零点在极点位置的左边
98’s Automatic Control Midterm_2 總分120
2009/12/16
1. Consider the system described in state variable form by
xtxtut()A()B(),, ytxt()C(),
where
010,,,, ABandC,,,,, , 11,,,,,,,,kk112,,,,
kk,kkand where and both and are real numbers. (10%) 1212
(a) Compute the roots of the characteristic polynomial. (5%)
(b) Use Routh-Hurwitz method to find the range of k, and k for stability. 12
(5%)
Ans:
2sksk,,(a) Characteristic polynomial= 21
(b) K>0 and K>0 12
2. A unity feedback system has the loop transfer function. (15%)
K10s,,, GsGs()(),css,5,,
(a) Determine the breakaway and entry points of the root locus and
sketch the root locus for K>0. (5%)
(b) Determine the gain K when the two characteristic roots have a
1damping ratio of . (5%)
2
(c) Calculate the roots. (5%)
3. A regulator system with unit step input is shown below. The
disturbance signal D(s) is modeled as a unit step function. The K is the
gain of a tunable proportional controller. Use mathematical analysis
and discuss the following questions when K=1, and K=10,
respectively. (20%)
(a) Set point regulation by analyzing settling time and steady-state error. (b) Steady-state error by disturbance.
T(c) The sensitivity of the closed-loop system T(s) with respect to K, . SK
(d) The total steady-state error due to both set point and disturbance.
disturbanceD(s)
20
1++KR(s)1-+(2)(1)ss,,set,point
1
Ans:
21T,2.67T,2.67(a) K=1 , K=10 e,,e,,()()ssrr36
20205(b) K=1 e,,, K=10 e,,, ()()dr1233
22ss,,32ss,,32TTS,S,(c) K=1 , K=10 KK22ss,,312ss,,33
eee()()(),,,,,(d) total steady totalrd
22022 K=1 e,,,, ()total333
1202211e,,,,, K=10 ()total612126
4. By graphics or setting and determine the phase behavior of a sj,,
sa,transfer function, , (10%) Gs(),sb,(a) when is a phase lead compensator.(5%) Gs()
(b) when is a phase lag compensator.(5%) Gs()
Ans:
(a)
(b)
零點在極點位置的左邊。
5. Explain the following terms(15%) (1) PID controller. (5%)
(2) Non-minimum phase system. (5%)
對於一個穩定的轉移函數,若零點在s平面的左半平面,則此系統稱為極小
相位系統(minimum phase system),反之若至少有一個零點在s平面的右半平
面,則稱為非極小相位系統(Non-minimum phase system)。非極小相位系統以
極座標表示系統的轉移函數,其大小(Magnitude)與相位(Phase)沒有任何關係
存在,以大小與相位作圖表示,則可看出大小的曲線雖已收斂(穩定),然而
相位仍舊繼續下降(改變),因此,非極小相位系統有可能會使系統發生不穩
定,通常系統有運輸延遲(Transportation lag)會出現非極小相位。 (3) Offset error. (5%)
Regulator響應的穩態誤差,以溫,調節控制系統為,,調節器在穩態時的溫,響應與期望的溫,值存在有固定偏差。
nd6. Derive that the percentage overshoot (P.O.)of a -order system to a 2
,,,21,,step input is where is the damping ratio. (15%) ,P.O.100,e
Ans:
KK7. For a DC-servo with position and tacho feedbacks, find and 12
to satisfy the control specs: settling time3 sec, dominant poles with ,
, ramp-input steady-state error 10%. (b)sketch root locus ,,0.707,
Kwith as parameter. (20%) 2
R(s)KC(s)+E(s)1a-ss(2),
1+Ks2
8. Use mathematical analysis to support the statement “The behavior of a
closed-loop system is dominated by its sensor when it is with high
gain open-loop transfer function.” (15%)
R(s
Y(s)+)G-
H
G1閉迴路系統可寫為 當很大時,,系統可寫為 HGHH1,GH
所以當系統sensor增益很大時,系統特性由sensor領導。
標準差19.7 分
平均56.8 分
龔泰宇 60.4 詹鎮宇 56.3 劉茂宏 64.6 黃世豪 42.1 顏瑋璘 77.1 黃啟哲 57.1 王祥賓 45 趙敏閔 12.5 楊宗瑋 47.1 陳俠安 44.2 彭忠驥 60.4 許博翔 50.8 林彬儀 55.8 黃昫璁 12.5 朱晃民 52.1 謝奇昇 47.5 黃軒蕙 65 蔡紹妤 14.2 林秉毅 39.2 朱証裕 45 吳嘉珮 76.7 盧天駿 54.2 鄔澤民 29.2 蘇少筠 病 呂致翰 46.3 王愷寧 25 劉奕甫 37.9 林佑安 62.1 蘇冠華 缺考 何恭愷 20.8 陳柏宇 79.2 陳健平 47.5 王文呈 50.4 蔡育安 81.3 楊敦榮 60 蔡沛吾 75 陳冠華 62.9 藍德鄰 52.1 江書豪 68.8 梁寶庭 65.8 莊子淵 77.5 廖紹宇 69.2 雷漢博 55 王誠正 44.2 吳咏翰 41.7 蘇牧寰 48.3 徐翊清 75.4 李政逸 79.2 曾俊諺 71.3 林雋哲 74.6 曾俊怡 93.8 謝任峰 75.4 吳健銘 79.2 張譽馨 92.1 黃景暉 66.7 吳宗翰 35.4 方曉涵 95.8 林睿奕 缺考 黃致堯 60.4