Which one of the following transfer functions represents the Bode plot as shown in the figure (where K is constant)?

F2 U.B Madhu 1.11.19 D 3

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  1. \(\frac{{K{s^2}}}{{{{\left( {1 + \frac{s}{{10}}} \right)}^3}}}\)
  2. \(\frac{{K{s^2}}}{{{{\left( {1 + \frac{s}{{10}}} \right)}^4}}}\)
  3. \(\frac{{K{s^2}}}{{{{\left( {1 + \frac{s}{{10}}} \right)}^5}}}\)
  4. \(\frac{{K{s^2}}}{{{{\left( {1 + \frac{s}{{10}}} \right)}^2}}}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{{K{s^2}}}{{{{\left( {1 + \frac{s}{{10}}} \right)}^5}}}\)
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Detailed Solution

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Concept:

In a bode magnitude plot, a pole adds a slope of -20 dB/decade whereas a zero adds a slope of 20 dB/decade.

The initial slope of the magnitude plot is ±20n dB/decade

+ for n zeros at origin

- for n poles at the origin

The standard transfer function of a Bode magnitude plot is:

\(TF = \frac{{K\left( {1 + \frac{s}{{{\omega _1}}}} \right)\left( {1 + \frac{s}{{{\omega _2}}}} \right) \ldots }}{{{s^n}\left( {1 + \frac{s}{{{\omega _3}}}} \right)\left( {1 + \frac{s}{{{\omega _4}}}} \right) \ldots }}\)

Here, ω1, ω2, ω3, ω4, … are the corner frequencies.

n is the number poles at the origin.

Calculation:

The initial slope of Bode plot is +40 dB/ decade, so the transfer function will have two zeros 2 at origin.

The slope of Bode plot changes to -60 dB/ decade

Change in slope = -100 dB/decade

So, it will have 5 poles at ω = 10 rad/sec.

Now, the transfer function is \(\frac{{K{s^2}}}{{{{\left( {1 + \frac{s}{{10}}} \right)}^5}}}\)

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