Consider the following statements: For a type-1 and a unity feed

SOLUTION

Concept:

Error constant and e_ss

Below table gives all information about the e_ss and error constant.

Type of input	Input: r(t)	ess	Error constant
Step	A.u(t)	\(\frac{A}{{1 + {k_p}}}\)	Position error constant \({k_p} = \mathop {\lim }\limits_{s \to 0} G\left( s \right)\)
Ramp	A.t.u(t)	\(\frac{A}{{{k_v}}}\)	Velocity error constant \({k_v} = \mathop {\lim }\limits_{s \to 0} {s}G\left( s \right)\)
Parabolic	A.u(t).t2/2	\(\frac{A}{{{k_a}}}\)	Acceleration error constant \({k_a} = \mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)\)

 

Calculation:

Let the type 1 system be assumed as:

\(G\left( s \right) = \frac{1}{{s\left( {1 + s\tau } \right)}}\)

Positional error constant

\({k_p} = \mathop {\lim }\limits_{s \to 0} \frac{1}{{s\left( {1 + s\tau } \right)}} = \infty \)

Statement 1 is wrong.

Acceleration error constant

\({k_a} = \mathop {\lim }\limits_{s \to 0} \frac{{{s^2}}}{{s\left( {1 + s\tau } \right)}} = 0\)

Statement 2 is correct.

Given input is Unit step

\({e_{ss}} = \frac{A}{{1 + {k_p}}}\)

We have calculated k_p and the value is ∞

\({e_{ss}} = \frac{1}{{1 + \infty }} = 0\)

Statement 3 is correct.

Tips & Tricks:

Input is taken as:

For a Step input, the input is taken as 0.

For Ramp input, the input is taken as 1.

For Hyperbolic input, the input is taken as 2.

Here we can conclude that input is taken as the power of ‘t’

Step input = t⁰u(t)

Ramp input = t¹u(t)

Hyperbolic input = u(t)t²/2

If Type > input then e_ss = 0

If Type < input then e_ss = ∞

If Type = input then e_ss is calculated as mentioned in the table which is in concept.