5. 圖 2 回授系統的 r 為定值,當 G(s)為何時?可使穩態誤差為零:
【用戶】is hello
【年級】幼兒園下
【評論內容】先解穩態誤差: T(s) = G(s)/(1 + G(s)) Y(s) = T(s) R(s)終值定理: (可設 r 為步階響應 R(s) = 1/s) ySS(t) = lim( t -> infinity) y(t) = lim(s -> 0) sY(s) = lim(s -> 0) s*T(s)*R(s) = lim(s -> 0) T(s) 得 ySS(t) = T(0), 對步階響應 (廣義: ySS(t) = T(s = 0)*(<s = 0>*R(s = 0)) )解穩態誤差: eSS = (1 - T(0))A: T(s) = (s + 2)/[(s + 1)2 +1] T(0) = 2/2 = 1 eSS = 0B: T(s) = (s + 2)/(s2 + 3s - 1) T(0) = 2/(-1) = -2 eSS = 3C: T(s) = (-4s + 3)/(s - 2)2 T(0) = 3/(-2) = - 3/2 eSS = 2.5D: T(s) = (1 - 2s)/(s2 - s +1) T(0) = 1/1 = 1...