The open-loop transfer function of the system is: \(G\left( s \r
The open-loop transfer function of the system is:
\(G\left( s \right)H\left( s \right) = \frac{{10{{\left( {s + 5} \right)}^2}}}{{s\left( {s + 2} \right)\left( {s + 10} \right)}}\)
With respect to the Bode magnitude plot, which of the following is / are correct?A. Magnitude at ω = 0.1 rad / sec is 20 dB
B. Initial slope of the line is -20 dB / decade
C. Slope of the line between frequency 5 rad / sec to 10 rad / sec is +40 dB / dec
D. Slope of the line at high frequency is -20 dB / dec
Please scroll down to see the correct answer and solution guide.
Right Answer is:
SOLUTION
\(G\left( s \right)H\left( s \right) = \frac{{10 \times {5^2}{{\left( {1 + \frac{s}{5}} \right)}^2}}}{{s \times 2\left( {1 + \frac{s}{2}} \right)\left( {10} \right)\left( {1 + \frac{s}{{10}}} \right)}}\)
\( = \frac{{12.5{{\left( {1 + \frac{s}{5}} \right)}^2}}}{{s\left( {1 + \frac{s}{2}} \right)\left( {1 + \frac{s}{{10}}} \right)}}\)
Calculation:
\(M{\left. \right|_{\omega = 0.1}} = 20\log 12.5 - 20\log \omega \)
= 20 log 12.5 – 20 log 0.1
≈ 42 dB
Observations:
1) Initial Slope is -20 dB/dec
2) Slope of the line between 5 rad / sec to 10 rad / sec is 0 dB / dec.
3) At high frequency, the slope of the line is -20 dB / dec.