Correct Answer - Option 2 : Decrease the time constant and decrease the steady state error
Concept:
LTI system in time constant form is defined as:
\(C\left( s \right) = \frac{{K\left( {1 + s{τ _1}} \right)\left( {1 + s{τ _2}} \right) \cdots }}{{{s^n}\left( {1 + s{τ _a}} \right)\left( {1 + s{τ _b}} \right) \cdots }}\)
τ1, τ2 ⋯ , and τa, τb ⋯ are time constants.
Steady-state error:
Kp = position error constant
= \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
Kv = velocity error constant
= \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
Ka = acceleration error constant
= \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)
Steady-state error for different inputs is given by
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
|
∞
|
∞
|
\(\frac{1}{{{K_a}}}\)
|
Calculation:
Given:
\(G(s)H(s)=\frac{K}{(1+sτ)}\)
GC(s) = KP
The close loop transfer function is given as:
\(CLTF=\frac{KK_P}{1+sT+KK_P}\)
\(CLTF=\frac{\frac{KK_P}{1+KK_P}}{1+\frac{T}{KK_P}s}\)
Comparing with the standard forms we get time constants as:
\(τ = \frac{T}{1+KK_P}\)
\(K_p=\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
\(K_p=\mathop {\lim }\limits_{s \to 0} \frac{KK_P}{1+sT}\)
\(K_p=KK_P\)
Steady-state error is calculated as:
\(e_{ss}=\frac{1}{{1 + {K_p}}}\)
\(e_{ss}=\frac{1}{{1 + {KK_P}}}\)
Hence, If KP is increased then both time constant and ess decreases.
Note: Don't get confused with KP and Kp.
KP: Transfer function of a proportional controller.
Kp: position error constant