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LMRC Assistant Manager Electrical : 2018 Paper

Option 4 : 12 dB per octave

CT 1: Current Affairs (Government Policies and Schemes)

54560

10 Questions
10 Marks
10 Mins

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

Application:

The given system is second order system. So, the number of poles is 2.

The slope of the asymptote = 2 × 20 dB/decade = 40 dB/decade

20 bB/decade = 6 dB/octave

Therefore, the slope of the asymptote = 12 dB/octave