Solutions t o B Problems
CHAPTER 2 B-2-1. Note that
&rtj =
+S
R e f erring to Egua tion (2-2 1, we obtain
Note t h a t
3 sin (5t
+
45 ") = 3 sin 5t cos 45'
-- - 3
Jz
So we have
fl
-+ 3
cos
s i n 5t + 3 cos 5t
5t sin 45'
Note that
Referring to Equation (2-21, we obtain
f(t) =
B-24.
t
0
=cos2Wtros3LLIt
t 1 0
Noting that cos 2 W t cos 3wt = + ( m s 5 d t + c Q s & J t ) w e have
F(S)
B-2-5.
=
L [ f ( t =) ~
~ + r tn s r ~ t ) ]
[%ms 5
The function f I t ) can be written as
f(t) = (h
- a)
l(t
- a)
The Laplace transform of f(t) is ~ ( s =)
[~(t)] =
J [(t - a )
~ ( -t a)] =
+ -as
6
B-2-6. -
- a)
f[t) = c l[t
- c I(t
- b)
The Laplace transform oE f ( t ) is ~ ( a = ) c
e-a S - ce-bs - 2 (,-as S S
S
B-2-7.
The Function f ( t ) can be written as
So the Laplace transform of f ( t ) becomes
As
a approacl~eszero, t h e l i m i t i n g value of F(s ) becomes as follows: F ( s ) = lim ad0 a+O
lim
= lim
= lirn a+O
B-2-8.
10
- 12.5
e-(a/5)s
+
2 5 =-as
a 2s
da
( 2 . 5 e-(a/5)s
-0.5 s e-(a/5)s 2
The function f(t) can be w r i t t e n as
So the Laplace transform of f(t) becomes
+
2 . 5 .-as)
2 . 5 s e-as
The limiting ~ l u 05 e F(s) as a approaches zero is l i m F(s) = l i r n ad0
a30
24(1
- as
e-*= a3s2
3-2-11.
Define y = x
Thw y(O+) = $(O+)
The i n i t i a l value of y can be obtained by use of the initial value theorem as follows:
Since y(s1
we obtain
B-2-12.
Note that
= J'+[y!t)l
= d;rGct,1
= sX(S)
- x((U)
Referring to Problm A-2-12,
we have
B-2-14. f ( t ~emst dt =
Referring to Problem A-2-12,
e-st at = 1 0-
we obtain
where
%(s) can thus be written as
and the inverse Laplace transform of Fl(s) is
Cbl
where
FZ(s) can thus be written as
and the inverse Laplace transform of F Is) is 2
me inverse Laplace transform of F (s) is 1
(bl
where
E (s) can thus be written as 2
and the inverse Laplace transform of F,(s)
.
is
L-
i p ) = -3 .-t
+
3 .-2t
B t .-2t
--2 + - + - -2 - - 2 - 5 s + l s
5 s + L
-
3 s + l
Rre inverse Laplam transform of f ( s ) is
The inverse Lapla-
transform of F(S) is
f { t )=
S ( t ) 'F
2
Hence f(t) = e-t cos 3t
4-
4t
1 e-t --
3
sin 3t
+- 5
s
Hence
The inverse Laplace transform of F(s) is
Hence
The inverse Laplace transform of F l s ) is
The inverse Uplacre transform of ~ ( s )is fit) =
1 1 sin (t - -
ld2
B-2-23. ~ ( s =)
L L J ~ ~
W
b e-as C (1 - e-as) , s2 S
a >O
The i m r s e Laplace transform of F(s) is f(t) = ct
3-2-24.
- c(t - a)f(t - a) - b
l(t - a)
A MATUG program to obtain p r t i a l - f r a c t i o n expansions of the given function F(s) is given b l e w .
From this eamputer output we obtain F(s) =
-- -0.25
I
+
2
f(t) = -0.25 e-*t + e-t
-
s4 + 333 +
2s2
s
+
+ s + l
-0.75
s
0.5
+-
s2
The inverse Laplace transform of F(s) is 0.75 + 0.5t
A possible MATLIE! program to obtain partial-fraction expansions B-2-25. of the given function F(s:l is given b l o w . f
I
num = [O 0 3 4 I]; d e n = [ l 2 5 8 101; [r,p,kJ = residue(num,den) r= 0.3661 - 0.4881i 0.3661 + 0.4881i -0.3661 0.0006i -0.3661+ O.0006i
-
P= 0.2758 + 1 .go81i 10.2758 1.9081i
-1 -2758 + I.0309i - 1.2758 - 1.0309i k=
0 L
From this cmputer output we obtain
Since the poles are complex quantities, we may rewrite F ( s ) as follows:
Then, the imrerse Laplace transform of F(s) is obtained as
f(tl = 0.7322 e092758t m s 1.9082t + 0.9762 e00275Btsin 1.9081 t - 0.7322 e-1-2758t cos 1.0309t
+ 0.001204 e-1*2758t sin 1.0309t
'Ihe Laplace transform of the given differential equation is
Substitution of the initial conditions into this l a s t equation gives
SoJ=ving for X ( s ) , we obtain
The inverse Laplace transform oE K(s) is x(t) = 5
COS
2t
This is the solution of the given differential equation. ---
---
..
B-2-27,
X+LJ
n
2x=t,
X(O)
;(a)
= 0,
= o
The Laplace transform of this differential equation is
Solving this q a t i o n for X ( s 1 , we obtain
The inverse Laplace transform of X ( s ] is
This is the solution of the given differential equation,
-
P
B-2-28
.
d ; + & + x = l ,
x(0) = 0,
210) = 2
The Uplace transform of this differential equation is 2[s2x(s)
-
sx(0) -
1 G(o)] + 2[sx(s) - ~ ( 0 )+) x ( s ) = s
Substitution of the initial conditions into this equation gives
Solving this last equation for X ( s ) , we get
The inverse Laplace transform of ~ ( s giws ) x ( t ) = 4 e-005t s i n 0.5t
+
1
= l + 3 e-0*5t sin 0.52
-e
cos 0.5t
-
a s 0.5t
- e-0*5t sin 0.5k
Taking the Laplace transform of b h i s differential equation, we obtain
-
2[sZ~(s) sxIO)
- s(03 1 + 7 ~ a i s -) x(0) 1 + 3X(s) = 0
By substituting the given initial conditions into this last equation,
-
~ [ s ~ x ( s )3s]
+
7[sX(s)
-
31 + 3XCs) = 0
Solving for X [ s ) yields
Finallyf taking the inverse laplace tmnsform of X(s), we obtain -0.5t x ( t ) = 3.6 e
-
B-2-30,
.. * x = sin 3t,
x
- 0.6
e-3t
x(0) = 0,
k(0) = 0
The Laplace transform of this differential equation is
Solving this equation for Xis), we get
The invezse. Laplace transform of X(s) gives
x(t) =
--
-'3
sin t
-I sin 8
3t
CHAPTER 3
B-3-2. Assume that the W y of known mmnt of i n e r t i a J through a small angle @ akrout the v e r t i c a l axis and then equation of motion far the oscillation is
where k is the torsfanal spring constant oE the string. be m i t t e n as
' is turned released.
The
This equation can
The period TO of this oscillation is
Next, we attach a rotating m the period T of oscillation.
y of unknown mmnt of inertia 9 and measure The equation for the perid T is
By eliminating the trimown torsional spring constant Ir from Equations (1) and ( 2 ) , we obtain
The unknown moment of inertia J can therefore be determined hy measuring t h e perid af oscillation T and s u b s t i t t t t i n g it i n t o Equation ( 3 ) .
B-3-3. Define the vertical displacement of t h e b a l l as x ( t ) with x(0) = 0. The positive direction is downward. The equation of motion for the system
is with initial conditimsx(O) = O m a n d G ( 0 ) = 2 0 m / s .
I\snrme that at t = tl the ball reaches t h e ground.
Sowe have
Then
frm which w e obtain
The ball reaches the ground in 2,915 s.
B-3-4. Define the torque applied to tfie flywheel as Tmotion for the system is
The equation of
from which we obtain
%y substituting n m r i c a l values into this equation, we have
Thus
T = 1256 N-III **
JQ=-T
(T = braking torque)
Integrating this equation,
Substituting the given numerical values,
Solving for TJJ, we obtain
Hence, the deceleration given by the brake is 5.33 rad/s2. The total angle rotated in 15-semnd period is obtained from
B-3-6.
Assume t h a t we apply force F to t h e spring system. F = k x + k ( x - y ) 1. 2
~liminatingy f r a the preceding equations, we obtain
Hence the equivalent spring constant
-3-7,
is given by
The equations for the system are
Eliminating y from the two equations gives
Then
The equivalent spring constant k
@3
is then o b b i n e d as
Next, consider the figure s h m below.
similar.
Note that A A B D and d C B E are
So we have
which can be rewritten as
--
K(OB +
+ OR) = -OA(OB - 4 El
Solving for OC, we obtain
B-3-8, (a)
The force f due to the dampers is
In terms of the ermivalent viscous friction coefficient beq, force f is given by Hence
(b)
The force f due to the dampers is
where z is t h e displacemnt of a point &tween damper bl and damper (Note t h a t the same force is transmitted through the shaft.) b2. From Equation (I), we have
In terrns of the equivalent viscous friction coefficient is given by f = hq(y
k tforce f
- );
By substituting Equation (2) i n t o Equation (13, we have
Hence,
B-3-9.
Since the sane force transmits t h e shaft, we have
where displacement z is defined in the figure Below.
In t e r n of the equitalent viscous friction coefficient, the force 5 is given by f= - 4) (21 From Equation (1) we have bl& + b2; + bgi = bl; + b2G + b3;
By substituting Equation (3) i n t o Equation
1 1 ) ~ we
have
Hence, by comparfng Equations ( 2 ) and ( 4 1 , we obtain
8-3-10.
The equation For khe system is
$+
(lc
1
+ k +Ic ) x = O 2 3
The natural frequency of the system is
B-3-ll.
?'he density
p of
the l i q u i d is
where A is the cross-sectio~lala r e a of the inside o f the g l a s s tube. mathernakical ~ d e for l the sysken is
The natural frequency is
The
B-3-12.
For a mll displacement x, the torque balance equation for the
system is
** mx(2a) =
- k(TI
x)a
or
Tire natural frequency is
A modified diagram for the systim s h m in Figure 3-55 is -3-14. given below.
mathemtical model for the system is given by the following t y ~ equations:
A
13;~
= -k,e,
I*
J$32 = k2(91
- k2(e1 - 82) - 82)
B-3-15. The following t w a equations describe the nation of the system and they are a mathematical model of the system.
-
I.
- k(G - $1 p(t) - kl(y - x ) - q(j.- i )
mlx = - ~ ( x ~3
m2$ = -kzy
+
Rewriting, we obtain
B-3-15.
A mathematical. m o d e l for the system is
B-3-17.
The equations of motion for the system a r e
Noting that x = 2 y , R8 = x - y = y r and J = the three equations can b rewritten as -I M R @2" = - @ =1 2 2 ( T~ T~)R
2
*. a = -
~ j *; icy = TI + T* Eliminating T2 f m t h e preceding equations gives
1 ~+N?+]CY -M 2
By changing y into xt
=
ml
I - 2nrj;
T2
The natural frequency 4s I
XE mass m is pullecl down a distance xo and released with zero i n i t i a l velmity, the motion of mass rn is
B-3-18.
Referring to the figure below, we have
(Note that since x is m a s u r d where T is the tension i n the wire. from the static equilibrium position, the term mq does not enter the equation.) For the rotational motion of the pulleyr WE have
Eliminating T frm Equetions (1) and (2), we obtain
Since x = Rlflr we have
Ill
This Last equation is a mathematical model of the system. the system is
The natural frquency of
3-3-19.
The equation of motion for the system is
Substituting the given numerical values for m, b, and k into t h i s equation, we obtain
wherex(0) = 0.1 and ?(o) = 0. The respofise ta the given initial condition can be o b t a i n e d by taking the Laplace transform of t h i s equation, solving the resulting equation for ~ ( s ) , a n dfinding the inverse Laplace transform o f The LapPaee transform of Equation ( 1 ) is X(s)
.
By subsbituting the given initial conditions ink0 this l a s t equation, w e get
Solving t h i s equation for X l s ) gives
-
The inverse Laplace transform of this last equation gives
x ( t ) = 0.1(J--
3
e-t s i n 3t + e l t
3t) ---
B-3-20.
The equation of notion for the system is
where Fk = &(mg
-F
s i n 30').
= 0.066 F
~ewrittingthis equation,
- 0.3(mg - 0.5 F)
For a constant speed motion, $ = 0 and the last equation becomes 1.016 F
or
- 0.3
mg
=b
m%
..---
.--.
3-3-21,
Tfie equations of mtion for the system are
.. Mx=T-
rx
kMs
0.
m=mgl-T
Elimination of T frm khese two equations gives
By substituting M = 2, m = I, and
JLK= 0.2
i n t o this l a s t equation, we get
Noting t h a t z ( 0 ) = 0, w e have
P s s w e that a t t = t
1
.
x(tl) - x(O) = 0.5 m.
?hen
Thus, the velocity of the block when it has maved 0.5 m can be found as
B-3-22.
The equations of motion for t h e system are
where x = R8 and J = $ mR 2
.
So we obkain
The natural frequency of the system is
B-3-23. Assume that the direction of the static friction force F is to S the left as shown in t h e diagram below. The equations for the system are
where J = $ .'~m Since the cylinder rolls without sliding, we have x = I?@. Consequently,
By eliminating
from Equations (I) and (2), we have
<
Since is found to be equal to -(1/3)F, the magnitude of Fs is one third of F and its direction is opposite to that assumed in the solution.
B-3-24.
The equation of motion for the system is **
ntx = F
-q
sin 3 0 -
- ppg
cos 30°,
x(0) = 0,
;(o)
By substituting the given numerical values into this equation, we obtain *b
x =F
or
and
- 1 x 9.81
x 0.5
- 0.2
x 1 x 9.81 x 0.866
= 0
Assume that a t t = tl, x ( t ) = 6 rn and
I
6 = (F
i(tl) =
- 6.604)
T¶x?n -
5 ds.
1 tI2 2
F m the last two equations, tl and F are fomd to h
Tfieref ore, Work done by force F = F x 6 = 8.69 x 6 = 52.14 N-m Work done by the grav3btional force =-mg
-
-
9.81 x 0.5 x 6 =
s i n 30mx 6
-
29-43 N-m
Work done by the sliding friction force =
8-3-26.
- 0.2 x mg m s 30'
x 6
The kinetic energy T is T = 1 ~ H2 e'2 ~ +'
d~
-
= + ( M + + ) I ~ ~ ~
The p t m t i a l energy U is U =t@f(1
-
- cos 0 ) +
(M+-+)gj?(l
- m s B)
Since the system f s conservative, we have
- m s 0) dJ
= constant Noting that d(T + U)/dt = 0, we obtain
[(M
12;
+
!gl+
+ (M +
sin 81
j) =
0
=
~
Since 0 is not identically zero, we have
Rewriting,
+
"
8 +
q
M+m 3
1
s
i
n
~
For small values of 0 ,
So the natural frequency is
B-3-27.
T"he k i n e t i c energy T of the system is
and the potential energy U of the system is
u
= 'j
r1R1
+ % k p l
- e2) 2
Using the law of conservation of energy, we have
+ )i k2o2Z
Noting that d ( T + ~ ) / d " , = 0, we obtain
Since
61 and
i) are not i d e n t i c a l l y zero, we must have 2
B-3-28. A t t = 0 (the instant the mss M is released' to m e ) t h e kinet energy T and the potential energy U of the system are I 1
The potential energy I s measured relative to the floor. A t the i n s t a n t mass m h i t s t h e fnmr t h e kinetic energy T and the 2 of the system are 2
plkential energy U
v2 is t h e velocity of the hanging mass m and G2 is the angle of rotakion of the pulley both at the instant the mass hits t h e f l o o r ; re2 - v2, and J = % mpr 2
&re
-
.
Using the law of consenration of energy, we obtain
By substituting ret = v2 into this l a s t equation, , = + M V ~ ~ + + I 2T T V ~ . . ) ~ ~ ~ V ~
~2
Solving this equation for v
E-3-29.
2
The force F necessary to move t h e weight is F = nl9 = 10OO x 9-81 = 1962 pJ 5 5
The p o w e r P is given by
where tJ = m~fx.
-3-30.
So we obtain
From the figure shmm to the right, we obtain
To keep the bar AB horizontal when prlling the weight n g , t h e rnament a b u t p i n t P must balance. Thus,
Solving this equation f o r x, we obtain
B-3-31.
Note that
where Gear ratio = 1/30
Since the power T L W l 0 2 the motor is
the torque T2 of the driven shaft is
B-3-32. Assume that the stiffness of the shafts of the gear nite, t h a t there is neither backTash nor elastic defamation, nt.mber of teeth on each gear is proportional to the radius of the angular displacement of slraf t 1 and shaft 2.as 0 and Q2, I By applying Newton's second law to 'the system, we obtain shaft (shaft 1 )
t r a i n is i n f i and t h a t the gear. Ikf i n e respectively. far the mtbr
where Tm is the torque developed by the motor and TI is the laad torque an 5@ar 1 due to the rest of the gear t r a i n .
For s h a f t 2 , we have
where T2 is the torque transmitted to gear 2. is equal to that of gear 2,
Sin-
e2/Q1
= n /n
1
2
, Equation
Since the work done hy gear 1
( 2 ) can be written as
Substituting Equation ( 3 ) into ~quatlon (I), we get
The equivalent moment of inertia of the gear t r a i n referred to the motor s h a f t is
Notice that if the r a t i o n /n
is very ttntch smaller than unity, then the 1 2 is negligible. effect of J2 an t h e equivalent mwnent of inertia J eq &-3-33. The equivalent mament of inertia J of mass m referred to the motor m s h a f t axis can be obtained frcm
where oE is the angular acceleration of the motor shaft and 5 is the linear acceleration of mass rn. Since d r = 2, we have
The equivalent mcment of inertia Jb of the belt is obtained from
S i n c e there is no slippage b e t w e e n the belt and the pulleys, the work: done by the belt and the right-side pulley ( ~ ~ and 0 ~that ) by the belt and t h e
left-side pulley (T2Q2)must he equal, or
where T is the load torque on the motor shaft and T is the torque trans1 2 rnitted to the left-side p u l l e y s h a f t , is t h e angular d i s p l a c m n t of the motor s h a f t , and G2 is the amy~lardisplacement of the left-side pulley shaft. Since the two pulleys are of t h e same size, we have BI = B2. Hence
For the motor s h a f t , we have
Also, for the left-side pulley shaft, we have
Since TI =
5' we ham
Since 0 = B2. this Last e q u a t i o n becomes 1
The equivalent moment of inertia J of the system with reeq s h a f t axis is
to the mtoi
CHAPTER 4
So we obtain
B-4-3.
figure 4-55 can be r e a w n as s
h below-
From the right side diagram we obtain the follaring equations: 10 i
20
1
+ 20(i 1
- i3) = E
i2 + 10(i2 + ig) = E
100 i, + 1O(i
2
+ i 1 = ZO(i 3
I
- i31
whi&
yield
So w e obtain
Solving f o r i2, ve get
Figure ( a )
Figure (b)
Fran Figure ( a ) abwe we find R 1 = ZR2.
Figure I c )
Referring to Figure (b) above, w e
set
So R
1
is obtained as
By substituting R1 = 2R2 into this l a s t equation, w e obtain
!t%en, frm Figure ( c ) a m , current i is obtained as
B-4-5. When switch and we have
S
is open, resismm between points A and B is l60R
When switch S is closed, resistanm between points A and B is
So we haw
Cansequently ,
Solving for Rj we obtain
R = 25.R M-6. -
F m the circuit diagram we obtain di,
Each of the p r d i n g two sets of equations constitutes a rrratharatifal model f o r the circuit.
Fran Figure
3-4-7.
4-55,
we have for t > 0
Taking Lapface transfom of Equations (1) and ( 2 ) , we get 1
1 1
where i -'(o) 2
(5)
* R2[%(s) - T(s)I'
0
is g i m by
t = 0
F m Equation ( 3 ) we have
'men Equation ( 4 ) can be rewritten as
Substituting Equation (5) into Equation (6)we obtain
Tbe inwrse Laplam transform of this last equation gives
Referring to Equation 15 1
B-4-8, -
W E have
A t steady stake (t
< 0) we haw
and
For t 2 0 the equation for the circuit is
By d i f f e r e n t i a t i q this l a s t quation, we obtain
Th-e Laplace transform o f this q u a t i o n is
Rl[al(s)
- i(O)]
+
L I(s) = 0 C
where
Hence
or
The inverse Laplace ttansfonn of I(s) gives
B-4-9.
The equations for t h e system are
where Jeq is the equi~lentmoment of inertia of the system referred to the motor rotor axis and is the equivalent viscous friction coefficient of the system referred
By taking Idplace transform of Equations (1) and ( 2 ) , w e obtain
Elimination of Eb(s) from the above two equations yields
(R, + L,s)I,(s)
+ l(bsB1(5) = E , ( s )
The Laplace transform of Equation (3) is J
shl(s) + b SB eq
1
(5)
= KIa(s)
Hence
By substituting Equation (5) into Equation ( 4 ) , we obtain
The numerical values of the equivalent moment of inertia Jq viscous friction coefficient b are, respectively, eq
and equivalent
Substituting these numerical v a l u e s into m a t i o n (61, we get
Since B2/B1= N /N = n = 0.1, ue have the transfer function B~(s)/E,(s) as 1
2
fallows:
R-4-10.
From the circuit shown to
t h e sight, we obtain
Since we have
X S O ,since we obtain
Equation (1) san be written as
Upla-
transforming this equation and simplifying, we get
Lapla-
transforming Equation ( 2 ) , we obtain
fran which w e obtain
By substituting Equation ( 4 )
i n t o Equation (31, we have
frm which the transfer function Eo(s)/Ei (s) can be obtained as
334-11.
From the circuit diagram shown, we obtain
Since we have
Also, since
i3 = 14 we get
Thus, (2)
Substituting Equation (2) i n t o Tquation (1)and simplifying, w obtain
Hence the transfer function Eo(s)/?Zi(s) is given by
The transfer function can given in t e r n of camplw impedances Z1 and Z 2 as follows:
3-4-12.
E,(s)/E~(s)
where
Hence
-14.
ZL = Ls,
- I- z
2
-
RCs + 1
R
Hence
Hence
Note that
Similarly,
The transfer function Eo(s)/Ei(s) can be given by
B-4-17. Then
Define the voltage at point A
as ek.
1
Define the voltage at point B as
eg.
Noting that and K 3> 1, we must have EA(s) =
EB(s)
Hence
frm w h i c h we obtain
8-4-18.
The voltage at point A is
or
The voltage a t point B is
Since
e =- 1 ~
(e + 2
@Dl
i
Thus, equating Equations (1) and ( 2 ) we obtain
Define the displacenent of midpoint between kg and b as x3. W-19. Then the equations for the system are
Using the force-voltage analogy, t h e preceding equations may k CcmVerted to
Rtle last three equations can be modified to di,
ti, -
1 dt
i2)dt +
C3
di2
+
R(i2 ~
L
2
R(ig
- i2)
=
-
i3) +
-
-
(i,
( i 2 - il)dt = 0
- ig)dt
From these thee equations we ean obtain the analogous electrical system as shown to the right,
(il
-
i
3
)dt = e(t)
B-s-20. Define the cyclic current in the left loop as il and that in t h e r l g n t Loop as i2. Then t h e equations for the circuit are
which can be rewritten as
Using the force-voltage analogy, we can convert the last two equations as follows:
From these equations an analogous mechanical system can be obtained as shuwn to the r i g h t .
////i=~/////
CHAPTER 5
By substituting t h e given numerical v a l u e s i n t o
B-5-1.
WE!
obtain
R-5-2. We s h a l l solve t h i s problem by using ttm different approaches: one based on t h e exact method and the other based on the use of an average resistance. (1)
Solution by the exact method.
For t h e liquid-level system we ha-
Py substituting C = 2 m 2 , Pi = 0.05 n 3/s, and p = 0.02 equation, we obtain
k t
As-
fi = x.
Then H = x 2 and dII = 2x &,
that at t = t
1
=\
So we
t k level reaches 2.5 rn.
*, dk
(
have
Then, t
1
is obtained as
.1
* -&
= - ZM)xly+1WO
Jfi I
=
lfi into t h i s last
- 200( 6- 1) t moo
dx 5-Zx
1
'"(5
- 2x)l /
Solution by use o f an averaqe resistance. (21 average resistance I? is obtained from
~ e f i n e h = H -1. system,
lhenqi
= a i - 0.02
Since Q = 0.02
and qo=h/R.
5, the
For the liquid-level
Which can be rewritten as
2
substituting C = 2 m , R = 129 s/m2, this l a s t equation yields
and qi = 0.05
- 0.02
3 = 0.03 m / s i n t o
Taking Laplace transforms of both sides of this l a s t equation, w e obtain
258[sH(s)
- h(O)]
3. B 7 + H(s) = 5
or
Solving t h i s q u a t i o n for H ( s ) ,
The inverse l a p l a c e transform of this last equation gives
Assume that at t = t
1'
Rewriting,
h ( t ) = 1.5.
1
The value of tl can be determined frm
So
we have
his solution has $een obtained by use of an average resistance. B-5-3.
The equations for the Piquid-level system are C dh
1 1
Since Rl = hl/ql
=
(6 + q - 6 - q,)dt
and R 2 = h2/q2, the system equations
can be rewritten as
From Equations (1) and (2) we obtain
By differentiating Equation ( 2 ) with r e s w t to t, we get dS2
+ - -1 - - - dh2
-
R2 d t
I
RL d t
By eliminating dh /dt from Equations ( 3 ) and (41, we obtain 1 dZh2 R C R C 1 1 2 2 &2
t
(RICI + R C
2 2
&2 dt
+h 2
=liZq
Substitution of h2 = RZq2 into this last equation y i e l d s
R C R C 1 1 2 2
dZq2 + dt2 +(RIC1
Hen= the transfer function of the the output is g i m by
R C ) -dq2 + q 2 = 4 2 2 dt
system when q is the input and q2 is
E-5-4.
The equations for the system are
Thus, we have
From Equation (1) we obtain
ClsHl(s) =
2 [H2(s1 R1
From Equation ( 3 ) we get
H~(s)1
By adding Equatfons (I), (21, and (31, and taking the Laplace transform of the resulting equation, we obtain
By substituting Equations ( 4 ) and ( 5 ) i n t o Ekqaticm (6), we get
Since H3(s)
= A 3 f ~ 0 ( ~ ) ,this last equation
can tR written as
This is k ? transfer ~ function r e l a t i n g Q O ~ S )and Q~(s) =
21-5-5.
For t h i s system
Hence
(+=27c
Q1
= -0.005Jii dt
Assume that .the head moves d m frm H = 2m to x for the 60 sewrid m o d . Then
5:
~ . ' H
dR = 4.005
-
which can be rewritten
as
Taking logarithm of both sides of this last q u a t i a n , we obtain 2.5
3.5089 log x = 1wI0 10
x = 1.652 m
B-5-6.
Frm Figure 5-32 we obtain
Using the e1ect;rical-liquid-1-1 analogy given below, e q a t i d n s for an analogous electrical system can be obtained.
C (capacitance)
R (resistance)
Analogaus equations for the electrical system are
Based on Equations ( 5 ) through (81, we obtain the analogous electrical system shuwn below,
B-5-7.
The equations for the liquid-level. system of Figuse 5-20 are
Using t h e table of electrical-liquid-level analogy shown in the s~lutian of Problem B-5-6, we can obtain an analogous electrical system. The analogous electrical equations are
Based on Equations 15) through ( 8 ) " r= system shown on n e t t page.
03tzin t h e analogous electrical
B-5-8. -
pv = mR,irT
In this problem p = 7 X 205
+ 1.0133 x 105
= 8.0133 x
lo5 ~
/ mabs~
The mass m of the air in the tank is
If ttre temperature of canpressed air is raised to 4bec, then T = 273 + 40 = 313 K and the pressure p beccanes
= 7.547 x 105 ~ / m 2gage = 7.695 kgf/m2 gage
= 109.4 1bf/in.*
B-5-9.
gage
mte t h a t
where q is t h e flow sate through the ~ 1 ~ and x 3 is given by
Hence
from which we obtain
PJs)
Pil s ) For t h e bell-
-
-
1 RCS + 1
and spring, we haw the following equation:
Ap,
= lot
The transfer function x(s)/Pi (s) is then given by
B-5-10.
Note t h a t P1
= 0.5 x 105 NJm 2 gage = 1.5133 x 105 N/m 2 abs
= 0 N/m 2 gage = 1.0133 x 105 N/m 2 a b s P2 If p2 3 0.528pl, the speed of air flaw is subsonic. So the flaw throughout the system is subsonic. The flow rate through the inlet wive is
The flow rate through the outlet valve is
Since bokh valves have identical flow characteristics, we have TT = K = R. 1 2 The equation for the system is
A t steady state, ve have d p i d t = 0 and t h i s last equation k c o m s
5 2 5 2 = 1.2633 x 10 N/m abs = 0.25 x 10 N/m gage
B-5-11.
For the toggle j o i n t shown in Figure 5-37, w e have
Hence
Neglecting the Mgher-clrder terms, a linearized equatlcm for the system can $e written as
Q
where
f(E1 =
- f(E) = a(H - A) C
f ( 4 ) = 0,2
Thus, a linearized m a t i o n becames
A linearized equatian for the system is
z
- z- = a ( x - 2)
where 2 = 2, E = 20, and
Thus, a linearized equation becomes
A linearized mathematical d e l is
where Z = 11,
y
- = 356, and
= 5, z
Thus* the l i n e a r l z d equation is
Define the radius and angle of rotation of the pinion as s and 8, respectively, Then, relative displacement between rack C and pinion B is re. 5
Relative displacement between rack A and rack C is 2re and this must q u a 1 displacement x, Therefore, we have
Since x = r0 + y, we obtain
B-5-17.
The heat balance equations for the system are
c1ae1 =
(U
C2dG2 = (q,
- ¶,)at
- s,)dt
Noting t h a t
Equations (1) and (2) can be modified to
from which we get
By eliminating Bl(s) frcm the preceding two equations, we obtain
Thus the transfer function
E b Z ( s ) f l l ( s ) can be given by
(1 (2)
B-6-1. &fine the current in the circuit as i (t) where t q u a t i a n for t h e circuit f o r t 9 0 is
Since the capacitor is not charged for t equation becaes
< 0, the
Since
we obtain
The i m f s e Laplace transform or E,(o)
B-6-2. -
gives
The equation for the circuit is
2 0.
The
laplace transform of this
Since q(0) = 0 and i10) = {(o) = 0, the Laplace transform of this la* -ation gives
S
LS~Q(S)+ RsQ(s) + -1Q ( s ) = E C
or
Since khe'current itt) is d q ( t )/atr we have
The current i ( t ) w i l l be o s c i l l a t o q if t h e t w o roots of the characteristic
equation
are cmmple~mnjugate.
If two roots are real, then the current is not osci-
Ilatory. Case 1
(Two roots of the characteristic equation are complex conjugate):
For this case' define
Then
The inverse Laplace transform of I ( s ) gives
The current i(t) approaches zero as t approaches infinity. Case 2 (Two rmts of the characteristic equation are r e a l ) :
For this case define
Then
L
Its) = E
(S
+
1
+ b)
The inverse Laplace transform OF I ( s ) gives
N o t i e that
Hence -E
-
1=
L b-a
The current i ( t ) can thus be given by
B-6-3, Referring to the circuit diagram shown t o the right, we have
Z1 = R1
"
Z
= R
4.-
Z1
7
C2s
Hence
Eo(s 1 Ei(5)
1'
Z2 +
2'
-
KP2s + 1 ( E +~ R ~ ) C ~+ S1
Next, we shall find the response eo(t ) when the input e . (t) is the unit step Z function of magnitude E. Since 1
.
the Inverse Laplace transform of E0(s) is
which gives t b response to t h e s t e p input of magnitude E
i'
~ 6 - 4 , The system equations are -
which can be rewritten as 0
bly
t
+ kly = k1xi + blxo
Noting that x,(O-) = O and y ( O - ) = 0 , by t a k i n g the two equations we obtain
b; -
transform of these
By eliminating Y ( s ) from the preceding two equations, we get
Simplifying ,
or
?he response
of the systm to xi(t) = Xil(t) can be obtained by taking the
inverse Laplace transform of
as fellows:
m.
The equations of motion for the system are k (x. 1
1
-
xo) = b (& 2
0
- y)
Rewriting these equations,
Noting that x ( 0 - ) = 0 and also yl(0-1 = 0, 2-transforms of these t w o equations becane
Elintinatfng Y ( s ) from the l a s t two equations, we obtain
which can be simplified as
Hence
Since the i n p u t xi(t) is given as
we have
Tne response Xo(s) is t h e n obtained as
Since
we have
Hence
B-6-6. -
First note that
Since R2 = 1.5 Rlr C = C 2
1
, and
R C = 1, we obtain 1 1
and the transfer function %(s)/Ei(s)
becomes
Since the input; ei(t) is given by
=0
elsewhere
we have
Hence, the response eo(t) can be obtained as follow:
Since
we obtain the respanse eoIt) as follows: e (t) = Ei (0.4 e-2t 0
- 0.4
e- (L/3)t
+
1)
lqarithmic d e c r e n t = In
'to = l I Xl
Solving for
By
xa
n
Xn
-
5C.dnT
5,
substituting n = 4 and xo/xq = 4 into t h i s last equation, we obtain
Noting t h a t Wn =@
and 2
5% =
b/m. w e find
and
E-6-8.
The system equation is
(rn+ 2 ) G + & + ) ~ = p
Substituting the given ntm#rical val~esm = 20 kg and p = 29 = 2 x 9.81 N into this l a s t equation, w e obtain 2 2 2 + b j t + h =2~9.81
At steady state l a S S = 2 x 9.81
From Figure 6-48 (b)p
,x
= 0.08 m.
Xss
--
'Thus
2 x 9-81 = 0.08 k'
Solving for k, we obtain
Since R
=&
=@=
3.34 r a d / s ,
b 2 2 U n -3
we obtain
E-5-9. -
FOP the x direction, the equation of motion is
For the rotational motion of t k pendulum,
Rewriting the preceding tm mationst
+ r n g(cos 0
Simplifying,
iZ + sin % 51 1 s i n 8 = - m2g 1sin
0
.*
x +
9.e ml
+
coseb3-
m2
sin 8
m2 ml
+
b2 +
m2
x=O
2k
"1
+
m2
Thus the equations of motion for t h e system are
;;+
mz +
1
0.
m21
cosQ0-
m2
"1
+
sin O
i2 + 2k
m2
"1
+
x = O
m2
The equation of mtian for t\e s y s t e m is
=.a
1 kg, k = 100 ~ / m ,p ( t ) = 10 &(t) N, x(0-1 = 0.1 m, and G(0-) = 1 By substituting the given n m ~ i c a vl a l v e s into the system equation, we obtain
w
e
rn =
m/s.
Taking the
a- transfarm of
this l a s t equation gives
Solving for X(s) gives
The inverse Laplace transform of X ( s ) gives
x(t1
sin =lo
10t+ 0.1 cos 10t
-
L e t us d e f i n e another impulse force to stop the nmtion as A 8 (t T), *ere A is the undetermined magnitude of the impulse force and t = T is the un'fhm, determined i n s t a n t that this impulse is to be given to t h e system.
t h e equation of motion for the system when the is
The
L- transform of
tkfo
impulse forces are given
this last equation gives
-
[ m s 2 + k ) ~ F s ) 1 + A e-ST
Solving for X(s) ,
me inverse taplace transform of X(s) is
If the mation of mass m is to be stopped at t = T, then x ( t ) mst be identically zero for t k T. Note t h a t x ( t ) can be made i d e n t i c a l l y zero for t > T if we choose
Thus, t k Xmti0n o f mass m can be stopped by anokher impulse force, such as
1
2
The system equation is
2 + b;t The
&-
=
d(t),
X(O-)
transform 0 5 this q u a t i o n is (ms2 + bs)X(s) = 1
Salving for X ( S ) ~we get
me response x ( t ) of the system is
= 0,
G(o-I
=
o
The velacity G(t ) is
I e-(b/m)t $(t) = m
The initial velocity can also be obtainl;d by use of the initial value thwrm.
'1Zlc nwrnenL u1 itjestla of t . 1 ~wndulum about The a n g l e o f rotation of the the pivot is J = m i 2 . pendulum is 8 rad. D e f i n e the force t h a t a c t s on the pendulum a t the time of sudden stop as F(t ) Then, k h e torque that a c t s on the pendulum due to the force F(t) is F ( t ) K cos 8. The equation for the pendulum system can be given by JJ-6-13.
.
We shall linearize this nonlinear ajvaiirn 4 erruning angle 9 is small. (Although @ = 20" is not quite small, the resulting linearized q u a t i m i will give an approximate solution.) By approximating cos Cl f 1 and s i n 0 f 8, Equation (1) can be written as
F(f)
3 J
p9
Since the velocity of the car at t = 0- is 10 m / s and the car stops in 0.3 s, the deceleration is 33.3 rn/s2, By a s d n g a constant acceleration of magnitude 33.3 m/s2 to act on the mass for 0.3 seconds, F ( t ) may & g i m by
Then, Equation ( 2 ) may be written as
Since
1 = 0.05
m, this l a s t equation becmes
1
Taking Laplace transforms of bath sides o f this last q u a t i o n , we obtain
b
where we used t h e i n i t i a l conditions t h a t 8(0-) = 0 and B(0-) = 0. Solving Equation ( 3 ) for @ ( s ) ,
The irrverse Laplace
transform of Qls) gives
-
<
(4)
Note t h a t l(t 0-3) = O for 13 ,< t 0.3. Let us a s s m e t h a t at t = tl, 0 = 2 0 8 = 0.3491 rad.
Tbenbytent a t i w l y assuming t h a t tl occurs before t = 0.3, we solm the following equation for tl:
which can
3 x simplified to
The result is
Since tl = 0.0327 < 0.3 our aasrenptim was correct. The terms involving l(t - 0 - 3 ) i n E q u a t i o n ( 4 ) d o n o t a f f e c t t h e v a l u e o f tl. It takes approximately 33 ms for the pendulum to swing ZOO.
B-6-14.
The equation of motion for t h i s system is
By substituting t h e numerical values for M, b, k, and g i n t o this equation, we obtain
By taking the Laplace transform of t h i s l a s t equation assuming zero initial conditions, we obtain
The inverse Laplace transfo m of X I S ) gives xdt) = 0-04905(l
- e-1*6666t cos 5.5277t - 0.3015
N e x t , we shall obtain the response Note that: X ( s ] can be written as
clurve
e-l-666fit s i n 5.5277t)
x ( t ) versus t with MATUB.
Now d e f i n e
nm=[Q 0
1.5351
R possible MATLAB program to obtain the response curve is given belcrw.
% ***** MATLAB program to solve Problem B-6-14
num = [O O 1.6351; den = [ I 3.3333 33.33331; step(num,den) grid title('Respoase ~(t)') Idabel('t sec') ylabel('x(t)')
*****
The resulting response curve x ( t ) versus t is sham k I m .
R-6-15.
The equation of motion f o r the system is
By substituting the numerical values of m, kl, k2, and bl i n t o this equation, we obtain
Laplace transforming this equation, we get
Solving this equation for X(s), X(s)
=
sx(0) + s2
$10)
+ 4s
+ 4x(O) 4
Since x(0) = 6-05 and g(0) = 1, X ( s ) becanes
16
The inverse Laplace transform of X ( s ) gives
x ( t ) = 0.05 e-2t m s 2
6t
+ 0.3175
e-2t sin 2. f3 t
.
This equation gives the time xespQme x(t) The response m m e x(t1 versus t can be obtained easily by use of NATLAE. Noting t h a t X(s) can be written as X(s) = 0.05~2 + 1.2s s2+4s+16
2 s
we may define
den = [I
4
163
and use a step camand. "ke following MRTLAB program will generate the reqmnse x(t) versus t as s h m below. % ***** PvlATltA3 pro_nfam to solve Problem B-6-15*****
mrn = L0.05 I .2 01; den=[I 4 161; st ep(num,den) ~d tit1e(lriesponse x(t)3 xlabelrt secq> ylabel(k(t)')
In this system F is the input and x2 is the output. B-6-17. Figure 6-57, we obtain the following equations:
Laplace transforming these tm -atimsr we obtain
By eliminating XI(s3
From
assuming zero i n i t i a l conditions,
f r a these two equtltionsr we get
simplifying this last equation, we get
from which xe obtain
By substituting numerical ~ l u e sfor ka, k2, blr and b2 into this l a s t equation, we obtain
Since the input F is a step formof ZM, we have ~ ( s = ) 21s. be obtained from
x2(s) can
The inverse Laplace transform of X ~ ( S ) gives
The response curve x 2 ( t ) versus t can be obtained with MATLAB as follows: F i r s t note t h a t
Then, define
and use a step comnand. The following MATLAB program will yield the response curve x 2 ( t ) - The resulting response curve is shown in the
figure below. % ***** MATLAB program to solve Problem B-6-17 *****
num = [O 0 0.83; den=[l 6.4 81, step(num,den) grid titl$Response x(t)3 xlabel('t sect) ylabel('x(t)')
Referring to the figure showr to the B-5-18. right, w e ham =-J-+C25,
Zl(d
Z2(s) = Rl +
R2
1 -
I
ClS
Hence
R2
Zr(s1 =
R ~ C ~+ S1
Z~(S= )
RrCls + 1 CIS
Thus
By rmbstituting the given nrrmerical wlues for Rl, R 2 , C1, and C2, we obtain
men ei (t1 = 5 V ( S*p
The reqmnse curve e,(t) MA'lTAEl program
input )l is applied to the system, we have
versus t can be obtained by entering the follmj-lng
i n t o the mqmter.
The resulting response curve eo(t) versus t is shown below.
R e m e BdU
I Sec
Note that
Notice t h a t the response curve is a sum of two exponential curves and a step function of magnitude 5.
CHAPTER 7 B-7-1.
The equation of m t i o n for the system is
whesex1(0)= 0 and
G(OE = O .
The laplace transform of t h i s equation is
By substituting the given numefical
values of
k, P, and c3 into this l a s t
equation, we obtain
Solving for X ( s ) , X(s> =
-- -5
10
(s2 + 100)(s2 + 4 )
- -1
2
96 s2
+
4
10 s2
+ loo
The inverse Laplace tsansfarm of X(s) gives the response x ( t ) x(t) =
B-7-2. -
5
(sin 2t
-'5 sin
I&)
The equation of motion for the system is
By substituting the *ven
nwrtlerlcal va3ues into t h i s equation, we get
+ 24;; + 2 0 0 ~= 5 sin 6t
or
--
+ 123;+ fOOx = 2.5 s i n
6t
73king the Laplace transform aE t h i s last equation, we o b i n 2
(S
t 12s
+
100)X(s) = 2.5
6
s
Solving for ~ ( ) ,s
-1-6
The i m s e laplace transfarm of X ( s ) gives x(t) =
3 e4t 1856
EJS)
Ei(d
s i n 8t +
-C
eat 464
1 s --
R+-
1
txe
8t +
sin 6t
-
m s 6t
1 RCs + 1
Cs
Hence
For the input e. (t) = E sin L3 t, the steady-state output e ( t )is g i m by 1 i 0
*-4.
The equations of motion for the system are
which can be m i t t e n as
I
Since we are interested in the steady state behavior of the system, we can assume that all initial conditions are zero. Thus, by assuming zero initial conditions and taking the Izlpkace transforms of the last two equations, we obtain
(m2s2+ bs + K)x*(s) = (h+ k)x1(s)
Ran Equation (2) we have
Substituting Equation (3) into Equation (1) and simplifying, we obtain
The transfer function Gl(s) between X1( 3 ) and P(s) can thus b obtained as
Hence
fram Mch we obtain
The steady-state output i ( t )a n , therefore, be given by
[
xl(t) = G ( j w ) P s i n dt + 11
1
L l G (jW)
to Equation 131 we have the transfer function G2 (s) and P ( s l as fallows:
NgXt, referring
between X
2
(8)
Hence
The magnitude and angle of G ( j W ) are given by 2
The steady-state output x (t) can be given by 2
xz(t) =
I
I ~ , ( j w ) P sin
ktI&+
B-7-5.
tension =
2 mu s = 0.1
X 6.28
2
x
1 = 3.94 M
Tne tension in the card is 3.94 N. 'rhe maximum angular speed can be obtained by salving the f o l l d n g q u a t ion for #
.
?he
maxfrmrm angular speed that can bE? attained without breaking the card is
1.59 Hz-
73-74.
F m the d i a g m shown centrifuqal force gravitational force
Solving for W
B-7-7. -
, we
k l o k r , we
obtain
nUJ2r - 0.15
,
m9
0.2598
h
obtain
The equations of m t i o n for the system are
MZ + & + m = m t d r
sinut:
Tf 10 % of the excitation form is to 1 s transmitted to the foundation, the Thus tranmnissibility must be q u a 1 to 0.1.
Since 5 is desired to be 0.2, we substitute J = 0.2 into this last equation*
Solving for
', we find I
86
Nating t h a t & > O ,
velnuat have!*
men
= 22.28.
k = 17.7 x 103 N/m
The amplitude of force F transmitted to the foundation is t
Rewriting
By substituting x
-y
= z into this last equationr we obtain
.
W
mz+&+kz=-my
The Zaplace transform of this equation, assuming zero initial conditians, is
Far the sinusoidal input y = Y sin&)
t r
The steady-state amplitude ratio of z to y is
Thus, the amplttude of sinusoidal dinplacetnent y of the base is equal to t h e amplitude of the relative displacemer~tz. If w<< W , we have
n
So the acceleration
of the base is ptopartional. to z.
B-7-9. Define the displacement of spring k2 as y . motion for t h e system are
'Then the eqmations of
The f o r e f(t) transmitted to the foundation is
By taking laplace transforrrrs of Equations (1) and (21, assuming zero i n k t i a l conditions, WF obtain
By eliminating YCs] fran t h e last two equations amd simplifying, w e get
The Laplace transform af Equation (3 5 is
So
w e have
and
The farce transmissibility TR is
The amplitude of the force transmitted to the foundation is
B-7-10. Define the d i ~ p l a - ~ k o f the t o p e n d of springk as z. 2 the equations of mtim for the system ape
where p = P s i n w t ,
Then
Rewriting these equatfom,
Laplace transforming these t w o qtatims, assuming zero i n i t i a l m d i t i o n s , we obtain
Eliminating Z(s) fran the last tm equations and simplifying gives
So
the transfer function ~(s)/P(s)is obtained as
W E motion transmissibility TR is
I
I
?RIPvibration amplitude X( jW ) of '.he machine is
B-7-11.
The equations of matian for the system are d + & + k x + ~ ( x - y ) = p = ~ s i n ~ t
Laplace tsansfoming these t w o equations, assuming zero i n i t i a l wnditicms, we obtain
Eliminating Y ( s ) E m the Last t w o equations and simplifying, we obtain
Note t b t if/-=#,
then X ( j W ) = 0.
Since
we have
By substituting
%/m,
= U 2 Into this last equation, ve get
Hence
I-I
=--
(Y(JW)
ma LrlL
ka "a-
The amplitude of dbration
05 mass
p
p ka
m,
m, is P&.
8-7-12. Assuming ma11 angles Q1 and system rriay be obtained as fo Llm:
e2 the
equations of nation for the
Rewriting these equations, we obtain
which can be simplified to
To find the n a t u r a l frequencies of the free yibntionr we assume the motion to be harmonic. That is, we assume
€I2 =B
= A sin W t ,
u
1
sinWt
0-
= -A&
sin
t,
e2 =
-B&*
sinwt
Substituting the preceding expressions i n t o Equations (1) and (2), w e obtain
Since these equations must be satisfied at all times and s i n d t cannot be zero a t a l l times, the quantities in the brackets must be equal to zero. Thus
For constants A and B to be nonzero, " ,determinant of t h e coefficients of Equations (3) and ( 4 ) must be equal t o zero, or
This determinant equation determines the natural frequencies of the systemEquation (5). can be rewritten as
This last equation can be factored as follows:
Thus
The first natural frequency is w l (first W e ) and the second natural frequency is L3 ( second mode) . 2
=J=,
A t the first natural freqwncy W
we obtain from
Equation
(3) the following expression: kit2 -
[ ~ o t ethat w e obtain the same .result from Equation ( 4 ) . ] Thus, at the first nrode the amplitude ratio A D k a m e s unity, or A = B. This means that both masses move the same munt in the same direction. This mode is depicted in
Figure ( a ) below. At
the second natural frequency & = LJ2 we obtain from Equation (3) ka -
- A- E3
mll 2
ka*
-9-+kaLL---I mlt2 P m1t2
ka2
- - - m2 m1
m212
[we obtain the same result fran Equation (4). 1 At the second mode, the mis means t h a t amplitude ratio A D becomes - m /m or A = (m2/ml)~. 2 1 masses mave in the opposite direction. This mode is depicted in Figure (b) shown b e l o w .
-
3-7-13.
T h e ~ a t i o n s c E m t i m f o rthesystemare
*.
mx =
-
(x + f l s i n
For small angle B, we have sin 8 m a t i o n s becnne
O)kl
-
(x
-I2sin €Ilk2
+ 8 and cos +3
1 and the preceding two
Notice that if flkl = [2k2t then the coupling terms become zero and equations become independent. However, in this problem
~I?Q
Therefore, mupling exists between Equation (11 and Equation (2). To f i n d the natural frequencies f o r the system, assume the following harmonic motion: x =A s h u t r
B
=:
T3
sinWt
Then, frm Equations [I) and (2) we obtain
For amp1itudes A and 3 to be nonzero, the determinant of the coefficients of Equations 13) and ( 4 ) must be equal to zero, or
Thfs determinant equation determines t h e n a t u r a l frequencies of the system. Equation ( 5 ) can ke rewritten as
which can be simplified to
Notice that this last: equation determines the natural frequencies of the system. By substituting the given numerical values f o r l l , 1 2 , kl, k2, m, and J into Equation (6), we o l ~ t a i n
a4- 1 4 0 W 2 + 3920 = O Solving this q u a t i o n for W *, w e get
Hence
The first natural frequency is Cdl = 6.2205 rad/s and t h e second natural frequency is u = 10.065 rad/s.
2
To determine the modes of vibration, notice t h a t from Equations (3) and (4)
have
By substituting the given numerical values into this last equation, we obtain
For the first mode of vibration ( &J = 6.2205 rad/s) t h e amplitude ratio A h 1 becomes as follows:
Notice that the ratio of the displacements of springs k and k are 1 2
'Ihe f i r s t mcde of vibration is sham in Figure ( a ) on next page.
Figure ( a )
For the second mode of vibration ( W = 10.065 rad/s) the ampl&ude 2 ratio A/B k c m e s as
Hence the ratio of the displacements o= springs k and k becomes 1 2
The secand mode of vibration is sharn in Figure (b) below.
Figure (b)
B-7-14. The system shown in ? i p e 7-50 is a p i a l case of the system shown in Figure 7-32 (Problem A-7-14). By defining kl=k,
k2=2k,
k3=kr
ml=m,
m2=m
Equations (740) and (7-41) become as follows:
Also,
that satisfies Equations ( 7 - 3 8 ) and (7-39) bcmes as follows:
X)ef ine
By substituting Ld12 into Equation (I), we obtain
Similarly, by substituting uZ2 into Equation (13,
[We get the sarrte result if we substitute dl2or Hence we have
get
aZZ into
Equation ( 2 ) , ]
which means that: in the first mode of vibration (with frequenq dl), the ma5ses mE in the S a m? d i mtion by the s:u n e m In the second mod .brat ion ( w i t 1I E r e qency w 2) r tht2 msse in opposite dir ; by the same amount. Figlures ( i3 ) and l u ) s r i m on next page d e p l c t the first mode of vibration and second mode of vibration, respectively.
Figure ( a )
B-7-15.
The equations of motion for the system are
Substituting the given numerical mlues into these two equations and simplifying, we have
To find the natural frequencies of the free vibration, assume that the motion is harmonic, or
Then .*
x=-A~*sinaIt,
.I
y=-Bw2sindt
If the preceding expressions are substituted into Equations (1) and (2), we obtain (-A
a2
+ II.A
- B) sin
=
o
Since these two equations must be satisfied a t all times and s i n u t cannot be zero a t all times, w e must hsve
Rearranging,
For constants A and B to 'b? nonzero, the determinant of the coefficient matrix must be equal t;o zero, or
which yields
which can be rewritten as (
t~ 2 -
7.2985)C d2
-
13.7016) = 0
Hence r
The frequency of the first mode is 2.7016 rad/s and t h e frequency of t h e second mode is 3.7016 rad/s. Frun Equations ( 3 ) and (4!, we obtain
By substituting
uI2 = 7.2985
i n t o A D , we obtain
Similarly, by substftuting w 2Z = 13.7016 into A D , we get
in the same direction, while a t the second mode of vibration, two masses move i n opposite d i m -
n m Hence, at the first m o d e of vibration, two masses r
tions.
Next, we s h a U obtain the vibrations x ( t ) and y(t) subjected to the given initial conditions. Laplace transforming Equations (1) and ( 2 ) ,
Substituting t h e given initial condition3 into the preceeding t w o equations we get
~liminatingY(S) frm Equations ( 5 ) and ( 6 ) ,
which can be simplified t o
Similarly, we can obtain Y ( s ) as follows:
To obtain the responses x(t) and y ( t ) to the given initial conditions, as follows:
we rewrite X(s) and Y ( s )
Possible MATLAB programs to p l o t x ( t ) and y ( t ) , respectively, are given next. The resulting plots x ( t 1 versus t and y ( t ) versus t are shown on next page. % ***** MATLAB pro_mrnto obtain vibration x(t) num = [0.05 0 0.5 0 01.
d e n = [ l 0 21 0 1001; t = 0:0.05:30; x = step(num,den,t);
plot(t,x) grid title('Vibration x(t) due to initial conditionds') xlabel('t sec') ylabd('x(t)')
*****
****#IM A n A B progranl to obtain vibration y(t) *I***
,
num = [0 0 0.5 0 01; den=[l 0 21 0 1001; t = 0:0.05:30; y = step(nurn,den,t);
I
pIot(t,yl grjd title('Vibration y(t) dare to initial conditionds') xlabel('t sec') ylabel(ly(t)') t
-
B-7-16. All necessary derivations of quatians for the system are given in Problem A-7-16. The e q u a t i ~ n sfor the system are
Referring to Equation 17-54) we have
Case ( a ) :
For the initial condition..; x(0) = 0.2807,
G(o)
-
y ( O ) = I,
OI
~ C O = ) 0
Equation (33 becomes as fellows:
By substituting
Equation ( 4 ) into Equation (23 and solving for ~ ( s ) ,W
obtain
S u b t i t u t i n g y ( 0 ) = 1 i n t o the l a s t e y a t i o n and simplifying, we get
To obtain plots of x ( t ) versus t and
y(S)
versus t, w e may enter the The resulting plots are
following M4TLAE program into the m p u t e s . shown in Figure ( a ) .
% ***** MATLAB p r o p o l to obtain x(t) and y(t), case (a) *****
numl = [0.2807 0 01; n u d = [I 0 01; d e n = [ l 0 7.19221; step(num I ,den) hold Current plot held step(num2,den) text(2,-0.5,'x(t)') trn(3,0.3>'flt)')
I
title(Responses x(t) and y ( t ] ~due to initial conditions (a)') xlabel (It sec') ylabel (k(t) and y(t33
Case (b):
For the i n i t i a l conditions x(0) = 1.7808,
G(03 = 0 ,
y ( 0 ) = -1,
we obtain the following expressions for ~ ( s and ) ~(s):
;(o)
= 0
k MATMI3 pragram for obtaining plots of x( t ) versus t and y(t) v;ersus t is 5below. The resulting p l o t s are shown in Figure (b) below.
*A **** * MATLAB program to obtain x(t) and fit), case (b)
* * * **
numf =[1.7808 0 01; m = [ - l 0 01; den=[] 0 27.8081; step(num1,den) hold Current plot held step(num2,den) texf(l.7,l.S,"x(t)') text(3 -5,-1 .Slfi(t)? title(Responses x(t) and fit) due to initial conditions (b)") xlabel ('t sec') ylabel ('x(t) and fit)')
Case
(c]:
~ { o =r 0.5, WE
I
For the i n i t i a l conditions
;f(o)
= O,
y(o) =
-as,
obtain the followhq expressions for XIS] and Y ( s } :
+(o) = o
A
MATLAB program to obtain plots of x ( t ) versus t and y(t) versus t is The resulting plots are shown in fiqure (c]
s h u n below.
% ***** MATLAB progranr to obtain x(t) and fit), case (c)
nmml = [0.5 0 2.5 0 01 num2 = 1-0.5 O -7.5 0 Q]; den=[] 0 35 0 2003; t = 0:0.02:5;
x = stq(mm 1,den,t);
pfot(t,s'el) hold Current plot held y = step[nuM,den,t); P~o~~~,Y*~~') text(1.6,0.5,'x(t)')
text(l.1,-0.3,'y(t)3 title('Responses x(l) and y(t) due to initial conditions (c)? xlabel ('E sm') yfabel ('x(t) and y(t)')
.
*****
Rsswmes: x(l) and dl)due to hRlal oondinions (c)
1
CHAPTER 8 B-8-1. Simplified b l w k diagrms The are s h m to the right, transfer function ~ ( s ) / R ( sis )
B-8-2.
Simplified block diagrams for the system are shown $elm.
The transfer function ~ ( s ) h ( s is )
Define the i n p u t impedance and feedback impedance as Z and Z B-8-3, respxtively, as shown in the figure bel,>w. 1 2
Then
Since el
0, Laplace transKorms af voltages e.(t) and e(t) are obtained as 1
E ~ ( s ) = ZII(s) = R I(s) 3.
Hence
USO r
Therefore,
The Mntrol acbion is propartianal plus integral.
-108
-
B-8-5.
klow
.
Let us define d i s p l a m n t s e, x , and y as shown in the figure
Fur relatively small angle 8 we can construct a block diagram as s h m below.
F r a the block diagram we obtain t h e transfer function Y(s)/B(s) as folloucts:
1
I
Since in such a systm ~ a / o[ (a + b) 1 is designed to be very large -red to I, ~(s)/B1sImay be simplified to
see that the piston displa-nt y is proportional to d e f l e i o n angle 9 of the control lever. Also, from the system diagram we see t h a t fox each m L l Mlue of y , t h e r e is a corresponding value of angle dm Therefore, for each -12 angle e of the control lever, there is a mrresponding steady-state elevator angle 8. We
B-8-6
If the engine spsed increases, the sleeve of the fly-ball -mar
moves upward. This movement acts as the input to the hydraulic tontroller. A positive error signal (upward motion of the sleeve) causes the power p i s t a to move dawnward, reduces the fuel mlve apeningc and decreases the engine Referring to Figure ( a ) shorn below, a block diagram for t h e system speed. can be dram as shown i n figure (b) on n e e page.
Figure ( a )
Figure Cb) From Figure (b3 the transfer function Y(s)/E(s) is obtafned as
K 8
al
+
a2
I+-
R s
al aI t a2 bs
bs
+k
Since such a speed mtruller is usually designed such t h a t
the transfer f u n c t i o n ~ ( s ) @ ( s ) kcomes
Thus, the control action of this s p d controller is proportional-plus-intq-1.
E-8-7.
For the first-order system
the step response curve is an q n e n t i a l 50 the the constant T can k determined from such an exponential m ily. Erom Figure 8-99 the time c o n s t a n t T is 2 s . If this thermometer is placed in a bath, the temperature of f i c h is increasing at a rate of 10°C/rnin = 1/6 "C/s, or
where a is a constant, then the steady-state error can be determined as follows? Noting that
we obtain
where
Therefere,
Thus, the steady-state error is 1/3*~. For a second-order system:
A typical response c u m , when this thermwneter is p l a d in a bath held at a constant temperature, is shown below.
B-8-B.
The closed-loop transfer function of the system is
For the unit-step input, w e have
B-8-9. -
Since MP is specifimf as 0.05, we haw
or Rewriting,
S o l v i n g Ear the damping ratio
5
we
obtain
The s e t t l i n g time ts is specified as 2 seconds.
So we
have
or
Iherefore,
-8-10.
The closed-loop transfer f u n e i o n of the system is
c(s1 -
Rls)
loo s3 + 2s2 + 10s + 100
T h i s system is unstable m u s e two m l e x-conjuqpte closgd-loop poles are in t h e right half plane. To vi:malize the urtstable r e s p n s e , WE may enter t h e folrmring MATLAB program ilIt0 the cDmPut.er The resulting unstable response olrve is shown in L.h~- I C r LYU e: UIL next page. To make the system s t a b l e , it is necessary to reduce the gain o f the system or add an appropriate -nsator. am-.I=<-
--
.
num = [0 O 0 1001; d e n = [ l 2 10 lC101; t
-
0:o.1 5 ;
stvInunldqt) €?id t i t l m i t - S t e p Response of Unstable System') AabelCt sec') yla bel('c(t )')
From the chara&eristic polynomial, we find
The rise t h w tr is obtained frm
t, =
7t:-B W d
Since
Wa
=w,JG= 4 JCTZ- 3.46
we have
?he peak time
$ is obtained
as
The maximum overshoot Mp is
The settling time t, is
B-8-12.
The closed-loop transfer fun~tlonfor the system is
From this quation, we o b t s i n
Since the damping ratio
7
is specified as 0 . 5 , we get
Th;lref ore, we have 0.5(1 + KKh) =
JG
The settling tine is specified as
Since the feedfomard transfer function G I s ) is GIs)
K 2s + .L
= 1
I
-=
+3 2s
-
S
1
the s t a t i c velocity error constalt Kv is
1 28+f,+mh 3t
S
This value must be equal to or greater than 50,
Hentx,
Thus, the-conditionsto be satisfied can be surmaarized as follows:
O
From W a t i a n s (1) and (2),
wz gwk
Frm Equation ( 3 ) we obtain
If we choose K = 5000, then we get
Thus, w e determined a set of values of K and Kh a s follows: R = 5000,
Kh = 0.0198
With these values o f K and Kh, a l l specifications are satisfied.
----------
Since R ( s ) = l/s2, the output C ( s ) is obtained as
Since the system is underdampecl, C(s) can be written as
The i n v e r s e Laplace transform of C(s) gives
The steady-state error ,e,
= lim
-
e r r o r can also 6e obbined by use of the f i n a l value Since t h e error signal ~ ( s )is
?he steady-state
theorem.
for s unit ramp input is
we obtain the steady-state error eSS as
B-8-14.
The characteristic equation is
The Routh array for this equation is 3
5
6
R
For the systm to be stable, there should be no sign changes in the first column. This requires
30
- H>0,
K > o
Hence, we get the mge of gain K f o r stability to be 3O>R>O
8-8-15. Since the system is of higher order (5th order), it is easier to find the range of gain K f o r stability by first plotting the r m t laci and then finding critical p i n t s (for stability) on the soot loci. The open-loop transfer function G(s) can be written as
The MATLAB program given on next page w i l l generate a plot of the root loci for the system. The resulting root-locus plat is s h m also on n
page.
e
1
num= [O 0 Q I 2 43; d e n = [ ! 11.4 39 43.6 24 01; rIocus(num,den) Waning: Divide by zero v = [-8 2 -5 51; axis(v); (wis('squareq) gtid
title(Root-Locus Plot (Problem B-8-15]?
Based on this plot, it can. be s e m that the system is cbnditianally &able. Pill critical points for stability lie on the j# axis. To deternine the crossing p i n t s of the r m t laci with the jd axis, substitute s = ju3 i n t o the characteristic equation w h i c h is
Then
(ju)5+ 11.4(ju)ld + 39(ju1)3 + (43.6
*
~ ) ( j l d )+*I24
+
2R)jkJ
This equation can be rewritten as
By equating the
we obtain
real part; and imaginary part equal to zero, respectively,
Equation (2) can be written as
From Equation (3)
WE
obtain
By substituting Equation ( 4 ) into Equat.ion (11, we get
which can IE simplified to
The roots of this l a s t q u a t i o n can be easily obtained by use of the Ft4ZZAB prcqram given below.
The rmt-locus branch in the upper half plane t h a t goes to i n f i n i t y crosses t h e jul axis at MJ = 1.2130, icl = 2.1509, a n d w = 3.7553. The gain values at these crossing p i n t s are obtainel a s follows:
K =
-1.2130~ + 39 x 1.2130~ - 2% 2
= 15-61
for hJ = 1.2130
R = -2.1509~ + 39
X
2.1509~ - 24
= 67-51
for OJ = 2.1509
,163.56
for W = 3.7553
2
K Based
or1
follows:
13-8-16.
=
-3 . ~ 5 5 3+~39 x 3 - 7 5 ~ 3- ~24 2
the K values above, we obtain t h e range of gain K for s t a b i l i t y as me system is s t a b l e if
A MATLPlB program
to p l o t the root loci and asymptotes for the
following system
is given b l o w and the result5ng root-locus p l o t is shorn on next pageN o t e t h a t the equation for the asymptotes is
num=[O 0 0 0 I]; d e n = [ l 1.1 10.3 5 01; numa=[O 0 0 0 I]; dena = [I 2.1 0.4538 0.08319 0.0057193; r = rlocus(num,den);
plot(rsl-'1 hold Current plot 'held plot(r,'ol)
rlocus(nurna,ctena); v' [-5 5 -5 51; axis(v); axis('square7 titlerPlot ofRoot Loci and Asymptotes')
I
B-8-17.
The open-loop transfer function G(s) is
The fallawing MATLW pragram will generate a r m t - l m s plot. resulting plot is shown on n e page.
The
num=[O 0 0 I 11; den=[l 7 10 0 01; rlocus(num,den) v = 1-6 4 -5 51; axislv); &sCsquare')
grid titlflmt-Locus Plot (Problem B-8-1713 Frm the plot we find that the critical Mlue of gain R for stability corresponds to t h e crossing pint of t;le rwt I m s branch that goes to infinity and the imaginary axis. Hence, we first f i n d the crossing
frquency and then find the m r r e s p d i n g gain m l u e .
The characteristic equation for t h i s systent is
By substituting s = ju) into the characteristic equation, we obtain ( j d 1 4 + 3 ( j # 1 ~ +1 0 ( j d ) 2 +Z ~ ( j u )+ Z = O
which can be rewritten as
-
(d4
10&12
3.
2K)
4
j U (-7&J2
+
2 ~ =) 0
By equating the real part and imaginary part of this last equation to zero, respectively, we get
u 4-
C I. )
L O ~ J ~Z K += 0
Equation ( 2 ) can be rewritten as
By suketitutimg Equation ( 2 ) i n t o Equation (11, we find
-
t ~ 4 lo&
which yields
+
7k12
=0
5 stability .
Since W = is the crossing frequency vith the j W ads, by substituting w = E i n t o Equation ( 3 ) we obtain the critical value of gain K for
Hence tfie stability range for K is
R-8-18.
The angle deficiency is 180*
- 126' -
120° =
- 6oe
A lead compensator can contribute 60*.
L e t us chwse the zero of the l e a d compensator at s = -1. Then, to obtain phase Lead angle of 60 *, the pole of the cmpensator m u s t he locate3 at s = -4. Thus,
The gain K can be determined fram t h e nlagnitude condition.
A
m the lead rompensator beromes as f=rllows:
The f e d f o r w a r d transfer function is
?"he following MATLAB program w i l l generate a root-locus plot. resulting plot is shown on next page. C
num = 10 0 8 81; den=[! 4 0 01; rloms(num,den) v = [-6 2 4 41; axis(v) axis('squarel)
grid titlc('Root-Locus Plot (Problem B-8-18)') h
The
4
-
5
4
3
-
2
-
a
1
2
Real A*if
Note that the closed-loop trznsfer function is
The closed-loop poles are located at s = -1 2 j
5and s = -2.
B-8-19. The MATUlB program a i m below generates a root-locus p l o t for the given system. The :ing plot is shorn on next page.
nutn = 60 0 O I]; den=[[ 5 4 01;
, rlocus(t~urn,den) v = [-6 4 -5 51; axislv); axis("square") grid tit le('Root-Locus Plot (Problem B-8-T9)y
Note t h a t constant-7 points (0< 5<1)1 straight line having angle 8 frm the jW axis as shown in the figure k l a w , From the figure we obtain .-
sin Note also that
= 0.6
8 =
Qn
line can be defined by s = -0.7%
+ ja
.
where a is a variable (0 < a (oo ) To find the value of K such that the damping xatio J of the dominant closed-lcxn, mles is 0.6 .can b found by finding the intersection of the line s = -0.7Sa + j a and root locus, The intersection point can be determined by solving the following simtaneaus equations far a.
By substituting E q u a t i o n (1) into Equation (21,
which can be rewritten as
part and imaginary part of t h i s l a s t equation to zero, respctively, we obtain
~y equating t h e real
Equation (43 can bs rewritten as
which can b written as Of
Hence
From Equation ( 3 ) we f i n d K = -1.8281a3 + 2. 1875a2
4
3a = 2.0535
K = -1.8281a3 + 2.1875a2 + 3a = -1759.74
for a = 0.5623 for a = 10.3468 .
-
S i n the ~ K value is positive far a = 0.5623 and negative- .for a = 10.3468, we choose a = 0.5623. The requixd gain R is 2,0535. Since the characteristic equation with K = 2.0535 is
the closed-loop poles can be obtained by use of t h e following MATLdB prqram
.
Thus, the closed-Imp poles are located at
The unit-step response of the systm with K = 2.0535 can be obtained by entering the following MRTLAE program into the caputer. The resulting unit-step response curve is s h m an next page. -
-
m m = [O 0 0 2.0535]; den= [ I 5 4 2.05351; step(num,den) &rid ticEe(Vnit-Step Response (Problem B-8-19]? xlabelrt Sect) , ylabeE(Uu0utput9
B-8-20.
The closed-loop transfer funct,ion of the system is
~ ( s c 2 as + bs + ab)
-
s3 + s
+
~ ( s 2+ as
4
bs
+ abl
Since the dominant closed-1pales are located at s = -1 f jl, t h e characteristic equation must he divisib1.e by
Hence
where s = - W is the unknown t h i r d pole. Ey dividing the left side 05 this last equation by s2 + 2s + 2, we obtain 53 + K S + ~ (1 + aK + bR)s
+ (aK + bK
-
4
2R + 3)s + Kab
?he remainder of d i v i s i o n must be zero.
Kab
abK = ( s 2
-
2(K
- 23
-
Z(K
4
-
2s + 2 ) l s + K 2)
Hence we set
= 0
-
2)
Since a is specified as 0 = 5 r hy s u k t i t ~ t i n ga = 0.5 into these two equations, we obtain
By substituting Equation (1) i n t o Equation (21, w e h a w
Then, by substituting K = 2 i n t o Eguation
I l ) r
we get
2b=1.5X2 - 3 - 0
The PFD controller with K = 2 and b = 0 kcoms
Thus, t h e controller beccxnes a P'D controller. The open-loop transfer function bxanes
The closed-loop transfer fmction (with b = 0 and K = 2 ) becwnes as
follows:
The root-lacus plat for the designed system can be obtained by enterring the following MATLAB program :.nto the computer.
num = 10 1 0.52; den= [ l 0 11; r!ocus(num,den) Y=
-
[-2 1 -1.5
1-51; axistv); axis('sguare')
&rjd title(Root-locus Plot (Problem B-8-20]?
The resulting root-lmus plot i s shown on next page.
129
CHAPTER 9
8-9-3. -
A Bode diagram of the PI cont!t-oiler
A Bode diagram
is shown in Figure (a )
of the ml controller is shown in Figure (b).
F i g u r e (b)
.
59-5,
Lead ne.t:wor-k
Lag network
'Ihe
the q s k m is
The equation o f motion for
B-9-6,
A_ transfo m
of this equation, using zero i n i t i a l conditions, givfs
Hence
Notice that t h i s system is a differentfating system. For the unit-step input X ( s ) = l/s, the output 8 ( s ) becomes
The inverse Laplace transform of 01s ) gives
e(t)
=
1 ,-(k/bSt R
Note that if the mlue of l r b is large, the response B( t ) approaches a Fhlle signal as s h m in Figure ( a 3 b l o w Since 'W
" 'I
-1
we o m i n
Figure ( a ) /G(jd)
= 90O
- tan-' Wb k
The 6teady-state output es,(t)
eSS It]= -
1
is therefore given by d
X
sin( Crl t + 90'-
-1 W b -)
tan
k
Next, substituting [ = 0.1 m, k = 2 ~ / m , and b = 0-2 N-si'm into G ( j L J ) gives
A Bode diagram
of G ( j t d ) is shown below. Bode Dlaqmn (Problem0.96)
B-9-7.
Noting that
we have
B-9-8. A possible MATLAB progmn for obtaining a Bode diagram of the given GCS 1 is s h m on next page. The resul t i n g Bode diagram is ahown also art next page.
num=[O
den=[l
0 0 320 6401; 9 72 64 01;
w = logspace(-2,3,IOO);
bode(nupden,w) subplot(2,1,1); title('Rode D i a w (Problem B-9-8)')
3-9-9, A possible MBTLF_B program to obtain a Bode diagram of the given The r e s u l t i n g Bode diagram is shown on next page. ~ ( sis ) sham below.
num=[O 20 20 lo]; den=[l 11 10 03; w = lo_espace(-2,3,100]; bode(num,den,w) subplat{2,1,1); tit le(Sode Diagram (Problem 3-9-9)')
A pssible WfTI.AB program for obtaining a Nyquist plot of the g i w n ~ ( s is ) shown bzlow. Note Chat to plat G ( j W ) locus only fox w 0, we use the following command: '3-9-10.
The resulting Nyquist p l o t is shown an n.ext gage.
C
0 0 1J; d e n = [ l 0.8 1 01; w = O.I:O.1:100; [re,im,w] = nyquist(num,dqw); plot(re,im)
num = [O
Y = 1-3
3
-4
23; ~ s ( v ) axis('squate? ;
~d titlewyquist Plot (Problem B-9-10$':1 XIabeI('Rea1 Axis'] ylabelC1Lmagh i s ' )
-
Since none of the open-loop poles lie in the right-half s plane and the G I j cc) 1 locus encircles the -1 + jO p i n t twice clockwise if G( j ul ) locus is plokted from W = to d = , the closed-loop system is unstable.
-
----I------
-
--
B-9-11. A possible MATLAB program f o r obtaining a Nyquist plot o f the given ~ ( s is ) sham belowThe resulting N y q u i s t plot is shown on next Since none of t h e open-loop p l e s l i e in the right-haif s plane page. and frm the Nyquist plot it can be seen that the G { j W ) lorus daes n o t encircle the -1 + jO point, the system is stable.
n u m - [O 0 0 20 203 den= [I 7 20 50 01; w = 0.1:0.1:100; [re,im,w] = nyquist(num,detr,w);
plot(re,im) v=[-1.5
1.5
-2.5
0.5]:a~tis(v);axis~square')
gtjd titfe('Nyquist Plot (Problem B-9-1I)') d a b e l ( R d Axis') ylabel(7mag h i s ' )
A possible EaTLRB prqram for obtaining a Nyquist plot of the B-9-12. me resulting N y q u i s t plot is shown un next given GIs) is s h m below. page
num = [O 1 2 I]; den = [1 0.2 f 1 J; w = 0:0.005:10; [re,im,w] = nyquist(nuqden,v~); plot(re,im) v = [-3 3 -3 31; aXis(v 1; isfsquare') grid title("Nycluist PIot (ProbIem B-9- 12)') xlabel('Real A x i s ' ) ylabel(7rnag Axis')
-
Fran the plot, it is seen that the G ( j W ) lozus encircles t h e -1 + jO t w i c e a 8 a 1 f s =rid f r u n W = - w to W = 0 to W = & . Referring to t h e Nyquist stability criterion (see page 497). we hare
mint
N = n m k r of doclnrise encirclmmt of the -1 + jO p i n t = -2
P = n m h r of poles of G ( s ) In the right-half s plane = 2 Note that there are s3.
hn open-loop poles
0 . 2 ~ 2+ s + 1
in the right-half s plane, because
Then Z =
number of zeros of 1 + G ( s ) in the right-half s plane
Thus, there are no closd-loop p l e s in t h e right-half s plane and the closed-loop systan is stable.
B-9-13. function
A closed-loop
systm with t h e following open-loop transfer
K E(S)H(S) =
s2(~is* 1)
( T >~O )
113
is tmskable, w h i l e a closed-loop system with the follbwing open-laap transfer function is stable.
Ndte that Nyquist plots of t k s e two systems are shgvnm next page. G ( j w 1 H l j L . d ) laci s t a r t fram negative infinity on the real axis (&r = 0) and approach the origin I w = clo ) The system with the o p n - l o o p transfer function given by Equation (1) encircles the -1 + jO point twice clockwise. The system is unstable. The system with open-loop transfer funckion given by equation ( 2 ) does not encircle the -1 + jO point* Hence, this system is s t a b l e ,
.
-
C
!rT--e
B-9-14.
11
(Stable)
I
For t h i s system
~y setting K
=
1, we draw a Nyquist diagram as shown below.
Note that
The ~ ~ q i i locus st crosses t h e negative real axis at CT = -0.442. for stability, w e require
Hence
The same result can also €xobtained analytically.
Since
by getting the imaginary past of G ( j w ) equal to zero, we obtain
Solv;ing this equation for the smallest positive
=rue o f u r we obtain
L-d = 2.029 Substituting Cu = 2.029 i n t o G ( j w ) yields
~(j2-029) =
K 1 + 2.029~
( m s 2.029
- 2.029
sin 2.029)
The critical value of K for stability ran k obtained by letting G(jZ.029) = -I, or 0.4421 K = L Thus, t h e range of gain K for stability is
The q u a d n t t e term kn the denominator has the undamped natural frqumcy of 2 raa/s and t h e damping ratio of 0.25. Define the frequency corresponding to the angle OF -130 to be
Solving this last equation for dl,we f i n d W 1 = 1.491. n u s , the phase a n g l e becomes equal to -130* at C J = 1.491 rad/s. A t t h i s frequency, themagnitudemust b e u n i t y , or IG(jdl)I = 1. The required gain K can ke determined from
Setting I~(j1.49111= 0.2890K = I, we find
Note that the phase crossover frequency is at U = 2 rad/s, since f ~ ( j 2 )= - / j ~- / - 0 . 2 5
x 22
+ 0.25
:c
j2 + 1 = -90'-
9 0 .
-180~
The magnitude IG(j2)1 with R = 3.46 becomes
Thus, the gain margin is 1.26 dB. R = 3.46 is shown below.
The Bode diagram of G ( j W ) with
Note that
B-9-16.
j Cc' + 0.1 j ~ + o . 5
K We
--
TO
ZK(10j 4 + f 3 jW(~jJ+llJjLJ+l)
shall plot the W e diagram when 2R = 1.
That is, w e plot t h e Bade
diagram o f
~ o j +d I '(jd'
=
jw
( 2 j LO c l ) ( j d + 1)
The diagram is shown below. me phase crurve shows that the phase angle is -130' at w = 1 ~ d / s . Since we require the phase margin to be 50°, the magnitud j1.438 must be q a l to 1 or 0 dB- Since the Bode diagram irrdic .hat (G(j1.43811 is 5.48 dS, w e need to choose 2K = -5,48 dB, or
Since the phase r u m s lies above the -18~' line for all Ld margin is + ~ dB. s
, the gain
B-9-17. Let us use the following lead compensator:
%?.+I
G,(s)
= k O (
= R,
d ~ s + 1
Since ,K
is specified as 4.0 s-1, we have
I s + -T S
f
1 eT
IC, = l i m sK,d s+o,
Ts + 1
K = K,dK s(O.ls+l)(s+I)
-
U T s + 1 n
L e t us set K = 1 and define R c q = K.
=4
Then
N e x t , p l o t a Elode diagram of
The following MATLAB pryram prduces the W e diagram s h m klm.
num = I0 0
0 41:
den=IO.f 1.1 bodelnum,den)
1
01;
subplot(2,l , I 1; title('8ode Diagram of G(s)
= 4/{slO. Is+ J l(s + I)]')
From this plot, the phase and gain mrgins are 17 * and 8 , 7 dB, respectively. Since the specifications call for a phase margin of 45 ', let us choose
(Tkis means that 12' has been added to canpensate for the shift in the gain crossover frequency.) The maximum phase lead is 40'. Since
sin
&=
1 -u' 1
d
i s d e l ; e n ~ i r ~ cas d 0.2174-
f
(& = 40 O)
-FN
~ us t choose, inskead o f 0.2174,M t o b e
0.21, or
Next s k p is to dctcrmi rkc t l ~ emrncr rrcquencies @ = TJT and d = l/(dT) N o t c ? tlraI; t h e miximutn phase-lead angle &, or tlie f pad mrp?tisalor. occurs at ttie q e m t r i c man or thc t w o carnor frequencies, or W = ~ / ( G T ) . 'I'l~earwunt or tile n l o r l l i i c a L i o r ~ i n tile rnagrlitude curve at tu'= l / ( & T ) due to t h e inclusion or the term (Ts + l}/(dTs + 1) is
we need to find the frequency p i n t where, when the lead cmpnsator is The magnitude I ~ w ( ) jI is a d d d , the total magnitude kmnes 0 dB. we select this frequency to -6.7778 dB corresponds to &J = 2.81 rad/s. be the new gain crosswer frquency Q Then we obtain
,.
Hence
and
Thus
The open-loop transfer function becomes as
The CLosed-loop transfer function is
The following MRTLM!- program produces the unit-step response nwe as shown below, I
num=[O 0 O 3.1064 41; den = [0.0163 I 0.2794 1.263 P step(nu m,d en)
4.1064 41;
grid
titIe('Ur Jt-Step Response of Compensated System (Problem B-9- 17)') KEabel('1 wc'] ylabeI('0utput c(t)') I
Similarly, the following MATLA3 program produces the unit-ramp response
cvrve as s h m on next page.
c = step(num,den,t);
ensated System problem B-9-17)')
B-9-18. form
To satisfy the xequirments, try a lead compensator Gc(s) of the
Define
Since t h e s t a t i c v e l o c i t y error c o n s t a n t Kv isgiven as
where K = Kc&. 50 s-l, we have
We shall now plot a W e diagrmn of
The following MATUB program produces the Bade diagram shom on next page. C
num = [O 0 501; den=(I 1 01; w = lopspace(-1,2,100); bode(nrm,den,w); subpIot(Z,1,I); title(8odc Diagram of GI (5) (Problem B9-18)y
.
From this plot, the phase margin is found to & 7 . ~ ~ The . gain margin is + dB. Since the specifications call for a phase maruin of 50 O r the additional phase lead a n g l e nr 3tisfy se m r glin requirement is 42 .ZO. We may as: num pha:se leadI r q i'ed ~ to h 48'. This means t h a t 5.8 ' te for t h e :d to a shift in the gain crosswer frequency. Since Llt13
sin
-
&, =
.
*
l-ol 1 +oC
-
a
= 0.114735. F$, = 48 mrreqmnds to Note that ol = 0.15 -corresponds to&=47.65Jb.) Whetherwe c h m s e 8 , = 4 8 " o r & , = 4 7 . 6 5 7 " d c e s n o t make much difference in t h e final solution. H e n c e , w e choose o( = 0.15. step is to determine the corner frequencies W = 1SS and T) of the lead compensator. Note that the m a x d phase lead angle &, m s at the geometric mean of the two comer frequencies, or W = 1 The m u n t o f the modification in themagnitude curve at # .= l / ( L ~ T )due to the inclusion of the tern (Ts. + l ) J ( d T s+ 2 ) is The nc&
ti, = l/(o(
Note t h a t
We n e d to f i n d the Frequency point where,
when the lead rompensator
is added, the t o t a l magnitude beccmes O dB-
The frequency at which
t h e magnitude of G ~ ( ~ L L ' is S q u a 1 to -8.239 dB s kt& = 10 and 100 r a d / s . From t h e Rode diagram we find t h e frequency p i n t w h e r e N o t i n g t h a t this freI G l ( j ~ ) I = -8.239 dB m s a t k 3 = 11..4 rad/s. quency corresponds to 1 / ( K T ) , or
we obtain
The lead c a n p n s a t o r thus determined is
-
"
cc(s) = yC S f -
T 1
= Kc
s
-F
s
+ 29.4347
4.4152
where Kc is determined as
The folLowhg MATTAB program produces the W e diagram of the l e a d ccanpensator just designed. It is s h m on next p g e .
num = [ I I .325 501; den = 1 11; w = lo, 1,3,100); bode(num,aen,w); subplot(2,2,I); titIeC'Bode Diagram of G4s) = SO(O.2265~+ 1)/(0.03397s + 1)3 Tbe open-loop transfer function of the designed system is
150
The following MATTAR program produces the Bade diagram of G,(S)G(S which is shown on next page.
1
From this Bode diagram, it is clearly seen that the phase margin is approximately 5 2 " , the gain margin is + w dB, and ??, = 50 s-1; a13 specifications a m met. ~ ~ I I tS h e, designed s v s t m is satisfactory. ~d unit-responses of the Next, we shall obtain thle unitoriginal uncompensated systm and th rnsated system, The original uncompensatd system has the following closed-loop transfer function:
The closed-loop transfer function of the cmpnsated system is
The closed-lmp p l e s of the campensated system are as follows:
The MATLkB proqsam given belw produces t h e wit-step responses of the uncompensated and campensated systems. The resulting r e s p n s e cumes are shown on next page. -
nurn=[O 0 I]; den= 11 1 I]; nvmc=[O 0 1000 4415.21; denc = 13 97.3041 1088.3041 4415.21; t = 0:0.01:8;
cl = step(num,den,t); c2 = step(nurnc,denc,t); plot(t,cl,t,c2) title("Unit-Step Responses of Uncompensated and Compensated Systems')
xlabel('t Sec') ylabel.IDutputs') text( 1, I .25,'Compensated system') tm(2,0.5,'Uncompensatd system?
The MITLAB prqram given &low produces the unit-ramp response of the uncanpensated system and mmpnsated system. The response c m s are
shorn on next, page. -
nurn=[O
0 0
I]; dm=[l 1 1 01; nurnc=[O 0 0 1000 4415.25; denc = [3 91.3041 1088.jD41 4415.2 01; t = 0:0.01:8; cl = step(num,degt); c2 = step(nurnc,denc,t); plat(t,c17t,c2,t,t) titlHVnit-Ramp Responses of Uncompensated and Compensated Systems') xlabel("tSec') ylabel(lJnit-Ramp Input and System Outputs')
text(l,5,'Campensated system') rext(4,I . 5,Vncompensated system')
Notice frm the unit-ramp respanse cu f o l l m the input ramp very closely. error in fellowing t h e i n p u t can be s zero for 0 . 5 X t . (The steady s t a t e 0.C2. )
urvr
?at the rompensated systm ;he compensaW system the : 0 < t < 0.5, but it is almost in themit-rampresponse is
CHAPTER 10 R-10-1.
The differential equation for t h e systm is
This is a s-nd-rder
system.
Therefore,
WF
need t w o state variables.
Define state variables xl and x2 as fullom:
Then we obtain
The output
y for the system is simply xl.
Y =
Thus,
XI
'Ihe state space representation far this rptm is given by
B-16-2.
The system q a t i o n s are
Then we obtain
The standard skate space representation for the system is
B-10-2.
The equations for the system are k
1
(U
bl(i
- Z) = bl(k
- j)
= k2y
-
By taking laplace transforms of these t w o equal:ions, assuming the zero i n i t i a l conditions, w e obtain
By eliminating Z ( s 1 from the l a s t two equations,
which can k simplified to
J
Yo-
kAS
-
5
k2
U s i n g Equations 21) and (21, Z(s)/U(s) ran be obtaind as follows:
Define
where
Then, Equations (25 and (3) can b~ written as
Hence
Then
Equations 14) and ( 5 ) can be written as
Rewriting ,
from which we get:
1 x 2 = - -aT
This is the state q u a t i o n .
a - i X2
aT
U
F m Equation (6) we have
Frm Equation ( 7 ) we get
Hence
This is the output equation.
B-10-4.
Note t h a t
Hence
which can be rewritten as
which is w i n l e n t to
T h i s equation can be w r i t t e n as
where
13efine
men, m a t i o n (13
m
s
from which the state equation can IE obtained as
Noting that y = eo = x + b u , the output equation can k given by 0
Equations ( 2 ) and ( 3 ) give a s t a t e space representation f o r the givlen system.
Method 1:
The s y s t m differential equation can bE? written as
Canparing t h i s equation with a standard third-order equation:
we f i n d
Referring t o Problem A-10-12,
we obtain
T h ~ the r state equation and output equation are
Method 2:
Referring to Problem A-10-13,
~(s?/U(s) can be written as
where
Then we have
... z + 18;
+ 192; + 6402 = u
Ikfi n e
Then
Hence, the s t a t e space representation for the system by this approaeh fs
B-10-6.
The equations for the circuit are
~ubstitutionof ql = xlt
$ -- X ~ ' 9 2= x3 i n t o the last t w u equations yields
H e n c e we have
B-10-7.
A partial-fraction v s i o n of Y (s ]/u(s) gives
Notice that
Then, f r a the p d i n g equations for
~l(s). x ~ ( s ) ,and
x3(s) we obtain
Also,
By taking the
inverse Laplace transforms of the last four quatims, we get
and
In the standard state space xepr@sentation, we have
B-10-8.
Nuti-
The equations f o r the system are
tht from Equations (11 and ( 2 )
Also, frm m a t i o n (3)
- 3(x1 - U ) a
=XI
.. - i - U ) +
+ 3(x1
O
Thus, q u a t i n g m a t i o n s ( 4 ) and (5) and simpl!:.fying, w e obtain
Substituting xl = y into this l a s t equation, we get I.
..I
y-3i;+3jr-y=u-2G+u
['She same result can be obtained by use of Equation (10-51). ]
B-10-9.
The eigenvector
ziassociated with an eigenvalue Xi is a -or
that s a t i s f i e s the following equation :
which can be rewritten as
example of eigenvectors corresponding to eigerlvalue
[I:] --
For
Xi
=
A*
An example
B-10-lo.
Wfine
The eigenvedors for t h i s matrix can be determined by solving the following
-
equation for x.
which can be rewritten as
which, i n turn, gives x
1
= arbitrary constant
X* = 0
x3 = arbitrary ~ n s t a n t
Hence,
where a and b are arbitrary nonzero constant.
Notice that
Note that
a m linearly independent vectors. linearly independent eigenveetsrs.
The e i g e r n e o r s of matrix M B involve two
N a , define
The eigemeckors for this matrix can be determined by solving the following
equation for
2.
which can be s m i t t e n as
from which we obtain x = arbitrary canstant 1
3 = arbitrary constant x3 = arbitrary m n s t a n t
Hence,
where a, b, and c are arbitrary nonzero m s t a n t ,
N o t i e that
Thus, the eigenvectors of matrix C involve three linearly indepndent eigenvectors. L
A MATLAB pragram to obtain a state-space representation of this system is
given on next page. Based on the MATLA13 cbutput we get the EoPloKing state spa- equations:
num = [6 6 25.04 5.0081; d e n = [ ] 5.03247 25.1026 5.008]; [A,B,C,D] = tf2ss(num,dm)
A= -5.0325 -25.1026 -5.0080 1.OOOO 0 0 0 1.0000 0
B= I 0 0
C= 0 25.0400
5.0080
D= 0
B-10-12.
Referring to Equation (10-511, WP have
where
Noking that
B-10-15.
The solution of the s t a t e q u a t i o n is
-
Since x(0) = 0 and u(t) CH
-
l(t). we obtain x ( t ~=
Note that
If e ~ , ' ~ Yb
B dT
where
Thus At:
R.
=
& -l[(s1 - A ) - ~ ] Y
b.
Hence
-TI B d T = a. m b.
Alternate solution Referring to Example 10-9, we have for the unit-step i n p u t the following solution:
Since x(O) = w0 in the present case, we have w. x(tl = m jI-'(ett
W
where
and
-
and x ( t ) is given by
- IJB .*I-
3-10-16.
The solution to the s t a t e equation when u ( T ) = T can be g i m by
Note that
Hence
B-10-17.
The equation of motion for the system is
By substituting the given numerical values for al, b, and k i n t o this last equation, we obtain
where u is the input and y is the output. By t a k i n g the Laplace transform of Equation (11, assuming the ~m i n i t i a l conditions, we have
Since the input u ( t ) is specified as a step displacement of 0 . 5 m, we have
With this input, Equation ( 2 ) can be rewritten as
The following M A W program w i l l generate the desired r e p = . resulting respbnse curve is s h m belaw. I
num = [0 0.5 01; den=[[ 1 0.51; stepCnurqden)
Erid
titlerstep Response (Problem B-I 0-17)') xlabell't Sec') ylabel('y(t)")
The
This is because the Note that t h e respnse reaches zero as t increases. transfer function Y(s)/U(s) given by Equation ( 2 ) involves s in the nume-
sator.
Onit-step response: The following MATLAB prqram y i e l d s the u n i t d t e p response of the given system. The resulting unit-step respnse curve is shown kllou. r
A = [-5.03247 1
-25.1026
-5.008
0
0
E
0 01;
B = [l;b;O]; C=[O
25.04
5.0081;
D = [O]; IY*%tl=~W3CAB,C,D); pIot(t,y)
srid
titEe('Unit-Step Response (Problem B-10-IS)? xlabelrt sect) ylabel("Outputy(t)')
Unit-ramp response: Referring to pages 596 - 599, we can obtain the unit-ramp response of the system by entering a MATLAB p q a m similar to M A W Program 10-7 into the computer. Note that
x4 is the butput of the system and is the unit-ramp respnse.
ing M A W pregsam produces the unit-ramp response. ponse curve is shown on next page.
A = [-5.03247 1
-25.1626
-5.008
0
0
0 B = [1;0;0]1; C = 10 25.04
1
01;
5.0081;
0 = [O]; AA = [A zeros(3,I );C 01;
BB = [B;O];
cc = [C
03;
DD = [O]; t = 0:0.01:5; t4qtI = step(AtiBB,CCDD, Lt); x4= [O 0 0 I]*x'; pl~t(t,x4,5t)
grid t i t l e ( V n i t - h p Response (Problem 33- 10-18)') xlabelrt Set')
ylabel('lnput and Output')
The follow-
The resulting res-
CHAPTER 11 Since
8-11-1.
sin(tclRT + 0 ) = sinlrlm m s 0
+ cosd
sin 8
we obtain
~ [ x ( r c r ) =l cm e
[ s i n ~ +~ sin l
+ sin 8
1
--
-
1 - z -1 mslFT -1 22 msWT + z - ~
( m s O ) Z - ' I S ~ ~ W T ) + (sin 0 ) (1 1
-
sin 1
By defining m = k
Since x (KT
1 &zinged to
- hT) 00
X k = O
.
o 7, ~ Q ) S Y J ~ I
-
8
+
-
22-= cosLdT
2-I s i n ( W T
2f1 msWT +
*z
- z'l
msdT)
-2
- 8)
2-2
- h, we obtain
= 0 f o r k < h, the limit of sumation in
can be k = h
nus ad
d
!Zince X(z) = X l ( z ) X 2 ( z ) t
by mmparing E q w t i m s (1) and (21, we o m i n
Differentiating X(z) with
xespect
to z, ~e obtain
Multiplying lmth sides of the l a s t equation by -Tz gives ob
Thus ve have
Xo= z
(1
- e'aT
(z-1)Iz-e
-aT
3 ~ -
s - 1
Hence
The inverse z transform of X ( z ) gives ~ ( k= )L
B-11-7.
Then
X(Z)
-
can be rewritten as
-aTe-aw
Q-
1
-aT z - e -aT e
Hence
B-11-8.
Since (z
we have
X(z) = z
-
-T e ]z1)(2 eWT}
II -
x(z> =
-
-T 1-e I
-
-
Thus
B-11-9.
Referring to Table 11-1, we f i n d
By writing 6' = a , we obtain
H e n c e we have
-
I 2-1
-
I z - e-T
%king the inverse z transform of each tern of this last equatizm, we obtain
3 2 1:-1 a(k) = a k a
+
.3*.k-1
Hence
Since a = , l ' e
we ktave -k-2
~ ( k= )
B-11-10.
k(k
* 1)
k = 0 , I f 2,
...
The z transform of the given difference equation becomes
Suhstitwting. the initial data into this last equation, we get
Hence
The Ln-e
z transform of X(z) gives
B-11-11 ,
If a # 0, then = x(0)
+
1 = 1
x(2) = x(1)
+
a = 1 + a
xll)
x/31 = x(2) t a2 = 1
+
a
+
a2
x(k) = l + a + a ' + - . -
+
,k-1 = 1 - ak l - a
l a f 1, a # 0 )
=k
( a = 1)
...
(a = 0 )
If a = 0, then ~ ( k= )0
k = 0 , I,
Zr
2 The pulse transfer function for G I s ) including the z e m r d e r P hold is obtained as follaws?
-
3
.
The p u l s e t r a n s f e r function f o r t h e
systemcanberemittenas
The inverse z transform of t h i s l a s t equation gives
Clamparing this d i f fesence equation with the standard s m d 4 r d e r difference
quation:
Define state variables as follows: xl(k) = y ( l )
- h#(k)
h = b = O 0
h
0
= b
- alhO = 1
Also, define h2 = b2
- alhl - aZhO= 2
-I=I
Then the state variables; 5 I k ) = y(k) r(k)
Z X
(k
1.
+ l) - u(L)
Referring to Section 11-6, the s t a t e equation for the system can k given by
The output equation is
N o t e that a numkr of d i f f e r e n t state space represenbtions are possible far
t h e system.
~1-u-14.
Frm t;be given transfer function we obtain
Define a state variable x as
(me input Variable is u and the output ~ r i a b l eis y . ) Then we have t k following continuous-time s t a t e equation and a u t p u t equation: x = - a x + u
The discretized s t a t e epuation is o b b i n d as
where
Hence the discrete-time state space representation for the system is given by
The pulse t r a n s f e r function for the discretized system is obtained from Equation ( 1 1 - 5 8 ) as follows:
8-11-15.
Referring to Equation (11-501, the purse transfer functfm F ( z )
is given by
-
The discretized state equation is
6
where
GCT) C
and HIT) can be obtained as follows: .c
Hence
Since
we have
Thus, the discrete-time state equation is given by
where T = 0-1 s,
Theoutput q a t i o n is
Referring to Equation (11-581, the pulse t.raflsfer function for t h i s discretized system is given by
-
GITP L.
The discretized
7
and
HIT)
LI*
s-te
equation is
m be obtained as fellows:
Since
we have
Hence
Thus, the discretized state space representation o f the system becames as follaws :
where T = 0.2 s.
Referring to Equation (11-583, the pulse transfer fmctim is obtain& as fbllows:
8-11-18 The given s y s t m has no input function, but is subjected to initial conditions. It is similar to free vibrations of a mechanical system consisting of a mass, danrper, and spring. Cbnsider first. the following system:
a state
where
space representation of the system is
~ ( k =) m(k) +
m(k)
Notice that the cneffirients bg, bl, and b2 affect o n l y matrices C and
D. Next, consider the case where the system has no input (u = 0). For such a case, MATLAX? prduces a p u l s e transfer f u n c k i a n o f the fcllawing
form :
If the original system bad an input function u such that
MATLAB prcduces the pulse transfer function
The following FlA?ZAB program w i l l yield a discrete-time state-space rep1resentation wl-Len the sampling period T is 0.1 s. This p r q m w i l l alsc3 y i e l d t h e pc~Lsetransfer fmction,
num = 10 0 03; den=[] 2 101; [AB,C,DJ = tDss(num,den); [ G , q = c2d(A,B,O. 1)
G= 0.7753 -0.8913 0.08g1 0.9536
M= 0.089 1 0.0046
[n-denzl-
sQtf(G,YC,D)
num= 0
0
0 I
I
denz = 1.0000 -1.7288
0.8 I87
Eased an the MkTLAB output, the discrete-time s t a t e equation is
The pulse transfer function obtained frm the M A W output is
If the original system had an input function U, the numerator becomes nonzero. The difference q u a t i o n carresponding to Equation (1) is
The i n i t i a l data y ( 0 ) and y(l) are g i w n by
Note that the original differential equation system given by n
y + 2; + 10y = 0,
y(0) = I,
;(o)
=
o
has t h e response curve (free vibration curve) a s s h m in Figure ( a ) ,
Figure ( a ) F Sec
m e MRTLABprogram used to obtain this c u r e is given below.
num=[l 2 01; den=[1 2 101; stepCnuqden) grid title(Tontinuous-Time System with y(O) = 1 and ydot(6) 0 (Froblem B-I 1-18)') xlabel('t Sec') ylabel('0utput y(t) of Original Continuous-Time System')
-
I
The response curve (free vibration curve) of the discretized version of the system given by Equation ( 2 ) is shown in F i g w e { b ) .
k
The MATLA3 program used to obtain this
curve is given b l m .
= [I -0.7752 01; den1 = [l -1-7288 O.Sl87J; u = [ l zeros(I,50$]; k = 050; y = filter(mm1 ,den1,u); plotCky,"*,siy,'-l) uid title@iscretised System with y(0) = 1 and y(35- 10.9536 (Prob B-11-18)') xlabel('k') yEabel('Output y(k) when SampIing Period is 0.1 srx')
rmml
r-
The pulse transfer function f o r this system m y be obtained by use t h e MATUB prqram s h m on n& page.
of
-
,
num = [O 0 21; dm= [i 3 23; [rlB,c,DI = tf2ss(num,den); [G,W = c2d(&B,O I); [numz,demj = ssZtfIG,H,C,D) nllm =
0
0.0091 0.0082
den%=
E .0000 - 1.7236 0.7408 me pulse transfer function obtained here
n m G(2) =
denz
--
is
0.00912-1 + 0,0082~'2 1 - 1.72362-1
+ 0.74082-2
This result is identical w i t h the pulse transfer function obtained in PKQ~~B A-11-12. II
Taking the z transform of this difference equation, we obtain
Hence Of
The Fibonacci series can be generatd as the response of X ( z ) to the Kronecker delta input, where X ( z ) is given by Equation (1)and t h e ICronecker d e l t a input u ( k ) is defined by
following MATWIB prrqmm w i l l generate the Fibonacci series.
x = fiIter(num,den,u)
Columns 1 through 6
Columns 7 through 12
Columns 13 t h u g h 18
Columns I 9 through 24 2584
4181
6765
10946
17711
28657
Colums 25 through 30 46368
Note
30.
75025
121393
19H18
317811
514229
t h a t column 1 corresponds to k = 0 and c o l m 31 corresponds to k = Thus, the Fibonacci series is givw by
The following MATLMl program will generate a plot of t h e mit-step response up to k = 50. The resulting unit-step response is s h m b l o w .
num = [O
0 0 0.3205 -0.18&5]; den=[1 -1.3679 0.3679 0.3205 -0.18851; u = ones(l,51); k = 0:50; y = filter(nwqden,u);
~lot(ky,"~r v = [O 50 0 1.21; axis(v)
.@a
titlewd-Step Response (P~*oblem B-11-21)> I
xtabel('k') ylabel('y(k)')
The following MATLAB program will generate a plot of the unit-response of the system up to k = 16. (The sampling period is I s . ) The resulting plot is shown on next page.
I
mrm = [O 0.5 15 1 -0.1452 -0.2963 0,05283; den = [ 1 - 1.8528 1.5906 -0.6642 0.05281; k = 0:16; x=[k]; y = filter(nurqden,x);
plotCky,'~',k~,'-',k,k'-3 grid titIcfUnit-Ramp Response (Ftoblem B-1 1-22)') xlabel('kl) ylabel("y(k)3
B-11-23. below.
1
I
A MATLAB prQgram to discretize the given equation is presented (The sampling period T is 0.05 s. )
~ l i MAT& s
program produces the output as shown on next p a p .
From the MATLRB output, the discrete-time state equation is given &bow.
q ( k
+
13
1.0259
0.0504
0
1.0389
k .0259
0
-0.0006
6.0000
6.0247
-0.0006
1.10000
0
The END
-
0 0560
1.0000
0.0250
