精華區beta NTU-Exam 關於我們 聯絡資訊
課程名稱︰自動控制 課程性質︰必修 課程教師︰王富正 開課學院:工學院 開課系所︰機械系 考試日期(年月日)︰101/10/30 考試時限(分鐘):180 是否需發放獎勵金:是 (如未明確表示,則不予發放) 試題 : Automatic Control 1st Midterm Exam, 30/Oct/2012 1. (5%) Given f(t) = e^-at sin(ωt), derive its Laplace Transform L[f(t)]. (必須有推導過程) 2. (20%) Refer to Fig.2, (1) (5%) Derive the two dynamic equations of the system. (2) (5%) Find the transfer function of X(s)/U(s). (3) (5%) Define the system state z = [ x xdot y ydot ]^T , find the state-space model of the system. (4) (5%) Draw the corresponding block diagram of the state-space representation of (3). ┌─┤├─┐ │ C2 │ ██████ ├─══─┤ ︴ │ │┌┐R2 │ k1 ︴ ︴ ││└┐ │ M1 □ ︴k2 ●─══─┤├┴┤- └┐│ ︱ ︴ R1 C1 │ ├┴───● b1│┴│ ︴ V1 ┌┤+ ┌┘ └┬┘ │ ││┌┘ V2 ┌┴──┴┐ │└┘ └─┬──┘M2 ●──────┼────────● ↓u(t) ┴ Fig.2 Fig.3 3. (15%) Refer to Fig.3, (1) (5%) Find the transfer function T = V2(s)/V1(s). (2) (5%) Find the sensitivity of T to R1, i.e., S(T)(R1). (3) (5%) Suppose R1 = 1 kΩ, R2 = 2 kΩ, C1 = 1 mF, C2 = 0.1 mF, find V2(t) to a step input V1(t) = 1, t ≧ 0 4. (20%) Consider a transfer function G(s) = ( s - 2 ) / ( s^3 + 6s^2 + 11s + 6 ), (1) (5%) Derive its controller canonical form, and draw the corresponding block diagram. (2) (5%) Derive its observer canonical form, and draw the corresponding block diagram. (3) (5%) Derive its Jordan form, and draw the corresponding block diagram. (4) (5%) Find the state response of (3) the the initial state [ 1 -1 1 ]^T. 5. (20%) Consider the closed-loop system of Fig.5, with G(s) = ( s + 1 ) / ( s^2 + 5s + 6 ) (1) (5%) Using a proportional (P) control K = Kp = constant, find the steady-state error to a step input R. (2) (5%) Using a proportional-integral (PI) control K = Kp + Ki/s, where Kp, Ki are constant, find the steady-state error to a step input R, and the steady-state error to a step disturbance Td. (3) (5%) If the disturbance is a sinusoidal signal Td = sin(ωt), the output response will be oscillatory with the above controllers. However, we can eliminate the oscillation by putting the unstable mode at the controller (it is so-called internal model principle) as K = α / ( s^2 + ω^2 ), where α = constant. Find the steady-state error to Td = sin(ωt). (4) (5%) Design a controller such that (i) the steady-state error to a step input R is zero; and (ii) the steady-state error to a sinusoidal disturbance Td = sin(60t) is zero. ┌─────┐ │Controller│ Td(s) + ├─────┤ +↓ R(s) ─→○─→│ K ├─→○──┬→ Y(s) -↑ └─────┘ + │ └──────────────┘ Fig.5 6. (20%) Consider the closed-loop system of Fig.5, with G(s) = 1 / ( s^2 + 3s + 2 ) and a proportional-integral-derivative (PID) control K = Kp + Ki/s + Kds, where Kp, Ki, Kd are constant. (1) (5%) Find the sensitivity of the closed-loop system to Ki i.e., S(T)(Ki), where T = T(R→Y). (2) (5%) To make the steady-state error to a unit ramp input (R(t)=t,t≧0) to be less than 0.1, find the minimal value of Ki. (3) (5%) Set the desired closed-loop poles at s = -3, -4, -5, find the required Kp,Ki,Kd values. (4) (5%) Suppose Kp = 17 , Ki = 12 , Kd = 5 , find the output response Y(t) to a unit step R(t)=1 , t≧0. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.112.243.213
jimy00ex :推 11/01 16:48
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