P13.4. A closed loop system has a zero-order hold and process.
2
(a) Determine G(Z) with T=1s and 𝐺𝑝 𝑠 = 𝑠+2
(b) Let r(t) be a unit step input and calculate the response of the system by synthetic division
Answer:
(a) 𝐺 𝑍 = 𝒵
1−𝑒 −𝑠𝑇 2
𝑠
𝑠+2
= 1 − 𝑍 −1 𝒵
1
𝑠
1
− 𝑠+2 = 1 − 𝑍 −1
(b) The Z-transform of unit step input is 𝑅 𝑍 =
𝑇 𝑍 =
0.865
𝑍
𝑍
𝑍−1
𝑍
0.865
− 𝑍−𝑒 −2𝑇 = 𝑍−0.135
𝑍
𝑍−1
𝐺(𝑍)
0.865
=
1 + 𝐺(𝑍) 𝑍 + 0.37
0.865𝑍
So 𝑌 𝑍 = 𝑇 𝑍 𝑅 𝑍 = 𝑍+0.37 × 𝑍−1 = 𝑍 2 −0.27𝑍−0.73
𝑌 𝑍 = 0.865𝑍 −1 + 0.234𝑍 −2 +∙∙∙=
∞
𝑘=0 𝑦
𝑘𝑇 𝑍 −𝑘
⇒ 𝑌 0 = 0, 𝑌 𝑇 = 0.865, 𝑌 2𝑇 = 0.234
P13.10 A unit feedback loop system has a zero-order hold and a process 𝐺𝑝 𝑠 =
1
,
𝑠(𝑠+10)
the
sampling period is T=0.1s
(a)
(b)
(c)
(d)
(e)
(f)
Let D(z)=K and determine the transfer function 𝐷 𝑍 𝐺(𝑍)
Determine the characteristic equation of the closed loop system
Calculate the maximum value of K for a stable system
Determine K such that the overshoot is less than 30%
Based on the K got from (d), determine the closed loop transfer function
Determine the location of the closed loop roots and the overshoot if K is one-half of the
value determined in (c)
Answer:
1 − 𝑒 −𝑠𝑇
1
𝐴 𝐵
𝐶
×
= 1 − 𝑍 −1 𝒵 2 + +
𝑠
𝑠 𝑠 + 10
𝑠
𝑠 𝑠 + 10
1
1
1
1
1
1
= 1 − 𝑍 −1 𝒵
× −
+
×
10 𝑠 2 100 𝑠 100 𝑠 + 10
1
10𝑇𝑍
𝑍
𝑍
=
1 − 𝑍 −1
−
+
2
100
(𝑍 − 1)
𝑍 − 1 𝑍 − 𝑒 −10𝑇
𝐺 𝑍 = 𝐺𝑜 𝐺𝑝 𝑍 = 𝒵
(a) With T=0.1s
1
𝑍
𝑍
𝑍
1 − 𝑍 −1
−
+
2
100
(𝑍 − 1)
𝑍 − 1 𝑍 − 𝑒 −1
−1
−2
1
𝑒 𝑍 − 2𝑒 + 1
0.0037𝑍 + 0.0026
=𝐾×
× 2
=
𝐾
×
100 𝑍 − 1 − 𝑒 −1 𝑍 + 𝑒 −1
𝑍 2 − 1.368𝑍 + 0.368
𝐷 𝑍 𝐺 𝑍 =𝐾×
(b) The characteristic equation of the closed loop system is
1+𝐷 𝑍 𝐺 𝑍 =0
(c) Based on the characteristic equation, we have
𝑞 𝑍 = 𝑍 2 − 1.368𝑍 + 0.368 + 𝐾0.0037𝑍 + 0.0026𝐾 = 0
Because the q(Z) is quadratic and has real coefficients, the necessary and sufficient
conditions for q(Z) to have all roots within the unit circle are
𝑞(0) < 1, 𝑞(1) > 0, 𝑎𝑛𝑑 𝑞(−1) > 0
⇒ 𝐾 < 240
(d)
𝐾
𝑠(𝑠+10)
=
𝐾/10
𝑠(0.1𝑠+1)
⇒ 𝜏 = 0.1,
𝑇
𝜏
= 1, 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝐹𝑖𝑔𝑢𝑟𝑒 13.19
We have K=75
(e) The closed loop transfer function is 𝑇 𝑍 =
0.2759𝑍+0.1982
𝑍 2 −1.092𝑍+0.5661
(f) K=119.5, the roots are 𝑧 = 0.4641 ± 0.6843𝑗. Based on the Figure 13.19, we have
overshoot is 0.55
10
AP13.4 A closed loop system has a zero-order hold and a process 𝐺𝑝 𝑠 = 𝑠+1. Determine the
range of sampling period T for which the system is stable.
Answer:
𝐺 = 𝐺𝑜 𝐺𝑝 𝑠 =
10(1 − 𝑒 −𝑠𝑇 )
𝑠(𝑠 + 1)
10 1 − 𝑒 −𝑠𝑇
10
10
𝐺 𝑍 =𝒵
= 1 − 𝑍 −1 𝒵
−
= 10 1 − 𝑍 −1
𝑠 𝑠+1
𝑠
𝑠+1
10(1 − 𝑒 −𝑇 )
=
𝑍 − 𝑒 −𝑇
The closed loop transfer function is
𝑍
𝑍
−
𝑍 − 1 𝑍 − 𝑒 −𝑇
𝑇 𝑍 =
For stability,
𝐺(𝑍)
10 − 10𝑒 −𝑇
10 − 10𝑒 −𝑇
=
=
1 + 𝐺(𝑍) 𝑍 − 𝑒 −𝑇 + 10 − 10𝑒 −𝑇 𝑍 − (11𝑒 −𝑇 − 10)
11𝑒 −𝑇 − 10 < 1 ⇒ 0 < 𝑇 < 0.2
𝐾
E13.14 A unity feedback system has a process 𝐺𝑝 𝑠 = 𝑠(𝑠+3) , with T=0.5
(a) Determine whether the system is stable when K=5
(b) Determine the maximum value of K for stable
Answer:
𝐾 1 − 𝑒 −𝑠𝑇
1/3 1/9
1/9
= 1 − 𝑍 −1 𝒵 2 −
+
2
𝑠 𝑠+3
𝑠
𝑠
𝑠+3
1
3𝑇𝑍
𝑍
𝑍
=𝐾× ×[
−
+
]
9
(𝑍 − 1)2 𝑍 − 1 𝑍 − 𝑒 −3𝑇
𝐺 𝑍 = 𝐺𝑜 𝐺𝑝 𝑍 = 𝒵
The characteristic equation of the closed loop transfer function is
𝑞 𝑍 =1+𝐺 𝑍 =0
𝐾
𝑞 𝑍 = 𝑍 − 1 𝑍 − 𝑒 −3𝑇 + 9 [3𝑇 𝑍 − 𝑒 −3𝑇 − 𝑍 − 1 𝑍 − 𝑒 −3𝑇 + (𝑍 − 1)2 ]
Because 𝑞(0) < 1, 𝑞(1) > 0, 𝑎𝑛𝑑 𝑞(−1) > 0, ⇒ 0 < 𝐾 < 6.32
So, when K =5, the system is stable.