Question

The Bode plot of a transfer function G(s) is shown in the figure below:

The gain $\left(20log|G\right(s\left)|\right)$ is 32 dB and -8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative for all $\omega$. The G(s) is.

• A. $\frac{39.8}{s}$
• B. $\frac{39.8}{s^2}$
• C. $\frac{32}{s}$
• D. $\frac{32}{s^2}$

Answer 1

The correct option is B $\frac{39.8}{s^2}$

Let,
$G\left(s\right)=\frac{K}{s^n}$

$G\left(j\omega \right)=\frac{K}{(j\omega {)}^n}$

$\because Gain=32dBat\omega =1rad/sec$

$\because 20log|\frac{K}{1^n}|=32$

$\Rightarrow 20logK=32$

$\Rightarrow K=39.8$

Now,
$\because Gain=-8dBat\omega =10rad/sec.$

$\therefore 20log|\frac{39.8}{{10}^n}|=-8$

$\Rightarrow {10}^n=100$

$\Rightarrow n=2$