# Consider the transfer function: $G(s)=\frac{5(s^2+10s+100)}{s^2 (s^2+15s+1)}$ The corner frequencies in Bode’s plot for this transfer function are a

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Consider the transfer function:

$G(s)=\frac{5(s^2+10s+100)}{s^2 (s^2+15s+1)}$

The corner frequencies in Bode’s plot for this transfer function are as

1. 10 r/s and 10 r/s
2. 100 r/s and 10 r/s
3. 10 r/s and 1 r/s
4. 100 r/s and 1 r/s

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Correct Answer - Option 3 : 10 r/s and 1 r/s

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.

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 zeros and P is the number of poles

Calculation:

$G(s)=\frac{5(s^2+10s+100)}{s^2 (s^2+15s+1)}$

$G(s)=\frac{5\times100(\frac{s^2}{100}+\frac{s}{10}+1)}{s^2 (s^2+15s+1)}$

Corner frequency in bode plot is defined for finite poles and zeros but in this given system poles and zeros are complex.

Hence, Corner frequencies = 10 r/s, 1r/s