The asymptotic Bode phase plot of \(\rm G(s)=\frac{{\rm{K}}}{{\left( {{\rm{s}} + 0.1} \right)\left( {{\rm{s}} + 10} \right)\left( {{\rm{s}} + {{\rm{p}}_1}} \right)}}\), with K and 𝑝_{1} both positive, is shown below.

The value of 𝑝_{1} is ________.

This question was previously asked in

GATE EC 2016 Official Paper: Shift 2

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

__Concept:__

**Corner Frequency**

The frequency at which slope changes from one level to another level.

It is nothing but the finite poles & zeroes locations in the form of magnitude.(1/τ)

**Asymptotic Bode Phase Plot**

Consider the transfer function with the single pole:

\(G(s) = \frac{1}{1+\frac{s}{ω _0}}\)

\(G(jω)=\frac{1}{1+\frac{jω}{ω _0}}\)

**At lower frequency (ω << ω _{0})**

G(jω) = 1 and** ∠G(jω) = 0° **

**At higher frequency (**ω >> ω0)

\(G(jω ) = \frac{1}{\frac{jω }{ω _0}}\)

∠G(jω) = - 90°

At the corner frequency (ω = ω0)

∠G(jω) = - 45°

NOTE: ω values are considered in log scale

__Calculation:__

The slope is calculated by:

\(slope = \frac{d\phi}{d(logω)}\)

Converting the transfer function into the time-constant form:

\(G(s)=\frac{\frac{K}{p_1}}{(1+\frac{s}{0.1})(1+\frac{s}{10})(1+\frac{s}{p_1})}\)

Consider the plot with the different points as shown:

The slope of OA is:

\(Slope = \frac{{ - 45^\circ - 0^\circ }}{{\log \left( {0.1} \right) - \log \left( {0.01} \right)}}\)

\(slope = \frac{{ - 45^\circ - 0^\circ }}{{\log \left( {\frac{{0.1}}{{0.01}}} \right)}}\)

**Slope = - 45°**

The slope of AB is:

\(Slope = \frac{{ - 225^\circ - (-45^\circ) }}{{\log \left( {10} \right) - \log \left( {0.1} \right)}}\)

\(slope = \frac{{ - 225^\circ +45^\circ }}{{\log \left( {\frac{{10}}{{0.1}}} \right)}}\)

\(Slope = \frac{-180^\circ}{2}\)

Slope = - 90°

The slope of BC is:

\(Slope = \frac{{ - 275^\circ - (-225^\circ) }}{{\log \left( {100} \right) - \log \left( {10} \right)}}\)

\(slope = \frac{{ - 275^\circ +225^\circ }}{{\log \left( {\frac{{100}}{{10}}} \right)}}\)

Slope = - 45°

If there is no pole between 0.1 and 10 then the slope will be - 45° but here the slope is - 90°. This implies that there is an existing pole between 0.1 and 10

From the given phase plot it is visible that the angle at the corner frequency 1 is - 135° (- 45° + (- 90°)). So, one pole is adding at the corner frequency 1 rad/sec.

By adding all the angles we get the phase plot which is given in the question

**NOTE: When systems are multiplied their angles are added.**

__Conclusion:__

Value of p_{1} = 1.