Question

The steady-state error for a Type 0 system for unit-step input is 0.2. In a certain instance, this error possibility was removed by the insertion of a unity gain block. Thereafter, a unit ramp was applied. The nature of the block and new steady-state error in this changed configuration will, respectively, be
1. integrator, 0.25
2. differentiator, 0.25
3. integrator, 0.20
4. differentiator, 0.20

Answer 1

Correct Answer - Option 1 : integrator, 0.25

Concept:

In type – 0 system, the steady-state error (e_ss) is defined for the step-input signal as:

\({e_{ss}} = \frac{A}{{1 + {k_p}}}\) 

Where,

A = magnitude of the step-input signal

K_p = positional error constant

K_p = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

Calculation:

Given that, Unit – step signal A = 1 and e_ss = 0.2.

\({e_{ss}} = \frac{1}{{1 + {K_p}}}\) 

\(\therefore {K_p} = \frac{1}{{0.2}} - 1\) 

= 5 -1

K_p = 4

So, the open-loop Transfer function will be:

\(= \frac{4}{{{s^0}}}\) type – 0 system.

\(\therefore {\rm{\;}}\mathop {\lim }\limits_{s \to 0} G\left( S \right)H\left( S \right) = {K_p} = 4\) 

Now, the error will be removed from the system if we increase the type of system form type – 0 to type – 1 system, i.e.

Type – 0 means \(\frac{1}{{{s^0}}}\) term in the open-loop Transfer function.

And type – 1 means \(\frac{1}{{{s^1}}}\) term in the open-loop Transfer function.

Hence we say that unity –gain block must be an integrator i.e \(\frac{1}{s}\) in the open-loop Transfer function.

Now, Open-loop Transfer function will be:

\(G'\left( s \right) = \frac{{G\left( S \right).H\left( S \right)}}{S}\) 

It is a type – 1 system.

Since, given that Unit ramp signal is applying to the system,

So, \(e_{ss}^1 = \frac{1}{{{K_v}}};\) for type – 1 system.

And K_v = velocity error Constant

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

\(= \mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right) \times \frac{{G\left( s \right)H\left( S \right)}}{s}\) 

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

= K_p

K_v = 4

New steady-state error will be:

\(e_{ss}^1 = \frac{1}{4} = 0.25\) 

Important Points:

Steady-state Error for different types of System is shown in the table:

Type	Step input	Ramp input	Parabolic input
Type – 0	\(\frac{A}{{1 + {k_p}}}\)	∞	∞
Type – 1	0	\(\frac{A}{{{K_v}}}\)	∞
Type – 2	0	0	\(A/{K_a}\)

 

e_ss is only valid for a closed-loop stable system and calculated for a closed-loop system by using the open-loop transfer function.