find the Laplace transform of the piecewise defined functions f(t) (illustrated below) by expressing the functions in terms of the piecewise function and the Heaviside step function, H(t). (a) Find L[f(t)]. Assume that 0 < a=””>< c=”” always=”” holds.=”” f(t)=”” (b,=”” 2)=”” (c,=”” 2)=”” (a,=”” 0)=”” (c,=”” 0)=”” t=”” none=””>
The Correct Answer and Explanation is :
To find the Laplace transform of a piecewise function using the Heaviside step function, we express the function in terms of the step function ( H(t) ), which is defined as:
[
H(t – c) = \begin{cases}
0 & \text{for } t < c \
1 & \text{for } t \geq c
\end{cases}
]
This allows us to describe functions that “turn on” at specific values of ( t ). We will assume the given function follows the form:
(a) Laplace Transform of the Piecewise Function
For the piecewise function ( f(t) ), we will assume that it is expressed in two parts:
- For ( t \in [0, a) ), the function ( f(t) = 0 )
- For ( t \in [a, \infty) ), the function ( f(t) = \text{some function of } t )
Expressing the Piecewise Function Using the Heaviside Step Function
Suppose the given function is ( f(t) = g(t) ) for ( t \geq a ), and ( f(t) = 0 ) for ( t < a ). Then we can write ( f(t) ) as:
[
f(t) = g(t) \cdot H(t – a)
]
Where ( H(t – a) ) is the Heaviside function that shifts the function ( g(t) ) to start at ( t = a ).
Laplace Transform of the Piecewise Function
To find the Laplace transform of this function, we apply the property of the Laplace transform for shifted functions:
[
\mathcal{L}{g(t) H(t – a)} = e^{-as} \mathcal{L}{g(t)}
]
Thus, the Laplace transform of ( f(t) ) becomes:
[
\mathcal{L}{f(t)} = e^{-as} \mathcal{L}{g(t)}
]
Example
If ( f(t) = t^2 \cdot H(t – 2) ), we first recognize that:
[
\mathcal{L}{t^2 H(t – 2)} = e^{-2s} \mathcal{L}{t^2}
]
The Laplace transform of ( t^2 ) is:
[
\mathcal{L}{t^2} = \frac{2}{s^3}
]
So, the Laplace transform of ( f(t) = t^2 H(t – 2) ) is:
[
\mathcal{L}{f(t)} = e^{-2s} \cdot \frac{2}{s^3}
]
Explanation in 300 Words:
To compute the Laplace transform of a piecewise function, we first express it in terms of the Heaviside step function, which effectively shifts the start of a function. The Heaviside step function ( H(t – a) ) is 0 for ( t < a ) and 1 for ( t \geq a ). This lets us describe a function that begins at a certain time ( t = a ) and behaves differently before and after that point.
Once the function is written as a product of the function and the Heaviside step function, we apply the known rule for the Laplace transform of shifted functions. Specifically, if ( f(t) = g(t) H(t – a) ), then the Laplace transform is given by:
[
\mathcal{L}{f(t)} = e^{-as} \mathcal{L}{g(t)}
]
This formula shows that the Laplace transform of a function that starts at time ( a ) is the Laplace transform of the original function multiplied by ( e^{-as} ), which shifts the Laplace transform in the complex frequency domain. This approach is particularly useful when working with piecewise functions that switch between different behaviors at specific times. By using this step function representation, we can handle these functions in a straightforward manner without needing to compute their transforms piece by piece.