Question

The transfer function of a system is \(\frac{1}{{1 + sT}}\) The input to this system is the ramp function, tu(t). The output would track this system with an error givne by
1. zero
2. \(\frac{T}{2}\)
3. T
4. \(\frac{{{T^2}}}{2}\)

Answer 1

Correct Answer - Option 3 : T

Concept:

Steady-state error for different inputs is given by

for unit ramp input \({e_{ss}} = \frac{1}{K_v}\)

Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

Calculation:

\(CLTF = \frac{1}{{1+sT}}\)

\(OLTF = \frac{1}{{1+sT-1}}=\frac{1}{sT}\)

Thus, the given system is a type-1 system and the input is ramp input

\({k_v} = \mathop {\lim }\limits_{s \to 0} s\left( {\frac{{1}}{{sT}}} \right) = \frac{1}{T}\)

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

Steady state error formulas are applicable only to open loop transfer function. Therefore if closed-loop transfer function is given, we must first calculate the open-loop transfer function to apply the formula.

Steady state error for different inputs is given by

for unit step input KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

for unit ramp input Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

for unit parabollic input Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)

Input	Type -0	Type - 1	Type -2
Unit step	\(\frac{1}{{1 + {K_p}}}\)	0	0
Unit ramp	∞	\(\frac{1}{{{K_v}}}\)	0
Unit parabolic	∞	∞	\(\frac{1}{{{K_a}}}\)