Question

2) a) Derie weayti for the critically damped case, 5-1. b) Determine the position, velocity and acceleration error constants for the forward-path 4(s+2 s" (s+5) transfer function given by G(s)-- c) Determine the steady-state errors of the unit feedback system with the forward-path paroc i ansfer funcioo given ru n

Answer 1

a)

Standard second order system:

$\small G(s) = \frac{w_n^2}{s^2+2\zeta w_n s+w_n^2}$

for critically damped case:

ZWnS

$\small G(s) = \frac{w_n^2}{(s+w_n)^2}$

Response to unit step input:

$\small C(s) = \frac{w_n^2}{(s+w_n)^2}*\frac{1}{s}$

Writing it as partial fractions:

$\small C(s) = \frac{1}{s}-\frac{1}{(s+w_n)}-\frac{w_n}{(s+w_n)^2}$

$\small c(t) = L^{-1}[\frac{1}{s}-\frac{1}{(s+w_n)}-\frac{w_n}{(s+w_n)^2}]$

c(t) = 1-e wnte unt

$\small c(t) = 1 - e^{-w_n t}(1+ w_n t )$

b)

G(s) = 4(8+2)

Position error

es,-lims *-(1-G(s))

ess = 1-lim G(s)

4(s + 2) ess = 1-1-0 s2(s + 5) im

$\small e_{ss} = \infty$

Position error constant:

$\small \Rightarrow K_p = \frac{1}{e_{ss}} = 0$

velocity error

$\small e_{ss} = \lim_{s\rightarrow 0} s*\frac{1}{s^2}(1-G(s))$

$\small e_{ss} = \lim_{s\rightarrow 0}\frac{1}{s}(1-G(s))$

$\small e_{ss} =\lim_{s\rightarrow 0}\frac{1}{s}( 1- \frac{4(s+2)}{s^2(s+5)})$

$\small e_{ss} =\lim_{s\rightarrow 0}\frac{1}{s}( \frac{s^2(s+5)-4(s+2)}{s^2(s+5)})$

$\small e_{ss} = \infty$

velocity error constant:

$\small \Rightarrow K_v = \frac{1}{e_{ss}} = 0$

acceleration error

$\small e_{ss} = \lim_{s\rightarrow 0} s*\frac{1}{s^3}(1-G(s))$

$\small e_{ss} = \lim_{s\rightarrow 0}\frac{1}{s^2}(1-G(s))$

$\small e_{ss} =\lim_{s\rightarrow 0}\frac{1}{s^2}( 1- \frac{4(s+2)}{s^2(s+5)})$

ess linm

$\small e_{ss} = \infty$

acceleration error constant:

$\small \Rightarrow K_a = \frac{1}{e_{ss}} = 0$

c)

G(s) = 4(8+2)

closed looped transfer function is:

$\small F(s) = \frac{G}{1+G}$

$\small F(s) = \frac{\frac{4(s+2)}{s^2(s+5)}}{1+\frac{4(s+2)}{s^2(s+5)}}$

$\small F(s) = \frac{4(s+2)}{s^2(s+5)+4(s+2)}$

$\small F(s) = \frac{4(s+2)}{s^3+5s^2+4s+8}$

Position error

$\small e_{ss} =\lim_{s\rightarrow 0} s*\frac{1}{s}(1- F(s))$

$\small e_{ss} = 1- \lim_{s\rightarrow 0}F(s)$

$\small e_{ss} = 1- \lim_{s\rightarrow 0} \frac{4(s+2)}{s^3+5s^2+4s+8}$

$\small e_{ss} = 0$

Position error constant:

$\small \Rightarrow K_p = \frac{1}{e_{ss}} = \infty$

velocity error

$\small e_{ss} =\lim_{s\rightarrow 0} s*\frac{1}{s^2}(1- F(s))$

$\small e_{ss} = \lim_{s\rightarrow 0} \frac{1}{s}(1- F(s))$

$\small e_{ss} = \lim_{s\rightarrow 0} \frac{1}{s}(1- \frac{4(s+2)}{s^3+5s^2+4s+8})$

$\small e_{ss} = \lim_{s\rightarrow 0} \frac{1}{s}(\frac{s^3+5s^2}{s^3+5s^2+4s+8})$

$\small e_{ss} = \lim_{s\rightarrow 0} (\frac{s^2+5s}{s^3+5s^2+4s+8})$

$\small e_{ss} = 0$

velocity error constant:

$\small \Rightarrow K_v = \frac{1}{e_{ss}} = \infty$

acceleration error

$\small e_{ss} =\lim_{s\rightarrow 0} s*\frac{1}{s^3}(1- F(s))$

$\small e_{ss} = \lim_{s\rightarrow 0} \frac{1}{s^2}(1- F(s))$

$\small e_{ss} = \lim_{s\rightarrow 0} \frac{1}{s^2}(1- \frac{4(s+2)}{s^3+5s^2+4s+8})$

$\small e_{ss} = \lim_{s\rightarrow 0} \frac{1}{s^2}(\frac{s^3+5s^2}{s^3+5s^2+4s+8})$

$\small e_{ss} = \lim_{s\rightarrow 0} (\frac{s+5}{s^3+5s^2+4s+8})$

$\small e_{ss} = 5/8$

acceleration error constant:

$\small \Rightarrow K_a = \frac{1}{e_{ss}} = 8/5$