written 7.7 years ago by | • modified 4.9 years ago |
Determine
(i) Error coefficient
(ii) Steady state error for input as $1+4t+\frac{t^2}{2}$.
written 7.7 years ago by | • modified 4.9 years ago |
Determine
(i) Error coefficient
(ii) Steady state error for input as $1+4t+\frac{t^2}{2}$.
written 7.7 years ago by | • modified 7.7 years ago |
G(S)=$\frac{10(S+1)}{S^2 (S+2)(S+10)}$
H(S)=1, unity feedback
Error coefficient,
Position error coefficient
$k_p$=$\lim_{s\to 0}$ G(S) H(S)=$\lim_{s\to 0}$ $\frac{10(S+1)}{S^2 (S+2)(S+10)}$ .1
=$\frac {10(1)}{(0)(2)(10)}$=10/0=∞
$k_p$=∞ Velocity error coefficient.
$k_v$=$\lim_{s\to 0}$ S G(S) H(S)=lim/(s→0) $\frac{S. (10(S+1)}{(S^2 (S+2)(S+10)}$ .1
=$\frac{10(1)}{(0(2)(10)}$=10/0=∞ $k_v$=∞
Acceleration error coefficient. $k_a$= $\lim_{s\to 0}$ $S^2$G(S) H(S)=$lim_{s\to 0}$ $\frac{(S^2)(10(S+1)}{S^2(S+2)(S+10)}$.1 =$\frac{10(1)}{(2)10}$ =$\frac{10}{20}$ $k_p$= 1/2
Steady state error is the combination of all the three types of inputs viz. step of magnitude $A_1$=1, Ramp of magnitude $A_2$=4 and parabolic of magnitude $A_3$=1.
Therefore, steady state error, $e_ss$=$e_ss$1+$e_ss$2+$e_ss$3
=$A_1$/(1+$k_p$ )+$A_2$/$k_v$ +$A_3$/$k_a$
=$1/(1+∞)+4/∞+1/0.5$
=$1/∞+4/∞+1/0.5$
=$0+0+1/0.5$
$e_ss$=2