laplacetransform e^{-4t}*cos(3t)

解答

拉普拉斯变换

e^{- 4 t} \cdot cos (3 t)

\frac{s + 4}{( s + 4 ) ^{2} + 9}

求解步骤

L {e^{- 4 t} cos (3 t)}

应用变换法则: if

L {f (t)} = F (s)

则

L {e^{a t} f (t)} = F (s - a)

对于

e^{- 4 t} cos (3 t) : f (t) = cos (3 t), a = - 4

= L {cos (3 t)} (s + 4)

L {cos (3 t)} : \frac{s}{s ^{2} + 9}

= \frac{( s + 4 )}{( s + 4 ) ^{2} + 9}

= \frac{s + 4}{( s + 4 ) ^{2} + 9}

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