课程名称︰控制系统 课程性质︰选修 课程教师︰陈永耀 开课学院：电资学院 开课系所︰电机系 考试日期（年月日）︰2013.04.25 考试时限（分钟）：180分钟 是否需发放奖励金：是 （如未明确表示，则不予发放） 试题 : 1. Given a DC motor speed control system as shown in Figure 1. The dynamics of the motor is described by the differential equation . y + 60y = 600v_a - 1500ω where y is the motor speed, v_a is the armature voltage, and w is the load torque. ○Ｗ │ ↓ ┌──┐ │1500│ └──┘ │ ↓－ ＋ e ┌─┐ v_a ┌──┐ ┌────┐ Ｒ○─→ Σ　──→│Ｄ│──→│600 │──→ Σ　─→│1/(s+60)│─┬─○Ｙ － └─┘ └──┘ ＋ └────┘　│ ↑ │ └─────────────────────────────┘ Figure 1 (a) Please describe the benefits of a closed-loop system over the open-loop system by using the above system as an example. (20%) (b) Assume that the armature voltage is given by the PI control law t v_a = k_p * e + k_I *∫ e dt 0 Please derive the transfer function from W to Y as a function of k_p and k_I. (10%) (c) Please determine the values for k_p and k_I so that the closed-loop system will have poles at -60±60j. (10%) 2. Given the open-loop transfer functions below, assume that it is configured in a unity-feedback configuration, s + 3 (a) KG(s) = ───────── s^3 + 2s^2 + s - 8 s + 1 (b) KG(s) = ────────────────────── s^6 + 2s^5 + 13s^4 + 10s^3 + 44s^2 + 7s + 31 5 * (s^2 + 2s + 1) (c) KG(s) = ──────────── s^4 + 2s^3 + 5s^2 + 10s Please determine the stability of the closed-loop system. (18%) 3. (a) Please describe the design process for a PID control system. (10%) (b) Assume that we start with P-control in the first place, with the design specifications the rise time t_r < 1 sec, the overshoot M < 10%, the settling time t_s < 3sec, the steady state error e_ss < 0.01, please determine the design steps in the following cases: (20%) Table 1 ┌────┬───┬───┬───┬───┐ │ │ t_r │ Ｍ │ t_s │ e_ss │ ├────┼───┼───┼───┼───┤ │ Case 1 │ 2 │ 12% │ 2 │ 0.1 │ ├────┼───┼───┼───┼───┤ │ Case 2 │ 1 │ 5% │ 4 │ 0.01 │ ├────┼───┼───┼───┼───┤ │ Case 3 │ 0.4 │ 30% │ 2 │ 0.002│ ├────┼───┼───┼───┼───┤ │ Case 4 │ 0.99 │ 10% │ 2.8 │ 0.009│ └────┴───┴───┴───┴───┘ 4. Given the following block diagrams, please determine the transfer functions of the systems. (12%) ┌───┐ ┌──┤ H2(s)│←────────┐ －│ └───┘ │ R(s) ┌─┐ ↓ ┌───┐ ┌───┐ │ ┌───┐ Y(s) ─→○→│Ｋ│→○─→○─→│ G1(s)│─┬→│ G2(s)│─┼→│ G3(s)│┬─→ ↑ └─┘ ↑ ↑ └───┘ │ └───┘ │ └───┘│ －│ －│ －│ ┌───┐ │ │ │ │ │ └──┤ H3(s)│←┘ │ │ │ │ └───┘ │ │ │ │ ┌───┐ │ │ │ └─────┤ H1(s)│←────────┘ │ │ └───┘ │ └────────────────────────────────┘ (a) ＋ ＋ ┌───────→○←───────┐ │ ↓ │ ┌┴┐ ┌─┐ │ │K2│ │K3│ │ └─┘ └─┘ │ ↑ ＋│ │ R(s) ┌───┐ │ ┌───┐ ↓ ┌───┐ │ Y(s) ─→○──→○─→│ G1(s)│──┴→│ G2(s)│─→○─→│ G3(s)│─┼─→ ↑ ↑ └───┘ └───┘ ↑ └───┘ │ －│ －│ ＋│ ┌───┐ │ │ │ └──┤ H1(s)│←┤ │ │ └───┘ │ │ │ ┌─┐ │ │ └───────────────┤K1│←────────┤ │ ┌───┐ └─┘ │ └───────────┤ H2(s)│←──────────────┘ └───┘ (b) --

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