系統(tǒng)的能控性能觀測性穩(wěn)定性分析-資料下載頁
2025-06-29 16:34本頁面
【正文】 3系統(tǒng)可觀a=0a = 10 27 18 1 0 0 0 1 0b = 1 0 0c = 0 1 0d = 0n = 3Qc = 1 10 73 0 1 10 0 0 1nc = 3系統(tǒng)可控Qo = 0 1 0 1 0 0 10 27 18no = 3系統(tǒng)可觀a=1a = 10 27 18 1 0 0 0 1 0b = 1 0 0c = 0 1 1d = 0n = 3Qc = 1 10 73 0 1 10 0 0 1nc = 3系統(tǒng)可控Qo = 0 1 1 1 1 0 9 27 18no = 2（c）已知系統(tǒng)矩陣為，，判別系統(tǒng)的能控性與能觀測性；n = 3Qc = 0 nc = 3系統(tǒng)可控Qo = 0 no = 3系統(tǒng)可觀（d）求系統(tǒng)的最小實現(xiàn)。Am = Bm = Cm = Dm = 0（2）穩(wěn)定性（a）代數(shù)法穩(wěn)定性判據(jù)z = 2p = 0 20 1k = 100（b）根軌跡法判斷系統(tǒng)穩(wěn)定性selected_point = + k =selected_point = +03poles = + + + （c）Bode 圖法判斷系統(tǒng)穩(wěn)定性已知兩個單位負反饋系統(tǒng)的開環(huán)傳遞函數(shù)分別為用Bode 圖法判斷系統(tǒng)閉環(huán)的穩(wěn)定性。G1（s）G2（s）（d）判斷下列系統(tǒng)是否狀態(tài)漸近穩(wěn)定、是否BIBO穩(wěn)定。z = p = + k = [P,D]=eig(A)P = + + D = 0 0 0 + 0 0 0 第 頁 共 頁
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