9.1.2. Standard Result

ℒ{u(t−a)} = \frac{e^{-as}}{s}, \text{ for } a  0

The Laplace Transform of the unit step function defined as u(t−a) is given by the formula ℒ{u(t−a)} = e^{-as}/s for any non-negative constant a. In this context, u(t−a) is a function that switches on at t=a. When a=0, this simplifies to the basic unit step function, denoted as u(t).

Imagine a switch that lights up a lamp only after a certain time (let's say 3 seconds). Before that time, the light remains off (value is 0), and once the time reaches 3 seconds, the light turns on (value is 1). The function that describes this behavior is similar to u(t−3), encapsulating when the light starts shining.

Using the Laplace Transform definition:

ℒ{u(t−a)} = ∫[u(t−a)e^{-st}] dt from 0 to ∞

Since u(t−a) = 0 for t < a, the lower limit can be changed to a:

= ∫[e^{-st}] dt from a to ∞ = [ \frac{e^{-st}}{-s} ] from a to ∞ = \frac{e^{-as}}{s}

To derive the transform, we start with the definition of the Laplace Transform. The unit step function is zero for all time t less than a, allowing us to change the lower limit of integration from 0 to a. When we perform the integration from a to infinity, we find that the result simplifies to e^{-as}/s.

Think of the integration process as analyzing the flow of water in a pipe that only opens at a certain time. Before it's opened, no water flows (zero contribution to the integral). Once you reach the opening time (here, time a), the flow begins, and you can measure how much water is flowing thereafter.

ℒ{f(t)u(t−a)} = e^{-as} ⋅ ℒ{f(t+a)}

When you have a function f(t) multiplied by the unit step function u(t−a), the Laplace Transform can be expressed using the second shifting theorem. This theorem states that you can shift the time a and factor out an exponential decay term e^{-as}, indicating that the function f(t) is delayed by a seconds.

Imagine a delivery truck that starts delivering packages only after 3 hours. The delivery process can be represented by a function. The fact that it starts later can be thought of as the truck being delayed: before 3 hours, there's no delivery (the truck isn’t in action), and after that, it starts delivering at a specific rate. This delayed action is modeled with the unit step function.

Example 1: ℒ{u(t−3)} = \frac{e^{-3s}}{s} Example 2: ℒ{(t−2)u(t−2)} = e^{-2s}⋅\frac{1}{s^2}

In Example 1, the transform of the delayed unit step function u(t−3) results in the expression e^{-3s}/s, indicating it starts at t = 3. Example 2 shows how to apply the second shifting theorem; ℒ{(t−2)u(t−2)} results in e^{-2s} times the Laplace Transform of the function adjusted by the shift (t = t - 2). Thus, you get an output of e^{-2s}/s^2.

Consider the delayed truck example again. If the truck starts working at 3 hours, the formula gives a measure of how “active” the truck becomes from that moment onward, likewise in the second example when considering a different start time but the same operational effect. These instances illustrate when different functions begin to operate.

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