}}{s^2} \), reflecting the delay and ramp characteristic of the function that starts at \( t = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heaviside Step Function
The Heaviside step function, often denoted as \( u(t-a) \), is a crucial tool in signal processing and control theory. It represents a signal that "turns on" at a certain point, effectively shifting the start of a function in the time domain. When working with Laplace Transforms, it helps in expressing piecewise functions in an integrated form.
Here's how it works: the Heaviside step function is defined such that \( u(t-a) = 0 \) for \( t < a \) and \( u(t-a) = 1 \) for \( t \geq a \). This simple characteristic makes it very useful for turning on a function at \( t = a \).
Here's how it works: the Heaviside step function is defined such that \( u(t-a) = 0 \) for \( t < a \) and \( u(t-a) = 1 \) for \( t \geq a \). This simple characteristic makes it very useful for turning on a function at \( t = a \).
- The Heaviside step function allows the clean expression of functions that do not start at \( t=0 \).
- It's pivotal in describing real-world signals that naturally have a delayed start.
- It enables the treatment of piecewise functions, like the one in our exercise, by expressing them as a single formula using only step functions.
Piecewise Functions
Piecewise functions are mathematical expressions defined by multiple sub-functions, each applying to a specific interval of the domain. They are integral in representing systems and signals with different behaviors over time.
In our example, the piecewise function is defined as \( f(t) = 0 \) for \( t < 1 \), and \( f(t) = t \) for \( t \geq 1 \). Such a function depicts a scenario where there is no value until a particular point, and then it behaves like a ramp function with a slope of 1.
In our example, the piecewise function is defined as \( f(t) = 0 \) for \( t < 1 \), and \( f(t) = t \) for \( t \geq 1 \). Such a function depicts a scenario where there is no value until a particular point, and then it behaves like a ramp function with a slope of 1.
- Piecewise functions are versatile as they model various real-life phenomena.
- They allow us to describe functions that cannot be captured by a single formula over their entire domain.
- These functions are vital for analytically handling systems that have different operational states.
Shift Theorem
The Shift Theorem, also known as the time-shifting property, is a fundamental concept in the domain of Laplace Transforms. It provides a means to handle the translation of functions along the time axis. By harnessing this theorem, we can simplify computations and compare delayed systems accurately.
Simply put, if you have a function \( f(t-a) \) with a delay \( a \), you can find its Laplace Transform by incorporating the exponential shift \( e^{-as} \) into the transform of \( f(t) \). Specifically, \( \mathscr{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) \), where \( F(s) \) is the Laplace Transform of \( f(t) \) with no delay.
Simply put, if you have a function \( f(t-a) \) with a delay \( a \), you can find its Laplace Transform by incorporating the exponential shift \( e^{-as} \) into the transform of \( f(t) \). Specifically, \( \mathscr{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) \), where \( F(s) \) is the Laplace Transform of \( f(t) \) with no delay.
- The Shift Theorem explains how a time-delay translates into the s-domain (Laplace domain) as an exponential factor.
- It is essential for managing systems or signals with inherent delays, ensuring precise modeling.
- This theorem streamlines the calculation of Laplace Transforms, especially for piecewise functions and systems with step inputs.