# The Bode plot of the open-loop transfer function of a system is described as foLlws: Slope – 40 dB/decade      ω < 0.1 rad/s Slop – 20 dB/decade        0.1 < ω < 10 rad/s Slope     0                          ω > 10 rad/s The system described will have

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ESE Electronics 2016 Paper 2: Official Paper
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1. 1 pole and 2 zeros
2. 2 poles and 2 zeros
3. 2 poles and 1 zero
4. 1 pole and 1 zero

Option 2 : 2 poles and 2 zeros
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## Detailed Solution

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:

As per the given details bode plot is:

DIAGRAM

The initial Slope of -40 dB indicates 2 poles at Origin.

The final slope of 0 dB indicates that 2 more zeros are there in the system.

Hence,

P = Z = 2