Question

Используя преобразование Лапласа найти изображение f(t)=cos3t

Answer 1

$cos(wt)*H(t)= \frac{s}{s^2+w^2} ==\ \textgreater \ \\ \\ cos3t= \frac{s}{s^2+9}$

Answer 2

Using Laplace Transform to Find the Image of \( f(t) = \cos(3t) \)

To find the Laplace transform of the given function \( f(t) = \cos(3t) \), we can use the following formula:

\[ F(s) = \int_0^\infty e^{-st} f(t) \, dt \]

where \( F(s) \) is the Laplace transform of \( f(t) \).

Now, let's proceed with finding the Laplace transform of \( f(t) = \cos(3t) \).

The Laplace transform of \( f(t) = \cos(3t) \) is given by:

\[ F(s) = \int_0^\infty e^{-st} \cos(3t) \, dt \]

To solve this integral, we can use the following trigonometric identity:

\[ \cos(a) = \frac{e^{ia} + e^{-ia}}{2} \]

Using this identity, we can express \( \cos(3t) \) in terms of exponentials and then proceed with the Laplace transform.

The Laplace transform of \( f(t) = \cos(3t) \) is:

\[ F(s) = \int_0^\infty e^{-st} \cos(3t) \, dt = \int_0^\infty e^{-st} \left( \frac{e^{i3t} + e^{-i3t}}{2} \right) \, dt \]

\[ F(s) = \frac{1}{2} \int_0^\infty e^{-st} e^{i3t} \, dt + \frac{1}{2} \int_0^\infty e^{-st} e^{-i3t} \, dt \]

\[ F(s) = \frac{1}{2} \int_0^\infty e^{-(s-3i)t} \, dt + \frac{1}{2} \int_0^\infty e^{-(s+3i)t} \, dt \]

\[ F(s) = \frac{1}{2} \left( \frac{1}{s-3i} \right) + \frac{1}{2} \left( \frac{1}{s+3i} \right) \]

\[ F(s) = \frac{1}{2} \left( \frac{1}{s-3i} + \frac{1}{s+3i} \right) \]

So, the Laplace transform of \( f(t) = \cos(3t) \) is:

\[ F(s) = \frac{s}{s^2+9} \]

This is the Laplace transform of the given function \( f(t) = \cos(3t) \).