Chapter 2 MATHEMATICAL FOUNDATION 2-1 (a)
Poles: s = 0, 0, Zeros: s =
−1, −10;
(b)
Poles: s =
−2, ∞, ∞, ∞.
−2, −2;
Zeros: s = 0. The pole and zero at s =
(c)
Poles: s = 0,
−1 + j, −1 − j;
Zeros: s =
(d)
Poles: s = 0,
−1, −2, ∞.
−2.
2-2 (a)
(b) G (s) =
5
( s + 5)
(c) G ( s) =
2
(d)
(s
4s 2
(e) G ( s)
=
−1 cancel each other.
1 s
2
∞
G (s)
+4
+4
=
)
∑e
+
1
G (s)
s+ 2
kT ( s + 5 )
1
= 1
k =0
−e
2-3 (a) g ( t ) = u s ( t ) − 2u s (t − 1) + 2 u s( t − 2) − 2 u s ( t − 3) + L G (s ) =
1 s
(1 − 2e − s + 2e−2 s − 2e −3s + L ) =
gT (t ) = u s (t ) − 2us (t − 1) + us (t − 2) GT (s ) =
1 s
(1 − 2e − s + e −2s ) = ( 1 − e − s ) 1
1−e
(
−s
s 1+ e
−s
0≤ t ≤ 2 2
s
1
)
−T ( s+5 )
=
4 s
2
+ 4s +8
∞
g(t )
∞
∑
=
g
T
− 2k )us (t − 2k )
(t
G (s)
∑s
=
k =0
1
−e
(1
−s
2
) e
−2 ks
=
k =0
−s
1− e
s (1 + e
−s
)
(b) g ( t) = 2tu s ( t ) − 4(t − 0.5) u s (t − 0.5) + 4(t − 1) us (t − 1) − 4(t − 1.5)us (t − 1.5) + L G ( s) = g
T
(1 − 2e
2 2
s
−0.5 s
+ 2e
−s
− 2e
−1.5 s
− 0.5 s ( ) + L) = 2 −0.5 s s (1 + e )
2 1−e
= 2 tu s ( t ) − 4 ( t − 0 . 5) u s ( t − 0 . 5) + 2( t − 1 ) u s ( t − 1 )
(t )
(1 − 2e−0.5 s + e− s ) = s 2 (1 − e−0.5 s )
2
GT ( s ) =
s
2
2
∞
g (t ) =
∑
k=0
G(s ) =
≤ t ≤1
2
∞
g T ( t − k )us ( t − k )
0
∑ s2 ( 2
1 −e
−0.5 s
k=0
)
2
e
− ks
=
( −0.5 s ) 2 −0.5s s (1 + e ) 2 1−e
2-4 g(t )
= ( t + 1 ) u s ( t ) − ( t − 1 ) u s ( t − 1 ) − 2 u s ( t − 1 ) − ( t − 2 ) u s ( t − 2 ) + ( t − 3) u s ( t − 3) + u s ( t − 3)
G ( s) =
s
2-5 (a)
(1 − e − s − e −2 s + e −3 s ) + s (1 −2e − s + e −3 s )
1
1
2
Taking the Laplace transform of the differential equation, we get 1 1 2 ( s + 5s + 4) F ( s) = F (s) = s +2 ( s + 1)( s + 2 )( s f (t )
=
1
e
−4 t
+
6
(b)
sX ( s ) 1
1
e
−t
−
3
− x1( 0 ) =
1
e
−2 t
t
+4)
=
1 6( s
+ 4)
=
X (s)
=
2
2
s
X (s)
=1
x (0)
2
s( s
+ 3 s +1
+ 1 )( s + 2 ) −1
(s
+ 1 )( s + 2 )
1
= 0 .5 + e
−t
− 0 .5 e
3( s
+ 1)
−
1 2( s
+ 2)
≥0
1
sX
2
(s)
− x 2 ( 0 ) = −2 X 1 ( s ) − 3 X 2 ( s ) +
1
= =
2s
+
−1 s
+1
1 s
+
+1
−
1 2( s
+ 2)
1 s
+2
Taking the inverse Laplace transform on both sides of the last equation, we get x (t )
1
2
Solving for X1 (s) and X2 (s), we have X 1 ( s)
+
−2 t
t
≥0
x (t ) 2
= −e
2
−t
+e
−2 t
t
≥0
1 s
x (0) 2
=0
2-6 (a) G (s)
(b) G (s)
=
(c) G (s ) =
=
1 3s
−2 . 5
(
50
5
−
s
(s
20
=
1 s
−
s s
2
g(t )
= 0 .5 t
2
e
−0.5 t
2 .5
+
s
2
=
1 s
g(t )
+3
s +4
−1
g (t )
+ 3)
3( s
30s + 20
+s+2
g (t ) = 1 + 1.069e
(e)
−
s +1
(d) G (s)
+ 1)
2
1
+
+ 2)
2( s
+
+1
s
1
−
+
)
e
s
1
−
1
e
3
2
= −2 . 5 e
−t
[
−s
−2 t
+s +2
−
+
1
e
−3 t
t
≥0
3
+ 5 te
g (t ) = 50 − 20e
1 2
=
−t
− (t −1)
+ 2 .5 e
−3 t
t
≥0
] us (t − 1)
− 30cos2(t − 1) − 5sin2(t − 1)
s s
2
Taking the inverse Laplace transform,
+s+2
[ sin1.323t + sin (1.323t − 69.3o ) ] = 1 + e−0.5 t (1.447sin1.323t − cos1.323t )
−t
t
≥0
2-7
−1 2 0 A = 0 −2 3 −1 −3 −1
2-8
0 0 B = 1 0 0 1
u (t ) =
(a)
u1( t) u ( t) 2
(b) Y (s )
=
R (s )
3s + 1 3
Y (s)
2
s + 2 s +5s + 6
=
R (s )
(c)
5 4
2
s + 10 s + s + 5
(d)
Y (s ) R (s )
=
s ( s + 2) 4
3
Y (s )
2
s + 10 s + 2 s + s + 2
R (s )
3
=
1+ 2e 2
−s
2s + s + 5
t≥0
4
5
6
7
8
9
10
11
12
13
Chapter 4 MATHEMATICAL MODELING OF PHYSICAL SYSTEMS
4-1 (a) Force equations: 2
f ( t) = M 1
d y1 dt
dy1 − dy 2 + K ( y − y ) 1 2 dt dt
dy1
+ B1
2
+ B3
dt
2 dy1 dy2 d y2 dy2 B3 − + K ( y 1 − y 2 ) + M 2 2 + B2 dt dt dt dt
Rearrange the equations as follows: 2
d y1 dt
=−
2
2
d y2 dt (i) State diagram:
2
(B
1
+ B 3 ) dy1
M1
B3 dy1
=
dt −
(B
M 2 dt
Since y
1
− y2
2
B3 dy2
+
M 1 dt
+ B3 ) dy2
M
1
dt
dx2 dt
=
K
x1 −
M2
(ii) State variables:
x
1
(B
=
2
+ B3 )
x2 +
M2 y , 2
x
2
=
dy
B3 M2 2
dx3
x3 x
,
1
dt
1
dt dx dt
3
= x2 =
x
dx dt dx
4
2
dt
4
=− =
K M
K
=
M
x
− y2 )
y ,
x
1
1
−
B
1
+
1
State diagram:
14
B M
2
+ B3
M 3
1
x
2
dy
=
4
1
x
2
K M
2
x
,
dt
B3 M1
x2 −
=
3
(B
1
dy
1
.
dt
+ B3 )
M1
x3 +
.
dt
+
2
−
dy
=
2
x1 +
M1
2
x
K
=−
dt
3
x
2
State equations: dx
M1
appears as one unit, the minimum number of integrators is three.
1
= − x2 + x3
f
− y2 ) +
1
(y
M2
State equations: Define the state variables as x = y − y ,
dx1
(y
M1 K
+
dt
2
K
−
x 1
3
K M
−
x 2
B
1
3
+
B
+ B3 M
1
3
M x
x
4
2
4
+
1 M
f 1
1 M
f
Transfer functions:
Y1 ( s ) F (s ) Y2 ( s ) F ( s)
M 2 s + ( B2 + B3 ) s + K 2
= =
{
3
{
3
s M 1 M 2 s + [( B1 + B 3 ) M 2 + ( B2 + B3 ) M 1 ] s + [K ( M 1 + M 2 ) + B1 B2 + B2 B3 + B1 B3 ] s + ( B 1 + B2 ) K 2
B3 s + K
s M1 M 2 s + [( B1 + B3 ) M 2 + ( B2 + B3 ) M1 ] s + [ K ( M1 + M 2 ) + B1 B2 + B2 B3 + B1 B3 ] s + ( B1 + B2 ) K 2
(b) Force equations: 2
d y1 dt
2
=−
(B
+ B2 ) dy1
1
M
+
dt
B2 dy2
+
M dt
1
dy2
f
M
dy1
=
dt
dt
K
−
B2
(i) State diagram:
Define the outputs of the integrators as state variables, x
1
=
y ,
x
2
=
2
dy
1
.
dt
State equations: dx
1
dt
=−
K B
x
1
dx
+ x2
dt
2
(ii) State equations: State variables: dx dt
1
=−
K B
x
1
+ x3
2
2
dx dt
2
x
1
=
=−
K M
y ,
x
2
dx
= x3
dt
Transfer functions:
15
x
3
2
1
−
=
=−
B
1
x
M
y ,
x
1
K M
2
x
1
3
−
1
+ = B
f
M
1
M
dy
1
.
dt x
3
+
1 M
f
y2
} }
Y1 ( s ) B2 s + K = 2 F (s ) s MB2 s + ( B1 B2 + KM ) s + ( B1 + B2 ) K
Y2 ( s) F ( s)
(c) Force equations: dy1
dy2
=
dt
2
1
+
dt
d y2
f
B1
dt
=−
2
(B
1
B2
=
M B2 s + ( B1 B2 + KM ) s + ( B1 + B2 ) K 2
+ B2 ) dy2 M
B1 dy2
+
dt
B1 dy1
+
M dt
M dt
(i) State diagram:
State equations: Define the outputs of integrators as state variables. dx
= x2
1
dt
dx
=−
2
dt
(ii) State equations: state variables: dx dt
1
= x3 +
1 B
dx
f
2
dt
1
x
1
K
x
M
1
=
y , 1
−
B
M x
dx
= x3
x
2
=
2
3
dt
2
+
1
f
M
y ,
x
2
=−
K
x
M
2
3
= −
dy
2
.
dt
B
2
M
x
3
State diagram:
Transfer functions:
Y1 ( s ) F (s )
Ms + ( B1 + B 2 ) s + K 2
=
(
B1 s Ms + B2 s + K 2
Y2 ( s )
)
F (s )
4-2 (a) Force equations:
16
=
1 Ms + B 2 s + K 2
+
1 M
f
−
K M
y2
y
=
1
1 K
+ Mg ) + y 2
( f
2
y
dt
2
d
2
=−
2
B dy M
−
2
dt
K
1
+ K2
y
M
+
2
K
2
M
y
1
State diagram:
State equations: Define the state variables as: x = y , 1
dx
dx
= x2
1
dt
=−
2
dt
K
1
x
M
x
2
1
−
B
x
M
2
2
=
+
dy
2
.
dt 1
( f
+ Mg
Ms
2
M
)
Transfer functions: Y (s) 1
F ( s)
s
=
2
+ Bs + K 1 + K 2
K ( Ms 2
2
Y (s) 2
+ Bs + K 1 )
F (s)
1
=
+ Bs + K 1
(b) Force equations: dy1 dt
=
1 B1
[ f ( t) + Mg] +
dy2 dt
−
K1 B1
(y
1
− y2 )
2
d y2 dt
2
=
B 1 dy 1
dy K B B dy − + ( y − y ) − ( y − y )− M dt dt M M M dt 2
1
2
1
2
2
1
2
State diagram: (With minimum number of integrators)
To obtain the transfer functions Y ( s ) / F ( s ) and Y ( s ) / F ( s ), we need to redefine the state variables as: x
1
=
y , 2
x
2
= dy 2
1
/ dt , and x
3
=
2
y . 1
State diagram:
17
2
Transfer functions:
Y1 ( s )
Ms + ( B1 + B 2 ) s + K 1 2
=
F (s )
s
2
[MBs + ( B B 1
4-3 (a) Torque equation: 2 d θ B dθ =− + 2 dt
1
2
Y2 ( s )
+ MK1 )]
=
F (s )
Bs + K 1
s [M B1 s + ( B1 B2 + MK1 ) ] 2
State diagram: 1
J dt
T (t )
J
State equations: dx
dx
= x2
1
dt
2
dt
=−
B
x
J
+
2
1
T
J
Transfer function:
Θ( s )
1
=
T (s)
+ B)
s ( Js
(b) Torque equations: d θ1 2
dt
2
=−
K J
(θ
1
−θ 2 ) +
1
dθ 2
K (θ 1 − θ 2 ) = B
T
J
dt
State diagram: (minimum number of integrators)
State equations: dx
1
dt
=−
K
x
B
1
dx
+ x2
dt
State equations: Let x = θ , 1
dx dt
1
=−
K B
x
1
+
K B
=−
2
x
2
x dx
2
dt
2
2
K
x
J
= θ1, =
x
1
+
1
T
J
and
x
3
dx 3
dt
State diagram:
18
= 3
dθ
1
.
dt
=
K J
x
1
−
K J
x
2
+
1 J
T
Transfer functions:
Θ1 ( s ) T ( s)
=
Bs + K
(
s BJs + JKs + BK 2
Θ2 (s )
)
T ( s)
=
K
(
s BJs + JKs + BK 2
)
(c) Torque equations: d θ1 2
T ( t ) = J1
dt
2
+ K (θ 1 − θ 2 )
d θ2 2
K (θ 1 − θ 2 ) = J 2
2
dt
State diagram:
State equations: state variables: dx dt
1
dx
= x2
dt
2
=−
K J
x
x
1
2
1
+
=θ2, K J
x
x
2
= dx
3
2
dt
dθ
2
x
,
dt 3
=
x
3
= θ1, dx
4
dt
4
x
=
=
4
K J
x
dθ
1
.
dt
1
1
−
K J
1
x
3
+
1 J
T
1
Transfer functions:
Θ1 ( s ) T ( s)
J 2s + K
Θ 2 (s )
2
=
s
2
J1 J2 s + K ( J1 + J 2 ) 2
T ( s)
=
K 2
s
J1 J 2 s + K ( J1 + J 2 ) 2
(d) Torque equations: d θm 2
T ( t) = J m
dt
2
+ K 1 (θ m − θ 1 ) + K 2 (θ m − θ 2 )
K 1 (θ m − θ 1 ) = J 1
19
d θ1 2
dt
2
K2 (θm − θ2 ) = J2
d θ2 2
dt
2
State diagram:
State equations: x = θ 1
dx
1
dt
dx
= −x2 + x3
2
dt
=
K J
1
x
m
− θ1,
dx 1
dt
1
3
x
=−
K J
2
1
dθ
=
x
1
x
,
dt
−
1
m
K J
2
x
3
+
4
m
=
dθ
x
,
dt
1 J
m
T
m
dx dt
4
4
= θm −θ2 ,
=
x
3
− x5
x
dx dt
5
5
=
=
dθ
2
.
dt
K J
x
2
4
2
Transfer functions:
Θ1 ( s ) T ( s) Θ 2 ( s) T (s )
K 1 (J 2 s + K 2 ) 2
=
s
2
s 4 + ( K1 J2 J m + K2 J 1 J m + K1 J 1 J 2 + K 2 J 1 J 2 ) s 2 + K1 K 2 ( J m + J 1 + J 2 ) K2 ( J1 s + K1 ) 2
=
s
2
s 4 + ( K1 J2 J m + K 2 J 1 J m + K1 J 1J 2 + K 2J 1J 2 ) s 2 + K1 K 2 ( J m + J 1 + J 2 )
(e) Torque equations: d 2θ m dt
2
=−
K1 Jm
( θ m − θ1 ) −
K2 Jm
( θ m − θ2 ) +
1 Jm
d θ1 2
T
dt
2
K1
=
J1
State diagram:
20
(θ
m
− θ1 ) −
B1 d θ1 J1 dt
d 2θ 2 dt
2
=
K2 J2
(θ
m
−θ 1 ) −
B 2 dθ 2 J 2 dt
State variables:
x
1
= θ m − θ1,
x
2
=
dθ
1
x
,
dt
dθ
=
3
m
x
,
dt
= θm −θ2 ,
4
x
dθ
=
5
.
2
dt
State equations: dx 1 dt
= −x 2 + x 3
dx
2
dt
K1
=
J
x
1
B1
−
J
1
dx 3
x
2
dt
1
Transfer functions:
Θ1 ( s )
(
K1 J 2 s + B2 s + K 2
=
K1
=−
2
J
x
1
−
K
m
J
2
)
J
Θ2 ( s )
dx
T
dt
m
=
(
dx
= x3 − x5
4
=
5
dt
K 2 J1 s + B1 s + K1 2
K2 J
x
4
−
2
B2 J
d θm 2
dt
2
Output equation:
+ Bm e
=
o
dθ m dt Eθ 20
∆( s )
3
m
2
1
1
2
m
1
2
2
1
2
2
1
2
1
1
m
1
1
2
m
2
1
1
2
1
2
m
2
+ K (θ m − θ L )
K ( θm − θ L ) = J L
d θL 2
dt
2
+ Bp
dθ L dt
L
π
State diagram:
Transfer function:
Θ L (s )
=
Tm ( s) Eo ( s )
=
Tm ( s)
(
)
K
(
)
s J m J L s + Bm J L + Bp J m s + Jm K + J L K + Bm B p s + Bm K 3
(
)
2
KE / 2 0π
(
)
s J m J L s + Bm J L + Bp J m s + Jm K + J L K + Bm B p s + Bm K 3
2
4-5 (a) d θ1 2
Tm ( t) = Jm
θ3 =
N1 N3 N2N4
dt θ1
2
+ T1
T2 =
T1 = N3 N4
N1 N2
T4 =
T3 =
T2
N3 N4
2
)
4-4 System equations: Tm ( t) = Jm
x
(B + B ) s + [ ( K J + K J ) J + ( K + K ) J J + B B J ] s + B K ) J + B K J + B K J ] s + K K ( J + J + J )} 4
+ [( B1K 2
1
+
T (s )
∆ ( s ) = s { J 1 J2 Jm s + J 2
4
m
∆( s )
T ( s)
x
d θ3 2
JL
dt
2
N3 N4
d θ3 2
T4 = J L
T4
dt
2
T2 = T3
θ2 =
N1 N2
θ1
2 N1 N3 d 2θ1 Tm = J m 2 + T4 = J m + JL 2 dt N 2N 4 N 2 N 4 dt
d θ1 2
21
N1 N 3
2
5
(b) d θ1 2
Tm = Jm θ2 =
dt
N1
2
+ T1
2
θ1
N2
d θ2
T2 = J2 N 1N 3
θ3 =
N2N4
dt
T4 = ( J3 + J L )
+ T3
2
d θ2 2
θ1
T2 = J 2
dt
2
2
N1
dt
T4 = J 2
N4
2 d θ3 d 2θ 2 N3 Tm ( t) = Jm + J 2 dt 2 + N ( J3 + J 4 ) dt 2 = Jm 2 dt N2 4
d θ1
2
T1 =
2
d θ2 2
N3
+
d θ3
dt
2
+
N3 N4
(J
N1 N2
T3 =
T2
+ JL ) 3
N3 N4
d θ3 2
dt
2
2 2 d 2θ1 N1 N1 N3 + J2 + N N ( J3 + J L ) dt 2 N2 2 4
4-6 (a) Force equations:
dy1 − dy 2 dt dt
2 dy1 − dy 2 = M d y 2 + B dy 2 t 2 dt dt dt dt
f ( t) = K h ( y1 − y2 ) + Bh (b) State variables:
x
1
=
y
1
− y2,
x
dy
=
2
K h ( y1 − y2 ) + Bh 2
dt
State equations: dx
=−
1
dt
K B
h
x
1
1
+
B
h
dx
f (t )
2
dt
h
=−
B
t
x
M
2
+
1
f (t )
M
4-7 (a) T
m
= Jm
2
d
θm
dt
2
d θm
+T1
2
Tm = Jm
Set
∂α L ∂n
dt
2
= 0.
T
=JL
d θL
2
2
θL
dt
2
d
(T
m
dt
2
T
N
1
T
2
= nT 2
(
2
)
(
θm N 1 = θ L N 2
2
nTm − n TL 2
Thus, α L =
)
− 2 nTL ) J m + n J L − 2nJL nTm − n J L = 0
n
∗
=−
J T m
2
L
2J T
L
2
Or, n
Jm + n JL 2
2
+
J T m
L
J T
+
2
J mTL
+ 4 J m J LT m
Torque equation about the motor shaft:
∗
=
L m
L
Relation between linear and rotational displacements:
22
−
J J
m L
where the + sign has been chosen.
2J T
Jm / J
n
2
= 0 , the optimal gear ratio is n
4-8 (a)
N
+ nTL =
L m
When T
=
L m
Optimal gear ratio:
(b)
1
J m + nJ α + nT L L L n
2
+ nJL
+TL
=0
T4
T
(b)
m
2
d
= Jm
θm
dt
2
+ Mr
2
d
2
θm
dt
dθ
+ Bm
2
m
= r θm
y
dt
Taking the Laplace transform of the equations in part (a), with zero initial conditions, we have
(
Tm ( s) = Jm + Mr
)sΘ
2
2
m
( s) + Bm sΘ m (s )
Y ( s) = rΘ m ( s)
Transfer function:
Y ( s) Tm ( s)
=
r
(
s J m + Mr
r
)s + B
m
4-9 (a) d θm 2
Tm = Jm
dt
2
(
+ r ( T1 − T2 )
2
T1 − T2 = M
dt
d θm 2
d y
Thus, Tm = J m
2
(c) State equations: dx1 = rx3 − x2 dt
dx2
=
Tm ( s)
=
dt
2
K1 + K 2
dt
(d) Transfer function: Y ( s)
)
T1 = K2 rθ m − rθ p = K 2 ( rθ m − y )
M
+ r ( K1 + K2 )( rθ m − y )
dx3
x1
=
dt
T2 = K1 ( y − rθ m ) 2
M
− r ( K1 + K 2 ) Jm
d y dt
x1 +
= ( K1 + K2 )( rθ m − y )
2
1 Jm
Tm
r ( K1 + K 2 )
s
2
Jm Ms + ( K1 + K2 ) ( Jm + rM ) 2
(e) Characteristic equation:
s J m Ms + ( K1 + K2 ) ( Jm + rM ) = 0 2
2
4-10 (a) Torque equations: d θm 2
Tm ( t) = Jm
dt
2
+ Bm
dθ m dt
+ K (θ m − θ L )
K ( θm − θ L ) = J L
State diagram:
23
d θL 2
dt
2
+ BL
dθ L dt
(b) Transfer functions: Θ L ( s) Tm ( s)
=
K
Θ m ( s)
∆ ( s)
Tm ( s)
=
J L s2 + BL s + K
∆ ( s) = s J mJ L s3 + ( Bm J L + BL J m ) s 2 + ( KJ m + KJ L + Bm BL ) s + Bm K
∆ (s )
(c) Characteristic equation:
∆( s ) = 0
(d) Steady -state performance:
T (t ) m
= T m = consta
T (s)
nt.
m
=
T
m
.
s
J L s + BL s + K 2
lim ω m ( t) = lim sΩ m ( s) = lim t →∞
(e)
s →0
s →0
Thus, in the steady state,
ω m = ω L.
The steady-state values of
ωm
4-11 (a) State equations: dθ L =ωL
dω
dt
dθ m dt
= ωm
L
dt
dω m dt
=
and
K J
=−
2
J m J L s + ( Bm J L + BL J m ) s + ( KJ m + KJ L + Bm BL ) s + Bm K
ωL
θm −
L
Bm Jm
3
2
do not depend on J
K J
ωm −
2
dθ
θL
L
(K
1
dt
+ K2 ) Jm
(b) State diagram:
(c) Transfer functions:
24
m
t
and J . L
= ωt
θm +
K1 Jm
θt +
dω dt
K2 Jm
t
=
θL +
K J
1
θm −
t
1 Jm
Tm
K J
1
t
θt
=
1 Bm
ΘL ( s )
=
(
K 2 J t s + K1 2
)
Θ t(s )
∆ ( s)
Tm ( s)
=
(
K1 J L s + K 2 2
)
Θ m (s )
∆ ( s)
T m( s)
J t J L s + ( K1 J L + K 2 J t ) s + K1 K 2 4
=
2
∆ ( s)
Tm ( s )
∆ ( s ) = s [ J mJ Ls + B mJ LJ t s + ( K1 J L J t + K 2J L J t + K 1Jm J L + K 2J m J t ) s 5
4
3
+ Bm J L ( K1 + K 2 ) s + K1 K2 ( J L + J t + J m ) s + BmK 1K 2 ] = 0 2
(d) Characteristic equation: ∆ ( s ) = 0 .
4-12 (a) K H (s ) −K 1 H i (s ) 1 + K1 H e ( s ) + 1 i H e (s) + B + Js Ra + L a s B + Js Ra + La s = ≅ =0 −1
Ω m (s ) TL ( s)
∆( s)
ωr = 0
∆( s)
Thus,
H e (s ) = −
H i (s )
H i (s)
Ra + La s
H e (s)
Ω m (s )
(b)
=
Ω r ( s) ∆ ( s ) = 1 + K1 H e ( s ) + = 1+
(R
a
Ω m (s ) Ω r ( s)
= TL =0
(R
(R
(R
+ La s ) ( B + Js )
K1 K b
K1 K b
+ La s )( B + Js )
+ La s )( B + Js )
+
R a + La s
d
2
θ 2
+
(R
K1 K i K b H e( s) a
+ La s ) ( ( B + Js )
K 1 Ki
(R
+ La s ) ( B + Js )
a
+ La s ) ( B + Js ) + Ki Kb + K1 Ki Kb H e ( s)
a
dt
F d a
1
= T s d 2 δ + K F d 1θ
= J α α1 = J
d
2
dt
θ 2
K
F
1
≅
Kb H e ( s)
d
1
θ
sin
δ≅δ
− K F d 1θ = T s d 2δ
Θ( s ) − K F d 1 Θ( s ) = T s d 2 ∆ ( s )
(b)
Js
(c)
With C and P interchanged, the torque equation about C is:
2
K1 H i ( s)
+
K1 K i
dt
J
a
∆( s)
4-13 (a) Torque equation: (About the center of gravity C) 2 d θ J = T s d 2 sin δ + F d d 1 2 Thus,
K1 K i
TL =0
a
= − ( Ra + L a s )
Ts ( d1 + d 2 ) δ + Fα d 2 = J
d θ 2
dt
2
Ts ( d 1 + d 2 ) δ + K F d 2θ = J
Js Θ(s ) − K F d 2 Θ( s ) = Ts ( d1 + d 2 ) ∆( s ) 2
25
Θ( s ) ∆ (s )
=
d θ 2
dt
Ts ( d1 + d 2 ) Js − K F d 2 2
2
θe = θr −θo
4-14 (a) Cause-and-effe ct equations: dia
Ra
=−
dt
d θm
La
dt
d θo
=−
2
dt
2
(θ
JL
a
1
+
J m dt
KL
=
(e
La
Bm dθ m
2
2
1
ia +
J
− eb )
e
= Ke
a
Tm = K ii a nKL
Tm −
= K sθ e
e
( nθ
Jm
m
− θo )
Tm
T2 =
θ 2 = nθ m
n
− θo )
2
State variables:
x
1
= θo,
x
= ωo ,
2
x
=θm,
3
x
4
= ωm,
x
5
= ia
State equations: dx
1
dt dx
4
dt
dx
= x2
=−
=−
2
dt nK J
K J
L
x
x
1
−
n KL J
m
x
3
nK J
L
2
L
+ 1
Bm
−
J
m
m
x
4
L
x
dx 3
dt
L
+
Ki J
x
3
= dx
5
x
5
dt
m
4
=−
KK L
s
x
1
−
a
Kb L
x
4
a
−
Ra L
x
5
+
KK
a
L
s
a
(b) State diagram:
(c) Forward-path transfer function: Θ o ( s) KK s K i nK L = 4 3 2 2 Θe ( s ) s J mJ L La s + J L ( Ra J m + Bm J m + Bm La ) s + n K LL a J L + K L J m La + Bm R a J L s +
(
(n R K J 2
a
L
L
)
)
+ R a KL J m + B mK L L a s + K i K bK L + R a B mK L
Closed-loop transfer function:
Θo ( s) Θr ( s )
=
L
)
(R J
(n R K J
L
+ Ra KL J m + Bm KL L a s + ( K i K b K L + Ra BmK L ) s + nKK s K i K L
a
K
(
L
2
(d)
KK s K i nK L
J mJ LL a s + J 5
L
a
= ∞, θ o = θ 2 = n θ m .
+ Bm J m + B m La ) s + n K LL a J L + K L J m L a + B mR a J L s + 4
m
)
J
L
2
2
is re flecte d to m otor s ide so
State equations:
26
J
T
=
J
m
+n
2
J . L
3
θr
dω dt
m
=−
B J
m
ωm +
T
K J
i
i
dθ a
dt
T
di
= ωm
m
a
dt
=−
R L
a
i
a
a
+
KK L
s
θr −
a
KK L
s
nθ
m
a
−
K L
b
ωm
a
State diagram:
Forward-path transfer function:
Θ o ( s) Θe ( s )
=
KKs K i n s J TL a s + ( Ra J T + Bm L a ) s + Ra Bm + K i K b 2
Closed-loop transfer function:
Θo ( s) Θr ( s )
=
KK sK in
JT L a s + ( R a J T + Bm La ) s + ( Ra Bm + Ki Kb ) s + KK s K i n 3
2
From part (c), when K
L
= ∞,
all t he ter ms wit hout K
L
in
Θo (s ) / Θe ( s )
and
Θ o ( s ) / Θr ( s )
can b e negl ected.
The same results as above are obtained.
4-15 (a) System equations:
f = K ii a = M T (b)
dv dt
+ B Tv
ea = Ra ia + ( La + Las )
dia dt
− Las
di s dt
0 = Rsis + ( L s + L as )
+ eb
di s dt
− L as
di a dt
Take the Laplace transform on both sides of the last three equations, with zero initial conditions, we have
Ea ( s ) = [ Ra + ( La + Las ) s ] I a ( s) − Las sI s ( s ) + K b V( s)
Ki I a ( s ) = ( MT s + BT ) V ( s )
0 = − Las sI a ( s) + [ Rs + s ( Ls + Las ) ] I s ( s ) Rearranging these equations, we get
V (s ) = Ia (s ) =
Ki M T s + BT
Y (s ) =
I a (s )
1 Ra + ( La + L as ) s
[E
a
V (s ) s
=
Ki
s ( M T s + BT )
( s ) + Las sI s ( s ) − KbV ( s ) ]
Block diagram:
27
Ia (s)
Is (s ) =
L as s
Ra + ( L a + L as ) s
I a ( s)
(c) Transfer function: Y ( s)
K i [R s + ( L s + L as ) s ]
=
s [R a + ( L a + L as ) s ][ R s + ( L s + L as ) s ]( M T s + BT ) + K i K b [R s + ( L a + L as ) s] − Lass 2
E a (s )
(M
2
T
s + BT )
4-16 (a) Cause-and-effect equations:
θe = θ r − θ L
e = K sθ e dω m
Tm = K i ia
dt
=
1
Bm
Tm −
Jm K
K s = 1 V/rad
b
Jm
ω−
= 15 . 5 V
KL Jm
(θ
m
ea = Ke dω L
−θ L )
=
dt 15 . 5
=
/ KRPM
ia =
× 2 π / 60
1000
KL
(θ
JL
m
ea − eb Ra
−θ L )
= 0 .148
V / rad / sec
Bm
KL
eb = Kbω m
State equations:
d θL dt
= ωL
d ωL
=
dt
KL JL
θm −
KL JL
θL
d θm dt
d ωm
= ωm
=−
dt
Jm
ωm −
Jm
θL+
1 Ki Jm Ra
( KK θ s
e
− K bω m )
(b) State diagram:
(c) Forward-path transfer function: G ( s) =
J m Ra J L
K i K Ks KL s J mJ L Ra s + ( Bm Ra + K i K b ) J L s + R a K L ( J L + J 3
= 0 .03 × 1 .15 × 0 . 05 = 0 . 001725
2
Bm Ra J L
= 10 × 1 .15 × 0 .05 = 0 . 575
28
m
)s
+ K L ( Bm Ra + K i K b )
Ki K bJ L
=
21 × 0 .148
× 0 . 05 = 0 .1554
R K J a
L
L
= 1.15 × 50000 × 0 . 05 =
R K J
2875
a
L
m
= 1 .15 × 50000 × 0 . 03 = 1725
K KK K i
K L ( Bm Ra + K i K b ) = 50000(10 × 1.15 + 21 × 0.148) = 730400
s
=
L
608.7 × 10 K
21
× 1 × 50000 K = 105000
0K
6
G ( s) =
(
s s + 423.42 s + 2.6667 × 10 s + 4.2342 × 10 3
2
6
8
)
(d) Closed-loop transfer function:
M (s ) =
Θ L ( s) Θ r ( s)
=
G( s) 1 + G ( s) M (s)
K i K Ks K L
=
=
J mJ LR a s + ( B m R a + K i K b ) J L s + R a K L ( J L + J m ) s2 + K L ( Bm Ra + K iK b ) s + K i K Ks K L 4
3
6 . 087 s
+ 423
4
.42 s
3
+ 2 .6667 × 10
Characteristic equation roots: K
4-17
=1
K
s
= − 1.45
s
= − 159
s
= − 131 . 05 ±
. 88
6
s
× 10 2
8
K
+ 4.2342 × 10
8
= 2738
s
+ 6 . 087 × 10
K
s
= ± j 1000
s
= −211 . 7 ±
j 1273 . 5
s
=
s
= −617
405
±
j1223 .4
.22
(a) Block diagram:
Tr ( s )
=
KM KR
( 1 + τ s ) (1 + τ s ) + K c
s
4-19 (a) Block diagram:
29
= m
KR
3.51 20 s + 12 s + 4.51 2
K
= 5476
j 1614. 6
(b) Transfer function: TAO ( s )
8
±
j 1275
(b) Transfer function: Ω( s ) α ( s)
=
K1 K 4 e
−τ D s
Js + ( JK L + B ) s + K 2 B + K3 K 4 e 2
− τ Ds
(c) Characteristic equation:
Js + ( JK L + B ) s + K 2 B + K 3 K 4e 2
(d) Transfer function: Ω( s ) α ( s)
− τD s
=0
K1 K 4 ( 2 − τ D s )
≅
∆ ( s)
Characteristic equation:
∆ ( s ) ≅ J τ D s + ( 2 J + JK 2τ D + Bτ D ) s + ( 2 JK 2 + 2B − τ D K 2 B − τ D K 3 K4 ) s + 2 ( K 2 B + K 3 K 4 ) = 0 3
2
4-19 (a) Transfer function: Ec ( s)
G ( s) =
=
E (s )
1 + R2C s
1 + ( R1 + R 2 ) Cs
(b) Block diagram:
(c) Forward-path transfer function: Ωm ( s)
=
E (s )
[1 + ( R
1
K (1 + R2C s )
+ R 2 ) Cs ] ( K b Ki + Ra JL s )
(d) Closed-loop transfer function: Ωm ( s) Fr ( s ) (e)
=
Gc ( s) =
[1 + ( R
1
E c ( s)
=
Kφ K (1 + R2C s )
+ R2 ) Cs] ( K b K i + Ra J L s ) + Kφ KK e N (1 + R2C s )
(1 + R C s )
E (s )
Forward-path transfer function:
Ωm ( s) E (s )
2
RCs 1
=
K (1 + R2C s )
RCs ( K b K i + Ra J L s ) 1
Closed-loop transfer function:
30
Ωm ( s) Fr ( s ) (f)
Ke
= 36
pulse s / rev
f
= 120
pulse s / sec
=
e
f
r
ω
NK
ω m = 120
Thus, N = 1. For
K φ K (1 + R2C s )
=
R1C s (K b K i + Ra J L s ) + Kφ KKe N (1 + R2C s )
= 36
/ 2 π pulse s / rad
ωm = pulse s / sec
ω m = 1800
=
= 5 . 73 pul
200 RPM
= 200(
ses / rad. 2 π / 60 ) rad / sec
N ( 36 / 2 π ) 200( 2 π / 60 )
RPM,
120
=
= 120
N pulse s / sec
N ( 36 / 2 π ) 1800( 2 π / 60 )
= 1080
N.
Thus,
N
= 9.
4-20 (a) Differential equations:
dθ m − dθ L dt dt dt dt 2 dθ dθ L = J d θ L + B dθ L + T K (θ m − θ L ) + B m − L 2 L L dt dt dt dt d θm 2
Ki ia = Jm
(b)
2
+ Bm
dθ m
+ K ( θm − θ L ) + B
Take the Laplace transform of the differential equations with zero initial conditions, we get
(
)
Ki I a ( s ) = J m s + Bm s + Bs + K Θm ( s ) + ( Bs + K ) Θ L ( s )
( Bs + K ) Θ Solving for
Θm ( s )
and
2
m
ΘL ( s )
Θ m (s ) = Θ L (s ) =
(
)
( s ) − ( Bs + K ) Θ L ( s ) = J L s + BL sΘ L (s ) + TL ( s) 2
from the last two equations, we have
Ki
J m s + ( Bm + B ) s + K 2
Bs + K J L s + ( BL + B ) s + K 2
I a (s ) + Θ m (s ) −
Bs + K J m s + ( Bm + B ) s + K 2
Θ L (s )
TL ( s )
J L s + ( BL + B ) s + K 2
Signal flow graph:
(c) Transfer matrix: 2 1 Ki J L s + ( BL + B ) s + K Θ m ( s ) = Θ (s ) ∆ (s ) Ki ( Bs + K ) L o
31
Ia ( s ) 2 Jm s + (B m + B ) s + K −TL ( s ) Bs + K
∆ o ( s ) = J L Jm s + [ J L 3
(B
+ B) + J m ( BL + B ) ] s + [ BL Bm + ( BL + BM
) B + (J
3
m
+J
m
) K]s
L
+ K ( BL + B ) s
2
4-21 (a) Nonlinear differential equations: dx ( t ) dt With R
a
= 0,
φ (t ) =
=
dv ( t )
v(t )
= −k ( v ) − g ( x ) +
dt e(t )
=K
K v(t )
f
=
i (t ) f
=
K i (t ) f
f
K
= − Bv
f (t )
i (t )
(t )
f (t )
i (t )
Then,
f a
+ a
=
b
=
K
φ ( t ) ia ( t ) = i
(b) State equations:
K e (t ) i
K
2
b
K
.
2
f
dv ( t ) Thus,
v (t )
dt
= − Bv
+
(t )
K 2
K K b
i
2
e (t )
2
f
v (t )
i ( t ) as input. a
dx ( t ) dt
dv ( t )
= v (t )
= − Bv
dt
(t ) + K K
2
i (t )
i
f a
(c) State equati ons: φ ( t ) as input. f (t )
= K iK
dx ( t ) dt
2
i (t )
ia ( t )
f a
dv ( t )
= v (t )
= if
= − Bv
dt
(t )
(t ) +
φ( t )
=
K
K K
f
φ
i
2
(t )
f
4-22 (a) Force and torque equations: Broom: vertical direction:
horizontal direction:
f
f
x
=
M
Car: horizontal direction:
Eliminating f
x
=
Define the state variables as x
1
v 2
d
x (t )
f
x
d b
2
θ
= θ,
x
dx1 dt dx3 dt
= x2
dx2
= x4
dx4
=
dt
dt
(M
( L cos
b
y
θ−
x (t )
=
≅
,
f L cos x
J
2
1
x
=
dt x and
cos x
1
b
=
1
u ( t) + M b Lx2 x1 − 3 M b gx1 / 4
(M
32
b
+ M c ) − 3M b / 4
2
b
x
4
+ M b ) gx1 − M b Lx2 x1 − u ( t )
L [ 4 (M b + M c ) / 3 − M b ]
(c) Linearization:
M L
≅ 1.
2
c
θ
3
x , and
3
2
=
θ)
dt
f L sin
dt dθ
2
and f y from the equations above, and sin x
2
d
+ Mc
2
d
2
=
2
dt
M
+ L sin θ
dt J
u(t )
− M bg =
b
rotation about CG:
(b) State equations:
K K b
2
f (t )
e(t 0
=
dx . dt
f
v (t )
∂ f1
∂ f1
=0
∂ x1 ∂ f2
∂ x2
=
c
b
∂ f2
∂f 3 ∂x 3
∂f 3
=0
M b Lx2 − 3 M b g / 4
∂ f4
∂ x1 ∂ f4
(M
+ M c ) − 3M b / 4
b
∂ f4
=0
∂ x3
∂ x2
(M
∂ f4
=0
∂ x4
=
∂ x3
=0
∂u
=0
2 M b Lx1 x2
+ M c ) − 3M b / 4
b
1
=
∂u
∂f 3
=0
∂x 4
2
=
L ( M b + M c − 3M b / 4 )
L [ 4 (M b + M c ) / 3 − M b ]
=0
∂x 2
∂ f2
=0
−1
=
∂u ∂f 3
=0
∂u
−2 M b x1 x2
=
∂ x2
∂ f1
=0
∂ x4
∂ f2
2
b
∂ f2
=0
∂ f4
b
c
=0
∂ x4 ∂x 1
∂ x3
(M + M ) g − M x = 0 L ( M + M − 3M / 4) b
∂ f1
=0
2
∂ x1
∂f 3
∂ f1
=1
(M
b
+ M c ) − 3M b / 4
Linearized state equations:
0 ∆x& 1 3 ( M b + M c ) g ∆x& L ( M + 4 M ) b c 2 = ∆x& 3 0 & − 3M b g ∆x 4 M + 4M b c
=
(b)
i
eq
E
=
Ki
2
y (t )
1 eq
dt dy
eq
=0
dt x
1
eq
= i, x
R
x
2 eq
2
=
y
=
y , and x
E
eq
R
=
3
E
=
eq
+
dt di ( t )
eq
E
dy
At equili brium,
R
=
dL ( y ) dy ( t )
Ri ( t ) + i ( t )
(t )
2
Define the state variables as
Then, x
=
dt
= Mg −
My ( t )
Thus,
d L ( y ) i( t )
+
Ri ( t )
0
0
L( y) = L y
4-23 (a) Differential equations: e(t )
0 ∆x −3 1 0 0 0 ∆x2 L ( M b + 4 M c ) + ∆u 0 0 1 ∆x 3 0 ∆x 4 4 0 0 0 M + 4M b c 1
L di ( t ) y
dt
dy ( t )
= 0,
=
dt
−
Ri ( t ) d
= 0,
2
L y
2
y (t )
dt
2
i( t )
dy ( t ) dt
=0
K
eq
R
Mg
dy . dt
K
x
Mg
3 eq
=0
The differential equations are written in state equation form: dx dt
1
=−
R L
x x 1
2
+
x x 1
x
2
3
+
x
2
L
e
=
f
dx 1
2
dt
(c) Linearization:
33
=
x
3
=
f
dx 2
dt
2
3
=
g
−
K x1 M x
2 2
=
f
3
+
L di ( t ) y
dt
∂f 1 ∂x 1
=−
∂f 1 ∂e
R
x
L
=
∂f 3 ∂x 1
x
2 eq
+
2 eq
x
=
L
=−
x
3 eq
E
1
K
L
Mg
x
∂f 2
eq
∂x 1
x
L
E
∂x 2
2 K x 1 eq M
x
x x 1
x
3
2
2 eq
K
∂e
2 Rg
Mg
Eeq
K
x
1 eq
Mg
=
K
2 eq
=0
∗
K 0 0
0
x
∆x& = A ∆x + B ∆e ∗
Eeq K RL Mg ∗ B = 0 0
Mg
0
=
=0
∂e ∂f 3
The linearized state equations about the equilibrium point are written as:
Eeq K − L Mg ∗ A = 0 2 Rg − E eq
∂x 3
∂f 2
Mg
eq
∂f 1
=0
=1
∂x 3
E
eq
L
2
2 Rg
=
E
+
3
∂f 2
=0
∂x 2
=
−
1 eq
2
∂f 3
eq
R
∂f 2
=0
2 Rg
2 eq
=−
∂x 2
Mg
R
=−
2
∂f 1
K
eq
L
2 eq
2 K x 1 eq M
E
=−
4-24 (a) Differential equations: d
M1
M
2
y1 (t ) 2
dt d
2
2
=
y 2 (t ) 2
dt
Define the state variables as x
1
=
=
M 2g
y ,
x
1
dy 1 ( t )
−B
M 1g
2
dt dy
−B
=
2
−
dy
(t )
2
dt 1
x
,
dt
3
2
Ki
(t )
2
y1 ( t )
2
1
(t ) y2 (t )
Ki
− =
+ Ki 2
2
x
4
2
(t )
y2 (t ) − y1 (t ) y ,
− y1 ( t )
=
dy
2
2
.
dt
The state equations are:
dx1 dt
= x2
M1
dx2 dt
Ki
= M 1 g − Bx2 − dx
At equilibrium,
1
dt
= 0,
M 1g − Solving for I, with X
1
dx
2
2
x1
dt
(x
3
= 0,
2
Ki
+
dx
2
KI
2
X1
+
3
dt
KI
(X
3
2
dx3
− x1 )
2
dt
dx
= 0,
4
dt
= 0.
= x4
Thus , x
M2
2 eq
=0
2
− X 1)
=0
2
M 2g −
dx4 dt
3
4 eq
=0
2
= 1 , we have
M + M2 Y2 = X 3 = 1 + 1 M2
1/2
(b) Nonlinear state equations:
34
= 0.
2
− X1)
Ki
(x
3
and x
KI
(X
= M 2 g − Bx4 −
( M1 + M 2) g I= K
1/2
2
− x1 )
2
dx1 dt
dx2
= x2
B
= g−
dt
M1
(c) Linearization: ∂ f1 =0 ∂ x1 ∂ f2
=
∂ x1
∂ f2 ∂i
∂ f4 ∂ x1
=
=
2 KI
x2 −
3
M 1 x1
2
∂ x2
2 KI
M2 ( X3 − X1 )
3
∂ f1 ∂ x3
∂ x2
dx3
∂ f3
∂ f4
=0
∂ x3
Linearized state equations:
M
1
B
∂ f2
M1
∂ x3
∂ x2
=
= 2,
∂ f3
=0
2 KI
∂ x3
= 1,
2
g
∂ f1 ∂i
=
−2 KI
= 32 .2,
32.2(1 + 2) X = 96.6 X = 9.8285X 1 1 1 1
B
= 0 .1,
1/2
I =
(
X1 =
)
X 3 = 1 + 1 + 2 X 1 = 2.732 X 1 = Y2 = 2.732 0 2 1 2 KI 1 + 3 3 M1 X1 ( X 3 − X 1 ) ∗ A = 0 2 −2 KI 3 M 2 ( X 3 − X1 )
1
0 −2 KI
M1
M 1 ( X 3 − X1 )
0
0 2 KI
2
M 2 ( x3 − x1 )
∂ f2 ∂ x4
3
∂ f3
= 1
∂i
∂ f4
M2
∂i
=
=0
= 0
−2 KI M2 ( X3 − X1)
= 1.
1 9.8285
2 3
2
M 2 ( X 3 − X 1)
3
=1
X 3 − X 1 = 1.732
1 0 0 115.2 −0.05 = 1 0 0 0 − B − 37.18 M 2
0 1 0 2 KI −1 + M 1 X 12 ( X 3 − X 1 ) 2 − 6.552 ∗ B = = 0 0 − 6.552 −2 KI 2 M 2 ( X 3 − X 1)
4-25 (a) System equations:
35
2
=0
B
K
Ki
0
−B
0
∂ x4
=−
x4 −
2
∂ f3
= 0
∂ x4
3
M2
M1 ( X3 − X1 )
∂ f4
2
M 2 ( X 3 − X1 ) M
dt
B
= g−
=0
∂ x4
∂ f3
=0
∂ x1
dt
dx4
= x4
∂ f1
=−
∂ x2
3
2
=0
∂ f2
M1 ( X 3 − X 1 )
∂ f4
2
M 1 ( x 3 − x1 )
2
−1 1 2+ 2 M 1 X1 ( X 3 − X 1 ) 2
Ki
=0
2 KI
−2 KI
i + 2
M 1x1
∂ f1
2
+
K
−18.59 0 0 1 37.18 − 0.1 0
0
2
Tm = K i ia = ( J m + JL ) T
D
=
dω m dt
d (sec) V
+ B mω m e
ea = Ra ia + La
= r −b
b
dia
= Ksy
dt
+ K b ωm
E ( s) a
=
y = nθ m
KG ( s ) E ( s ) c
Block diagram:
(b) Forward-path transfer function: Y (s )
=
E (s )
K Kin G c ( s ) e
− TD s
s {( Ra + La s ) [( Jm + J L ) s + Bm ] + K b Ki }
Closed-loop transfer function:
Y (s ) R (s )
=
− TD s
K Kin G c ( s ) e
s ( Ra + La s ) [( Jm + J L ) s + Bm ] + K bK i s + KG c ( s )K i ne
36
− TD s
y = y ( t − TD )
Chapter 5 STATE VARIABLE ANALYSIS OF LINEAR DYNAMIC SYSTEMS 5-1 (a) State variables:
x
1
=
y,
x
2
=
dy dt
State equations:
Output equation:
dx1
dt 0 1 x1 0 = + r dx2 −1 −4 x2 5 dt (b) State variables:
x
1
=
y,
x
2
=
y = [1
dy
x
, dt
3
2
d
=
dt
x1 = x1 x2
0]
y 2
State equations:
Output equation:
dx1
dt 1 0 x1 0 dx 2 = 0 0 1 x2 + dt dx −1 −2.5 −1.5 x3 3 dt x1 =
(c) State variables:
∫
t
0
y (τ ) d τ ,
0 0 r 0.5
x2 =
dx1
x& 2 = x&3 x& 4
0
dt
0
0
x1
0
x
1
=
y,
x
2
=
dy ,
x
dt
3
State equations:
x& 2 = x&3 x& 4 5-2
dt
d y dt
2
x1 x 2 y = [1 0 0 0] = x1 x 3 x 4
0 0 1 0 x 0 2+ r 0 0 0 1 x3 0 − 1 − 1 − 3 −5 x 1 4
0
2
x4 =
,
Output equaton:
1
(d) State variables:
x&1
dy
x3 =
,
State equations:
x&1
x1 y = [1 0 0 ] x2 = x1 x3
=
2
d
dt
3
y 2
x
,
4
=
d y dt
3
Output equation:
1
0
0
x1
0
x1 x 2 y = [1 0 0 0] = x1 x 3 x 4
0 0 1 0 x2 0 + r 0 0 1 x3 0 0 −1 −2.5 0 −1.5 x 1 4
We shall first show that
Φ ( s ) = ( sI − A ) = We multiply both sides of the equation by
I
A
1 A
2
+ 2 + +L 2 s s 2! s ( sI − A ) , and we get I = I. Taking the inverse Laplace transform −1
37
on both sides of the equation gives the desired relationship for
5-3 (a) Characteristic equation: Eigenvalues:
s
∆( s ) =
= −0 . 5 +
− A = s2 + s + 2 = 0
sI
− 0 .5 −
j 1. 323 ,
φ( t ) .
j 1. 323
State transition matrix:
cos1.323t + 0.378sin1.323t
φ ( t) =
(b) Characteristic equation:
−1.512sin1.323 t
∆( s ) =
sI − A
−0.5t e −1.069sin (1.323t − 69.3 ) 0.756sin1.323t
o
= s 2 + 5s + 4 = 0
Eigenvalues:
= − 4,
s
−1
State transition matrix:
1.333 e− t − 0.333e−4 t
φ ( t) =
−t
−1.333 e − 1.333 e
(c) Characteristic equation:
−0.333e + 1.333e −t
0.333 e − 0.333e
−4 t
−t
∆ ( s ) = ( s + 3) = 0 2
−4 t
−4 t
Eigenvalues:
= −3, − 3
s
State transition matrix:
e −3 t φ ( t) = 0 (d) Characteristic equation:
∆( s ) =
−3 t
− 9 = 0 Eigenvalues:
2
s
e 0
s
= −3 ,
3
s
=
−
State transition matrix:
e3 t φ ( t) = 0 (e) Characteristic equation:
e 0
−3 t
∆ ( s ) = s + 4 = 0 Eigenvalues: 2
j2,
j2
State transition matrix:
cos2 t − sin2 t
φ ( t) = (f) Characteristic equation:
∆( s ) =
s
3
s i n 2t
cos2t
+ 5 s + 8 s + 4 = 0 Eigenvalues: 2
s
= − 1, − 2 ,
−2
State transition matrix:
e− t φ ( t) = 0 0 (g) Characteristic equation:
∆( s ) =
s
3
0 e
0
+ 15
e φ ( t) = 0 0
−5 t
5-4 State transition equation: x (t ) = φ (t )x( t ) +
−2 t
+ 75 s + 125 = 0
2
s
te −2t e 0
−2 t
te e
−5 t
−5 t
0
Eigenvalues:
s
= − 5,
− 5,
−5
te −5 t e 0
−5 t
∫ φ (t − τ )Br (τ )d τ t
0
(a)
38
φ (t )
for each part is given in Problem 5-3.
∫
t
0
1 s + 1 1 0 1 1 1 ( sI − A) −1 BR( s ) = L−1 ∆ ( s ) −2 s 1 0 1 s s+2 2 s(s + s + 2) 1 + 0.378sin1.323 t − cos1.323t −1 = =L t≥0 s−2 −1 + 1.134sin1.323 t + cos1.323t s s2 + s + 2 ) ( −1
φ ( t − τ ) Br (τ ) d τ = L
(b)
∫ φ ( t − τ ) Br (τ ) dτ = L ( sI − A) t
−1
0
−1
BR ( s ) = L
−1
1 s + 5 1 1 1 1 ∆( s) −4 s 1 1 s
s+6 1.5 − 1.67 + 0.167 1.5 − 1.67 e− t + 0.167e −4 t s ( s + 1)( s + 2) −1 = L−1 s s + 1 s + 4 = =L −t s−4 −1 + 1.67 − 0.667 −1 + 1.67 e − 0.667 e − 4t s( s + 1)( s + 4) s s + 1 s + 4 (c)
1 t −1 −1 −1 s + 3 ∫0 φ ( t − τ ) Br (τ ) dτ = L ( sI − A) BR( s) = L 0 0 0 −1 = =L 1 t≥ −3t 0.333 (1 − e ) s ( s + 3 )
0 1 1 1 s s + 3 0
0
(d)
1 t −1 −1 −1 s − 3 φ ( t − τ ) B r ( τ ) d τ = L ( s I − A ) B R ( s ) = L ∫0 0 0 = =L 1 s ( s + 3 ) −1
0 0.333 (1 − e −3t )
(e)
39
0 1 1 1 s s + 3
t≥ 0
0
t≥0
1 2 t −1 −1 −1 s + 4 φ ( t − τ ) B r ( τ ) d τ = L s I − A B R ( s ) = L ( ) ∫0 −2 s 2 + 4 2 s 2 −1 =L = t ≥0 1 0.5sin2t ( s 2 + 4 )
0 1 s 1 s 2 s + 4 2
(f)
1 s + 1 t −1 −1 −1 ∫0 φ ( t − τ ) Br (τ ) dτ = L ( sI − A) BR( s) = L 0 0 0 0 1 −1 −2 t = 0.5 (1 − e ) =L t≥0 s( s + 2) 0 0
0 1 s+ 2 0
0 1 1 1 2 (s + 2 ) s 0 1 s +2 0
(g)
1 s + 5 t −1 −1 −1 ∫0 φ ( t − τ ) Br (τ ) dτ = L ( sI − A) BR (s ) = L 0 0
1
( s + 5) 1 s +5 0
2
0 1 1 0 2 ( s + 5) s 1 1 s+5 0
0 0 0 1 0.04 0.04 0.2 −1 −1 −5 t −5 t =L =L 2 s − s + 5 − ( s + 5 ) 2 = 0.04 − 0.04 e − 0.2 te u s ( t) −5t s ( s + 5) 0.2 − 0.2 e 1 0.2 0.2 − s ( s + 5 ) s s +5 5-5 (a) (b)
Not a state transition matrix, since
φ( 0 ) ≠ I
(identity matrix).
Not a state transition matrix, since
φ( 0 ) ≠ I
(identity matrix).
(c) φ ( t )
is a state transition matrix, since
φ ( 0 ) = I and
40
1 [φ ( t ) ] = − t 1 − e (d) φ ( t )
is a state transition matrix, since
0 e
t
= φ ( − t)
φ ( 0 ) = I , and
e2 t [φ ( t) ]−1 = 0 0 5-6 (a) (1) Eigenvalues of A:
−1
1 = −t 1 − et e 0
−1
− te
2 2t
2t
− te
0
e
e
2 .325 ,
t e / 2
2t
− 0 . 3376 +
= φ ( − t)
2t
2t
j 0 . 5623 ,
− 0 .3376 −
j 0 . 5623
(2) Transfer function relation: −1
−1
s 1 −1 X( s) = ( sI − A ) B U ( s) = 0 ∆ (s ) 1
s 2 + 3s + 2 0 U ( s) = 1 −1 ∆ (s ) −s 1
0 0
−1 s + 3
s 2
s +3
1 0
1 1 s ( s + 3 ) s 0 U ( s) = s U (s ) ∆ (s ) 2 2 −2 s − 1 s 1 s
∆( s ) = s + 3 s + 2 s + 1 3
2
(3) Output transfer function:
1 1 s = = C ( s ) ( sI − A) B = [1 0 0 ] 3 2 U (s ) ∆( s ) 2 s + 3 s + 2 s + 1 s Y ( s)
(b) (1) Eigenvalues of A:
1
−1
− 1,
− 1.
(2) Transfer function relation:
1 ( s + 1) 2 1 0 1 s + 1 −1 U (s ) X ( s ) = ( sI − A) BU ( s ) = U (s ) = ∆ ( s ) 0 s + 1 1 1 ( s + 1)
∆ ( s ) = s + 2s + 1 2
(3) Output transfer function:
1 ( s + 1) 2 Y ( s) 1 1 s+2 −1 = = C ( s ) ( sI − A) B = [1 1] + = 2 2 U (s ) 1 ( s + 1) s + 1 ( s + 1) s + 1 (c) (1) Eigenvalues of A:
− 1,
0,
− 1.
(2) Transfer function relation:
s + 2s = 1 2
1 −1 X( s) = ( sI − A) BU ( s ) = ∆ (s )
0 0
s+2
1 0
1 1 s ( s + 2) ) s 0 U ( s) = s U ( s) ∆ (s ) −s s 1 s 2
41
2
(
)
∆( s ) = s s + 2s + 1 2
(3) Output transfer function:
1 s +1 1 = C ( s ) ( sI − A) B = [1 1 0 ] s = = 2 U (s ) s ( s + 1) s ( s + 1) s 2 Y ( s)
5-7
dy We write dt
=
dx
1
dt
+
dx
−1
2
dt
dx1 dt d x dy = = dt dt 2 d y dt2
2
d
= x 2 + x3
dt
y
=
dx
+
2
dt
dx dt
3
= −x1 − 2 x2 − 2 x3 + u
0 1 0 x1 0 0 1 1 x + 0 u 2 −1 −2 −2 x3 1
x1 1 0 0 x = y = 1 1 0 x y& 0 1 1
(1)
1 0 0 x = − 1 1 0 x 1 − 1 1
(2)
Substitute Eq. (2) into Eq. (1), we have
−1 1 0 = A 1 x + B1u = 0 0 1 x = dt −1 0 −2
dx
5-8 (a) s sI
− A = −1 1
a1 M = a2 1
s
−2
0
−2
0
0
a2 1 0
1 0 0
s
0 0 u 1
= s − 3 s + 2 = s + a 2 s + a1 s + a0 3
2
3
2
a
0
= 2,
a
1
= 0,
−1
0 −3 1 = − 3 1 0 1 0 0
S = B
−2 2 0 P = SM = 0 −1 1 −4 −2 1
42
AB
0 2 4 A B = 1 2 6 1 1 − 1 2
a
2
= −3
(b) s sI
− A = −1
−2
0
−1
0
s
−1
1
a1 M = a2 1
(c) s sI
−A =
+2
0
sI
−A =
0 0
1 0
+1
3
2
3
1 0
a
2,
s
2
s
a
3
+ a 2 s + a1 s + a0 2
a
0
3 3 1 = 3 1 0 1 0 0
s
−1 2
= 0,
−1 s
= 12 ,
= 1,
S = B
0 1 −1 AB A B = 1 0 − 1 1 − 1 1 0 1 1 2
(e) −A =
0
+1
2 1 P = SM = 2 3 1 2
sI
1
a
= −3
2
a
1
= 16 ,
a
2
1 −1 0 S = B AB A B = 1 −2 4 1 −6 23 9 6 1 P = SM = 6 5 1 −3 1 1
3
0
0 0
=
2
−1 = s + 3 s + 3 s + 1 =
1
0
2
16 7 1 = 7 1 0 1 0 0
=1
a2
a
= s + 7 s + 16 s + 12 = s + a 2 s + a 1 s + a 0
0
0
2
2
−1 s
0
a1 M = a2 1
1
a2
3
+3
s
2
a1 M = a2 1
2
1 2 6 S = B AB A B = 1 3 8 0 0 1 0 −1 1 P = SM = −1 0 1 1 0 0
0
s
− 3 s + 2 = s + a 2 s + a1 s + a 0
3
0 −3 1 = − 3 1 0 1 0 0
−1
1
s
0 0
0
0
(d)
1
1
+2
s
−1
s
a2
=
+3
=
s
2
+ 2 s −1 = s + a1 s + a 0 2
43
a
0
= − 1,
a
1
=2
a
1
= 3,
a
2
=3
=7
M=
a1 1
1
2 1 0 1 0 P = SM =
5-9 (a)
S = [B
=
AB ] =
0 1 1 −3
1 0 −1 1
From Problem 5-8(a),
0 −3 1 M = − 3 1 0 1 0 0 Then,
C V = CA = 2 CA (b)
0.5 1 3 Q = (MV ) = 0.5 1.5 4 −0.5 −1 −2 −1
From Problem 5-8(b),
C V = CA = 2 CA (c)
1 0 1 − 1 2 1 1 2 1
16 7 1 M = 7 1 0 1 0 0 1 0 1 −1 − 1 3 1 Q = (MV ) = 2 5 1
0.2308 0.3077 1.0769 0.1538 0.5385 1.3846 −0.2308 −0.3077 −0.0769
From Problem 5-8(c),
C V = CA = 2 CA
1 0 0 − 2 1 0 4 −4 0
Since V is singular, the OCF transformation cannot be conducted.
(d)
From Problem 5-8(d),
3 3 1 M = 3 1 0 1 0 0 Then,
C V = CA = 2 CA (e)
1 0 1 − 1 1 −1 1 −2 2
−1 1 0 Q = ( MV ) = 0 1 − 2 1 − 1 1 −1
From Problem 5-8(e),
44
M=
5-10 (a)
2 1 1 0
Eigenvalues of A:
T = [p1
Eigenvalues of A:
2
T = [ p1
1
2
Q = ( MV ) =
p are the eigenvectors. 3
−0.7321
1, 2.7321,
where p , p , and
1 0 1 1
0 0.5591 0.8255 p3 ] = 0 0.7637 −0.3022 1 −0.3228 0.4766
p2
where p , p , and
(b)
C CA 2 =
−0.7321
1, 2.7321,
1
V=
Then,
p2
0 0.5861 0.7546 p3 ] = 0 0.8007 −0.2762 1 0.1239 0.5952
p are the eigenvalues. 3
(c)
Eigenvalues of A:
-3, -2, -2. A nonsingular DF transformation matrix T cannot be found.
(d)
Eigenvalues of A:
−1, −1, −1
The matrix A is already in Jordan canonical form. Thus, the DF transformation matrix T is the identity matrix I.
(e)
Eigenvalues of A:
0.4142,
−2.4142 T = [p2
p2 ] =
0.8629 −0.2811 − 0.5054 0.9597
5-11 (a)
1 −2 0 0
S = [B
AB] =
S = B
1 −1 1 2 AB A B = 2 −2 2 3 −3 3
S = [B
AB] =
S is singular.
(b) S is singular.
(c)
2 2
2 +2 2 2+
2
S is singular.
(d)
45
−1
0 1 1 −3
1 −2 4 AB A B = 0 0 0 1 −4 14
S = B 5-12 (a)
2
S is singular.
Rewrite the differential equations as:
d θm 2
dt
2
B d θm 2
=−
J dt
2
−
K J
θm +
State variables: x = θ ,
x
m
1
Ki
2
J =
dia
ia dθ
=−
Kb dθ m
dt m
x
,
dt
3
La dt
−
Ra La
ia +
K a Ks
(θ
La
r
−θm )
= ia
State equations:
Output equation:
dx1 dt 0 dx2 = − K dt J dx K K 3 − a s dt La
1 −
B J
−
Kb La
0 x1 Ki x2 + 0 θ r J x K K Ra 3 a s − La La 0
y
=
1
0
0 x
(b) Forward-path transfer function:
s Θm ( s ) K G ( s) = = [1 0 0 ] J E (s ) 0
−1 s+
B J
Kb La
0 Ki − J Ra s+ L a
−1
0 KiK a 0 = K ∆ o ( s) a La
∆ o ( s ) = J La s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + KRa = 0 3
2
Closed-loop transfer function:
s Θm ( s ) K M ( s) = = [1 0 0 ] J Θr ( s ) KaKs La =
−1 s+
B J
Kb La K i Ka K s
0 Ki − J R s+ a La
−1
0 K s G( s ) 0 = K K 1 + K s (s ) a s La
JLa s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + K i K a K s + KRa 3
2
5-13 (a)
46
= x1
A=
0 1 −1 0
A = 2
− 1 0 0 −1
A = 3
0 − 1 1 0
A = 4
1 0 0 1
(1) Infinite series expansion: 3 5 t t t2 t4 1 − + − L t − + − L 1 2 2 2! 4! 3! 5! φ ( t) = I + At + A t + L = 3 5 2 4 2! −t + t − t + L 1 − t + t L 3! 5 ! 2! 4 !
cos t = − sin t
sin t cos t
(2) Inverse Laplace transform: −1
1 s 1 s −1 Φ ( s ) = ( sI − A ) = 1 s = s 2 + 1 −1 s −1
φ (t ) =
cos t − sin t
sin t cos t
(b) A=
−1 0 0 −2
A = 2
1 0 0 4
A = 3
−1 0 0 − 8
A = 4
−1 0 0 16
(1) Infinite series expansion: 2 3 4 t t t 1 − t + − + +L 0 1 2 2 2 ! 3! 4 ! φ ( t) = I + At + A t + L = 2 3 2! 4t 8t 0 1 − 2t + − +L 2! 3!
e −t = 0
e 0
−2 t
(2) Inverse Laplace transform: Φ ( s ) = ( sI − A ) = −1
s + 1 0
1 s +1 = s + 2 0
1 s + 2 0
−1
0
e− t
−2 t e 0
φ (t ) =
0
(c) A=
0 1 1 0
A = 2
1 0 0 1
A = 3
0 1 1 0
A = 4
1 0 0 1
(1) Infinite series expansion: 3 5 t t t2 t4 1+ + +L t+ + L 1 2 2 e− t + et 2! 4! 3! 5 ! φ ( t) = I + At + A t + L = = −t t 3 5 2 4 2! t + t + t L 1 + t + t + L −e + e 3! 5 ! 2 ! 4!
e +e −t
−e + e −t
(2) Inverse Laplace transform: −1
1 s 1 s −1 −1 Φ ( s ) = ( sI − A ) = −1 s = s 2 − 1 1 s =
47
0.5 − s +1 −0.5 + s + 1
s − 1 s + 1 s − 1 0.5 0.5 0.5 + s − 1 s + 1 s − 1 0.5
− 0.5
+
0.5
t
t
e− t + et φ ( t) = 0.5 − t t −e + e 5-14 (a)
e = K s (θ r − θ y ia = Solve for i
a
)
ea = e − es
eu − eb
eb = K b
Ra + Rs
in terms of
θy
dθ and
y
e +e −t
−e + e −t
t
es = Rs ia
dθ y
t
eu = Kea
Tm = K i ia = ( J m + J L )
dt
d θy 2
dt
2
, we have
dt
ia =
KKs (θr − θ y ) − Kb
dθ y dt
Rs + Rs + KRs
Differential equation:
d θy 2
dt
2
K ii a
=
Jm + J L
State variables: x = θ ,
x
y
1
dθ y − KK sθ y + KK sθ y −K b ( J m + J L ) (R a + R s + KRs ) dt Ki
=
2
=
dθ
y
dt
State equations:
dx1 0 1 0 dt x1 θ = + − KK s K i − K b Ki − KK s Ki r dx2 ( J + J ) (R + R + KR ) ( J + J ) (R + R + KR ) x2 ( J + J ) (R + R + KR ) m L a s s m L a s s s dt m L a s 1 x1 0 0 = + θr −322.58 −80.65 x2 322.58 We can let v ( t )
= 322
.58
θr,
then the state equations are in the form of CCF.
(b) −1
−1 1 s s + 80.65 1 = 2 322.58 s + 80.65 s + 80.65 s + 322.58 −322.58 s −0.014 0.014 −0.06 − 1.059 s + 76.42 s + 4.22 s + 76.42 + s + 4.22 = 1.0622 0.0587 4.468 − 4.468 s + 76.42 s + 4.22 s + 76.42 − s + 4.22
( sI − A ) = −1
For a unit-step function input, u ( t ) s
=1 /
s.
322.2 1 s ( s + 76.42)( s + 4.22) = ( sI − A)−1 B = s 322.2 s ( s + 76.42)( s + 4.22)
48
1 + 0.0584 − 1.058 s s + 76.42 s + 4.22 − 4.479 + 4.479 s + 76.42 s + 4.22
−76.42t −4.22t −0.014e + 0.01e −0.06e −76.42 t − 1.059e −4.22 t x (t ) = x (0) −76.42 t −4.22t −76.42t −4.22t − 4.468e 1.0622 e − 0.0587 e 4.468 e 1 + 0.0584 e −76.42 t − 1.058e −4.22 t = t≥0 −76.42 t −4.22t + 4.479e −4.479 e
(c) Characteristic equation: (d)
∆( s ) =
s
2
+ 80 . 65 s + 322
From the state equations we see that whenever there is to increase the effective value of
Ra by (1 + K ) Rs .
=0
. 58
Ra there is ( 1 + K ) Rs .
Thus, the purpose of R is s
This improves the time constant of the system.
5-15 (a) State equations: dx1
0 1 0 dt x1 θ + − KKs Ki − Kb Ki KKs Ki = r dx2 J ( R + R + KR ) J ( R + R + KR ) x2 J ( R + R + KR ) s s s s s s dt 1 x1 0 0 = + θ r −818.18 −90.91 x2 818.18
Let v
= 818
.18
θr.
The equations are in the form of CCF with v as the input. −1
s −1 1 s + 90.91 ( sI − A ) = = 818.18 s + 90.91 ( s + 10.128 ) ( s + 80.782) −818.18 −10.128t −80.78t 0.01415 e − 0.0141e 1.143 e−10.128t − 0.142 e−80.78t x1 (0) x (t ) = −10.128 t −80.78 t −10.128 t −80.78t + 0.1433e − 0.1433 e + 1.143e − 11.58 e x2 (0) 11.58 e−10.128 t − 11.58 e−80.78t + t≥0 −10.128 t −80.78t + 0.1433 e 1 − 1.1434 e −1
(b)
(c) Characteristic equati on:
2
. 18
=0
− 10.128, −80.782
Eigenvalues:
(d)
∆ ( s ) = s + 90 . 91 s + 818
Same remark as in part (d) of Problem 5-14.
5-16 (a) Forward-path transfer function: Y ( s) 5 ( K1 + K 2s ) G ( s) = = E ( s ) s [s ( s + 4)( s + 5) + 10 ] M ( s) =
Y (s ) R (s )
=
G (s) 1 + G(s )
=
Closed-loop transfer function:
5 ( K1 + K 2s )
s + 9 s + 20 s + (10 + 5 K 2 ) s + 5 K1 4
3
2
(b) State diagram by direct decomposition:
49
1 s
State equations:
x&1
x& 2 = x&3 x& 4
0 0 0 −5 K
(c) Final value:
Output equation:
x1 0 1 0 x2 0 + r 0 1 x3 0 −20 −9 x4 1
1
0
0 0 − (10 + 5 K 2 )
r(t )
= u s ( t ),
R( s)
0
1
=
5-17
s →0
=
5 K1
5K2
. s
lim y( t) = lim sY ( s ) = lim t →∞
y
s →0
5 ( K1 + K 2s )
s + 9 s + 20 s + ( 10 + 5 K 2 ) s + 5 K1 4
3
2
=1
In CCF form,
0 0 A= M 0 − a0
1
0
0
L
0
1
0
L
M
M
M
O
0
0
0
L
− a1
− a2
− a3
L
s 0 sI − A = M 0 a 0 sI
−A =
s
n
0 M 1 − an −1
−1
0
0
s
−1
0
M
M
M
0
0
0
a1
a2
a3
+ a n −1 s
n −1
0 0 B = M 0 1
0
+ a n−2 s
L
L 0 O M 0 −1 L s + a n n−2
0
+ L + a1 s + a 0
Since B has only one nonzero element which is in the last row, only the last column of going to contribute to the last row of
adj (s I − A ) B
( sI − A ) .
. The last column of
Thus, the last column of
adj (s I − A )
adj (s I − A ) B
50
is 1
adj (s I − A )
is
is obtained from the cofactors of s
s
2
L
s
n −1
'
.
0 x
5-18 (a) State variables:
x
1
=
y,
x
2
=
dy , dt
x
3
=
d
2
y 2
dt
x& ( t ) = Ax ( t ) + B r ( t )
State equ ations:
0 1 0 A= 0 0 1 −1 −3 −3
0 B = 0 1
(b) State transition matrix:
s2 + 3 s + 3 s + 3 1 s −1 0 1 −1 2 Φ ( s ) = ( sI − A ) = 0 s −1 = −1 s + 3s s ∆ (s ) 2 − s 1 3 s + 3 −3 s − 1 s 1 1 1 1 2 1 + + + 2 3 ( s + 1) 2 ( s + 1)3 ( s + 1 )3 s + 1 ( s + 1) ( s + 1) −1 1 1 2 s = + − 2 s + 1 ( s + 1) ( s + 1 )3 ( s + 1) 3 ( s + 1) 3 2 −s −3 2 s + ( s + 1 )3 ( s + 1) 2 ( s + 1)3 ( s + 1)3 −1
∆ (s ) = s3 + 3 s2 + 3s + 1 = ( s + 1)
(1 + t + t 2 / 2 ) e− t 2 −t φ ( t) = −t e / 2 ( −t + t 2 / 2 ) e −t
(t + t ) e (1 + t − t ) e −t
2
2
2
t e
3
( t − t / 2 ) e (1 − 2t + t 2 / 2 ) e −t 2 −t
t e /2
−t
−t
2
−t
(d) Characteristic equation: ∆ ( s ) = s + 3 s + 3 s + 1 = 0 3
− 1, −1, −1
Eigenvalues:
5-19 (a) State variables:
2
x
1
=
y,
x
2
=
dy dt
State equations:
dx1 (t ) dt 0 1 x1 ( t) 0 = + r ( t) dx2 ( t) − 1 − 2 x2 ( t) 1 dt State transition matrix:
s+ 2 ( s + 1) 2 s −1 −1 Φ ( s ) = ( sI − A ) = 1 s + 2 = − 1 ( s + 1) 2 −1
( s + 1) s ( s + 1)2 1
2
51
(1 + t ) e−t
φ (t ) =
− te
−t
(1 − t ) e te
−t
−t
Characteristic equation:
=
y,
−y=
x
(b) State variables:
x
1
∆ (s ) = ( s +1) = 0 2
x
2
=
y
dy
+
dt
State equations: dx dt
1
=
dy dt
=
x
2
dx
− x1
2
2
dt
d
=
2
y
dt
dy
+
2
dt
= −y −
dy
+r = −x2 +r
dt
dx1 dt −1 2 x1 0 = + r dx2 0 −1 x2 1 dt State transition matrix:
1 s + 1 −2 s + 1 Φ (s ) = = 0 s + 1 0
( s + 1) 1 s +1 −2
−1
2
(c) Characteristic equation: ∆ (s ) = ( s +1) = 0
φ (t ) =
e −t 0
− te e
−t
−t
2
5-20 (a) State transition matrix: ω s − σ sI − A = −ω s − σ
( sI − A )
−1
which is the same as in part (a).
s − σ ∆( s) ω 1
=
s − σ −ω
2
cos ω t − sin ω t σ t ( sI − A )−1 = e sin ω t cos ω t
−1
φ ( t) = L
σ + jω , σ − jω
(b) Eigenvalues of A:
5-21 (a) Y (s) 1
U ( s)
=
1
Y ( s) 2
U (s) 2
s 1+ s
−1
+ 2s s
= 1
+s
−1
(b) State equations [Fig. 5-21(a)]:
0 1 0 A1 = 0 0 1 −3 −2 −1 State equations [Fig. 5-21(b)]:
0 0 −3 A 2 = 1 0 −2 0 1 −1
−3 −2
+3s
−3
=
1 s
3
+s +2s+3 2
−3
+2s
−2
+ 3s
−3
=
1 s
3
x& = A x + B u 1
Y ( s) 1
U (s) 1
Output equation: y = C x
1 1
1
0 B1 = 0 1
C1 = [1
x& = A x + B u 2
+ s + 2s +3 2
=
2
1 B2 = 0 0
2
0
1
0]
Output equation: y
C2 = [0
52
0
(
∆ ( s ) = s − 2σ + σ + ω
1]
2
= C 2x
2
2
)
A
Thus,
= A1 '
2
5-22 (a) State diagram:
(b)
State diagram:
5-23 (a) G (s) X (s)
=
Y (s) U ( s)
=
10 s 1 + 8 .5 s
= U ( s ) − 8 .5 s
−1
−1
−3
+ 20 . 5 s
X ( s)
X (s) −2
− 20 . 5 s
+ 15
−2
s
−3
X ( s ) − 15 s
State diagram:
State equation:
x& ( t ) = Ax ( t ) + B u ( t )
53
Y ( s)
X (s) −3
X (s)
= 10
X (s)
1 0 0 A= 0 0 1 −15 −20.5 −8.5 (b) G (s) Y ( s)
=
Y (s) U ( s)
= 10
s
−3
=
10 s
−3
+ 20
1 + 4. 5 s
−1
+ 20
−4
X ( s)
s
0 B = 0 1 s
−4
+ 3 .5 s
A and B are in CCF
X (s) −2
X (s)
X (s)
X ( s)
= −4. 5 s
−1
X ( s ) − 3 .5 s
−2
X ( s) +U ( s)
State diagram:
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
0 0 A= 0 0
0 1 0 0 0 1 0 −3.5 −4.5 1
0
0 0 B= 0 1
0
(c) G (s)
Y (s)
=
Y (s) U ( s)
= 5s
−2
State equations:
=
5( s s(s
X (s)
+ 1)
+ 2 )( s + 10 )
+ 5s
−3
X (s)
=
5s
−2
1 + 12 s
+ 5s
−1
X (s)
A and B are in CCF
−3
+ 20
s
X (s) −2
X (s)
= U ( s ) − 12
x& ( t ) = Ax ( t ) + B u ( t )
54
s
−1
X (s)
− 20
s
−2
X ( s)
0 0 1 A= 0 0 1 0 −20 −12
0 B = 0 1
A and B are in CCF
(d) G ( s) = Y ( s)
=
Y (s ) U (s ) s
−4
=
(
1
)
s ( s + 5) s + 2s + 2
X ( s)
2
X (s)
=
= U ( s) − 7s
s
−4
X (s )
−1
−2
1 + 7 s + 12 s + 10 s −1
−2
X ( s ) − 12 s
X (s)
−3
− 10
X (s ) s
−3
State diagram:
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
0 0 0 1 0 0 1 0 A= 0 0 0 1 0 −10 −12 −7
0 0 B= 0 1
A and B are in CCF
5-24 (a) G (s)
=
Y (s) U ( s)
=
10 s
3
+ 8 . 5 s + 20 . 5 s + 15 2
=
5 . 71 s
55
+ 15
−
6 . 67 s
+2
+
0 . 952 s
+5
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
−1.5 0 0 A= 0 −2 0 0 −5 0
5.71 B = −6.67 0.952
The matrix B is not unique. It depends on how the input and the output branches are allocated.
(b) G (s)
Y (s)
=
=
U ( s)
10( s 2
s (s
+ 2)
−4. 5
=
= 1 )( s + 3 . 5)
0 .49
+
s
+ 3 .5
s
4
+ s
State diagram:
x& ( t ) = Ax ( t ) + B u ( t )
State equation:
0 0 1 0 0 0 0 0 A= 0 0 −1 0 0 0 0 −3.5 (b) G (s)
=
Y ( s) U (s)
State equations:
=
5( s s( s
+ 1)
+ 2 )( s + 10 )
=
0 1 B= 1 1
2 .5 s
+
0 . 313 s
+2
x& ( t ) = Ax ( t ) + B u ( t )
0 0 0 A = 0 −2 0 0 0 −10
1 B = 1 1
(d)
56
−
0 . 563 s
+ 10
+1
+
5 . 71 s
2
Y (s )
G ( s) =
U (s )
=
(
1
s ( s + 5) s + 2s + 2 2
)
=
0.1 s
−
0.0118 s+ 5
−
0.0882s + 0.235 s + 2s + 2 2
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
0 0 0 −5 A= 0 0 0 0
0 1 −2
0
1 1 B= 0 1
0
0 0 −2
5-25 (a) G (s)
=
Y (s) U ( s)
=
10
+ 1. 5)( s + 2 )( s + 5)
(s
State diagram:
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
−5 1 A = 0 −2 0 0
1 −1.5 0
0 B= 0 10
(b) G ( s) =
Y (s ) U (s )
=
10 s + 2 1 2 s ( s + 1)( s + 3.5) s s + 1 s + 3.5 10(s + 2)
2
=
State diagram:
57
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
0 0 1 0 0 0 1 1 A= 0 0 −1 1 0 0 0 −3.5
0 0 B= 0 10
(c) G ( s) =
Y (s )
=
U (s )
5 s + 1 1 s( s + 2)( s + 10) s s + 2 s + 10 59 s + 1)
=
State diagram:
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
1 0 0 A = 0 − 10 −1 0 0 −2
0 B = 0 5
(d) G ( s) =
Y (s) U (s )
=
(
1
s ( s + 5) s + 2s + 2 2
)
State diagram:
58
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
0 1 0 0 A= 0 −2 0 0
0 1 −5
0
0 0 B= 0 1
0
1 −2 0
5-26 (a) G (s)
=
Y (s) E (s)
=
10 s(s
+ 4 )( s + 5)
=
10 s 1+ 9 s
−1
−3
+ 20
X (s) s
−2
X (s)
(b) Dynamic equations:
x&1 x& = 2 x&3
1 0 0 0 −10 −20
x1 0 1 x2 + 0 r −9 x3 1 0
y = [10
0
0] x
(c) State transition equation:
s −1 (1 + 9s −1 + 20 s −2 ) s −2 (1 + 9 s −1 ) s −3 x1 (0) s −3 X 1 (s ) −3 −1 −1 −2 X (s ) = 1 1 s −2 1 − 10 s s (1 + 9 s ) s x2 (0) + 2 ∆ (s ) ∆ ( s ) −1 s −2 −2 −1 s − 10 s − 20 s s x3 (0) X 3 ( s ) 1 2 s s + 9 s + 20 s+9 1 x1 (0) 1 1 = − 10 s( s + 9) s x2 (0) + 1 ∆ c (s ) ∆ c (s ) 2 s − 10 s − 10 ( 2 s + 1 ) s x3 (0) ∆( s ) = 1 + 9 s
−1
+ 20
s
−2
+ 10
s
−3
∆c( s) =
59
s
3
+ 9 s + 20 s + 10 2
1.612 0.946 0.114 −0.706 −1.117 −0.169 −0.708t x(t ) = −1.14 −0.669 −0.081 e + 1.692 2.678 4.056 e −2.397t 0.807 0.474 0.057 −4.056 −6.420 −0.972
0.0935 0.171 0.055 + −0.551 −1.009 − 0.325 e −5.895t x(0) 3.249 5.947 1.915
0.1 − 0.161e−0.708t + 0.0706 e−2.397t − 0.00935e−5.895 t −0.708 t −2.397 t −5.895 t + − 0.169e + 0.055e 0.114 e −0.087 e−0.708t + 0.406 e− 2.397t − 0.325e− 5.895t
t ≥0
(d) Output:
(
y ( t ) = 10 x1 (t ) = 10 1.612e
−0.708t
(
+ 10 1.141 e
− 0.706e
−0.708 t
−2.397t
− 0.169 e
+ 0.0935e
−2.397 t
−5.897 t
+ 0.0550 e
) x (0) + 10 ( 0.946e − 1.117e + 0.1711e ) x (0) ) x (0) + 1 −1.61e −0.708t + 0.706e −2.397 t − 0.0935e −5.895t −0.708 t
−2.397t
−5.895t
1
2
−5.895 t
3
5-27 (a) Closed-loop transfer function: Y (s)
10
=
R( s)
s
3
+ 9 s + 20 s + 10 2
(b) State diagram:
(c) State equations:
x&1 x& = 2 x&3
(d) State transition equations: [Same answers as Problem 5-26(d)]
1 0 0 0 −10 −20
x1 0 1 x2 + 0 r −9 x3 1 0
(e) Output:
5-28 (a) State diagram:
(b) State equations:
60
[Same answer as Problem 5-26(e)]
t≥0
x&1 x& 2 = x&3 x& 4
−2 20 −1 0 x1 0 −1 0 −10 1 0 x2 0 0 u + −0.1 0 −20 1 x3 0 0 T D 0 0 0 − 5 x4 30 0
(c) Transfer function relations: From the system block diagram, −0.2s 0.3 30 e U ( s) 90U( s) −1 T ( s ) + T ( s ) + + D D ∆( s) s + 2 ( s + 2)( s + 20) ( s + 2)( s + 5)( s + 20) (s + 5)( s + 20)
1
Y ( s) =
∆( s ) = 1 + Y ( s)
−0 . 2 s
+ 2 )( s + 20
(s
− ( s + 19 . 7 )
=
+ 2 )( s + 20 ) + 0 .1e
(s
−( s + 20
Ω( s ) =
−0 . 2 s
)
+ 2 )( s + 20 ) + 0 .1e
(s
5-29 (a)
0 . 1e
−0 . 2 s
=
+ 2 )( s + 20 ) + 0 .1 e
(s
(s
) T
D
30 e
(s)+
+ 5)
(s
T (s) D
+ 2 )( s + 20 −0 . 2 s
(s
(s
+ 5)
(s
)
+ 90(
s
+ 2 )U ( s )
+ 2 )( s + 20 ) + 0 . 1e 30 e
+
− 0. 2 s
−0 . 2 s
−0 . 2 s
U ( s)
+ 2 )( s + 20 ) + 0 .1e
There should not be any incoming branches to a state variable node other than the s
−1
−0 . 2 s
branch. Thus, we
should create a new node as shown in the following state diagram.
Notice that there is a loop with gain −1 after all the s
(b) State equations: dx
1
dt
=
17
x
2
1
+
1
x
2
dx 2
2
dt
=
15 2
x
−
1
1 2
x
2
+
1
r
R (s )
= 2.
Output equation:
Ks + 5 s + 1
(
( s + 1) s + 11s + 2 2
( s + 1) ( s
)
2
y
= 6 .5 x 1 + 0 .5 x 2
K = 1: Y (s) R( s)
=
s s
3
2
+ 12
+ 5s +1 s
2
+ 13 s + 2
State diagram:
61
)
+ 11s + 2 = 0
Roots of characteristic equation: −1, −0.185, −-10.82. These are not functions of K.
(c) When
∆
(b) Characteristic equation: 2
=
branches are deleted, so
2
5-30 (a) Transfer function: Y (s )
−1
(d) When
K = 4:
Y (s ) R (s )
4s + 5s + 1
(s + 1)(4 s + 1)
2
=
( s + 1) ( s
2
+ 11s + 2
=
) ( s + 1) (s
2
+ 11s + 2
)
=
4s + 1 s + 11s + 2 2
State diagram:
(e) Y (s) R( s)
=
Ks (s
2
+ 5s =1
+ 1 )( s + 0 .185)(
s
+ 10 .82
When K = 4, 2.1914, 0.4536, pole-zero cancellation occurs. )
5-31 (a) Gp ( s) =
Y ( s) U (s )
=
1
( 1 + 0.5s ) (1 + 0.2s + 0.02s
2
)
=
100 s + 12 s + 70 s + 100 3
2
State diagram by direct decomposition:
State equations:
x&1 x& = 2 x&3
0 0 −100
x1 0 0 1 x2 + 0 u −70 −12 x3 1 1
0
(b) Characteristic equation of closed-loop system:
Roots of characteristic equation:
62
s
3
+ 12
s
+ 70 s + 200 = 0
2
− 5. 88 ,
− 3 . 06 +
j 4. 965 ,
− 3 . 06 −
5-32 (a) Gp ( s) =
Y ( s) U (s )
≅
1 − 0.066 s
(1 + 0.5s ) (1 + 0.133s + 0.0067s
2
)
=
−20( s − 15) s + 22 s + 190 s + 300 3
2
State diagram by direct decomposition:
State equations:
x&1 x& = 2 x&3
0 0 −300
x1 0 0 1 x2 + 0 −190 −22 x3 1 1
0
Characteristic equation of closed-loop system: s
3
+ 22
s
2
+ 170
5-33 (a) State variables:
x
s
+ 600 = 0
1
= ωm
State equations: dω dt
m
=−
K K b
i
and x
−12, −5 + j5, −5 −j5
2
+ K bRa
JR
a
Roots of characteristic equation:
= ωD
+
K
D
J
ωD +
KK JR
i
e
a
(b) State diagram:
63
dω dt
D
=
K J
D R
ωm −
K J
D R
ωD
j 4. 965
(c) Open-loop transfer function: Ωm ( s)
=
E (s )
KK i ( J R s + K D )
JJ RR a s + ( K b J R Ki + K DR a J R + K D JRa ) s + K DK b Ki 2
Closed-loop transfer function:
Ωm ( s) Ω r ( s)
=
Ks KKi ( J R s + KD )
JJ R Ra s + ( Kb J R Ki + KD R a J R + KD JRa + Ks K Ki J R ) s + KD K b Ki + Ks K Ki KD 2
(d) Characteristic equation of closed-loop system: ∆ ( s ) = JJ R R a s + ( K D J R K i + K D R a J R + K D JRa + K s KK i J R ) s + K D K b K i + K s K Ki KD = 0 2
∆( s ) = Characteristic equation roots:
−19.8,
s
2
+ 1037
+ 20131
.2
=0
−1017.2
5-34 (a) State equations: x& ( t ) = Ax ( t ) + B r ( t ) A=
s
−b d −2 1 c − a = 2 − 1
B=
0 1
S = [B
AB ] =
0 1 1 −1
Since S is nonsingular, the system is controllable.
(b) S = [B
AB ] =
0 d 1 − a
AB
1 −1 1 A B = 1 −1 1 1 −1 1
The system is controllable for d
≠ 0.
5-35 (a) S = B
2
S is singular. The system is uncontrollable.
(b)
64
1 − 1 1 AB A B = 1 − 2 4 1 −3 9
S = B
2
S is nonsingular. The system is controllable.
5-36 (a) State equations: x& ( t ) = Ax ( t ) + B u ( t )
A=
−2 3 1 0
B=
Output equation:
V = C
'
'
AC
'
y
=
1 1
S = [B
= Cx
0 x
1
1 −2 = 0 3
1 1 1 1
C = 1
S is singular. The system is uncontrollable.
0
V is nonsingular. The system is observable.
(b) Transfer function: Y (s)
AB ] =
s
=
U ( s)
s
2
+3
+2s−3
1
= s
−1
Since there is pole-zero cancellation in the input-output transfer function, the system is either uncontrollable or unobservable or both. In this case, the state variables are already defined, and the system is uncontrollable as found out in part (a).
5-37 (a) α = 1, (b)
2 , or 4
.
These values of
α will cause pole-zero cancellation in the transfer function.
The transfer function is expanded by partial fraction expansion, Y (s) α −1 α R( s)
=
3( s
+ 1)
−
2( s
By parallel decomposition, the state equations are: x& ( t )
−1 0 0 A = 0 −2 0 0 −4 0 The system is uncontrollable for
(c)
α = 1,
or
−2 + 2)
+
α −4 6( s
+4)
= Ax ( t ) + B r ( t ) ,
α −1 B = α − 2 α − 4
output equation: y ( t )
D=
= C x ( t ).
1 − 1 3 2
1 6
α = 2, or α = 4.
Define the state variables so that
−1 0 0 A = 0 −2 0 0 0 − 4
The system is unobservable for
1 3 −1 B= 2 −1 6
α = 1, or α = 2, or α = 4.
65
D = [α − 1
α −2
α − 4]
5-38 S = [B
AB] =
b 1 b ab − 1
S = ab − 1 − b ≠ 0 2
The boundary of the region of controllability is described by ab
− 1 − b = 0. 2
Regions of controllability:
5-39 S = [B
AB] =
b1 b1 + b 2 b b 2 2
The system is completely controllable when b
V = C
'
'
AC
'
S = 0 when b1 b2 − b1 b2 − b2 = 0,or b2 = 0 2
≠ 0.
2
d2 d1 = d2 d 1 + d 2
The system is completely observable when d
≠ 0.
2
5-40 (a) State equations: dh 1 K nN K = ( qi − qo ) = I θm − o h dt A A A State variable:
x
1
= h,
x
=θm,
x
x& = Ax + B e
State equations:
−K o A A= 0 0
2
K I nN A 0 0
dθ m
=ω
dt 3
=
dθ dt
m
d ωm m
=−
dt
Ki Kb JRa
ωm +
Ki K a JRa
ei
J
= Jm +n
= ωm
i
1 = KK − i b JRa 0
V = 0 when d 1 ≠ 0 .
0 −1 0.016 0 0 1 0 0 −11.767
State diagram:
66
0 B = 0 = K K i a JRa
0 0 8333.33
2
J
L
(b) Characteristic equation of A: s+ sI − A =
Ko
− K I nN
A
A
0
s
0
0
Eigenvalues of A:
0,
0
s+
= s s +
−1 Ki K b
s + Ki K b = s ( s +1)( s + 11.767) A JRa
Ko
JRa
−1, −11.767.
(c) Controllability: S = B
0 133.33 0 A B = 0 8333.33 − 98058 8333.33 −98058 1153848 2
AB
S ≠ 0.
The system is controllable.
(d) Observability: (1) C =
1
0
V = C
(2) C =
0
(3) C =
0
'
1
V = C
'
'
'
'
'
'
−1 1 1 ( A ) C = 0 0.016 −0.016 0 0.016 0 ' 2
'
V is nonsingular. The system is observable.
:
AC
1
'
'
AC
0
0
V = C
:
0
0 0 0 ( A ) C = 1 0 0 0 1 −11.767 ' 2
'
V is singular. The system is unobservable.
:
AC
5-41 (a) Characteristic equation:
0 0 0 ( A ) C = 0 0 0 1 −11.767 138.46 ' 2
'
∆( s ) = s I − A
∗
Roots of characteristic equation: −5.0912,
=
s
4
− 25 . 92
s
5.0912, 0, 0
(b) Controllability:
67
2
=0
V is singular. The system is unobservable.
S = B
∗
∗
AB
∗
∗2
A B
− 0.0732 0 − 1.8973 0 −0.0732 0 −1.8973 0 ∗3 ∗ A B = 0 0.0976 0 0.1728 0.0976 0 0.1728 0
∗
∗
S is nonsingular. Thus, A ,
B
∗
is controllable.
(c) Observability: (1)
C
∗
=
1
0
V = C
∗'
0
0
∗'
A C
∗'
∗' 2
(A ) C
∗'
1 0 ∗' 3 ∗' ( A ) C = 0 0
0
25.92
1
0
0
0
0
0
25.92 0 0 0
S is singular. The system is unobservable. ∗
(2) C = 0
1
V = C
0
∗'
0
∗'
A C
∗'
∗'
2
(A ) C
∗'
0 671.85 0 25.92 1 0 25.92 0 ∗' 3 ∗' (A ) C = 0 0 0 0 0 0 0 0
S is singular. The system is unobservable.
∗
(3) C = 0
0
1
V = C
∗'
0
∗'
A C
∗'
∗'
2
(A ) C
∗'
0 0 ∗' 3 ∗' (A ) C = 1 0
S is nonsingular. The system is observable.
∗
(4) C = 0
0
0
1
68
0
− 2.36
0
0
0
0
1
0
− 2.36 0 0 0
V = C
∗'
∗'
A C
∗'
∗' 2
(A ) C
0 −61.17 0 −2.36 0 0 −2.36 0 ∗' 3 ∗' ( A ) C = 0 0 0 0 1 0 0 0
∗'
S is singular. The system is unobservable.
5-42
The controllability matrix is
0 − 384 0 −1 0 −16 −1 0 −16 0 −384 0 0 0 0 16 0 512 S= 0 512 0 0 0 16 0 1 0 0 0 0 0 0 0 1 0 0
Rank of S is 6. The system is controllable.
5-43 (a) Transfer function: Θv ( s ) R (s )
=
J vs
2
(J
KI H s + K Ps + K I + K N 2
G
)
State diagram by direct decomposition:
x& ( t ) = Ax ( t ) + B r ( t )
State equations:
0
1
0 0 A = 0 0 0 0
0 1 − KP J G
0
0
1 0 − ( KI + KN
(b) Characteristic equation:
JG
Jvs
2
(J
)
0 0 B= 0 1
)
s + K P s + K I + KN = 0 2
G
69
5-44 (a) State equations: x& ( t ) = Ax ( t ) + B u 1 ( t )
A=
−3 1 0 − 2
B=
S is nonsingular. A ,
0 1
S = [B
(b)
With feedback, u 2
= − kc 2 ,
'
0 1 1 −2
B is controllable.
Output equation: y 2 = C x
V = C
AB ] =
'
AC
'
C = −1
−1 3 = 1 −3
the state equation is: x& ( t )
−3 − 2 k A = 1+ g 0
=
V is singular. The system is unobservable.
Ax ( t )
1+ k − 2 1
1
+ B u1 ( t ) . 1 0 S= 1+k 1 −2
0 B= 1
S is nonsingular for all finite values of k. The system is controllable. State diagram:
y2 = Cx
Output equation:
V = D '
−1 1 + k −1 1 + K ' ' A D = 1 1 + k
1
C=
1 + k
(1 + k ) 3 + 2k − 2 (1 + k ) 3 + 2k
2
V is singular for any k. The system with feedback is unobservable.
5-45 (a) S = [B V = C (b) u = − k 1
1 2 2 − 7 1 − 1 ' ' A C = 1 −2
AB ] = '
S is nonsingular. System is controllable.
V is nonsingular. System is observable.
k2 x
0 1 k1 − −1 −3 2 k1
A c = A − BK =
70
1 − k2 −k1 = 2 k2 − 1 − 2 k1 − 3 − 2 k 2 k2
S = [B
A cB ] =
For controllabillity, k 2
V = C For observability, V
'
≠−
1 − k1 − 2 k2 + 2 2 − 7 − 2k − 4k 1 2
11 2 '
A cC
'
− 1 − 1 − 3 k1 = 1 −2 − 3 k 2
= −1 + 3k 1 − 3 k 2 ≠ 0
71
S = − 11 − 2 k2 ≠ 0
Chapter 6 6-1 (a)
STABILITY OF LINEAR CONTROL SYSTEMS = 0 , − 1. 5 +
Poles are at s
j 1. 6583 ,
− 1. 5 −
One poles at s = 0. Marginally stable .
j 1. 6583
(b) Poles are at s = − 5, − j 2 , j 2 (c) Poles are at s = − 0 .8688 , 0 .4344 + j 2 . 3593 , 0 .4344 − j 2 . 3593 (d) Poles are at s = − 5, − 1 + j , − 1 − j (e) Poles are at s = − 1.3387 , 1. 6634 + j 2 . 164, 1. 6634 − j 2 .164 (f) Poles are at s = − 22 . 8487 ± j 22 . 6376 , 21 . 3487 ± j 22 . 6023 6-2 (a)
s
+ 25
3
s
+ 10 s + 450 = 0
2
Two poles on j ω axis. Marginally stabl e . Two poles in RHP. Unstable . All poles in the LHP. Stable . Two poles in RHP. Unstable . Two poles in RHP. Unstable .
Roots: − 25 . 31 , 0 .1537 + j 4.214, 0 .1537 − 4.214
Routh Tabulation: s
3
s
2
s
1
1
10
25
450
− 450
250
Two sign changes in the first column. Two roots in RHP.
= −8
0
25
(b)
s
s
0
3
+ 25
450
+ 10 s + 50 = 0
2
s
Roots: − 24. 6769 , − 0 .1616 + j 1.4142 , − 0 .1616 − j 1.4142
Routh Tabulation: s
3
s
2
s
1
s
0
1
10
25
50
− 50
250
No sign changes in the first column. No roots in RHP.
=8
0
25
(c)
s
50
+ 25
3
s
2
+ 250
s
+ 10 = 0
Roots: − 0 . 0402 , − 12 .48 + j 9 . 6566 , − j 9 . 6566
Routh Tabulation: s
3
s
2
s
1
s
0
1
250
25
10
− 10
6250
= 249
No sign changes in the first column. No roots in RHP. .6
0
25
(d)
2s
4
10
+ 10
s
3
+ 5 . 5 s + 5 . 5 s + 10 = 0 2
Roots: − 4.466 , − 1.116 , 0 .2888 + j 0 . 9611 , 0 .2888 − j 0 . 9611
Routh Tabulation:
71
s
4
s
3
s
2
55
1
10 24.2
s
2
5 .5
10
5 .5
− 11
= 4.4
− 100
10
10
= − 75 . 8
4.4 s
0
10
Two sign changes in the first column. Two roots in RHP.
(e)
s
6
+ 2 s + 8 s + 15 5
4
s
3
+ 20
s
2
+ 16 s + 16 = 0
Roots: − 1.222 ± j 0 .8169 , 0 . 0447 ± j 1.153 , 0 .1776 ± j 2 .352
Routh Tabulation: s
6
s
5
s
4
s
3
s
2
s
1
16
1
8
20
2
15
16
− 15
40
= 0 .5
− 16
2
2
− 33
− 48
−396 + 24
= 11 .27
−33 − 541 .1 + 528
= −1.16
16
= 12
16
0
11 .27 s
0
0
Four sign changes in the first column. Four roots in RHP.
(f)
s
4
+ 2 s + 10 3
s
2
+ 20 s + 5 = 0
Roots: −0 .29 , − 1. 788 , 0 . 039 + j 3 .105 , 0 . 039 − j 3 .105
Routh Tabulation: s
4
s
3
s
2
20
1
10
2
20
− 20
=0
5
5
2
6-3 (a)
s
4
s
2
s
1
s
0
ε 20
5
ε − 10 ε
≅−
Replac e 0 in last row by
10
ε
Two sign changes in first column. Two roots in RHP.
5
+ 25 s + 15 3
s
2
ε
+ 20 s + K = 0
Routh Tabulation:
72
s
4
s
3
s
2
1
K
15 25
20
− 20
375
= 14.2
K
25 s
− 25
284
1
K
= 20 − 1. 76
K
20
− 1. 76
K
>0
>0
K
or K
< 11 . 36
14.2 s
0
K
Thus, the system is stable for 0 < K < 11.36. When K = 11.36, the system is marginally stable. The auxiliary equation equation is A ( s ) = 14.2 s frequency of oscillation is 0.894 rad/sec.
(b)
s
2
+ 11 . 36 = 0 .
The solution of A(s) = 0 is s
2
= − 0 .8 .
The
+ Ks + 2 s + ( K + 1) s + 10 = 0
4
3
2
Routh Tabulation: s
4
s
3
s
2
1
2
K
K
− K −1
2K
K
=
K
−1
10
+1
>0
K K
10
K
−9 K − 1
>1
2
s
1
s
0
−9 K −1 > 0 2
−1
K 10
The conditions for stability are: K > 0, K > 1, and − 9 K − 1 > 0 . Since K is always positive, the last condition cannot be met by any real value of K. Thus, the system is unstable for all values of K. 2
(c)
s
2
+ ( K + 2 ) s + 2 Ks + 10 = 0
3
2
Routh Tabulation: s
3
s
2
s
1
K 2K
1
2K
+2
10
2
+ 4 K − 10 K
s
0
+2
K
> −2
K
2
+ 2 K −5 > 0
10
The conditions for stability are: K > −2 and K + 2 K − 5 > 0 or (K +3.4495)(K − 1.4495) > 0, or K > 1.4495. Thus, the condition for stability is K > 1.4495. When K = 1.4495 the system is 2
marginally stable. The auxiliary equation is A ( s ) The frequency of oscillation is 1.7026 rad/sec.
(d)
s
3
+ 20
s
2
+ 5 s + 10
K
= 3.4495
s
=0
Routh Tabulation:
73
2
+ 10 = 0 .
The solution is s
2
= −2 .899
.
s
3
s
2
s
1
1
5 10 K
20
− 10
100
K
= 5 − 0 .5 K
5
− 0 .5 K > 0
K
>0
or K
< 10
20 s
0
10 K
The conditions for stability are: K > 0 and K < 10. Thus, 0 < K < 10. When K = 10, the system is
= 20
marginally stable. The auxiliary equation is A ( s ) equation is s
(e)
s
= −5 .
2
3
2
+ 100 = 0 .
The solution of the auxiliary
The frequency of oscillation is 2.236 rad/sec.
+ Ks + 5 s + 10 s + 10
4
s
2
K
=0
Routh Tabulation: s
4
s
3
s
2
5
K
10
K
10 K
5K
− 10
5K
10 K
1
>0 − 10 > 0
K
− 100
50 K s
K 5K
1
− 10
K
>2
2
=
− 10
or K
50 K
− 100 − 10 5K
K
3
5K
− 10
− 10 − K > 0 3
K s
0
10 K
K
The conditions for stability are: K > 0, K > 2, and 5 K
β K
+ 2 . 9055
γε
K
2
− 2 . 9055
K
+ 3 .4419
ϕ
<0.
Thus, the conditions for stability are: K > 2 and K < is unstable for all values of K.
(f)
s
4
− 10 − K > 0 . 3
>0 The last condition is written as
The second-order term is positive for all values of K.
−2.9055.
Since these are contradictory, the system
+ 12 . 5 s + s + 5 s + K = 0 3
2
Routh Tabulation: s
4
s
3
s
2
1
1
12 . 5
5
−5
12 . 5
= 0 .6
K
K
12 . 5 s
1
3
− 12 . 5 K
= 5 − 20 .83
K
5
− 20 . 83 K > 0
or K
< 0 .24
0 .6 s K K >0 The condition for stability is 0 < K < 0.24. When K = 0.24 the system is marginally stable. The auxiliary 0
equation is A ( s ) = 0 . 6 s oscillation is 0.632 rad/sec.
2
6-4
The characteristic equation is Ts
3
+ 0 .24 = 0 .
The solution of the auxiliary equation is s
+ ( 2T +1) s + ( 2 + K ) s + 5 K = 0 2
Routh Tabulation:
74
2
= − 0 .4.
The frequency of
s
3
T
s
2
2T
s
1
( 2T
+1
0
T
>0
5K
T
> −1 /
+ 1 )( K + 2 ) − 5 KT 2T
s
+2
K
K (1 − 3 T )
+1
5K
T > 0, K > 0, and K
4T
+2
3T
−1
+ 24
Ks
<
T-versus-K parameter plane is shown below.
6-5 (a)
Characteristic equation: s
5
+ 600
s
4
+ 50000
s
3
+ Ks
2
+ 4T + 2 > 0
>0
K
The conditions for stability are:
2
. The regions of stability in the
+ 80
K
=0
Routh Tabulation: s
5
s
4
s
3
3
s
2
214080
1
50000
24 K
600
K
80 K
× 10
7
−K
14320 K
600
3 s
1
0
< 3 × 10
7
600 00 K
−K
2
80 K
× 10 − K 7
− 7 .2 × 10
+ 3 .113256 × 10
16
600( 214080 s
K
00
11
− 14400
K
−K
K
< 214080
K
00
2
K
2
− 2 .162 × 10
)
80 K
K
Conditions for stability:
75
>0
7
K
+ 5 × 10
12
<0
3
From the s
2
From the s
< 3 × 10 7 K < 2 .1408 × 10 7 12 − 2 .162 × 10 K + 5 × 10 < 0 7
K
row: row:
1
From the s row:
K
2
Thus, 0
From the s
2 . 34
× 10
When K
(b)
= 2 . 34 × 10 7 = 2 .1386 × 10
< K < 2 .1386 × 10
5
)( K
− 2 .1386 × 10
7
)
7
s
3
× 10
2 . 34
5
< K < 2 .1386 × 10
7
ω = 10 . 6 rad/sec. ω = 188 . 59 rad/sec.
5
Characteristic equation:
− 2 . 34 × 10
K>0
row:
Thus, the final condition for stability is: When K
5
(K
or
+ ( K + 2 ) s + 30 2
Ks
+ 200
=0
K
Routh tabulation: s
3
s
2
s
1
s
0
30 K
1
+2
K
2
− 140
K
+2
30 K
200 K K
200 K
Stability Condition:
Characteristic equation: s
3
> −2
K
> 4. 6667
K
>0
K > 4.6667
When K = 4.6667, the auxiliary equation is A ( s ) The frequency of oscillation is 11.832 rad/sec.
(c)
K
+ 30
s
+ 200
2
= 6 . 6667
s
2
+ 933
. 333
=0.
The solution is s
2
+K =0
s
Routh tabulation: s
3
s
2
s
1
1
200
30
K
−K
6000
K
< 6000
K
>0
30 s
0
K
Stabililty Condition:
0
<
< 6000
K
When K = 6000, the auxiliary equation is A ( s ) The frequency of oscillation is 14.142 rad/sec.
(d)
Characteristic equation:
s
3
= 30
s
2
+ 6000 = 0 .
The solution is s
2
= − 200
.
+ 2 s + ( K + 3) s + K + 1 = 0 2
Routh tabulation: s
3
s
2
s
1
K
+3
1
K
2
K +1
+5
K
> −5
K
> −1
30 s
0
K +1
Stability condition:
K>
−1.
When K =
−1 the zero element occurs in the first element of the
76
= − 140
.
<0
s row. Thus, there is no auxiliary equation. When K = −1, the system is marginally stable, and one of the three characteristic equation roots is at s = 0. There is no oscillation. The system response would increase monotonically. 0
6-6 State equation:
x& ( t ) = Ax ( t ) + B u ( t )
Open-loop system:
A=
1 −2 10 0
x& ( t ) = ( A − BK ) x ( t )
Closed-loop system:
A − BK =
0 1
B=
1 10 − k 1
−2
−k 2
Characteristic equation of the closed-loop system:
sI − A + BK =
s −1
2
−10 + k1
s + k2
Stability requirements: k
2
20
−1 > 0
= s + ( k 2 − 1 ) s + 20 − 2 k1 − k 2 = 0 2
or k
2
>1
− 2 k1 − k 2 > 0
or
k
< 20 − 2 k 1
2
Parameter plane:
6-7 Characteristic equation of closed-loop system: s −1 0 sI − A + BK = 0 k1
= s + ( k 3 + 3 ) s + ( k 2 + 4 ) s + k1 = 0
s
−1
k2 + 4
s + k3 + 3
3
2
Routh Tabulation:
s
3
s
2
s
1
s
k2 + 4
1 k3 + 3
(k
3
k3 +3>0 or k3 > − 3
k1
+ 3) ( k 2 + 4 ) − k 1
(k
k3 + 3 0
+ 3 )( k + 4) − k > 0 2
1
k >0
k
1
1
Stability Requirements:
k 3 > − 3, 6-8 (a)
3
(k
k 1 > 0,
3
+ 3) ( k 2 + 4 ) − k 1 > 0
Since A is a diagonal matrix with distinct eigenvalues, the states are decoupled from each other. The second row of B is zero; thus, the second state variable, x is uncontrollable. Since the uncontrollable 2
77
state has the eigenvalue at −3 which is stable, and the unstable state x with the eigenvalue at −2 is 3
controllable, the system is stabilizable.
(b) 6-9
Since the uncontrollable state x has an unstable eigenvalue at 1, the system is no stabilizable. 1
The closed-loop transfer function of the sysetm is
Y (s)
=
R (s )
1000 s + 15.6 s + ( 56 + 100 K t ) s + 1000 3
s
The characteristic equation is:
2
3
+ 15 . 6 s + ( 56 + 100 2
K )s t
+ 1000 = 0
Routh Tabulation: s
3
s
2
s
1
1 15 . 6 873 . 6
+ 100
56
+ 1560
K
t
1000 Kt
− 1000
1560 K t
15 .6 s
0
.4
>0
1000
Stability Requirements:
6-10
− 126
K
t
> 0 . 081
The closed-loop transfer function is Y (s) R( s)
=
K(s s
3
s
The characteristic equation:
+ Ks 3
+ Ks
+ 2 )( s + α )
+ ( 2 K + αK − 1 ) s + 2 αK
2
2
+ ( 2 K + α K − 1) s + 2 αK = 0
Routh Tabulation: s
3
s
2
s
1
2 αK
K (2
+ αK − 1
2K
1
K
+ α ) K − K − 2 αK
>0
2
(2 + α )K
K s
0
2 αK
Stability Requirements: α > 0 , K -versus- α Parameter Plane:
α >0 K
> 0,
K
>
1
+ 2α
2
+α
78
.
− 1 − 2α > 0
6-11 (a)
Only the attitude sensor loop is in operation: K
Θ( s) Θr ( s ) If KK
s
If KK
s
t
= 0. The system transfer function is:
G (s) p
=
1+ K G ( s) s
=
p
K s
2
− α + KK
s
>α,
the characteristic equation roots are on the imaginary axis, and the missible will oscillate.
≤α,
the characteristic equaton roots are at the origin or in the right-half plane, and the system
is unstable. The missile will tumble end over end.
(b)
Both loops are in operation: The system transfer function is
Θ( s) Θr ( s ) KK
For stability: When K
=0
t
G (s) p
1 + K sG t
> 0,
t
and KK
=
KK
>α,
s
(s)
p
s
+ K sG p ( s )
=
K s
2
+ KK
t
s
+ KK s − α
−α > 0 .
the characteristic equation roots are on the imaginary axis, and the missile
will oscillate back and forth. For any KK − α if KK < 0, the characteristic equation roots are in the right-half plane, and the system s
If KK
t
t
is unstable. The missile will tumble end over end. > 0 , and KK < α , the characteristic equation roots are in the right-half plane, and the system is t
unstable. The missile will tumble end over end.
6-12
Let s
1
= s + α,
then when s
= −α,
s
1
= 0.
This transforms the
s = −α axis in the s-plane onto the imaginary
axis of the s -plane. 1
(a)
F (s)
=
Or
s
+ 5s +3 = 0
s
2
2
+ 3 s1 − 1 = 0
1
Le t s
= s1 − 1
2
s1
Routh Tabulation:
s s
We get
1
−1) + 5 ( s1 −1) + 3 = 0 2
−1
1
1
(s
3
1
−1
0 1
Since there is one sign change in the first column of the Routh tabulation, there is one root in the region to the right of s = −1 in the s-plane. The roots are at −3.3028 and 0.3028.
(b)
F (s)
=
s
3
+ 3s + 3s +1 = 0 2
Let s
=
s
1
−1
We get
79
( s1 −1)3
+ 3( s1 −1) + 3 ( s1 −1) + 1 = 0 2
(c)
Or
s
F (s)
=
= 0.
3 1
s
Or
3
s
3 1
The three roots in the s -plane are all at s 1
+ 4 s + 3 s + 10 = 0 2
=
Let s
s
1
1
= 0.
Thus, F(s) has three roots at s =
We get
1
−2
1
10
(s
−1
1
−1.
−1) + 4 ( s1 −1) + 3 ( s1 −1) +10 = 0 3
2
+ s 1 − 2 s1 + 10 = 0 2
s s
Routh Tabulation:
3 1 2 1
− 12
1
s1 s
0
10
1
Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = −1 in the s-plane. The roots are at −3.8897, −0.0552 + j1.605, and −0.0552
(d)
F (s) Or
=
s
+4s +4s +4 = 0
3
s
− j1.6025. 2
3 1
Let s
=
s
1
−1
(s
We get
1
−1) + 4 ( s1 −1) + 4 ( s1 −1) + 4 = 0 3
2
+ s1 − s1 + 3 = 0 2
s s
Routh Tabulation:
3 1 2 1 1
s1 s
1
−1
1
3
−4
0
3
1
Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = −1 in the s-plane. The roots are at −3.1304, −0.4348 + j1.0434, and −0.4348
−j1.04348.
6-13 (a) Block diagram:
(b) Open-loop transfer function: G ( s) =
H ( s)
=
E (s )
K a K i nK I N
s (R a Js + K i K b ) ( As + K o )
=
16.667 N s( s + 1)( s + 11.767)
Closed-loop transfer function: H (s) R( s)
G (s)
= 1
+G ( s)
16 .667 N
= s
3
+ 12 . 767
(c) Characteristic equation:
80
s
2
+ 11 . 767
s
+ 16 . 667
N
s
+ 12 . 767
3
2
s
+ 11 . 767
s
+ 16 . 667
=0
N
Routh Tabulation: s
3
s
2
s
1
1
11 . 767 16 . 667 N
12 .767 150 .22
− 16 . 667
N
− 16.667
1 50.22
N
>0
or
N
NA
>0
12 . 767 s
0
N
16 .667N
Stability condition:
0
<
N
>0
<9.
6-14 (a) The closed-loop transfer function: H (s)
250 N
=
R( s)
s ( 0 .06 s
+ 0 . 706
+ ( 0 . 706
A
)( As
+ 50 ) + 250
N
The characteristic equation: 0 . 06 As
3
+ 3) s + 35 . 3 s + 250 2
N
=0
Routh Tabulation: s
3
s
2
s
1
0 . 06 A 0 . 706 A 24. 92 A
+ 105
07 . 06 A s
0
250 N
− 15 NA
+3
1
>0
0 . 706 A
+3 > 0
24.92 A
+ 105
.9
− 15
N > 0
250N
From the s row,
(b)
+3
.9
A
35 .3
N < 1.66 + 7.06/A
When A
→∞
N
max
→ 1. 66
For A = 50, the characteristic equation is
3 s + ( 35.3 + 0.06 K o ) s + 0.706 Ko s + 250 N = 0 3
2
Routh tabulation
81
Thus, N
max
= 1.
<9
s
3
s
2
s
1
0706 K
3
+ 0 . 06
35 . 3
2
0 . 0424 K o
(c)
0
250 N
o
+ 24.
92 K o
+ 0 . 06
35 . 3 s
K
o
− 750
K
N
> − 588
o
. 33
2
N < 0.0000 5653 K o
K
250 N
N
+ 0 . 03323
>0
N = 10, A = 50. The characteristic equation is
s + ( 35.3 + 0.06 K o ) s + 0.706 Ko s + 50 KI = 0 3
2
Routh Tabulution: s
3
s
2
s
1
0 . 706 K
1 35 . 3
+ 0 . 06 2
0 . 04236 K o
0
50 K
6-15 (a) Block diagram:
50 K
o
+ 24. 92
Ko
+ 0 . 06
Ko
35 .3 s
K
− 50
o
K
I
KI
I
o
> −588
. 33
KI
< 0 . 000847
K
>0
I
2
2Ko
+ 0 .498
(b) Characteristic equation: 2 Ms + K s + K + K = 0 D s P 500 s
82
2
+ K D s + 500 + K P = 0
Ko
K
o
(c)
For stability, K
D
> 0,
0 .5
+ K P > 0.
Thu s,
K
P
> − 0 .5
Stability Region:
6-16 State diagram:
∆ = 1 + s + Ks
2
Characteristic equation: s
Stability requirement:
2
+s+K =0
K>0
83
Chapter 7
TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS
7-1 (a) ζ ≥ 0 . 707
(c) ζ ≤ 0 . 5
7-2 (a)
Type 0
7-3 (a)
K
(b)
K
(c)
K
(d)
K
(e)
K
(f)
K
ωn ≥ 2
1
rad / sec
≤ ω n ≤ 5 rad
(b)
/ sec
Type 0
(c)
(d)
Type 1
Type 2
≤ ζ ≤ 0 . 707
(b)
0
(d)
0 .5
(e)
Type 3
ωn ≤ 2
≤ ζ ≤ 0 . 707
(f)
ω n ≤ 0 . 5 rad
Type 3
= lim
G ( s)
= 1000
Kv
= lim
sG ( s )
=0
Ka
= lim
s G (s)
p
= lim
G ( s)
=∞
Kv
= lim
sG ( s )
=1
Ka
= lim
s G (s)
p
= lim
G ( s)
=∞
Kv
= lim
sG ( s )
=K
Ka
= lim
s G (s)
p
= lim
G ( s)
=∞
Kv
= lim
sG ( s )
=∞
Ka
= lim
s G ( s)
p
= lim
G ( s)
=∞
Kv
= lim
sG ( s )
=1
Ka
= lim
s G (s)
p
= lim
G ( s)
=∞
Kv
= lim
sG ( s )
=∞
Ka
= lim
s G (s)
p
s→0
s→0
s→0
s→0
s→0
s→0
s →0
s →0
s→0
s →0
s →0 s →0
84
s→0
s→0
s→0
s→0
s→0 s→0
rad / sec
2
=0
2
=0
2
=0
2
=1
2
=0
2
=
K
/ sec
7-4 (a)
Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )
K
tu ( t )
K
2
K
s
s
t u (t ) / 2 s
(b)
p
= 1000
v
=0
a
=0
1 1001
∞ ∞
Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )
K
tu ( t )
K
2
K
s
s
t u (t ) / 2 s
p
=∞
0
v
=1
1
a
=0
∞
(c) Input
Error Constants Steady -state Error ________________________________________________________________________________ u (t )
K
tu ( t )
K
2
K
s
s
t u (t ) / 2 s
p
=∞
0
v
=K
1/ K
a
=0
∞
The above results are valid if the value of K corresponds to a stable closed-loop system.
(d)
The closed-loop system is unstable. It is meaningless to conduct a steady-state error analysis.
(e)
Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )
K
tu ( t )
K
2
K
s
s
t u (t ) / 2 s
(f)
p
=∞
0
v
=1
1
a
=0
∞
Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )
K
tu ( t )
K
2
K
s
s
t u (t ) / 2 s
p
=∞
0
v
=∞
0
a
=K
1/ K
The closed-loop system is stable for all positive values of K. Thus the above results are valid.
7-5 (a)
K
H
=
H (0)
=1
M (s) a
0
=
= 3,
G (s) 1+ G ( s)H ( s) a
1
= 3,
a
2
s
=
=
3
s
+1
+ 2 s + 3s +3 = 1, b 1 = 1. 0 2
b
2,
Unit-step Input:
ess = Unit-ramp input: a
(b)
K
H
=
H (0)
2 =3
0
− b0 K H = 3 − 1 = 2 ≠ 0.
0
− b0 K H = 2 ≠ 0
Unit-parabolic Input: a
b0 KH 1 − a KH 0 1
and
a
Thus
1
e
ss
= ∞.
− b1 K H = 1 ≠ 0.
Thus
e
ss
= ∞.
=5 M (s)
=
G (s) 1+ G ( s)H ( s)
=
1 s
2
+5s + 5
85
a
0
= 5,
a
1
= 5,
b
0
= 1,
b
1
= 0.
Unit-step Input:
ess = Unit-ramp Input:
b0K H 1 − a KH 0
= 0:
i e
ss
=
a a
− b1 K H
1
e
(c)
K
H
=
H (0)
ss
1
5 1 − = 0 5 5
− b0 K H = 0
0
a K
5
=
i
= 1:
s
+5
a
1
− b1 K H = 5 ≠ 0
1
=
25
H
0
Unit-parabolic Input:
=
1
5
=∞
=1/5 M (s)
=
G (s) 1+ G ( s)H ( s) a
0
= 1,
a
=
+ 15 s + 50 s + s + 1 a = 50 , a = 15 , b = 5 , 2 3 0
s
= 1,
1
4
3
2
The system is stable. b
1
=1
Unit-step Input:
ess = Unit-ramp Input:
b0K H 1 − a KH 0
= 0:
i e
ss
=
a a
− b0 K H = 0
0
− b1 K H
1
a K
e
(d)
K
H
=
H (0)
ss
=
= 1:
i
1−1 / 5
a
1
− b1 K H =
4 /5
≠0
=4
1/5
H
0
Unit-parabolic Input:
5/5 = 5 1 − 1 = 0
1
=∞
= 10 M (s)
G (s)
=
1+ G ( s)H ( s)
a
0
= 10 ,
a
=
= 5,
1
1 s
+ 12 s + 5 s + 10 a = 12 , b = 1, 2 0 3
The system is stable.
2
b
= 0,
1
b
=0
2
Unit-step Input:
ess = Unit-ramp Input:
b0K H 1 − a KH 0
= 0:
i e
ss
=
a a
1
− b0 K H = 0
0
− b1 K H a K 0
Unit-parabolic Input: e
7-6 (a)
M (s)
=
s s
4
+ 16
s
3
+4
ss
=
5
i
= 1:
a
1
− b1 K H = 5 ≠ 0
= 0 . 05
100
H
=∞
+ 48 s + 4 s + 4 a = 4, a = 4, 0 1 2
1 10 = 10 1 − 10 = 0
1
K a
=1
H
2
= 48 ,
The system is stable. a
3
= 16 ,
b
0
= 4,
b
1
= 1,
b
2
= 0,
Unit-step Input:
ess = Unit-ramp input: i
= 0:
b0K H 1 − a KH 0 1
a
0
4 = 1 − 4 = 0
−b0 KH = 0
i
= 1:
86
a
1
− b1 K H = 4 − 1 = 3 ≠ 0
b
3
=0
e
=
ss
a
− b1 K H
1
a K 0
Unit-parabolic Input: e
(b)
M (s)
s
3
3
=
4
H
4
=∞
+ 3)
K(s
=
ss
−1
4
=
K
+ 3s + ( K + 2)s + 3K a = 3K , a = K + 2, 0 1 2
a
=1
H
The system is stable for K
= 3,
2
b
0
=3K,
> 0.
=K
b
1
Unit-step Input:
ess = Unit-ramp Input:
b0K H 1 − a KH 0
= 0:
i e
=
ss
3K = 1 − 3 K = 0
1
a a
0
− b0 K H = 0
− b1 K H
1
a K 0
Unit-parabolic Input: e
ss
+2 −K
K
=
= 1:
i
=
3K
H
a
1
− b1 K H = K + 2 − K = 2 ≠ 0
2 3K
=∞
The above results are valid for K > 0.
(c)
M (s)
=
s s
4
+ 15
s
3
+5 + 50 s + 10 s a = 0 , a = 10 , 0 1 2
a
2
10 s
=
H ( s)
s+5 50 , a
=
K
3
H
= lim
H (s)
s →0
= 15 ,
b
0
= 5,
s b
1
=2 =1
Unit-step Input:
ess =
a 2 − b1K H a KH 1
Unit-ramp Input: e
ss
=∞
ss
=∞
Unit-parabolic Input: e
(d)
M (s)
=
s
4
+ 17
s
3
K(s
+ 5)
+ 60
s
a
2
K
+ 5 Ks + 5 K
= 5K ,
0
1 50 − 1 × 2 = 2 10 = 2.4
1
a
1
= 5K ,
a
2
H
=1
The system is stable for 0 < K < 204.
= 60 ,
a
3
= 17 ,
b
0
= 5K,
b
1
=
K
Unit-step Input:
ess = Unit-ramp Input: i
= 0:
e
=
ss
b0K H 1 − a KH 0 1
a a
1
0
−b0 KH = 0
− b1 K H a K 0
5K = 1 − 5 K = 0
=
H
Unit-parabolic Input: e
ss
5K
i
−K
5K
= 1:
=
4 5
=∞
The results are valid for 0 < K < 204.
87
a
1
− b1 K H = 5 K − K = 4 K ≠ 0
7-7 G (s)
=
Y (s)
=
E ( s)
KG 1
(b) (c)
r(t )
r(t )
r(t )
K
= u s ( t ):
e
= tu s ( t ): =t
2
e
u ( t ) / 2:
e
s
ss
=
=
+ K tG p( s)
Error constants:
(a)
( s ) 20 s
p
1
p
= ∞,
=
ss
=
ss
K
1 1+ K 1
20 s ( 1 + 0 .2 s
=
5K 1 + 100 K
+ 100 K
,
Type-1 system. K ) t
a
=0
t
=0 p
1 + 100 K
=
K
v
100 K
t
5K
v
=∞
K
a
7-8 G p (s)
G (s)
100
= (1
+ 0 .1 s )( 1 + 0 . 5 s )
(b) (c)
r(t )
r(t )
K
= u s ( t ):
e
= tu s ( t ): =t
2
KG
=
E (s)
+ 0 .1 s )( 1 + 0 . 5 s ) + 100
Error constants:
r(t )
Y (s)
=
20 s 1
p
(s)
+ K tG p ( s )
100 K
= 20 s ( 1
(a)
G (s)
e
u ( t ) / 2:
e
s
ss
=
1 K
p
ss
ss
= ∞,
= =
K
1 1+ K 1 K
=
5K 1 + 100 K
,
K
a
=0
t
=0 p
= v
v
Kt
1 + 100 K
t
5K
=∞ a
Since the system is of the third order, the values of K and K
must be constrained so that the system is
t
stable. The characteristic equation is
s + 12 s + ( 20 + 2000 K t ) s + 100 K = 0 3
2
Routh Tabulation: s
3
s
2
s
1
1
20
+ 24000
K
t
100 K
12 240
+ 2000
Kt
− 100
K
12 s
0
100 K
Stability Conditions:
K>0
12 ( 1+ 100 K t ) − 5 K > 0 or
1 + 100 K t
Thus, the minimum steady-state error that can be obtained with a unit-ramp input is 1/12.
88
5K
>
1 12
7-9 (a)
From Figure 3P-19,
Θ o ( s)
=
Θr ( s )
Θo ( s) Θr ( s )
=
K1 K 2
1+ 1+
K1 K 2 Ra + La s
Ra + La s
+
K i K b + KK1 K i K t
(R
a
+ La s ) ( Bt + J t s )
K i K b + KK1 K i K t
+
(R
a
+ La s ) ( Bt + J t s )
+
KK s K 1K i N
s ( Ra + La s )( Bt + J t s )
s [( Ra + La s ) ( Bt + Jt s ) + K1 K2 ( Bt + Jt s ) + Ki Kb + KK1 K i K t ]
L a J t s + ( L a Bt + Ra J t + K 1 K 2 J t ) s + ( Ra Bt + K i K b + K Ki K1K t + K 1 K 2 Bt ) s + KK s K 1K i N 3
2
θ r ( t ) = u s ( t ),
Θr ( s ) =
1
lim s Θ ( s ) e
s →0
s
=0
Provided that all the poles of s Θ ( s ) are all in the left -half s-plane. e
(b)
For a unit-ramp input,
Θr ( s ) = 1 / e
ss
=
lim
t →∞
2
s .
θ e (t ) =
lim s Θ ( s ) e
s →0
R B
=
a
t
+ K 1 K 2 B t + K i K b + KK KK
s
1
K K i
t
K K N i
1
if the limit is valid.
7-10 (a) Forward-path transfer function:
[n(t) = 0]:
K (1 + 0.02s ) G ( s) =
Y ( s)
=
E (s )
Error Constants:
K
For a unit-ramp input, r ( t )
2 K (1 + 0.02s ) s ( s + 25) = 2 K Kt s s s + 25 s + KKt 1+ 2 s ( s + 25)
(
= ∞,
p
= tu s ( t ),
K
R( s)
=
1
=
v
1 s
2
K ,
K
,
a
)
Type-1 system.
=0
t
e
ss
=
lim e ( t )
t→ ∞
= lim
s →0
sE ( s )
=
1 K
=
K
t
v
Routh Tabulation:
s s
3
2
1
s
1
KK t + 0.02 K
25
K
25 K ( Kt + 0.02) − K 25
s
0
K K >0
Stability Conditions:
(b)
With r(t) = 0, n ( t )
= u s ( t ),
N (s)
=1/
25 ( Kt + 0.02 ) − K > 0 or K t > 0.02
s.
System Transfer Function with N( s) as Input:
K Y (s ) N (s )
K s ( s + 25) = 3 2 K (1 + 0.02 s) K Kt s s + 25 s + K ( K t + 0.02 ) s + K 1+ 2 + 2 s ( s + 25) s ( s + 25) 2
=
89
Steady -State O utput due to n ( t):
=
y ss
7-11 (a)
n(t )
= 0,
r (t )
lim y ( t )
t →∞
= lim
=1
sY ( s )
s→0
if the limit is valid.
= tu s ( t ).
Forward-path Transfer function:
G ( s) =
Y ( s) E (s )
K ( s + α )( s + 3)
=
(
2
Ramp-error constant:
Kv
= lim
Steady -state error:
e
=
Characteristic equation: s + Ks Routh Tabulation: 3
s
3
s
2
s
1
s
0
ss
3K
s →0
1 K
sG ( s )
=− v
= −3 K α
1 3K
v
+ αK − 1 3 αK
K K (3K
Type-1 system.
+ [ K ( 3 + α ) − 1] s + 3α K = 0
2
1
)
s s −1
n =0
+ αK − 1 ) − 3 αK K 3α K
+ αK − 1 − 3α > 0
3K
Stability Conditions:
K
or
1+ 3K
>
3
+α
αK > 0 (b)
When r(t) = 0, n ( t )
= u s ( t ),
N (s)
=1/
s.
K ( s + 3) Y (s )
Transfer Function between n ( t) and y( t):
N (s ) Steady -State Output due to n ( t):
=
y ss
lim y ( t )
t →∞
= 1+
r =0
= lim
s→0
2 Ks ( s + 3) s −1 = K ( s + α )( s + 3) s 3 + Ks 2 + [ K ( s + α ) − 1]s + 3α K
(
s s −1 2
)
=0
sY ( s )
7-12
if the limit is valid.
− πζ
Percen t maxi mum ov
=e
ershoo t = 0 .25
1 −ζ
2
Thus
πζ Solving for
ζ
Peak T ime
1−ζ
2
= − ln0.25 = 1.386
from the last equation, we have t
max
=
π ωn
1−ζ
ζ
= 0 .01
(
π ζ = 1.922 1 − ζ 2
2
2
)
= 0.404. sec.
π
ωn =
Thus,
2
1 − ( 0 .404 )
0 . 01
Transfer Function of the Second-order Prototype System: Y (s) R( s)
ωn 2
=
s
2
+ 2ζω n s + ω n 2
7-13 Closed-Loop Transfer Function:
=
117916 s
2
+ 277
.3 s
+ 117916
Characteristic equation:
90
= 343 2
.4
rad / sec
Y (s )
=
R (s )
25 K
s + ( 5 + 500 Kt ) s + 25 K = 0 2
s + ( 5 + 500 Kt ) s + 25 K 2
For a second-order prototype system, when the maximum overshoot is 4.3%,
ωn =
2 ζω
25 K ,
n
= 5 + 500
K
ζ = 0 . 707
= 1.414
t
.
25 K
Rise Time: [Eq. (7-104)] t
r
=
1 − 0 .4167
ζ + 2 . 917 ζ
K K
With
= =
=
ωn ωn 2
Thus,
2
( 10 . 82 )
=
25 4. 68
K
= 0 .2
ωn
sec
ω n = 10 . 82
Thu s
rad / sec
2
= 4. 68
25 and
2 .164
t
5
= 0 . 0206
,
+ 500
K
t
= 1.414 ω n = 15 . 3
the sy stem t ransfe
Y (s)
t
=
10 . 3
= 0 . 0206
500
r func tion i s
117
=
R( s)
K
Thus
s
2
+ 15 . 3 s + 117
Unit-step Response: y = 0.1 at t = 0.047 sec. y = 0.9 at t = 0.244 sec. t = 0 .244 − 0 . 047 = 0 .197 r
y
7-14 Closed-loop Transfer Function: Y (s )
=
R (s )
25 K
2
πζ 1−ζ
, we get
ζ
( 4. 32% max.
s + ( 5 + 500 Kt ) s + 25 K = 0
s + ( 5 + 500 Kt ) s + 25 K
ζ
= 0 . 0432
Characteristic Equation:
2
When Maximum overshoot = 10%,
Solving for
max
= − ln0.1 = 2.3
(
π ζ = 5.3 1 − ζ 2
2
2
2
)
= 0.59.
The Natural undamped frequency is
ωn =
25 K
Thus,
5 + 500 K
t
= 2 ζω n = 1.18 ω n
Rise Time: [Eq. (7-114)] r
=
K
=
t
1 − 0 .4167
ζ + 2 . 917 ζ ωn
ωn
2
= 0 .1 =
1. 7696 sec.
ωn
Th us
ω n = 17 . 7
rad / sec
2
= 12 . 58
25 With K = 12.58 and K
t
=
sec.
K
t
=
15 . 88
= 0 . 0318 500 0 . 0318 , the system transfer function is Thus
Y (s) R( s)
313
= s
2
+ 20 .88 s + 314.
5
Unit-step Response: y = 0.1 when t = 0.028 sec. y = 0.9 when t = 0.131 sec.
91
overs hoot)
t
y
7-15
= 0 . 131 − 0 . 028 = 0 .103
r
= 1.1
max
( 10%
sec.
max.
overs hoot )
Closed-Loop Transfer Function:
Characteristic Equation:
Y (s)
s + ( 5 + 500 Kt ) s + 25 K = 0
R( s)
=
25 K s
2
+ ( 5 + 500
2
+ 25 K
K )s t
πζ
When Maximum overshoot = 20%,
1−ζ Solving for
ζ
, we get
ζ = 0 .456
The Natural undamped frequency
= − ln0.2 = 1.61
(
π ζ = 2.59 1 − ζ 2
2
2
2
)
.
ωn =
25 K
5
+ 500
K
t
= 2 ζω n = 0 . 912 ω n
Rise Time: [Eq. (7-114)] t
r
=
1 − 0 .4167
ζ + 2 . 917 ζ ωn
2
= 0 . 05 =
1.4165 sec.
ωn
Thus,
ωn
ωn =
1.4165
=
28 .33
0 . 05
2
= 32 .1 5 + 500 K = 0 . 912 ω = 25 . 84 Thus, t n 25 With K = 32.1 and K = 0 . 0417 , the system transfer function is K
=
K
t
= 0 . 0417
t
Y (s) R( s)
802 . 59
=
s
2
+ 25 . 84 s + 802
. 59
Unit-step Response: y = 0.1 when t = 0.0178 sec. y = 0.9 when t = 0.072 sec. t = 0 . 072 − 0 . 0178 = 0 . 0542
sec.
r
y
7-16 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K From Eq. (7-102), Delay time t
d
≅
When Maximum overshoot = 4.3%,
1.1
max
= 1.2
With K = 20.12 and K
max.
overs hoot )
Characteristic Equation:
s + ( 5 + 500 Kt ) s + 25 K = 0 2
+ 0 .125 ζ + 0 .469 ζ
2
= 0 .1
ωn
ζ = 0 . 707
t
.
=
d
1.423
ωn
sec.
= 0 .1
sec.
Thus
ω n = 14.23 = 8.1 5 + 500 K = 2ζω = 1.414 ω = 20.12 K = t n n 5 5 2
( 20%
ω n = 14.23
rad/sec.
2
t
= 0 . 0302
, the system transfer function is Y (s) R( s)
202 . 5
= s
2
+ 20 .1 s + 202
Unit-Step Response:
92
.5
Thus
K
t
=
15 .12 500
= 0 . 0302
When y = 0.5, t = 0.1005 sec. Thus, t = 0 .1005 sec. d
y
7-17 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K From Eq. (7-102), Delay time t
≅
d
1.1
( 4. 3%
max.
overs hoot )
s + ( 5 + 500 Kt ) s + 25 K = 0 2
+ 0 .125 ζ + 0 .469 ζ
2
= 0 . 05 =
ωn
1. 337 Thus,
ωn
ωn =
1. 337
= 26 . 74
0 . 05
2
K =
With K = 28.6 and K
= 1. 043
Characteristic Equation:
ω n = 26.74 = 28.6 5 5 2
max
t
= 0 . 0531
,
5
+ 500
K
t
= 2 ζω n = 2 × 0 . 59 × 26 . 74 = 31 . 55
the sy stem t ransfe
Y (s)
K
t
= 0 . 0531
r func tion i s
715
=
R( s)
Thus
s
2
+ 31 . 55 s + 715
Unit-Step Response: y = 0.5 when t = 0.0505 sec. Thus, t = 0 . 0505 sec. d
y
7-18 Closed-Loop Transfer Fu nction: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K For Maximum overshoot = 0.2, ζ = 0.456 From Eq. (7-102), Delay time t
d
=
1.1
max
= 1.1007
( 10 . 07%
max.
overs hoot )
Characteristic Equation:
s + ( 5 + 500 Kt ) s + 25 K = 0 2
.
+ 0 .125 ζ + 0 .469 ζ
2
=
ωn
1.2545
= 0 . 01
ωn
sec.
ω n = 15737.7 = 629.5 25 5 2
Natural Undamped Frequency 5
+ 500
K
t
ωn =
1.2545
t
= 0.2188
.45
rad/sec. Thus,
0 . 01
= 2 ζω n = 2 × 0 .456 × 125
With K = 629.5 and K
= 125 .45
= 114.41
K =
Thus, K
t
= 0.2188
, the system transfer function is Y (s) R( s)
=
15737 . 7 s
2
+ 114.41
Unit-step Response:
s
+ 15737
.7
y = 0.5 when t = 0.0101 sec.
93
Thus, t y
7-19 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K ζ = 0 .6 2 ζω = 5 + 500 n From Eq. (7-109), settling time t
K
t
=
3 .2
≅
s
1.2 ω
n
−5
= 1.2
sec.
( 20% ma
x. overs hoot )
Characteristic Equation:
s + ( 5 + 5000 Kt ) s + 25 K = 0 2
K
=
ζω n
max
= 0 . 0101
d
t
= 1.2 ω n 3.2
0 .6 ω
= 0 .1
sec. Thus,
ωn =
n
ωn
3 .2
= 53 . 33
rad / sec
0 . 06
2
= 0 .118
=
K
500
= 113
. 76
25
System Transfer Function: Y (s) R( s)
=
2844 s
2
+ 64 s + 2844
Unit-step Response:
y(t) reaches 1.00 and never exceeds this value at t = 0.098 sec. Thus, t = 0 . 098 sec. s
7-20 (a) Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K For maximum overshoot = 0.1, ζ = 0 . 59 . Settling time:
t
s
=
K
3 .2
ζω n t
=
3 .2
=
1.18
0 .59
ωn
ωn −5
Characteristic Equation:
s + ( 5 + 500 Kt ) s + 25 K = 0 2
5
+ 500
= 0 . 05
K
t
= 2 ζω n = 2 × 0 . 59 ω n = 1.18 ω n ωn =
sec.
ωn
3 .2 0 . 05
× 0 . 59
= 108
.47
2
= 0 .246
K
=
500
= 470
. 63
25
System Transfer Function: Y (s) R( s)
=
11765 . 74 s
2
+ 128
s
+ 11765
. 74
Unit-Step Response: y(t) reaches 1.05 and never exceeds this value at t = 0.048 sec. Thus, t = 0 . 048 sec. s
94
(b)
ζ = 0 .456
For maximum overshoot = 0.2, Settling time t
s
=
3 .2
ζω n
.
3 .2
=
0 .456 K
t
=
5
+ 500
= 0 . 01
ωn
0 . 912
K
t
= 2 ζω n = 0 . 912 ω n ωn =
sec.
ωn −5
3 .2
× 0 . 01
0 .456
= 701 . 75
rad / sec
= 1.27
500
System Transfer Function: Y (s) R( s)
=
492453 s
2
+ 640
s
+ 492453
Unit-Step Response: y(t) reaches 1.05 and never exceeds this value at t = 0.0074 sec. Thus, t = 0 . 0074 sec. This is less s
than the calculated value of 0.01 sec.
7-21 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K Damping ratio
ζ = 0 . 707
Characteristic Equation:
s + ( 5 + 500 Kt ) s + 25 K = 0 2
Settling time t
.
s
=
4. 5ζ
ωn
=
3 .1815
ωn
= 0 .1
sec.
Thus,
ω n = 31 .815 ωn 2
5
+ 500
K
t
= 2 ζω n = 44. 986
Thus , K
t
= 0 . 08
K
=
2ζ
= 40 .488
System Transfer Function: Y (s) R( s)
Unit-Step Response:
=
1012 .2 s
2
+ 44.
986 s
+ 1012
.2
The unit-step response reaches 0.95 at t = 0.092 sec. which is the measured t . s
95
rad/sec.
7-22 (a)
When
ζ = 0 . 5 , the rise time is t
r
≅
1 − 0 .4167
ζ + 2 . 917 ζ ωn
2
=
1. 521
=1
ωn
sec.
Thus
ω n = 1. 521
rad/sec.
The second-order term of the characteristic equation is written s
2
+ 2 ζω n s + ω n = 2
The characteristic equation of the system is
For zero remainders,
28 .48 a
=
2
+ 1. 521 s + 2 . 313 = 0
s
3
+ ( a + 30
)s
+ 1. 521 s + 2 . 313
2
s
Dividing the characteristic equation by
s
45 . 63
Thus,
a
+ 30
2
as
+K =0
, we have
= 1. 6
K
= 65 .874 + 2 . 313
a
= 69 . 58
Forward-Path Transfer Function: G (s)
69 . 58
= s( s
+ 1.6 )( s + 30
)
Unit-Step Response: y = 0.1 when t = 0.355 sec. y = 0.9 when t = 1.43 sec. Rise Time: t = 1.43 − 0 . 355 r
(b)
seconds
The system is type 1.
(i)
For a unit-step input, e
(ii)
For a unit-ramp input,
= 1. 075
ss
= 0. K
= lim
v
s→0
sG ( s )
=
K 30 a
=
60 . 58 30
× 1. 6
7-23 (a) Characteristic Equation: s
3
+ 3s + (2 + K ) s − K = 0 2
Apply the Routh-Hurwitz criterion to find the range of K for stability.
Routh Tabulation:
96
= 1.45
e
ss
=
1 K
= 0 .69 v
sec.
s
3
s
2
s
1
s
0
6
+K
1
2
3
−K
+4K 3
−K
Stability Condition:
-1.5 < K < 0 This simplifies the search for K for two equal roots. When K = −0.27806, the characteristic equation roots are: −0.347,
−0.347, and −2.3054. (b) Unit-Step Response: ( K = − 0.27806)
(c) Unit-Step Response ( K = − 1)
The step responses in (a) and (b) all have a negative undershoot for small values of t. This is due to the zero of G(s) that lies in the right-half s-plane.
7-24 (a)
The state equations of the closed-loop system are: dx
= − x1 + 5 x 2
dx
= −6 x 1 dt dt The characteristic equation of the closed-loop system is 1
∆=
2
s +1
−5
6 + k1
s + k2
− k1 x 1 − k 2 x 2 + r
= s + ( 1 + k2 ) s + ( 30 + 5 k1 + k 2 ) = 0 2
97
For
(b)
For
ω n = 10
ζ = 0 . 707
rad / sec,
, 2ζω
2
For
ω n = 10
1
2
k
2
1
+ k 2 = 70
.
1.414
ζ = 0 . 707 an d
k 2 = 59 + 10 k1 2
Thus
, 1
+ k 2 = 2ζω n = 14. 14
k
Th us
2
= 13 .14
The closed-loop transfer function is
Y (s)
=
R (s )
5 s + ( k 2 + 1) s + ( 30 + 5 k1 + k 2 ) 2
lim y ( t )
For a unit-step input,
t →∞
(e)
k
= 30 + 5k1 + k 2
= 11 . 37 .
1
Thu s 5 k
.
2
2
and
1
ωn = 1 +
Th us
+ k 2 = 100
Solving for k , we hav e
(d)
2
(1 + k )
rad / sec 5k
+ 5 k 1 + k 2 = ω n = 100
= 1+ k2.
n
ωn = (c)
30
= lim
sY ( s )
s→0
5
=
=
5 s + 14.14 s + 100 2
= 0 . 05
100
For zero steady-state error due to a unit-step input, 30
+ 5 k1 + k 2 = 5
Thus
5k
1
+ k 2 = −25
Parameter Plane k versus k : 1
2
7-25 (a) Closed-Loop Transfer Function Y (s) 100 ( K P + K D s ) = 2 R ( s ) s + 100 K D s + 100 K P The system is stable for K > 0 and P (b)
For
ζ = 1,
2ζω
ω n = 10
n
= 100 K
K
P
(c)
See parameter plane in part (g) .
(d)
See parameter plane in part (g) .
D
(b) Characteristic Equation: s K
D
> 0.
n
= 100
2
+ 100
K
.
T hus
2ω
KD
= 20
98
KP
K
D
= 0 .2
K
P
D
s
+ 100
K
P
=0
(e)
K
Parabolic error constant
= 1000
a
−2
sec
K a = lim s G ( s ) = lim100 ( K P + K D s ) = 100 K P = 1000
Thus K P = 10
2
s →0
(f)
s →0
Natural undamped frequency
ω n = 50
ω n = 10 (g)
When K
P
K
rad/sec.
= 50
P
K
Th us
=
P
25
= 0, G (s)
=
100 K s
D
s
2
100 K
=
D
(pole-zero cancellation)
s
7-26 (a) Forward-path Transfer Function: Y ( s)
G ( s) = When r ( t )
(b)
= tu s ( t ),
=
E (s ) K
= lim
v
KKi
s [ Js(1 + Ts ) + K i K t ]
s→0
sG ( s )
=
K K
e
=
ss
10 K
(
s 0.001 s + 0.01 s + 10 K t =
t
2
1 K
=
K
)
t
K
v
When r(t) = 0
Y (s ) Td ( s) For T ( s ) d
(c)
=
1 s
=
1 + Ts
=
s [ Js(1 + Ts) + Ki Kt ] + KKi lim y ( t ) t→ ∞
=
lim sY ( s ) s →0
=
1 + 0.1s
(
1 if the system is stable. 10 K
The characteristic equation of the closed-loop system is 0 . 001 s
3
+ 0 . 01 s + 0 .1 s + 10 2
K
=0
The system is unstable for K > 0.1. So we can set K to just less than 0.1. Then, the minimum value of the steady-state value of y(t) is 1 + =1 − 10 K K =0 .1 However, with this value of K, the system response will be very oscillatory. The maximum overshoot will be nearly 100%.
(d)
For K = 0.1, the characteristic equation is 0 . 001 s
3
+ 0 . 01 s + 10 2
K s t
For the two complex roots to have real parts of
+1 = 0
s
or
3
+ 10
s
2
+ 10
4
K s t
+ 1000 = 0
−2/5. we let the characteristic equation be written as
99
)
s 0.001 s + 0.01 s + 10 Kt + 10 K 2
