Question

If the function h() is [(4 sin(21) + 8 cos (21)) (], identify the Laplace transform H(s) of the given function. 2s + 8 S + 8 8s + 2 S2 + 4 8s + 8 S2 + 4 8s + 8 52 + 2

Answer 1

We can write the given function as

ht) - (4 sin 2t + 8 cos 2t)u(t) = (4 sin 2t + 8 cos 2tuít -0)

where u is the Heaviside step function.

Now, the Laplace transform of this is given by

$\small \frak L \{h(t)\}=\frak L\left \{(4\sin2t+8\cos2t)u(t-0)\right \}=e^{-s\cdot 0}\frak L \left \{4\sin2(t+0)+8\cos2(t+0) \right \}$

So we have

$\small =1\cdot \frak L \left \{4\sin(2t)+8\cos(2t) \right \}$

$\small =\frak L \left \{4\sin(2t) \right \}+\frak L \left \{8\cos(2t) \right \}$

$\small =4\frak L \left \{\sin(2t) \right \}+8\frak L \left \{\cos(2t) \right \}$..........(1)

Now, from standard Laplace transform formulas, we have

$\small \frak L\left\{\sin \left(2t\right)\right\}:\quad \frac{2}{s^2+4}$

$\small \frak L\left\{\cos \left(2t\right)\right\}:\quad \frac{s}{s^2+4}$

So, using these in (1) we have

$\small =4\cdot{2\over s^2+4}+8\cdot {s\over s^2+4}$

$\small ={8\over s^2+4}+ {8s\over s^2+4}$

$\small ={8+8s\over s^2+4}$

Thus, the answer is given by

$\small \frak L \{h(t)\}={8s+8\over s^2+4}$