The Bode asymptotic magnitude plot of a minimum phase system is shown in the figure.

If the system is connected in a unity negative feedback configuration the steady-state error of the closed-loop system, to a unit ramp input, is _______.

This question was previously asked in

GATE EC 2014 Official Paper: Shift 2

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

__Concept__**:**

Steady-state error due to ramp Input is given as:

\({e_{ss}} = \frac{1}{{{k_a}}}\)

Where:

\({k_a} = \mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

__Analysis__**:**

From the bode plot given, we can write:

Initial slope = -20 dB/dec which indicates 1 pole at origin.

At ω_{1} = 2, Slope = -20 + 20 which indicates 1 zero at ω_{1} = 2.

At ω_{2} = 10, slope = 0 – 20 which indicates 1 pole at ω_{2} = 10.

The open-loop transfer function of the system is:

\(G\left( s \right)H\left( s \right) = \frac{{k\left( {\frac{s}{2} + 1} \right)}}{{s\left( {\frac{s}{{10}} + 1} \right)}}\) ---(1)

26.02|_{ω = 0.1} = 20 log k – 20 log ω

20 log k = 6.02

k = 1.99

k ≈ 2

From equation (1), we can write:

\(G\left( s \right)H\left( s \right) = \frac{{2\left( {\frac{{s + 2}}{2}} \right)}}{{s\left( {\frac{{s + 10}}{{10}}} \right)}}\)

\(G\left( s \right)H\left( s \right) = \frac{{10\left( {s + 2} \right)}}{{s\left( {s + 10} \right)}}\)

Now, the steady-state error will be:

\(\left( {{e_{ss}}} \right) = \frac{1}{{{k_a}}}\)

\({k_a} = \mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

\( = \mathop {\lim }\limits_{s \to 0} s \times \frac{{10\left( {s + 2} \right)}}{{s\left( {s + 10} \right)}}\)

k_{a} = 2

\({e_{ss}} = \frac{1}{2} = 0.5\)