MID EXAM - ELECTRICAL - Linear System and Control - Muhammad Zubair - SU-20-02-048-004
MID EXAM - ELECTRICAL - Linear System and Control - Muhammad Zubair - SU-20-02-048-004
MID EXAM - ELECTRICAL - Linear System and Control - Muhammad Zubair - SU-20-02-048-004
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Registration No SU-20-02-048-004
Contents
No. Title Page
1 Answer no 1 4
2 Answer no 2 7
3 Answer no 3 part a 10
4 Answer no 3 part b 11
Page |4
Question #1:
Answer1:
Step 1
Step 2
Page |5
16(s4 −2 s 2+1)
1+(16 s4 −32 s 2+16)(1)
u(t) y(t)
16(s4 −2 s 2+1)
1+(16 s4 −32 s 2+16)(1)
16 (s 4 −2 s2 +1)
1+(16 s4 −32 s 2+16)
16(s4 −2 s 2+ 1)
16 ¿ ¿
y ( t ) s 4−2 s 2+ 1
=
u (t ) 17
s 4−2 s2 +
16
b 0=1 ,b 1=0 , b2=−2 ,b 3=0 , b 4=1
b4s 4 +¿
b3 s ¿
¿
G(s)= 3+¿
b
2s 2+¿
b
1s
1+¿
b0 ¿
¿
0s5 + a s4 +a s 3 +a s2 +a s 1+ a
4 3 2 1 0
0 1 0 0
0 0 1 0
A= 0 0 0 1
0 1 0 0
Or A= 0 0 1 0
0 0 0 1
-1.06 0 2 0
1
0
B= −2 ⌉
⌈ C= ⌈ 10 0 0 0 ⌉
0
1
Page |7
Question #2
Answer 2:
First we check Eigen value of Matrix A so we check that the
system is stable or unstable.
det ( λI −A )=0
[ 10 ]
I= 0 1
λ I= [ 0λ 0λ]
[ λ 0] [ 0.5 1]
( λI − A ¿= 0 λ ‒ 0.8 2
Or [ λ−0.5−1 ]
( λI − A ¿= −0.8 λ−2
Page |8
det ( λI −A )= λ−0.5−1
[ −0.8 λ−2 ]
det ( λI −A )=¿
λ=−b ± √b 2−4 ac
2a
λ=2.5± √ 6.25−0.8
2
λ 1=2.5 ± √ 5.45
2
λ 1=2.5 ± 2.33
2
λ 1=2.5+2.33
2
λ 1=2.415
2.5−2.33
λ 2=
2
Page |9
λ 2=0.085
Here it is clear that Eigen Values are positive so the system is unstable
And we need to stabilize it, following the pre-requisite steps for Matrix
C, and the matrix C must be Identity so
STEP 1:
P a g e | 11
C( s) G 1+G 2
=
R( s) 1+G 1+G 2(G 4−G3)
Q3(B)
G1
R(s) G2
C(s)
H1
H2
STEP 1:
G1
P a g e | 12
R(s) G2
C(s)
H1
H2
STEP 2:
G1
R(s)
G2
C(s)
H2-H1
STEP 3:
R(s) G2 C(s)
1+ G2(H 2−H 1)
STEP 4:
G2
R(s) 1+G1 1+ G2(H 2−H 1)
P a g e | 13
C(S)
STEP 5:
(1+ G1)G 2
R(s) 1+ G2(H 2−H 1)
C(S)
STEP 6:
¿¿
R(s)
C(S)
C( s)
=¿ ¿
R( s)