Questions: Consider the function f(t)= 0 if 0 ≤ t<4π sin(t−4π) if 4π ≤ t a. Use the graph of this function to write it in terms of the Heaviside function. Use h(t-a) for the Heaviside function shifted a units horizontally. f(t)= b. Find the Laplace transform F(s)=Lf(t). F(s)=Lf(t)=

Consider the function

f(t)=
0 if 0 ≤ t<4π
sin(t−4π) if 4π ≤ t

a. Use the graph of this function to write it in terms of the Heaviside function. Use h(t-a) for the Heaviside function shifted a units horizontally.

f(t)=

b. Find the Laplace transform F(s)=Lf(t).

F(s)=Lf(t)=

Transcript text: Consider the function \[ f(t)=\left\{\begin{array}{ll} 0 & \text { if } 0 \leq t<4 \pi \\ \sin (t-4 \pi) & \text { if } 4 \pi \leq t \end{array}\right. \] a. Use the graph of this function to write it in terms of the Heaviside function. Use $h(t-a)$ for the Heaviside function shifted $a$ units horizontally. \[ f(t)= \] $\square$ help (formulas) b. Find the Laplace transform $F(s)=\mathcal{L}\{f(t)\}$. \[ F(s)=\mathcal{L}\{f(t)\}= \] $\square$ help (formulas)

Solution

Solution Steps

Solution Approach

a. To express the given function $ f(t) $ in terms of the Heaviside function, we need to recognize that the Heaviside function $ h(t-a) $ is 0 for $ t < a $ and 1 for $ t \geq a $. We can use this property to rewrite $ f(t) $ by shifting the sine function using the Heaviside function.

b. To find the Laplace transform $ F(s) $ of $ f(t) $, we will use the properties of the Laplace transform, particularly the shifting theorem. The Laplace transform of $ \sin(t) $ is known, and we will apply the shifting property to account for the Heaviside function.

Step 1: Express $ f(t) $ in terms of the Heaviside function

Given the function: \[ f(t) = \begin{cases} 0 & \text{if } 0 \leq t < 4\pi \\ \sin(t - 4\pi) & \text{if } 4\pi \leq t \end{cases} \] We can use the Heaviside function $ h(t - a) $ to rewrite $ f(t) $. The Heaviside function $ h(t - 4\pi) $ is 0 for $ t < 4\pi $ and 1 for $ t \geq 4\pi $. Therefore, we can express $ f(t) $ as: \[ f(t) = \sin(t - 4\pi) \cdot h(t - 4\pi) \]

Step 2: Find the Laplace transform $ F(s) = \mathcal{L}\{f(t)\} $

To find the Laplace transform of $ f(t) $, we use the shifting property of the Laplace transform. The Laplace transform of $ \sin(t) $ is: \[ \mathcal{L}\{\sin(t)\} = \frac{1}{s^2 + 1} \] Using the shifting property $ \mathcal{L}\{f(t - a)h(t - a)\} = e^{-as}F(s) $, where $ a = 4\pi $, we get: \[ \mathcal{L}\{\sin(t - 4\pi)h(t - 4\pi)\} = e^{-4\pi s} \cdot \frac{1}{s^2 + 1} \]