text{Find the Laplace transform } F(s)=Lleft{f(t)right} text{of the function } f(t)=6+sin(3t) text{defined on the interval } tgeq0

Question
Laplace transform
asked 2021-02-16
\(\text{Find the Laplace transform }\ F(s)=L\left\{f(t)\right\}\ \text{of the function }\ f(t)=6+\sin(3t) \ \text{defined on the interval }\ t\geq0\)

Answers (1)

2021-02-17
\(\text{Step 1}\)
\(\text{From the given statement, the function is }\ f(t)=6+\sin(3t)\)
\(\text{Step 2}\)
\(\text{To find the Laplace transform of the function as follows.}\)
\(L(f(t))=L(6+\sin(3t))\)
\(=L(6)+L(\sin(3t))\)
\(\text{Known fact: }\)
\(L(1)=\frac{1}{s}\)
\(L(\sin(\omega t))=\frac{\omega}{s^2+\omega^2}\)
\(\text{Therefore, }\)
\(L(6)+L(\sin(3t))=6L(1)+L(\sin(3t))\)
\(=6\bigg(\frac{1}{s}\bigg)+\frac{3}{s^2+3^2}\)
\(=\frac{6}{s}+\frac{3}{s^2+9}\)
\(=\frac{6s^2+3s+54}{s^3+9s}\)
\(\text{Thus, the Laplace transform of the function is } \frac{6s^2+3s+54}{s^3+9s}\)
0

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