laplacetransform (sin(t))^2

解答

拉普拉斯变换

(sin (t))^{2}

\frac{2}{s ( s ^{2} + 4 )}

求解步骤

L {(sin (t))^{2}}

利用以下特性:

sin^{2} (x) = \frac{1}{2} - cos (2 x) \frac{1}{2}

= L {\frac{1}{2} - cos (2 t) \frac{1}{2}}

利用拉普拉斯变换的线性特性:

对于函数

f (t), g (t)

和常数

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {\frac{1}{2}} - \frac{1}{2} L {cos (2 t)}

L {\frac{1}{2}} : \frac{1}{2 s}

L {cos (2 t)} : \frac{s}{s ^{2} + 4}

= \frac{1}{2 s} - \frac{1}{2} \cdot \frac{s}{s ^{2} + 4}

化简

\frac{1}{2 s} - \frac{1}{2} \frac{s}{s ^{2} + 4} : \frac{2}{s ( s ^{2} + 4 )}

= \frac{2}{s ( s ^{2} + 4 )}

\int 2 t \cdot (- cos (2 t)) d t

t an g e n t f (x) = e^{x}, at x = 5

x \to 0 lim (\frac{2}{1 - 2 ^{\frac{1}{x}}})

d er i v a t i v e f (x) = \frac{2}{x - 1}

in v erse l a pl a ce - \frac{5}{s ( 50 + 20 s )}