For the Bode plot of the system

\(G(s)=\frac{10}{0.66s^2+2.33s+1}\) the corner frequencies are:

This question was previously asked in

ESE Electronics 2011 Paper 2: Official Paper

Option 4 : 0.50 and 3.00

CT 1: Ratio and Proportion

2536

10 Questions
16 Marks
30 Mins

**Concept:**

**Corner frequency: **

The point in Bode plot, where slope changes i.e. location of poles & zeros.

**Calculation:**

\(G\left( s \right) = \frac{{10}}{{0.66{s^2} + 2.33s + 1}}\)

\( = \frac{{15.15}}{{\left( {s + 0.5} \right)\left( {s + 3.03} \right)}}\)

From above open loop transfer function we get,

Poles = - 0.5, - 3.03

∴ No of corner frequencies = 2

∴ Given system have corner frequency = 0.5, 3.03

**Note:**

In general, bode plot will have transfer function in time constant form.

i.e.

\(G\left( s \right) = \frac{{10}}{{0.5\left( {\frac{s}{{0.5}} + 1} \right)3.03\left( {\frac{s}{{3.03}} + 1} \right)}} = \frac{{6.6}}{{\left( {\frac{s}{{0.5}} + 1} \right)\left( {\frac{s}{{3.03}} + 1} \right)}}\)

where, std. the open-loop transfer function for bode plot;

\(G\left( s \right) = \frac{K}{{\left( {\frac{s}{{{w_{{c_1}}}}} + 1} \right)\left( {\frac{s}{{{w_{{c_2}}} + 1}}} \right)}};\)

By comparing;

K = 6.6

\({w_{{c_1}}}\) & \({w_{{c_2}}}\) are corner frequencies.