Question

In each exercise, graph the function \(f(t)\) for \(0 \leq t<\infty\), and use Table \(5.1\) to find the Laplace transform of \(f(t)\). \(f(t)=\sin (t-2 \pi) h(t-2 \pi)\)

Short answer

Question: Determine the Laplace transform of the function \(f(t) = \sin(t-2\pi)h(t-2\pi)\) for \(0 \leq t < \infty\) after graphing it. Answer: The Laplace transform of the given function \(f(t) = \sin(t-2\pi)h(t-2\pi)\) is \(F(s) = \frac{1}{s^2 + 1}e^{-2\pi s}\).

Step-by-step solution

Analyze the unit step function

The unit step function \(h(t-2\pi)\) will "activate" or be equal to 1 when \((t-2\pi)\geq0\), or \(t\geq2\pi\). When \(t<2\pi\), \(h(t-2\pi)=0\). This will affect the overall behaviour of the given function \(f(t)\).

Graph the sine function for \(t \geq 2 \pi\)

With the unit step function active for \(t \geq 2 \pi\), we can plot the sine function for that range. The function \(\sin(t - 2 \pi)\) is equal to \(\sin(t)\) shifted 2 \(\pi\) units to the right. For \(0 \leq t < 2\pi\), \(f(t) = 0\) as the unit step function is not yet active. For \(t \geq 2 \pi\), we have \(f(t) = \sin(t - 2 \pi)\), which is a regular sine wave.

Consult Table 5.1 for Laplace transform of sine function

Looking up Table 5.1, we can find the Laplace Transform of the sine function, \(F(s) = L\{\sin(at)\} = \frac{a}{s^2 + a^2}\) where \(a = 1\) in our case. But as there is a time shift of 2\(\pi\), we'll need to modify this expression.

Apply time shift property to find Laplace transform

The time shift property of Laplace transforms states that if we have a function \(f(t-a)\) shifted by 'a' units, then its Laplace transform can be found by multiplying the Laplace transform of \(f(t)\) by \(e^{-as}\). In our case, \(f(t) = \sin(t-2\pi)h(t-2\pi)\) and as found earlier, the Laplace transform of the sine function is \(\frac{1}{s^2 + 1}\). To adjust for the time shift(\(a=2\pi\)), we can multiply \(e^{-as}\) with the Laplace transform of the sine function: \(F(s) = L\{f(t)\} = \frac{1}{s^2 + 1}e^{-2\pi s}\). So, the Laplace transform of the given function \(f(t)\) is \(F(s) = \frac{1}{s^2 + 1}e^{-2\pi s}\).

Key concepts

Unit Step Function

The unit step function, often denoted as h(t) or u(t), acts like a mathematical on-off switch. Its primary role is to turn on a function at a certain point in time. For times less than its activation point, the unit step function has a value of zero; it 'switches on' to one at the activation point and maintains that value for all later times.

For instance, consider the function h(t - 2π). This function stays at 0 for all t less than 2π and then suddenly jumps to 1 at t = 2π and remains at 1 indefinitely. This behavior can be used to model systems or signals that begin at a specific time. When paired with other functions, such as sine or cosine, the unit step function can 'activate' these other functions at desired times, making it a valuable tool in time-dependent processes.

Graphing Functions

Graphing functions is an essential skill in mathematics as it helps to visualize the behavior of functions over time. To graph a piecewise function like f(t) = sin(t - 2π)h(t - 2π), we first determine the intervals over which different pieces of the function are active.

In this case, for 0 ≤ t < 2π, f(t) = 0, as the unit step function h(t) has not activated. But, for t ≥ 2π, the function takes the form of a regular sine wave, since h(t - 2π) is now 1. Therefore, for graphing, we would plot a flat line at zero up to t = 2π and then a sine wave for times beyond 2π. This approach enables anyone reviewing the graph to quickly determine when a signal starts and how it behaves over time.

Time Shift Property

The time shift property is an invaluable concept when working with Laplace transforms, especially in the context of engineering and physics where time-delayed signals and processes are commonplace. Essentially, if a function f(t) is shifted to the right in time by a units, resulting in f(t - a), then its Laplace transform is modified by multiplying it by e^{-as}.

This property makes it straightforward to find the Laplace transform of functions that have been delayed in time. For example, the function sin(t - 2π)h(t - 2π) from the original exercise involves both a sine wave and a unit step function, with the entire function starting at t = 2π. By using the time shift property, we can determine that the Laplace transform of the given function f(t) is F(s) = L{f(t)} = (1/(s^2 + 1))e^{-2πs}, effectively capturing the behavior of the delayed sine wave mathematically.