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EE 4314 Homework2 Solutions
3.32
a). y + 2n
y +n
2y = 0 , y 0
( )= y
0, y 0
( )= 0 (1)
Take the laplace transform of equation (1), we get:
s2Y s( ) s y0 + 2n sY s( )+n
2Y s( ) = H s( )
then
G s( ) =sy
0
s2+ 2n s+n
2
=
k1
s s1
+
k1
*
s s1
*
where
s1= n + jn 1
2
s2= n jn 1
2
k1=
nej cos
1( )
2n 12e
j / 2
then
y t( ) =ent
2 12e
j n 12
t+
2cos1
+ e j n 1
2t+
2cos1
y t( ) = y0e
t
12sin dt cos
1( )
b).dy t( )
dt= 0 t=
n
d
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tMax
=
2
d
n
y t( )tMax
yn = y0e nd
1
2sin cos
1( )
yn =y
01
2
12
end (2)
For = lny
0
yn=d
From (2): y1= y
0e
d lny0
yn
= d
For lny
1
y1
lnyi
yi
We have yn= y
n1 y
n,
yn = y0end y
0e n1( ) d
= y0end 1 e
d( ) ,
yn
yn=
y0end
y0end 1 e
d
( )
yn
yn=
yi
yifor all i, n.
3.43
(a) s4 + 8s3 + 32s2 + 80s+100 = 0
The Routh array is,
s4 : 1 32 100
s3 : 8 80
s2 : 22 100
s1 : 80-800/22=43.6
s0 : 100
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No roots not in the LHP.
(b) s5 +10s4 + 30s3 + 80s2 + 344s+ 480 = 0
The Routh array is,
s5 : 1 30 344
s4 : 10 80 480
s3 : 22 296
s2 : -545 480
s1 : 490
s0 : 480
2 roots not in the LHP.
(c) s4 + 2s3 + 7s2 2s+ 8 = 0
The Routh array is,
s4 : 1 7 8
s3 : 2 -2
s2 : 8 8
s1 : -4
s0 : 8
2 roots not in the LHP.
(d) s3 + s2 + 20s+ 78 = 0
The Routh array is,
s3 : 1 20
s2 : 1 78
s1
: -58s0 : 78
2 roots not in the LHP.
(e) s4 + 6s2 + 25 = 0
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The Routh array is,
s4 : 1 6 25
s3 : 4 12
s2 : 3 25
s1 : 12-100/3=-21.3
s0 : 25
2 roots not in the LHP.
3.44
s5+ 5s
4+10s
3+10s
2+ 5s+K= 0
The Routh array is,
s5 : 1 10 5
s4 : 5 10 K
s3 : a
1 a
2
s2 : b
1 K
s1 : c
1
s0 : K
where
a1=
5 10( ) 1 10( )5
= 8
a2=
5 5( )1 K( )5
=
25K
5
b1=
a1( ) 10( ) 5( ) a2( )
a1
=
55 +K
8
c1=
b1( ) a2( ) a1( )K
b1
=
K2+ 350K1375( )5 55+K( )
For stability: all elements in the first column must be positive. The following
conditions must be satisfied.
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1. b1=
55+K
8> 0 K> 55
2. c1=
K2+ 350K1375( )5 55+K
( )
> 055 < K< 3.88
3. K> 0
By intersecting all 3 conditions, we have
0
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3.46
The transfer function for the above system is
LetK
1
=
1
K
Y s( )R s( )
=
K1
s+ z( )s+ p( )
K0
s2 a
2( )
1+K
1s+ z( )
s+ p( )K
0
s2 a
2( )
=
K1K
0s+ z( )
s3+ ps
2+ K
1K
0 a
2( )s+ K1K0z pa2
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Assume K0=1, we can construct Routh array as follow
s3
: 1K
1 a
2
s2 : p K
1z pa
2
s1 :
K1z + pa
2+K
1p pa
2
p=
K1z +K
1p
p
s0 : K
1z pa
2
For stability: all elements in the first column must be positive. The following
conditions must be satisfied.
p > 0,
K1pK
1z > 0 if K
1> 0 p > z
K1z pa
2> 0 ifK
1> 0 z >
pa2
K1
=Kpa2
3.47
(a)Y s( )R s( )
=
AesT
s s+1( )
1+Ae
sT
s s+1( )
=
AesT
s2+ s+ Ae
sT
The characteristic equation is: s2 + s+ AesT
(b) Using eTs 1Ts, the characteristic equation is
s2+ 1TA( )s+ A = 0
The Routh array is,
s2 : 1 A
s1 : 1-TA 0s0 : A
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For stability: all elements in the first column must be positive. The following
conditions must be satisfied.
A > 0,
TA
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scatter(real(r), imag(r), 'x')
end
hold offgrid on
3.22
Using block-diagram algebra.
Define dummy variables r1, r
2and r
inas shown in the figure below.
We have
r1= G
2rin (r
1+ r
2)H
2( )G4 (1)
r2= G
3rin r
1+ r
2( )H2( )G5 (2)
Letr
0
= r1
+ r2
(1)+ (2);r0=G
2G
4rinG
4H
2r0+G
3G
5rinG
5H
2r0
1+G4H
2+G
5H
2( )r0 = G2G4 +G3G5( )rin
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r0
rin
=
G2G
4+G
3G
5
1+G4H
2+G
5H
2( )
Define G7 =(G2G4 +G3G5)G1G6
1+G4H2 +G5H2( ) (3)
Define dummy variable e3
as shown in the figure below.
We now have,
e3= R s( ) G7H4e3 G7H3e3
e3=
R s( )1+ G
7H
4+ G
7H
3( )
Since
Y s( ) = e3G7
We now have,
Y s( )R s( )
=
G7
1+ G7H
4+ G
7H
3( )(4)
Substitute (3) into (4), we get
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Y s( )R s( )
=
G1G
6G
2G
4+ G
3G
5( )1+ G
2H
4+ G
2H
5( )
1+G
1G
6H
4G
2G
4+ G
3G
5( )1+ G
2
H4
+ G2
H5( )
+
G1G
6H
3G
2G
4+ G
3G
5( )1+ G
2
H4
+ G2
H5( )
=
G1G
6G
2G
4+G
3G
5( )1+G
2H
4+G
2H
5( )+G1G6H4 G2G4 +G3G5( )+G1G6H3 G2G4 +G3G5( )
Using Masons rule,
G =Your
Yin
=
Gk
k
k=1
N
Gk=1
=G1G
6G
2G
4
Gk=2
=G1G
6G
3G
5
k=1
=1
k= 2
=1
=1 Li + LiLj LiLjLk ++ 1( )m+
Li=1
=G4H
2
Li= 2
=G5H
2
Li=1Lj=1 =G1G2G4G6H4
Li=2Lj=2 =G1G3G5G6H4
Li= 3Lj= 3 =G1G2G4G6H3
Li=4Lj=4 =G1G3G5G6H3
therefore
=1G4H
2G
5H
2+G
1G
2G
4G
6H
4+G
1G
3G
5G
6H
4+G
1G
2G
4G
6H
3+G
1G
3G
5G
6H
3
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Y s( )R s( )
=
G1G
2G
4G
6
+
G1G
3G
5G
6
=
G1G
2G
4G
6+G
1G
3G
5G
6
1G4H
2G
5H
2+G
1G
2G
4G
6H
4+G
1G
3G
5G
6H
4+G
1G
2G
4G
6H
3+G
1G
3G
5G
6H
3
4.2
(a) ForK=10 and y = 10r , we have:
Case a:
Y
R
= 1K
3
1= 0.01
Case b:
Y
R=
K
1+ 2K
3
2= 0.364
Case c:
Y
R=
K3
1+ 3K
3 3 = 0.099
(b) Sensivity SK
G , G =Y
R
Case a:
dG
dK= 3
1K
2
SK
G=
K
G
dG
dK=
K
1K
33
1K
2( ) = 3
Case b:
SK
G= 0.646
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Case c:
SK
G= 0.03
Case c is the least sensitive.
(c)
Case b:
SK
G= 2.354
Case c:
SK
G= 0.99
The closed-loop system is much more sensitive to errors in the
feedback path than in the forward path.
4.4 G s( ) =A
s s+ a( )
(a) T s( ) =G s( )
1+ G s( )=
A
s s+ a( )
1+A
s s+ a( )
=
A
s2+ as+ A
dT
dA=
s2+ as+ A( ) A
s2+ as+ A( )
2
SAT =
A
T
dT
dA=
s s+ a
( )s s+ a( )+ A
(b)
dT
da=
sA
s2+ as+ A( )
2
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a
T
dT
da=
a s2+ as+ A( )
A
sA
s2+ as+ A( )
2
Sa
T=
as
s s+ a( )+ A
(c)
T s( ) =G s( )
1+ G s( )
dT
d=
G s( )2
1+ G s( )( )2
T
dT
d=
1+ G( )G
G2
1+ G( )2=
G
1+ G
ST=
A
s s+ A( )
1+A
s s+ a( )
=
A
s s+ a( )+ A
- If a = A =1, the transfer function is most sensitive to variations in a andA near=1 rad/sec
- The steady-state response is not affected by variations in A and a.- The steady-state response is heavily dependent on since ST 0( ) =1.0
4.9
(a)For a unity feedback system to be Type 1 the open loop transferfunction must have a pole at s = 0. Thus in this case, since G has nosuch pole, it is necessary for D to have a pole at s = 0.
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(b)Y s( ) =
1
s2+ D s( )+ K
W s( )
lims0
s1
s2+ D s( )+ K
1
s= 0
if and only if
lims0
s1
D s( ) =
if and only if =1 since D(s) has a pole at the origin. Therefore
the system will reject step disturbances with zero error.
4.19
(a) System (a):E= R Y =
R s( )1+ G s( )
E s( ) =s 4s+1( )
4s2+ s+ K
0K
1
R s( )
Using final value theorem:
ess,ramp =1
K1
=
1
Kv K
1=Kv =1
System (b):
E= R Y=1+G K
2G
1+GR
E s( ) =4s+1+ K
3K
01K
2( )4s+1+ K
3K
0
R s( )
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Using final value theorem:
ess,ramp =1+K3K0 1K2( )
1+K3
K0
= 0
forK0=11+K
31K
2( ) = 0
ess,ramp =4
1+K3
=
1
Kv
forKv=1 K
3= 3
K2=
4
3 ,K
3=
3
(b)
Let K0=K
0+ K
0, system (a):
ess,ramp = lims0
ss 4s+1( )
4s2+ s+K
0+ K
0( )1
s= 0
System (b):
ess,ramp =1+K3 K0 + K0( ) 1K2( )
1+ K3 K0 + K0( ) K0=1=
K0
1+ 3 1+ K0( ) 0
System type of system (b) is not robust. The system (a) is preferred over
system (b) because it is more robust to parameter changes.
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4.20
(a)
T s( ) =Y s( )R s( )
=
10 kp
s+ kI
( )s s+1( ) s+10( )
1+ 10 kp s+ kI( )
s s+1( ) s+10( )
E s( ) = 1T s( )( )R s( )
=
s s+1( ) s+10( )+10 kp s+ kI( )10 kp s+ kI( )s s+1( ) s+10( ) +10 kp s+ kI( )
1
s2
For=1,
ess = lims0
ss s+1( ) s+10( )
s s+1( ) s+10( ) +10 kp s+ kI( )
1
s2
=
10
10kI
=
1
kI
for characteristic equation: s3 +11s2 +10s+10 kps+ kI( ) = 0
The Rouths array is,
s3 : 1 10 1+ kp( )
s2 : 11 10k
I
s1 :
110 1+ kp( )10kI11
s0
:10k
I
For stability: all elements in the first column must be positive. The
following conditions must be satisfied;
kI> 0 and 111+ kp( ) kI > 0 .
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(b) = 0.9
E s( ) =s s+1( ) s+10( )+ 9 kp s+ kI( )10 kp s+ kI( )
s s+1( ) s+10( )+ 9 kp s+ kI( )
R s( )
=
s s+1( ) s+10( ) kp kIs s+1( ) s+10( )+ 9 kp s+ kI( )
R s( )
Using final value theorem, for R s( ) =1
s2
ess
(c) for R s( ) =1
s
lims0
sE s( ) = lims0
s s+1( ) s+10( ) kp s kIs s+1( ) s+10( ) + 9 kp s+ kI( )
= k
I
9kI
= 1
9
The system is type 0. Kp is defined such that ess =1
1+Kp . Kp = 8.
Without the magnitude an equivalent result is that Kp = 10 .