Question

For the transfer function, \(G\left( s \right) = \frac{{5\left( {s + 4} \right)}}{{s\left( {s + 0.25} \right)\left( {{s^2} + 10s + 25} \right)}}\), the values of the constant gain term and the highest corner frequency of the Bode plot respectively are:

1. 3.2, 5.0
2. 16.0, 4.0
3. 3.2, 4.0
4. 16.0, 5.0

Answer 1

Correct Answer - Option 1 : 3.2, 5.0

Concept:

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

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

ωc1, ωc2, … are corner frequencies.

k is the constant gain term

Calculation:

Given transfer function is 

\(G\left( s \right) = \frac{{5\left( {s + 4} \right)}}{{s\left( {s + 0.25} \right)\left( {{s^2} + 10s + 25} \right)}}\)

\(= \frac{{5 \times 4\left( {1 + \frac{s}{4}} \right)}}{{0.25 \times 25 \times s \times \left( {1 + \frac{s}{{0.25}}} \right)\left( {1 + \frac{{10s}}{{25}} + \frac{{{s^2}}}{{25}}} \right)}}\)

\(= \frac{{3.2\left( {1 + \frac{s}{4}} \right)}}{{s\left( {1 + \frac{s}{{0.25}}} \right)\left( {1 + \frac{{10s}}{{25}} + \frac{{{s^2}}}{{25}}} \right)}}\)

Constant gain = 3.2

Corner frequencies = 0.25, 4, 5

Highest corner frequency = 5 rad/sec