Question

The Laplace transform of a signal x(t) is given by , $X\left(s\right)=\frac{s^2+2s+1}{s^2-s+1}$
$ifx\left(t\right)=\delta \left(t\right)+2\sqrt{3}{e}^{0.5t}\left(cos(\frac{\sqrt{3}}{2}t+\theta )\right)u\left(t\right)$, then $\theta$ equal to ______degrees.

• A. -30

Answer 1

The correct option is A -30
$X\left(s\right)=\frac{s^2-s+1+3s}{s^2-s+1}=1+\frac{3s}{(s-\frac{1}{2}{)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$
$=1+\frac{3(s-\frac{1}{2})}{(s-\frac{1}{2}{)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}+\frac{\frac{3}{2}}{(s-\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}$

$\Rightarrow x\left(t\right)=\delta \left(t\right)+(3e^{0.5t}cos\left(\frac{\sqrt{3}}{2}t\right)+\sqrt{3}{e}^{0.5t}sin\left(\frac{\sqrt{3}}{2}t\right))u\left(t\right)$

$=\delta \left(t\right)+\left(\sqrt{12}{e}^{0.5t}cos(\frac{\sqrt{3}}{2}t-ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right))\right)u\left(t\right)$

$=\delta \left(t\right)+2\sqrt{3}{e}^{0.5t}cos(\frac{\sqrt{3}}{2}t-ta{n}^{-1}\left(\frac{\sqrt{3}}{2}t\right))u\left(t\right)$

$=\delta \left(t\right)+2\sqrt{3}{e}^{0.5t}cos(\frac{\sqrt{3}}{2}t-{30}^0)u\left(t\right)$

$\Rightarrow \theta =-{30}^o$