Answers-of-Homework Controle de sistemas

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Answers of Homework Questions Chapter 4 (2014/2015) 
Question 1. 
What pole locations characterize (1) the underdamped system, (2) the over-damped system, 
(3) the critically damped system? 
 
Answer: a. Complex conjugate pole locations, like −σ ± jω, 
b. Real and separated pole locations, like −𝜎1, −𝜎2, 
c. Real identical pole locations, like −𝜎1, −𝜎2, 
 
Question 2. Find the step response for each of the systems with the transfer functions below. 
 a. G(s) =
1
𝑠+2
 
 b. G(s) =
5
(𝑠+2)(𝑠+3)
 
Answer: Applying inverse Laplace transform of 𝐂(𝐬) =
𝟏
𝒔
𝑮(𝒔) gives the step response of the system 
 a. C(s) =
1
𝑠(𝑠+2)
=
𝐴
𝑠
+
𝐵
(𝑠+2)
=
(𝐴+𝐵)𝑠+2𝐴
𝑠(𝑠+2)
, 𝐴 = 1 2⁄ , 𝐵 = −
1
2⁄ 
 c(t) =
1
2
−
1
2
𝑒−2𝑡 
 b. C(s) =
1
𝑠(𝑠+2)(𝑠+3)
=
𝐴
𝑠
+
𝐵
(𝑠+2)
+
𝐶
(𝑠+3)
=
(𝐴+𝐵+𝐶)𝑠2+(5𝐴+3𝐵+2𝐶)𝑠+6𝐴
𝑠(𝑠+2)(𝑠+3)
, 𝐴 = 5 6⁄ , 𝐵 = −
15
6⁄ , 𝐶 =
10
6⁄ 
 c(t) =
5
6
−
5
2
𝑒−2𝑡 +
5
3
𝑒−5𝑡 
(Attention !! The answer is obtained by the inverse transform of C(s) not the transfer function G(s) itself) 
Question 3. (TEST 2 2013/2014) with answer 
 
Question 4. Given the transfer function below, find ζ and 𝜔𝑛. (See Example 4.3 Nise’s textbook) 
 
 a. G(s) =
25
𝑠2+3.6𝑠+25
 
 b. G(s) =
400
𝑠2+8𝑠+400
 
 
Answer: (Hint: Use following general form ) 
 a. 𝜔𝑛
2 = 25, 𝜔𝑛 = 5, 2𝜁𝜔𝑛 = 3.6 thus, ζ = 0.36 
 b. 𝜔𝑛
2 = 400, 𝜔𝑛 = 20, 2𝜁𝜔𝑛 = 8 thus, ζ = 0.2 
 
Question 5. For the unit step response shown below, find the transfer function of the system.(Nise book, Prob.29 
p208) 
 
 
Answer (Hint: Use %OS=40% and 𝑻𝒑 = 𝟒 second which are measured from the graph) 
 
 
 From the graph above %OS=
1.4−1.0
1.0
= 40%, 𝑇𝑝 = 4 second. Since the relationship between %OS and ζ is 
given by the graph below, ζ ≈ 0.28. 
 
4 
≈0.28 
(We may also solve this by using analytical relationship given by ζ =
−ln (%OS/100)
√𝜋2+ln2(%𝑂𝑆/100)
) 
Using the formula 𝑇𝑝 =
𝜋
𝜔𝑛√1−𝜁
2
 and ζ = 0.28, we get 𝜔𝑛 ≅ 0.818 
Thus, 
G(s) =
𝜔2
𝑠2 + 2𝜁𝜔𝑛𝑠 + 𝜔
2
=
0.669
𝑠2 + 0.458𝑠 + 0.669
 
 
Question 6. (See Skill-Assessment Exercise 4.6, Nise book) 
Determine the validity of a second-order approximation for each of these two transfer functions: 
 a. G(s) =
200
(𝑠+5)(𝑠2+3.6𝑠+25)
 
 b. G(s) =
320
(𝑠+11)(𝑠2+2𝑠+40)
 
Answer 
 a. Poles of this system are: 
 s + 5 = 0 s = −5 
 𝑠2 + 3.6𝑠 + 25 = 0 𝑠 =
−3.6±√(3.6)2−4×25
2
= −1.8 ± 𝑗√87.04 
From the five times rule of thumb, the 2-nd order approximation is not valid because that the real pole 
(s = −5) is less than five times farther to left than the real part (-1.8) of the complex poles above. 
 b. Poles of this system are: 
 s + 11 = 0 s = −11 
 𝑠2 + 2𝑠 + 40 = 0 𝑠 =
−2±√4−4×40
2
= −1 ± 𝑗√39 
From the five times rule of thumb, the 2-nd order approximation is not valid because that the real pole 
(s = −11) is less than five times farther to left than the real part (−1) of the complex poles above. 
END