Sbe final exam jan17 - solved-converted

1. Page1/9
DYNAMIC CONTROL SYSTEMS
(SBE_305A)
Final Exam
Spring 2015
Cairo University
Faculty of Engineering
Systemsand Biomedical Engineering Dept.
Answer ONLY FIVE Questions (Full Mark 75 points)____ Time Allowed : 3 Hours
Question 1 (15 points):
a) Using block diagram reduction:
Now 𝑮 𝟒(𝒔) =
𝑮 𝟑(𝒔)
𝟏+𝑮 𝟑(𝒔)𝑯 𝟑(𝒔)
𝑮 𝟓(𝒔) =
𝟏
𝑮 𝟐(𝒔)
𝑮 𝟔(𝒔) =
𝑮 𝟐(𝒔)
𝟏+𝑮 𝟐(𝒔)𝑯 𝟐(𝒔)
𝑮 𝟕(𝒔) =
𝑮 𝟏(𝒔)𝑮 𝟔(𝒔)
𝟏+𝑮 𝟏(𝒔)𝑮 𝟔(𝒔)𝑯 𝟏(𝒔)
=
𝑮 𝟏(𝒔)∙
𝑮 𝟐(𝒔)
𝟏+𝑮 𝟐(𝒔)𝑯 𝟐(𝒔)
𝟏+𝑮 𝟏(𝒔)∙
𝑮 𝟐(𝒔)
𝟏+𝑮 𝟐(𝒔)𝑯 𝟐(𝒔)
∙𝑯 𝟏(𝒔)
=
𝑮 𝟏(𝒔)𝑮 𝟐(𝒔)
𝟏+𝑮 𝟐(𝒔)𝑯 𝟐(𝒔)+𝑮 𝟏(𝒔)𝑮 𝟐(𝒔)𝑯 𝟏(𝒔)
=
𝑮 𝟏(𝒔)𝑮 𝟐(𝒔)
𝟏+𝑮 𝟐(𝒔)[𝑯 𝟐(𝒔)+𝑮 𝟏(𝒔)𝑯 𝟏(𝒔)]
𝑮 𝟖(𝒔) = 𝟏 + 𝑮 𝟓(𝒔) = 𝟏 +
𝟏
𝑮 𝟐(𝒔)
=
𝟏 + 𝑮 𝟐(𝒔)
𝑮 𝟐(𝒔)
∴ 𝑻(𝒔) =
𝑪(𝒔)
𝑹(𝒔)
= 𝑮 𝟕(𝒔)𝑮 𝟖(𝒔)𝑮 𝟒(𝒔) =
𝑮 𝟏(𝒔)𝑮 𝟐(𝒔)
𝟏 + 𝑮 𝟐(𝒔)[𝑯 𝟐(𝒔) + 𝑮 𝟏(𝒔)𝑯 𝟏(𝒔)]
𝟏 + 𝑮 𝟐(𝒔)
𝑮 𝟐(𝒔)
𝑮 𝟑(𝒔)
𝟏 + 𝑮 𝟑(𝒔)𝑯 𝟑(𝒔)
R(s) C(s)
G1(s) G2(s) G4(s)
H2(s)
H1(s)
+ + +
+--
G1(s) G2(s) G4(s)
G5(s)
H1(s)
H2(s)
R(s) C(s)
G5(s)
G1(s) G6(s) G4(s)
H1(s)
R(s) C(s)
G7(s)
G5(s)
G4(s)
R(s)
C(s)
G7(s) G8(s) (s)4G
R(s)
C(s)

2. Page2/9
=
𝑮 𝟏(𝒔)𝑮 𝟑(𝒔)[𝟏 + 𝑮 𝟐(𝒔)]
[𝟏 + 𝑮 𝟑(𝒔)𝑯 𝟑(𝒔)]{𝟏 + 𝑮 𝟐(𝒔)[𝑯 𝟐(𝒔) + 𝑮 𝟏(𝒔)𝑯 𝟏(𝒔)]}
𝑻(𝒔) =
𝑪(𝒔)
𝑹(𝒔)
=
∑ 𝑷 𝒊∆𝒊
∆
=
𝑷 𝟏∆ 𝟏+𝑷 𝟐∆ 𝟐
∆
𝑷 𝟏 = 𝒔 𝑷 𝟐 =
𝟏
𝒔 𝟐
𝑳 𝟏 = 𝑳 𝟑 = −𝒔 𝟐
𝑳 𝟐 = 𝑳 𝟒 = −
𝟏
𝒔
∴ ∆= 𝟏 − (𝑳 𝟏 + 𝑳 𝟐 + 𝑳 𝟑 + 𝑳 𝟒) = 𝟏 + (𝒔 𝟐
+
𝟏
𝒔
+ 𝒔 𝟐
+
𝟏
𝒔
) = 𝟏 +
𝟐𝒔 𝟑
+ 𝟐
𝒔
=
𝟐𝒔 𝟑
+ 𝒔 + 𝟐
𝒔
∴ 𝑻(𝒔) = (𝒔 +
𝟏
𝒔 𝟐
) ∙
𝒔
𝟐𝒔 𝟑 + 𝒔 + 𝟐
=
𝒔(𝒔 𝟑
+ 𝟏)
𝒔 𝟐(𝟐𝒔 𝟑 + 𝒔 + 𝟐)
=
𝒔 𝟑
+ 𝟏
𝒔(𝟐𝒔 𝟑 + 𝒔 + 𝟐)
c) Let 𝑮 𝟏(𝒔) = 𝒔 𝟐
𝑮 𝟐(𝒔) =
𝟏
𝒔
𝑮 𝟑(𝒔) = 𝑮 𝟏(𝒔) + 𝑮 𝟐(𝒔)
𝑮 𝟒(𝒔) = 𝒇𝒆𝒆𝒅𝒃𝒂𝒄𝒌(𝑮 𝟑(𝒔), 𝟏)
𝑮 𝟓(𝒔) = 𝑮 𝟒(𝒔) ∙
𝟏
𝒔
𝑻(𝒔) = 𝒇𝒆𝒆𝒅𝒃𝒂𝒄𝒌(𝑮 𝟑(𝒔), 𝟏)
Thus, the Matlab code is given by:

3. Page3/9
>>num1=[1 0 0]
>>den1=[1]
>>G1=tf(num1,den1)
>>num2=[1]
>>den1=[1 0]
>>G2=tf(num2,den2)
>>G3=G1+G2
>>G4=feedback(G3,[1])
>>G5= G4*G2
>>num6=[1 0]
>>den6=[1]
>>G6=tf(num6,den6)
>>T=feedback(G5,G6)
Question 2 (15 points):
a) The system poles are given by: s = -10, s = -3 ± j
b) The damping ratio and natural frequency of the dominant poles can be obtained by equating the
second factor of the denominator with zero. This takes the form:
𝒔 𝟐
+ 𝟐𝜻𝝎 𝒏 + 𝝎 𝒏
𝟐
= 𝟎
∴ 𝟐𝜻𝝎 𝒏 = 𝟔 ===> 𝜻𝝎 𝒏 = 𝟑
𝝎 𝒏
𝟐
= 𝟏𝟎 ===> 𝝎 𝒏 = √𝟏𝟎 𝒓𝒂𝒅/𝒔
∴ 𝜻√𝟏𝟎 = 𝟑 ===> 𝜻 =
𝟑
√𝟏𝟎
= 𝟎. 𝟗𝟒𝟗
c) For the unit impulse response R(s) = 1. Thus
𝒀(𝒔) =
𝟓𝒔 + 𝟏
(𝒔 + 𝟏𝟎)(𝒔 𝟐 + 𝟔𝒔 + 𝟏𝟎)
=
𝟒𝟗𝒔 + 𝟓𝟒
𝟓𝟎(𝒔 𝟐 + 𝟔𝒔 + 𝟏𝟎)
−
𝟒𝟗
𝟓𝟎(𝒔 + 𝟏𝟎)
=
𝟒𝟗(𝒔 + 𝟑) − 𝟗𝟑
𝟓𝟎[(𝒔 + 𝟑) 𝟐 + 𝟏]
−
𝟒𝟗
𝟓𝟎(𝒔 + 𝟏𝟎)
∴ 𝒚(𝒕) = 𝟎. 𝟗𝟖𝒆−𝟑𝒕
∙ 𝐜𝐨𝐬 𝒕 − 𝟏. 𝟖𝟔𝒆−𝟑𝒕
∙ 𝐬𝐢𝐧 𝒕 − 𝟎. 𝟗𝟖𝒆−𝟏𝟎𝒕
= 𝟐. 𝟏𝟎𝟐𝟑𝟖𝒆−𝟑𝒕
∙ 𝐜𝐨𝐬(𝒕 + 𝟔𝟐. 𝟐°) − 𝟎. 𝟗𝟖𝒆−𝟏𝟎𝒕
d) For the unit step response 𝑹(𝒔) =
𝟏
𝐬
. Thus
𝒀(𝒔) =
𝟓𝒔 + 𝟏
𝒔(𝒔 + 𝟏𝟎)(𝒔 𝟐 + 𝟔𝒔 + 𝟏𝟎)
=
𝟏
𝟏𝟎𝟎𝒔
+
𝟒𝟗
𝟓𝟎𝟎(𝒔 + 𝟏𝟎)
+
𝟖𝟑 − 𝟐𝟕𝒔
𝟐𝟓𝟎(𝒔 𝟐 + 𝟔𝒔 + 𝟏𝟎)
=
𝟏
𝟏𝟎𝟎𝒔
+
𝟒𝟗
𝟓𝟎𝟎(𝒔 + 𝟏𝟎)
−
𝟐𝟕(𝒔 + 𝟑) − 𝟏𝟔𝟒
𝟐𝟓𝟎[(𝒔 + 𝟑) 𝟐 + 𝟏]
∴ 𝒚(𝒕) = 𝟎. 𝟎𝟏 + 𝟎. 𝟎𝟗𝟖𝒆−𝟏𝟎𝒕
− 𝟎. 𝟏𝟎𝟖𝒆−𝟑𝒕
∙ 𝐜𝐨𝐬 𝒕 + 𝟎. 𝟔𝟓𝟔𝒆−𝟑𝒕
∙ 𝐬𝐢𝐧 𝒕
= 𝟎. 𝟎𝟏 + 𝟎. 𝟎𝟗𝟖𝒆−𝟏𝟎𝒕
− 𝟎. 𝟔𝟔𝟒𝟖𝟑𝒆−𝟑𝒕
∙ 𝐜𝐨𝐬(𝒕 + 𝟖𝟎. 𝟕°)
e) ess(∞) = 1 – T(0) = 𝟏 −
𝟏
𝟏𝟎×𝟏𝟎
= 𝟎. 𝟗𝟗 = 𝟗𝟗%

4. Page4/9
Question 3 (15 points):
a) Let the driving pressure be given by Pd = P1 – P2
∴ 𝑸 = 𝑲√𝑷 𝒅
∴ ∆𝑸 =
𝒅𝑸
𝒅𝑷 𝒅
∆𝑷 𝒅 =
𝑲
𝟐√𝑷 𝒅
∆𝑷 𝒅 =
𝑲
𝟐√𝑷 𝟏 − 𝑷 𝟐
∆(𝑷 𝟏 − 𝑷 𝟐)
Now if Pd = P1 – P2 = 0, then in that limiting case the slope of the Q – (P1 – P2) characteristic
approaches infinity. So, for an infinitesimal change in P1 – P2 the change in Q grows beyond all
limits.
b) 𝑺 𝑮
𝑻
=
𝟏
𝟏+𝑮(𝒔)𝑯(𝒔)
=
𝟏
𝟏+𝑮(𝒔)
=
𝟏
𝟏+
𝟒𝟎𝟎
𝒔 𝟐+𝟖𝒔+𝟒𝟎𝟎
=
𝒔 𝟐+𝟖𝒔+𝟒𝟎𝟎
𝒔 𝟐+𝟖𝒔+𝟖𝟎𝟎
𝑺 𝑯
𝑻
=
− 𝑮( 𝒔) 𝑯(𝒔)
𝟏 + 𝑮( 𝒔) 𝑯(𝒔)
= −
𝟒𝟎𝟎 × 𝟏
𝒔 𝟐 + 𝟖𝒔 + 𝟒𝟎𝟎
∙
𝒔 𝟐 + 𝟖𝒔 + 𝟒𝟎𝟎
𝒔 𝟐 + 𝟖𝒔 + 𝟖𝟎𝟎
= −
𝟒𝟎𝟎
𝒔 𝟐 + 𝟖𝒔 + 𝟖𝟎𝟎
c) 𝑺 𝑮
𝑻
=
𝟏
𝟏+𝑮(𝒔)𝑯(𝒔)
=
𝟏
𝟏+
𝟒𝟎𝟎
𝒔 𝟐+𝟖𝒔+𝟒𝟎𝟎
∙
𝟏
𝒔
=
𝒔( 𝒔 𝟐
+𝟖𝒔+𝟒𝟎𝟎)
𝒔( 𝒔 𝟐+𝟖𝒔+𝟒𝟎𝟎)+𝟒𝟎𝟎
=
𝒔 𝟑
+𝟖𝒔 𝟐
+𝟒𝟎𝟎𝒔
𝒔 𝟑+𝟖𝒔 𝟐+𝟒𝟎𝟎𝒔+𝟒𝟎𝟎
𝑺 𝑯
𝑻
=
− 𝑮( 𝒔) 𝑯(𝒔)
𝟏+𝑮( 𝒔) 𝑯(𝒔)
= − 𝟒𝟎𝟎
𝒔 𝟐+𝟖𝒔+𝟒𝟎𝟎
∙ 𝟏
𝒔
∙ 𝒔 𝟑+𝟖𝒔 𝟐+𝟒𝟎𝟎𝒔
𝒔 𝟑+𝟖𝒔 𝟐+𝟒𝟎𝟎𝒔+𝟒𝟎𝟎
= −𝟒𝟎𝟎
𝒔 𝟑+𝟖𝒔 𝟐+𝟒𝟎𝟎𝒔+𝟒𝟎𝟎
d) The circuit becomes
Y1(s) = R1·I(s)
𝑰(𝒔) =
𝑼 𝟏(𝒔)
𝒁(𝒔)
Z(s)= Z1(s)+ Z2(s)
Z1(s)= R1+sL1

5. Page5/9
𝒁 𝟐(𝒔) =
𝑹 𝟐(𝑹 𝟑+
𝟏
𝒔𝑪
)
𝑹 𝟐+𝑹 𝟑+
𝟏
𝒔𝑪
=
𝑹 𝟐(𝒔𝑪𝑹 𝟑+𝟏)
𝒔𝑪(𝑹 𝟐+𝑹 𝟑)+𝟏
∴ 𝒁(𝒔) = 𝑹 𝟏 + 𝒔𝑳 𝟏 +
𝑹 𝟐( 𝒔𝑪𝑹 𝟑 + 𝟏)
𝒔𝑪( 𝑹 𝟐 + 𝑹 𝟑) + 𝟏
=
[ 𝒔𝑪( 𝑹 𝟐 + 𝑹 𝟑) + 𝟏](𝑹 𝟏 + 𝒔𝑳 𝟏) + 𝑹 𝟐( 𝒔𝑪𝑹 𝟑 + 𝟏)
𝒔𝑪( 𝑹 𝟐 + 𝑹 𝟑) + 𝟏
=
𝒔 𝟐
𝑪𝑳 𝟏(𝑹 𝟐 + 𝑹 𝟑) + 𝒔[𝑪(𝑹 𝟐 + 𝑹 𝟑)𝑹 𝟏 + 𝑳 𝟏 + 𝑪𝑹 𝟐 𝑹 𝟑] + (𝑹 𝟏 + 𝑹 𝟐)
𝒔𝑪(𝑹 𝟐 + 𝑹 𝟑) + 𝟏
∴ 𝑰(𝒔)
=
𝒔𝑪(𝑹 𝟐 + 𝑹 𝟑) + 𝟏
𝒔 𝟐 𝑪𝑳 𝟏(𝑹 𝟐 + 𝑹 𝟑) + 𝒔[𝑪(𝑹 𝟐 + 𝑹 𝟑)𝑹 𝟏 + 𝑳 𝟏 + 𝑪𝑹 𝟐 𝑹 𝟑] + (𝑹 𝟏 + 𝑹 𝟐)
𝑼 𝟏(𝒔)
∴
𝒀 𝟏(𝒔)
𝑼 𝟏(𝒔)
=
𝑹 𝟏[𝒔𝑪(𝑹 𝟐 + 𝑹 𝟑) + 𝟏]
𝒔 𝟐 𝑪𝑳 𝟏(𝑹 𝟐 + 𝑹 𝟑) + 𝒔[𝑪(𝑹 𝟐 + 𝑹 𝟑)𝑹 𝟏 + 𝑳 𝟏 + 𝑪𝑹 𝟐 𝑹 𝟑] + (𝑹 𝟏 + 𝑹 𝟐)
Question 4 (15 points):
a) We can obtain the state space model from the transfer function as follows:
U(s) +
Let 𝐺2(𝑠) = 𝐺1(𝑠) ∙
1
𝑠
Now 𝐺1(𝑠) = 𝑎 +
𝑏
𝑠
=
𝑎𝑠+𝑏
𝑠
∴ 𝐺2(𝑠) =
𝑎𝑠 + 𝑏
𝑠2
∴ 𝐺(𝑠) =
𝑌(𝑠)
𝑈(𝑠)
=
𝐺2(𝑠)
1 + 𝐺2(𝑠)
=
𝑎𝑠+𝑏
𝑠2
1 +
𝑎𝑠+𝑏
𝑠2
=
𝑎𝑠 + 𝑏
𝑠2 + 𝑎𝑠 + 𝑏
This is of the form
𝐺(𝑠) =
𝑏 𝑜 𝑠2
+ 𝑏1 𝑠 + 𝑏2
𝑠2 + 𝑎1 𝑠 + 𝑎2
In that case the state model is given by:
𝐀 = [
0 1
−𝑎2 −𝑎1
] = [
0 1
−𝑏 −𝑎
]
G1(s)
𝟏
𝒔
_
Y(s)

6. Page6/9
𝐁 = 𝜷 = [
𝛽1
𝛽2
] = 𝐛 − 𝐋𝛃′
= [
𝑏1
𝑏2
] − [
𝑎1 0
𝑎2 𝑎1
] [
𝑏 𝑜
𝛽1
]
𝐂 = [1 0] 𝐃 = 𝑏 𝑜 = 0
𝛽1 = 𝑏1 − 𝑎1 𝑏 𝑜 = 𝑎
𝛽2 = 𝑏2 − 𝑎2 𝑏 𝑜 − 𝑎1 𝛽1 = 𝑏 − 𝑎2
b) The system eigenvalues can be computed from:
|𝜆𝐈 − 𝐀| = |
𝜆 −1
𝑏 𝜆 + 𝑎
| = 𝜆(𝜆 + 𝑎) + 𝑏 = 𝜆2
+ 𝑎𝜆 + 𝑏 = 0
∴ 𝜆 =
−𝑎 ± √𝑎2 − 4𝑏
2
c) The transfer function can be derived from the state model as follows:
𝐺(𝑠) = 𝐂(𝐬𝐈 − 𝐀)−𝟏
𝐁 + 𝐃
Now 𝐬𝐈 − 𝐀 = [
𝑠 −1
𝑏 𝑠 + 𝑎
]
(𝐬𝐈 − 𝐀)−1
=
1
𝑠2+𝑎𝑠+𝑏
[
𝑠 + 𝑎 1
−𝑏 𝑠
]
∴ 𝐺(𝑠) = [1 0]
1
𝑠2 + 𝑎𝑠 + 𝑏
[
𝑠 + 𝑎 1
−𝑏 𝑠
] [
𝑎
𝑏 − 𝑎2] =
𝑎𝑠 + 𝑏
𝑠2 + 𝑎𝑠 + 𝑏
d) The system poles are the same as the eigenvalues.
e) If a = 2 & b = 4, then:
𝐀 = [
0 1
−4 −2
] 𝐁 = [
𝑎
𝑏 − 𝑎2] = [
2
4 − 4
] = [
2
0
] 𝐂 = [1 0] 𝐷 = 0
The transfer function becomes:
𝐺(𝑠) =
2𝑠 + 4
𝑠2 + 2𝑠 + 4
f) >>A=[0 1;-4 -2]
>>B=[2;0]
>>C=[1 0]
>>D=[]
>>S=ss(A,B,C,D)
>>tf(S)

7. Page7/9
Question 5 (15 points):
a) 𝐺(𝑠) =
3𝑠+2
𝑠2+7𝑠+12
=
𝑏 𝑜 𝑠2+𝑏1 𝑠+𝑏2
𝑠2+𝑎1 𝑠+𝑎2
∴ 𝑎1 = 7 𝑎2 = 12 𝑏1 = 2 𝑏2 = 2 𝑏 𝑜 = 0
𝐀 = [
0 1
−𝑎2 −𝑎1
] = [
0 1
−12 −7
] 𝐂 = [1 0] 𝐃 = 𝑏 𝑜 = 0 = 𝛽 𝑜
𝐁 = 𝛃 = 𝐛 − 𝐋𝛃′
∴ [
𝛽1
𝛽2
] = [
𝑏1
𝑏2
] − [
𝑎1 0
𝑎2 𝑎1
] [
𝛽 𝑜
𝛽1
] = [
3
2
] − [
7 0
12 7
] [
0
𝛽1
]
∴ 𝛽1 = 3 − 7 × 0 − 0 × 𝛽1 = 3
𝛽2 = 2 − 12 × 0 − 7𝛽1 = 2 − 7 × 3 = −19
∴ 𝐁 = [
3
−19
]
b) The controllable state model is given by:
𝐀 𝐜 = [
0 1
−𝑎2 −𝑎1
] = [
0 1
−12 −7
]
𝐁 𝐜 = [
0
1
]
𝐂 𝐜 = [ 𝑏2 − 𝑎2 𝑏 𝑜| 𝑏1 − 𝑎1 𝑏 𝑜] = [2 − 12 × 0|3 − 7 × 0] = [2 3]
𝐃 𝐜 = 𝑏 𝑜 = 0
c) The observable state model is given by:
𝐀 𝐎 = 𝐀 𝐜
𝐓
= [
0 −12
1 −7
]
𝐁 𝐎 = 𝐂 𝐜
𝐓
= [
2
3
] 𝐂 𝐎 = 𝐁 𝐜
𝐓
= [0 1] 𝐃 𝐎 = 𝑏 𝑜 = 0
d) 𝐺(𝑠) =
3𝑠+2
𝑠2+7𝑠+12
=
3𝑠+2
(𝑠+3)(𝑠+4)
=
10
𝑠+4
−
7
𝑠+3
=
𝑐1
𝑠+𝑝1
+
𝑐2
𝑠+𝑝2
𝐀 = [
− 𝑝1 0
0 − 𝑝2
] = [
−4 0
0 −3
] 𝐁 = [
1
1
] 𝐂 = [ 𝑐1 𝑐2] 𝐃 = 𝑏 𝑜 = 0
e) >>num=[3 2]

8. Page8/9
>>den=[1 7 12]
>>[A,B,C,D]=tf2ss(num,den)
Question 6 (15 points):
a) 𝐺(𝑠) =
𝑠2+4𝑠+2
(𝑠+1)(𝑠2+4𝑠+4)
=
𝑠2+4𝑠+2
(𝑠+2)2(𝑠+1)
=
𝑠2+4𝑠+2
(𝑠+𝑝1)2(𝑠+𝑝3)
=
2
(𝑠+2)2
+
2
𝑠+2
−
1
𝑠+1
=
𝑐1
( 𝑠+𝑝1)
2 +
𝑐2
𝑠+𝑝1
+
𝑐3
𝑠+𝑝3
∴ 𝐀 = [
− 𝑝1 1 0
0 − 𝑝1 0
0 0 − 𝑝3
] = [
−2 1 0
0 −2 0
0 0 −1
] 𝐁 = [
0
1
1
]
𝐂 = [ 𝑐1 𝑐2 𝑐3] = [2 2 −1] 𝐃 = 𝑏 𝑜 = 0
b) We first compute the eigenvalues as follows:
|𝜆𝐈 − 𝐀| = |
𝜆 −1
3 𝜆 + 4
| = 𝜆(𝜆 + 4) + 3 = 𝜆2
+ 4𝜆 + 3 = (𝜆 + 1)(𝜆 + 3) = 0 ===> 𝜆1 = −1
𝜆2 = −3
Then we construct the transformation matrix P such that x = Pz, where z is an auxiliary vector.
𝐏 = [
1 1
𝜆1 𝜆2
] = [
1 1
−1 −3
]
The state model becomes:
𝐳̇ = 𝐏−𝟏
𝐀𝐏𝐳 + 𝐏−𝟏
𝐁𝑢
𝑦 = 𝐂𝐏𝐳 + 𝐃𝑢
Now 𝐏−𝟏
𝐀𝐏 = [
− 𝜆1 0
0 − 𝜆2
] = [
−1 0
0 −3
] 𝐏−𝟏
𝐁 =
1
2
[
1
−1
] 𝐂𝐏 = [1 1] 𝐃 = 0
c) 𝐆(𝐬) = 𝐂(𝐬𝐈 − 𝐀)−1
𝐁 + 𝐃 = 𝐂(𝐬𝐈 − 𝐀)−1
𝐁
𝐬𝐈 − 𝐀 = [
𝑠 −1
2 𝑠 + 4
]
∴ 𝐆(𝐬) =
1
𝑠2 + 4𝑠 + 2
[
1 𝑠 + 4
𝑠 −2
]
d) >>A=[0 1;-2 -4]
>>B=[0 1;1 0]
>>C=[1 0;0 1]
>>D=[]

9. Page9/9
>>S=ss(A,B,C,D)
>>tf(S)