Question

22. For the unity feedback system given in Figure P9.1 with G(S) = 5(+ 5)(s + 11) do the following: [Section: 9.4] a. Find the gain, K, for the uncompensated system to operate with 30% overshoot. b. Find the peak time and K, for the uncompensated system, c. Design a lag-lead compensator to decrease the peak time by a factor of 2, decrease the percent overshoot by a factor of 2, and improve the steady-state error by a factor of 30. Specify all poles, zeros, and gains. R(s) + E(s) C(S) G(s) FIGURE P9.1

Answer 1

ANSWER:

Given that, for the unity feedback system with G(S) - s (5+5)(5+1) a) Now, we have to find system the gain, K.for operate with the uncompensated to 30 / overshoot. Pormula is, fo ovess hoof -57 X100 e 1-52 Ymp e forom the given information lmp:30 ore et X 100 30 -Sī 100 Ni-sa 3 (12) septe

Squaring 2 on both sides then we qet: 1.44 - => 1.44(1-52), 55,2 => 144-1.446=(9.14)52 >> 1.44 - 1.44s?9.85952 1.44-1.445²-9-85953 0 => 1.44-11-89959=0 11.899537.44 5 1.44/ 11:299 s? 0.1274 >> S-0.35 18° degree yield -1.464 +53.816 witte again of K was "218"

b) NOW, formula for peak time Тр: ан : 3.818 3.143.818 Tp = 0. 8aas Kvasi ( from formula tos velocity euros constant) 218 55 Kv - 3.96 C) From the compensator given information in tead design, the percentage Over Shoot was 15% and peak time coas 0.4116 s - n (%100) 20.517 V2+12(4.100) aze : 7.634

: condo- TRX design point is located at - Scont joan de-sa -461 +jay, 634 The un compensated System Poles as well as the compensatece Zero al -5 yields .171.2 Compensator Pole Corylcobuce is 1771.2. 180's -8.8" ferom the fiqure, 1 7.634 , tan 8.8° Pc-4.61 7.634 = (Pc -1161 (0:1548) Pc-4.61 - 49.3152 Pcs 49.3152 +4.61 - 53.9252 The compensated open loop aanstcę function $($+11) (5+53-92)

Function at Point -9.61 +57.624 Poduce the 'k' value al 44zo ; Kve 216.6 5(11) Kyo 26:6 218.6 e 55 = 3.9-75 Kv was 30 (3,975) = 119-25 Then the lead compensated kv 4430 Kv 1(53.92) 7.469 = 119.25 d 7.469 =15.97 F! It is the required impsovement over the lead compensated system] Sto.01597 Geog(s) = St0.001 4430 (5+0.1597) s(s+11) (5+53.92) (S+0.001)