i. All static error co-efficients ($ K_p , K_v , K_a $)

ii. Steady state error of ramp input with magnitude 4.

Mumbai University > Electronics Engineering > Sem 4 > Linear Control Systems

Topic : Time Response Analysis

Difficulty : High

Marks : 10M

Given, G(s) = $\frac{20(s+1)}{s^2(s+2)(s+4)}$

let, H(s) = 1 (unity feedback)

=> G(s)H(s) = $\frac{20(s+1)}{s^2(s+1)(s+4)}$

Since, there is two pole at the origin, This is type 2 system.

Hence it would give us finite steady state error only for parabolic input. For step and ramp input, it would give us zero steady state error.

1. To find static error coefficient

$$K_p = \lim_{s\to0} G(s).H(s)$$

$$\lim_{s \to 0} \frac{20(s+1)}{s^2 (s+2)(s+4)}$$

$$ \therefore K_P = {\infty}$$

$$K_V = \lim_{s \to 0} S.G(s).H(s)$$

$$K_V = \lim_{s \to 0} \frac{20(s+1)}{s(s+2)(s+4)}$$

$$ \therefore K_V = \infty$$

$$K_a = \lim_{s \to 0} s^2G(s)H(s)$$

$$K_a = \lim_{s \to 0} s^2 \frac{20(s+1)}{s^2(s+2)(s+4)}$$

$$K_a = \lim_{s \to 0} \frac{20(s+1)}{(s+2)(s+4)}$$

$$K_a = \frac{20(0+1)}{(0+2)(0+4)}$$

$$K_a = \frac{20}{8}$$

$$K_a = 2.5$$

2. Steady state error of ramp i/p with magnitude 4

A = 4

$$ess = \frac{A}{K_V} = \frac{4}{\infty}$$

$$ess = 0$$