Question

For the Bode plot shown in figure, the transfer function is,

• A. $\frac{1-s}{1+s}$
• B. $\frac{1}{(1+s{)}^2}$
• C. $\frac{1}{s^2}$
• D. $\frac{1}{s(s+1)}$

Answer 1

The correct option is A $\frac{1-s}{1+s}$

The gain of the function is 1, given system is all pass system. Hence option (a) is correct answer.
We can cross check

$GH\left(s\right)=\frac{1-s}{1+s}$
$GH\left(j\omega \right)=\frac{1-j\omega }{1+j\omega }$
$\angle GH\left(j\omega \right)=-ta{n}^{-1}\left(\omega \right)-ta{n}^{-1}\left(\omega \right)=-2ta{n}^{-1}\left(\omega \right)$

By varying $\omega :0\to \infty$

The phase varies from $0^{o}to-{180}^o$ as shown in the plot.