課程名稱︰控制系統
課程性質︰選修
課程教師︰陳永耀
開課學院：電資學院
開課系所︰電機系
考試日期（年月日）︰2013.04.25
考試時限（分鐘）：180分鐘
是否需發放獎勵金：是
（如未明確表示，則不予發放）
試題 :
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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