The Bode magnitude plot of the transfer function $G\left(s\right)=\frac{K\left(1+0.5s\right)\left(1+as\right)}{s\left(1+\frac{s}{8}\right)\left(1+bs\right)\left(1+\frac{s}{36}\right)}$ is shown below. Note that – 6dB/octave = -20 dB/decade. The value of $\frac{a}{bK}$ is ___________

Concept:

Bode plot transfer function is represented in standard time constant form as

$T\left(s\right)=\frac{k\left(\frac{s}{{\omega }_{c_1}}+1\right)\dots }{\left(\frac{s}{{\omega }_{c_2}}+1\right)\left(\frac{s}{{\omega }_{c_3}}+1\right)\dots }$

ωc1, ωc2, … are corner frequencies.

In a Bode magnitude plot,

- For a pole at the origin, the initial slope is -20 dB/decade
- For a zero at the origin, the initial slope is 20 dB/decade
- The slope of magnitude plot changes at each corner frequency
- The corner frequency associated with poles causes a slope of -20 dB/decade
- The corner frequency associated with poles causes a slope of -20 dB/decade
- The final slope of Bode magnitude plot = (Z – P) × 20 dB/decade

Where Z is the number of zeros and P is the number of poles

Calculation:

By given magnitude plot

$G\left(s\right)=\frac{K\left[1+\frac{s}{2}\right]\left[1+\frac{s}{4}\right]}{s\left[1+\frac{s}{8}\right]\left[1+\frac{s}{24}\right]\left[1+\frac{s}{26}\right]}$

By comparing with the given transfer function,

$G\left(s\right)=\frac{K\left(1+0.5s\right)\left(1+as\right)}{s\left(1+\frac{s}{8}\right)\left(1+bs\right)\left(1+\frac{s}{36}\right)}$

$a=\frac{1}{4};b=\frac{1}{24};$

For finding K:



$K={\left({\omega }_1\right)}^n$ ; where n is no. of poles from the given plot

$K={\left(8\right)}^1$

$\Rightarrow K=8$

so, $\frac{a}{bK}=\frac{\frac{1}{4}}{\frac{1}{24}\cdot 8}=0.75$