SOLUTION

The frequency response function is obtained by simply substituting s = jω

$\begin{array}{l}G(s) = \dfrac{{s + 10}}{{(s + 1)(s + 2)}}\\\\G(j\omega ) = \dfrac{{j\omega + 10}}{{(j\omega + 1)(j\omega + 2)}}\\\\\angle G(j\omega ) = {\tan ^{ – 1}}\dfrac{\omega }{{10}} – \left( {{{\tan }^{ – 1}}\dfrac{\omega }{1} + {{\tan }^{ – 1}}\dfrac{\omega }{2}} \right)\end{array}$

For ω = 0

$\angle G(j\omega ) = {\tan ^{ – 1}}\dfrac{0}{{10}} – \left( {{{\tan }^{ – 1}}\dfrac{0}{1} + {{\tan }^{ – 1}}\dfrac{0}{2}} \right) = 0^\circ $

For ω = ∞

$\angle G(j\omega ) = {\tan ^{ – 1}}\dfrac{\infty }{{10}} – \left( {{{\tan }^{ – 1}}\dfrac{\infty }{1} + {{\tan }^{ – 1}}\dfrac{\infty }{2}} \right) = 0^\circ $

90º − (90º + 90º)

−90º