Q$3$:

given:$1$

$z_n=0\mathrm{.}2$

${}_n=4$

$\frac{1}{(s+2)(s+4)+K}$

$K=5$

Find:

a$)$ For the given $K$ find the $P$ hase angle and the magnitude

b$)$ for the given Zm fin the value of $k$

c$)$ Find $$

$\frac{501}{a}$

$1$

$s=jw$

$=>\frac{1}{(jw+2)(jw+4)+5}$

$=>\frac{1}{(ubrace(j^2{w}^2+4jw+2jw+8+5\text{u}\text{b}\text{r}\text{a}\text{c}\text{e}))}$

$=>\frac{1}{-1}$ubrace$(w^2+13+$ubrace$(6$wubrace${)}_b$jubrace${)}_a$

$2$

$M=\frac{1}{\sqrt[\phantom{2}]{a^2+b^2}}$

$M=\frac{1}{\sqrt[\phantom{2}]{(-w^2+13{)}^2+(6w{)}^2}}$

$3$

$=ta{n}^{-1}(\frac{a}{b})$

$=ta{n}^{-1}(\frac{-w^2+13}{6w})$

b$)$

$s=a+bi$

$a=-m_n,b=w_n\sqrt[\phantom{2}]{1-z_n^2}$

$1$

$a=-0\mathrm{.}2(4),\frac{2}{b}=4\sqrt[\phantom{2}]{1-{0\mathrm{.}2}^2}$

$a=-0\mathrm{.}8$

$b=3\mathrm{.}92$

$3\frac{?}{b}$

$ar(s)=-0\mathrm{.}8+3\mathrm{.}92$hat$()$

$(s+2)(s+4)=-K$

$=>(-0\mathrm{.}8+3\mathrm{.}92j+2)(-0\mathrm{.}8+3\mathrm{.}92j+4)=-k$

$K=11\mathrm{.}53-17\mathrm{.}24i$

$k=11\mathrm{.}53$

c$)$

$B=co{s}^{-1}(z_n)$

$B=co{s}^{-1}(0\mathrm{.}2)=78\mathrm{.}5$

For the k is it correct?