Analysis Of Engineering Systems - Homework 6

David Clark MAE 488 Homework #6

5.42) Input sys1=tf([18 10],[3,18,10]) step(sys1) sys2=tf([10],[3,18,10]) step(sys2) Output

Overshoot: 6.56%

Peak time: 0.924 Settling time: 3.02

Rise time: 0.298

David Clark MAE 488 Homework #6

Overshoot: 0% 3.6

Peak time: 8.91

8.59) Input program: sys1=tf([0.2 1],[1 24 225 900 2500]) sys2=tf([1 1],[1 24 225 900 2500]) sys3=tf([10 1],[1 24 225 900 2500]) step(sys1) step(sys2) step(sys3) Output

Settling time: 6.51

Rise time:

David Clark MAE 488 Homework #6

Overshoot : 29.6% 1.87

Peak time: 0.745 Rise time: 0.285

Settling time:

David Clark MAE 488 Homework #6

Overshoot : 190% 2.38

Peak time: 0.497 Rise time: 0.115

Overshoot: 2250%

Peak time: 0.469 Rise time: 0.0376 Settling time: 2.33

8.68 Input rlocus([1,8,8,0],[1,16,9,8,4]) sgrid(0.707,[]) K=3.69 roots([1,(16+K),(9+8*K),(8+8*K),4]) rlocus([1,8,8,0],[1,16,9,8,4]) sgrid([],0.25) sys1=tf([8,4],[1,17.51,21.08,14.04]) step(sys1) sys2=tf([1,9.51,5],[1,17.51,21.08,14.04,4]) step(sys2) Output

Settling time:

David Clark MAE 488 Homework #6

K= 3.6900 ans = -17.6245 -0.9725 + 0.9673i -0.9725 - 0.9673i -0.1206

David Clark MAE 488 Homework #6

David Clark MAE 488 Homework #6