Step Functions

In this discussion, we will investigate piecewise defined functions and their Laplace Transforms.  We start with the fundamental piecewise defined function, the Heaviside function.  

The Heaviside function y  =  u_c(t) and y = 1 - u_c(t) are graphed below.

Example

We can write the function 

In terms of Heaviside functions.

Solution

We tackle the functions in parts.  The function that is 1 from 0 to 2 and 0 otherwise is 

        1 - u₂(x)

Multiplying by 3 gives

        3(1 - u₂(x))  =  3 - 3u₂(x)

To get the function that is 1 between 2 and 5 and 0 otherwise, we subtract

        u₂(x) - u₅(x)

Now multiply by e^x to get

        e^x(u₂(x) - u₅(x))  =  e^x u₂(x) - e^x u₅(x)

Adding these together gives

        f(x)  =  3 - 3u₂(x) + e^x u₂(x) - e^x u₅(x)

        =  3 + (e^x - 3)u₂(x) - e^x u₅(x)

We can find the Laplace transform of u_c(t) by integrating

In practice, we want to find the Laplace transform of a more general piecewise defined function such as 

This type of function occurs in electronics when a switch is suddenly turned on after one second and a forcing function is applied.  We can write 

        f(x)  =  u_p(x) sin x

We will be interested in the Laplace transform of a product of the Heaviside function with a continuous function.  The result that we need is 

By taking inverses we get that if F(s)  =  L{f(t)}, then

        L^-1{e^-csF(s)} =  u_c(t)f(t - c)

Proof

We use the definition to get

Example

Find the Laplace transform of 

Solution

We use that fact that 

        f(x)  =  u_p(x)sin x  =  -u_p(x) sin[(x - p)]  

Now we can use the formula to get that 

        L{f(x)}  =  -L{u_p(x) sin[(x - p)]}  =  -e^-cs L{sin x}

By the table, we get

                -e^-cs
        =                     
               s² + 1

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