What is laplacetransform e^t*cos(t)+j*e^t*sin(t) ?

The answer to laplacetransform e^t*cos(t)+j*e^t*sin(t) is (s-1)/((s-1)^2+1)+j/((s-1)^2+1)

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laplacetransform e^tcos(t)+je^t*sin(t)

laplace transform

e^{t} \cdot cos (t) + j \cdot e^{t} \cdot sin (t)

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

Solution steps

L {e^{t} cos (t) + j e^{t} sin (t)}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

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

= L {e^{t} cos (t)} + j L {e^{t} sin (t)}

L {e^{t} cos (t)} : \frac{s - 1}{( s - 1 ) ^{2} + 1}

L {e^{t} sin (t)} : \frac{1}{( s - 1 ) ^{2} + 1}

= \frac{s - 1}{( s - 1 ) ^{2} + 1} + j \frac{1}{( s - 1 ) ^{2} + 1}

Refine

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

= \frac{s - 1}{( s - 1 ) ^{2} + 1} + \frac{j}{( s - 1 ) ^{2} + 1}

e^{x} d x = 2 y d y

\frac{d}{d y} (x + 3 y^{\frac{1}{3}})

\frac{\partial}{\partial y} (2 x y^{4})

f (x) = (x)^{(3)} e^{(3 x)}

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

What is laplacetransform e^t*cos(t)+j*e^t*sin(t) ?
The answer to laplacetransform e^t*cos(t)+j*e^t*sin(t) is (s-1)/((s-1)^2+1)+j/((s-1)^2+1)