Question

What is the laplace transform of the following?

2 sin (πt-3/5) u(t-7)
a. (-11.32s - 7.48) e⁻⁷ˢ/6.79 + 25s²
b. (30.52s - 24.88) e⁻⁷ˢ/9.87 + 25s²
c. (-8.31s - 4.52) e⁻⁷ˢ/1.53 + 25s²
d. this problem is not at all possible to convert to the frequency domain and none of these answer are correct
e. (5.92s - 4.26) e⁻⁷ˢ/6.79 - 25s²
f. (26.13s - 5.43) e⁻⁷ˢ/3.27 - 15s²
g. (-16.29s + 7.81) e⁻⁷ˢ/3.27 - 25s²
h. (12.36s - 17.29) e⁻⁷ˢ/5.48 + 15s²
i. (-7.49s - 13.26) e⁻⁷ˢ/25 + s²

Answer 1

The Laplace transform of 2 sin (πt-3/5) u(t-7) is (2πe^(-3s/5))/(s^2 + π^2).

Step-by-step explanation:

To find the Laplace transform of 2 sin (πt-3/5) u(t-7), we can use the Laplace transform property for sine functions. According to the property, if the Laplace transform of f(t) is F(s), then the Laplace transform of sin(at) is (a/s^2 + a^2).

In this case, the frequency of the sine function is π and the Laplace transform of sin(πt) is (π/s^2 + π^2). We also have a phase shift of -3/5, so we can multiply the Laplace transform by e^(-3s/5).

Thus, the Laplace transform of 2 sin (πt-3/5) u(t-7) is (2πe^(-3s/5))/(s^2 + π^2).