Question

The asymptotic log-magnitude curve for open loop transfer function is sketched below,


Open loop transfer function is

• A. $T\left(s\right)=\frac{100(s+8)(s+4)}{s^2(s+1.268)}$
• B. $T\left(s\right)=\frac{16(s+1.268)(s+4)}{s^2(s+8)}$
• C. $T\left(s\right)=\frac{10(s+1.268)(s+8)}{s^2(s+4)}$
• D. $T\left(s\right)=\frac{8(s+1.268)(s+8)}{s^2(s+4)}$

Answer 1

The correct option is B $T\left(s\right)=\frac{16(s+1.268)(s+4)}{s^2(s+8)}$

From the above Bode plot,
For section de, slope is -20 dB/dec

$\therefore$ $-20=\frac{y-0}{log8-log16}$
y = 6.02 dB
Now for section bc, slope is -20 dB/dec

$\therefore$ $-20=\frac{16-6.02}{log{\omega }_1-log4}$
${\omega }_1=1.268$ rad/sec

To find value of gain K

y = mx + c
16 = -40 log 1.268 + 20 log K
K = 10.14

From all the result, transfer function is,

$T\left(s\right)=\frac{10.14(\frac{s}{1.268}+1)(\frac{s}{4}+1)}{s^2(\frac{s}{8}+1)}$

$T\left(s\right)=\frac{16(s+1.268)(s+4)}{s^2(s+8)}$