system dynamics ogata

B-2-1.

Note that

Solutions to B Problems

CHAPTER 2

& r t j = +- S

R e f erring to Egua tion (2-2 1, we obtain

Note tha t

3 sin (5t + 45 ") = 3 sin 5t cos 45' -+ 3 cos 5t sin 45'

- 3 - - 3

Jz s in 5t + - cos 5t

f l So we have

Note that

Referring to Equation (2-21, we obtain

B-24. f( t) = 0 t <U

= c o s 2 W t r o s 3 L L I t 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) = L[f( t )~ = [%ms 5 ~ t + r n s r ~ t ) ]

B-2-5. The function f I t ) can be written as

f(t) = (h - a) l(t - a )

The Laplace transform of f(t) is

-as ~ ( s ) = [~(t)] = J [(t - a) ~ ( t - a)] = +

6

B-2-6. - f[t) = c l[t - a ) - c I(t - b) The Laplace transform oE f ( t ) is

e-a S ~ ( a ) = c - - c - e-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~es zero, t h e l imit ing value of F(s ) becomes as follows:

10 - 12.5 e-(a/5)s + 2 5 =-as l im F(s ) = lim a d 0 a+O 2 a s

( 2 . 5 e-(a/5)s - 2 . 5 .-as) = lim da

-0.5 s e-(a/5)s + 2.5 s e-as = lirn

a+O 2

B-2-8. The function f(t) can be wri t ten as

So the Laplace transform of f(t) becomes

The limiting ~ l u e 05 F(s) as a approaches zero is

24(1 - as e-*= l i m F(s) = l i r n a d 0 a 3 0 3 2 a s

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 = J'+[y!t)l = d;rGct,1 = sX(S) - x((U)

we obtain

B-2-12. Note that

Referring to Problm A-2-12, we have

B-2-14.

f(t~ emst dt = e-st at = 1 0-

Referring to Problem A-2-12, 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 + B t .-2t . 3 .-2t

- 2 5 5 - 3 5 - 2 + - + - - - - 2 - + - s + l s s + L s + l s

Rre inverse Laplam transform of f (s ) is

The inverse Lapla- transform of F(S) is

f{t) = S ( t ) 'F 2 4- 4 t

H e n c e 1 f(t) = e-t cos 3t - - e-t sin 3t 3

Hence

The inverse Laplace transform of F(s) is

H e n c e

The inverse Laplace transform of Fl s ) is

The inverse Uplacre transform of ~ ( s ) is

1 1 (t - - sin L L J ~ ~ fit) = - ld2 W

B-2-23. C

~ ( s ) = - b (1 - e-as) , - e-as s2 S

a > O

The imrse Laplace transform of F(s) is

f(t) = ct - c ( t - a ) f ( t - a ) - b l(t - a)

3-2-24. A MATUG program to obtain prtial-fraction expansions of the given function F(s) is given b l e w .

From this eamputer output we obtain

I - - -0.25 + -0.75 0.5 F(s) = + +-

s4 + 333 + 2s2 s + 2 s + l s s2

The inverse Laplace transform of F(s) is

f(t) = -0.25 e-*t + e-t - 0.75 + 0.5t

B-2-25. A possible MATLIE! program to obtain partial-fraction expansions

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 + 0.4881 i -0.3661 - 0.0006i -0.3661 + O.0006i

P =

0.2758 + 1 .go81 i 10.2758 - 1.9081 i

-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 e00275Bt sin 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 + L J 2 x = t , X ( O ) = 0, ;(a) = o

n

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

1 2[s2x(s) - sx(0) - 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 - e cos 0.5t - e -0*5t sin 0.5k

= l + 3 e-0*5t sin 0.52 - a s 0.5t

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 - 0.6 e-3t

.. - B-2-30, x * x = sin 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

sin t - - x(t) = -- I s i n 3t '3 8

--

CHAPTER 3

B-3-2. Assume that the W y of known mmnt of i n e r t i a J is turned through a small angle @ akrout the ve r t i ca l axis and then ' released. The equation of motion far the oscillation is

where k is the torsfanal spring constant oE the string. This equation can be m i t t e n as

The period TO of this oscillation is

Next, we attach a rotating m y of unknown mmnt of inertia 9 and measure the period T of oscillation. 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 . Sowe have

I\snrme that at t = tl the ball reaches the ground. 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 T- The equation of motion for the system is

from which we obtain

%y substituting nmrical values into this equation, we have

Thus

T = 1256 N-III

** J Q = - 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 tha t we apply force F to t he spring system. Then

F = k x + k ( x - y ) 1. 2

~liminating y f r a the preceding equations, we obtain

Hence the equivalent spring constant is given by

-3-7, The equations for the system are

Eliminating y from the two equations gives

The equivalent spring constant k is then obbined as @3

Next, consider the figure s h m below. Note that AABD and d C B E are similar. 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 the displacemnt of a point &tween damper bl and damper b2. (Note t ha t the same force is transmitted through the shaft.) From Equation (I), we have

In terrns of the equivalent viscous friction coefficient kt force f is given by

f = hq(y - ;) By substituting Equation (2) in to Equation (13, we have

Hence,

B-3-9. Since the sane force transmits the shaft, we have

where displacement z is defined in the figure Below.

In tern of the equitalent viscous friction coefficient, the force 5 is given by

f = - 4) (2 1

From Equation (1) we have

bl& + b2; + bgi = bl; + b2G + b3;

By substituting Equation (3) i n to Equation 1 1 ) ~ we have

Hence, by comparfng Equations (2) and (41 , we obtain

8-3-10. The equation For khe system is

$ + (lc + k + I c ) x = O 1 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~lal area of the inside of the g l a s s tube. The mathernakical ~ d e l for the sysken is

The natural frequency is

B-3-12. For a mll displacement x, the torque balance equation for the system is

** I mx(2a) = - k(T x)a

or

Tire natural frequency is

-3-14. A modified diagram for the systim s h m in Figure 3-55 is given below.

A mathemtical model for the system is given by the following t y ~ equations :

13;~ = -k,e, - k2(e1 - 82) I*

J$32 = k2(91 - 82)

B-3-15. The following twa equations describe the nation of the system and they are a mathematical model of the system.

I.

mlx = - ~ ( x - ~3 - k(G - $1 + p(t)

m2$ = -kzy - kl(y - x ) - q(j. - i ) 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

2 Noting that x = 2 y , R8 = x - y = y r and J = the three equations can b rewritten as

I 2" 1 - M R @ = - @ = 2 2 ( T ~ - T~)R

*. a = -

~ j ; * icy = TI + T*

Eliminating T2 f m the preceding equations gives

1 - M ~ + N ? + ] C Y = ml I - 2nrj; T2

2

By changing y into xt

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

where T is the tension i n the wire. (Note that since x is masurd 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 = Rlf l r we have

This Last equation is a mathematical model of the system. The natural f rquency of the system is

I l l

3-3-19. The equation of motion for the system is

Substituting the given numerical values for m, b, and k into th i s equation, we obtain

wherex(0) = 0.1 and ?(o) = 0. The respofise ta the given initial condition can be obtained by taking the Laplace transform of this equation, solving the resulting equation for ~ ( s ) , a n d finding the inverse Laplace transform of X(s) . The LapPaee transform of Equation (1 ) is

By subsbituting the given initial conditions ink0 this l a s t equation, we 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-- e-t s i n 3t + e l t - 3 t ) 3

--- . . - -- . - -.

B-3-20. The equation of notion for the system is

where Fk = &(mg - F sin 30'). ~ewritting this equation,

= 0.066 F - 0.3(mg - 0.5 F) For a constant speed motion, $ = 0 and the last equation becomes

1.016 F - 0.3 mg = b

or m%

3-3-21, Tfie equations of mtion for the system are rx .. M x = T -

kMs

0.

m = m g l - 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 that z ( 0 ) = 0, w e have

P s s w e that a t t = t . x(tl) - x(O) = 0.5 m. ?hen 1

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 the system are

2 where x = R8 and J = $ mR . 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 30- - ppg cos 30°, x ( 0 ) = 0 , ; ( o ) = 0

By substituting the given numerical values into this equation, we obtain

* b x = F - 1 x 9.81 x 0.5 - 0.2 x 1 x 9.81 x 0.866

or

and

Assume that a t t = tl, x(t ) = 6 rn and i(tl) = 5 d s . T¶x?n - I

1 6 = (F - 6.604) - 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 s i n 30m x 6

- - - 9.81 x 0.5 x 6 = - 29-43 N-m

Work done by the sliding friction force

= - 0.2 x mg m s 30' x 6

8-3-26. The kinetic energy T is

1 2 '2 T = ~ H ~ ' e + d~ = + ( M + + ) I ~ ~ ~ -

The ptmtia l energy U is

U = t @ f ( 1 - cos 0 ) + - m s 0) d J

- - ( M + - + ) g j ? ( l - m s B)

Since the system f s conservative, we have

= constant

Noting that d(T + U)/dt = 0, we obtain

[(M + 1 2 ; + (M + +lg! sin 81 j) = 0

Since 0 is not identically zero, we have

Rewriting,

" + 8 + q s i n ~ = ~

M+m 1 3

For small values of 0 ,

So the natural frequency is

B-3-27. T"he ki ne t i c energy T of the system is

and the potential energy U of the system is

2 u = 'j r R + % k p l - e2) + )i k2o2Z 1 1

Using the law of conservation of energy, we have

Noting that d ( T + ~ ) / d " , = 0, we obtain

Since 6 and i) are not i d e n t i c a l l y zero, we must have 1 2

B-3-28. A t t = 0 (the instant the mss M is released' to m e ) the 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 ins tant mass m h i t s the fnmr the kinetic energy T and the plkential energy U of the system are 2

2

&re v2 is the velocity of the hanging mass m and G2 is the angle of rota- - kion of the pulley both at the instant the mass hits the f loor; re2 - v2,

2 and J = % mpr . Using the law of consenration of energy, we obtain

By substituting ret = v2 into this las t equation,

, = + M V ~ ~ + + I T T V ~ . . ) ~ ~ ~ V ~ 2 ~2

Solving this equation for v 2

E-3-29. The force F necessary to move t he 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. So we obtain

-3-30. From the figure shmm to the right, we obtain -

To keep the bar AB horizontal when prlling the weight ng, the rnament a b u t p i n t P must balance. Thus,

Solving this equation for x, we obtain

B-3-31.

where

Note that

Gear ratio = 1/30

Since the power TLWl 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 t r a i n is i n f i - nite, t ha t there is neither backTash nor elastic defamation, and t h a t the nt.mber of teeth on each gear is proportional to the radius of gear. Ikf i n e the angular displacement of slraf t 1 and shaft 2. as 0 and Q2, respectively.

I

By applying Newton's second law to 'the system, we obtain far the mtbr shaft (shaft 1)

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 in . For shaf t 2 , we have

where T2 is the torque transmitted to gear 2. Since the work done hy gear 1 is equal to that of gear 2,

Sin- e2/Q1 = n /n , Equation ( 2 ) can be written as 1 2

Substituting Equation (3) into ~quatlon (I), we get

The equivalent moment of inertia of the gear t r a in referred to the motor sha f t is

Notice that if the ra t io n /n is very ttntch smaller than unity, then the 1 2

effect of J2 an the equivalent mwnent of inertia J is negligible. eq

&-3-33. The equivalent mament of inertia J of mass m referred to the motor shaft axis can be obtained f rcm m

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

Since there is no slippage between the belt and the pulleys, the work: done by the belt and the right-side pulley ( ~ ~ 0 ~ ) and that by the belt and the

left-side pulley (T2Q2) must he equal, or

where T is the load torque on the motor shaft and T is the torque trans- 1 2

rnitted to the left-side pulley s h a f t , is the angular d i s p l a c m n t of the

motor sha f t , and G2 is the amy~lar displacement of the left-side pulley shaft . Since the two pulleys are of the same size, we have BI = B2. Hence

For the motor shaf t , we have

Also, for the left-side pulley shaft, we have

Since TI = 5' we ham

Since 0 = B2. 1

this Last equat ion becomes

The equivalent moment of inertia J of the system with re- to the mtoi shaft axis is eq

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(i - i3) = E 1 1

20 i2 + 10(i2 + ig) = E

100 i, + 1O(i + i 1 = ZO(i - i31 2 3 I

whi& yield

So w e obtain

Solving for i2, ve get

Figure ( a ) Figure (b) Figure Ic)

Fran Figure (a) abwe we find R 1 = ZR2. Referring to Figure (b) above, we set

So R is obtained as 1

By substituting R1 = 2R2 into this l a s t equation, we obtain

!t%en, frm Figure ( c ) a m , current i is obtained as

B-4-5. When switch S is open, resismm between points A and B is l60R and we have

When switch S is closed, resistanm between points A and B is

So we haw

Cansequen tly ,

Solving for Rj we obtain

R = 2 5 . 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 for the circuit.

3-4-7. Fran Figure 4-55, we have for t > 0

Taking Lapface transfom of Equations (1) and ( 2 ) , we get

1 ( 5 ) * R2[%(s) - T(s)I ' 0 1 1

where i -'(o) is g i m by 2

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

Referr ing to Equation 15 1 WE have

B-4-8, 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 en t i a t i q this l a s t quation, we obtain

Th-e Laplace transform of this q u a t i o n is

Rl[al(s) - i ( O ) ] + L I(s) = 0 C

where

H e n c e

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~lent moment 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 ( 5 ) = KIa(s) eq 1

Hence

By substituting Equation (5) into Equation ( 4 ) , we obtain

The numerical values of the equivalent moment of inertia Jq and equivalent viscous friction coefficient b are, respectively,

eq

Substituting these numerical values 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

XSO, 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 we obtain

By substituting Equation ( 4 ) i n to 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

3-4-12. The transfer function E,(s)/E~(s) can given in tern of camplw impedances Z1 and Z2 as follows :

where

Hence

-14.

H e n c e

ZL = Ls, I - - - RCs + 1 z 2

- R

Hence

Note that

Similarly,

The transfer function Eo(s)/Ei(s) can be given by

B-4-17. Define the voltage at point A as ek. Then

1

Define the voltage at point B as eg.

Noting that

and K 3> 1, we must have

EA(s) = EB(s) Hence

frm which we obtain

8-4-18. The voltage at point A is

1 e = - (e + @Dl ~ 2 i

or

The voltage a t point B is

Since

Thus, equating Equations (1) and (2) we obtain

W-19. Define the displacenent of midpoint between kg and b as x3. 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 ,

1 dt t i , - i 2 ) d t + - (i l - i )dt = e(t)

3 C3

d i 2

L 2 ~ + R(i2 - i3) + - (i2 - il)dt = 0

R(ig - i2) = - (i, - i g ) d t

From these thee equations we ean obtain the analogous electrical system as shown to the right,

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 the equations for the circuit are


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 ight . / / / / i = ~ / / / / /

CHAPTER 5

B-5-1. By substituting the given numerical va lues i n t o

WE! obtain

R-5-2. We sha l l solve t h i s problem by using ttm different approaches: - one based on the exact method and the other based on the use of an average resistance.

(1) Solution by the exact method. For t he liquid-level system we ha-

2 3 Py substituting C = 2 m , Pi = 0.05 n /s, and p = 0.02 lfi into t h i s last equation, we obtain

2 k t fi = x. Then H = x and dII = 2x &, So we have

As- that at t = t t k level reaches 2.5 rn. Then, t is obtained as 1 1

*, = \ dk ( * - & 1.. = - Z M ) x l y + 1 W O Jfi dx

5 - Z x I

1 = - 200( 6 - 1) t moo '"(5 - 2x)l /

(21 Solution by use of an averaqe resistance. Since Q = 0.02 5, the average resistance I? is obtained from

~ e f i n e h = H - 1. lhenqi = a i - 0.02 and qo=h/R. For the liquid-level system,

Which can be rewritten as

2 3 substituting C = 2 m , R = 129 s/m2, and qi = 0.05 - 0.02 = 0.03 m /s i n t o this l a s t equation yields

Taking Laplace transforms of both sides of this l a s t equation, w e obtain

3. B 7 258[sH(s) - h(O)] + H(s) = - 5

or

Solving t h i s q u a t i o n for H ( s ) ,

The inverse laplace transform of this last equation gives

Assume that at t = t h ( t ) = 1.5. The value of tl can be determined frm 1' 1

Rewriting,

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 = (6 + q - 6 - q,)dt 1 1

Since Rl = hl/ql 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

d S 2 +---- - 1 dh2 - I R2 d t RL d t

By eliminating dh /dt from Equations (3 ) and (41, we obtain 1

dZh2 R C R C - &2

t (RICI + R C - + h =liZq 1 1 2 2 &2 2 2 dt 2

Substitution of h2 = RZq2 into this last equation y i e l d s

dZq2 R C R C - dq2 + R C ) - 1 1 2 2 dt2 +(RIC1 2 2 dt + q 2 = 4

Hen= the transfer function of the system when q is the input and q2 is the output is g i m by

E-5-4. The equations for the system are

Thus, we have

From Equation (1) we obtain

ClsHl(s) = 2 R1

[H2(s1 - H~(s) 1

From Equation (3) we get

By adding Equatfons (I), (21, and (31, and taking the Laplace transform of the resulting equation, we obtain

By substituting Equations ( 4 ) and ( 5 ) in to Ekqaticm (6), we get

Since H3(s) = A 3 f ~ 0 ( ~ ) , this last equation can tR written as

This is k ? ~ transfer function relating 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 mod. Then

5: H'.~ dR = 4.005 -


Taking logarithm of both sides of this last qua t ian , we obtain

2.5 log x = 1wI0 3.5089 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 he 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= 03tzin the analogous electrical system shown on n e t t page.

B-5-8. - pv = mR,irT

In this problem

p = 7 X 205 + 1.0133 x 105 = 8.0133 x l o 5 ~ / m ~ abs

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 2 gage = 7.695 kgf/m2 gage

= 109.4 1bf/in.* gage

B-5-9. mte tha t

where q is the flow sate through the ~ 1 ~ x 3 and is given by

Hence


PJs) - - 1

Pi l s ) RCS + 1

For the bell- 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 ha t

5 2 5 2 P1 = 0.5 x 10 NJm gage = 1.5133 x 10 N/m abs

2 5 2 P2

= 0 N/m gage = 1.0133 x 10 N/m abs

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. The equation for the system is 1 2

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 in 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

C

Q - f(E) = a(H - A) where

f(E1 = f ( 4 ) = 0,2

Thus, a linearized m a t i o n becames

A l inearized 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 = 5, z = 356, and

Thus* the l inea r l zd equation is

5 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.

Relative displacement between rack A and rack C is 2re and this must qua1 displacement x, Therefore, we have

Since x = r0 + y, we obtain

B-5-17. The heat balance equations for the system are

c ae = (U - ¶,)at 1 1 (1

C2dG2 = (q, - s,)dt (2)

Noting that

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 EbZ(s)f l l (s) can be given by

B-6-1. &fine the current in the circuit as i (t) where t 2 0. The q u a t i a n for t h e circuit for t 9 0 is

Since the capacitor is not charged for t < 0, the laplace transform of this equation becaes

Since

we obtain

The i m f s e Laplace transform or E,(o) gives

B-6-2. The equation for the circuit is -

Since q(0) = 0 and i10) = {(o) = 0, the Laplace transform of this la* -a- tion gives

E LS~Q(S) + RsQ(s) + -1 Q ( s ) = - C S

or

Since khe'current itt) is d q ( t ) /a t r we have

The current i(t) w i l l be osc i l l a toq if t he 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

E Its) = - 1 L (S + + 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 to the right, we have Z1

Z1 = R1 " 7 Z = R 4.- C2s

Hence

Eo(s 1 - - Z2 - - KP2s + 1

Ei(5) '1 + '2 ( E ~ + R ~ ) C ~ S + 1

Next, we shall f ind 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 the s t e p input of magnitude E i '

~ 6 - 4 , The system equations are -


0 t

bly + kly = k x + blxo 1 i

Noting that x,(O-) = O and y(O-) = 0, by t ak ing the b; - transform of these two equations we obtain

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 . - xo) = b (& - y ) 1 1 2 0

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 npu t xi(t) is given as

we have

Tne response Xo(s) is t hen obtained as

Since

we have

Hence

B-6-6. - First note that

Since R2 = 1.5 Rlr C = C , and R C = 1, we obtain 2 1 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 - 0.4 e- (L/3)t + 1) 0

'to lqarithmic d e c r e n t = In - = - xa l I n - - 5C.d T

Xl Xn n

Solving for 5 ,

By substituting n = 4 and xo/xq = 4 into this last equation, we obtain

Noting tha 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~es m = 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 . 8 1

At steady state

l a S S = 2 x 9.81

From Figure 6-48 (b)p x,, = 0.08 m. 'Thus

- 2 x 9-81 = 0.08 Xss - k '

Solving for k, we obtain

Since

=& =@= 3.34 r a d / s , b R 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

+ rng (cos 0 iZ + sin % 51 1 s i n 8 = - m2g 1 sin 0

Simplifying,

.* 9 .e x + c o s e b 3 - m2 sin 8 b2 + 2k x = O

ml + m2 ml + m2 "1 + m2

Thus the equations of motion for the system are

mz 1 0. ;; + c o s Q 0 - m21 s i n O i2 + 2k x = O + m2 "1 + m2 "1 + m2

=.a The equation of mtian for t\e s y s t e m is

w e rn = 1 kg, k = 100 ~ / m , p ( t ) = 10 &(t) N, x(0-1 = 0.1 m, and G(0-) = 1 m/s. By substituting the given n m ~ i c a l valves into the system equation, we obtain

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

sin 10t + 0.1 cos 10t x(t1 =lo

L e t us def ine 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 undetermined i n s t a n t that this impulse is to be given to the system. 'fhm,

the equation of motion for the system when the t k f o impulse forces are given is

The L- transform of 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 ident i ca l ly zero for t > T if we choose

Thus, t k Xmti0n of mass m can be stopped by anokher impulse force, such as

1 2 The system equation is

2 + b;t = d( t ) , X(O-) = 0, G(o-I = o

The &- transform 05 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

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.

JJ-6-13. '1Zlc nwrnenL u1 itjestla of t . 1 ~ wndulum about the pivot is J = m i 2 . The angle of rotation of the pendulum is 8 rad. Def ine the force that a c t s on the pendulum a t the time of sudden stop as F(t ) . Then, k h e torque that acts 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

We shall linearize this nonlinear ajvaiirn 4 erruning F(f) -3 1 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

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

Taking Laplace transforms of bath sides of this last qua t ion , we obtain

b

where we used the 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 tha t l(t - 0-3) = O for 13 ,< t < 0.3.

Let us assme that at t = tl, 0 = 2 0 8 = 0.3491 rad. Tbenbyten- t a t i w l y assuming tha t tl occurs before t = 0.3, we solm the following equation for tl:

which can 3x simplified to

The result is

Since tl = 0.0327 < 0.3 our aasrenptim was correct. The terms involving l(t - 0 - 3 ) inEquat ion ( 4 ) d o n o t af fect 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 the numerical values for M, b, k, and g i n to this equation, we obtain

By taking the Laplace transform of th i s l a s t equation assuming zero initial conditions, we obtain

The inverse Laplace transf o m of X I S ) gives

xdt) = 0-04905(l - e-1*6666t cos 5.5277t - 0.3015 e-l-666fit s i n 5.5277t)

N e x t , we shall obtain the response clurve x(t ) versus t with MATUB. Note that: X(s] can be written as

Now define

n m = [ 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 kIm.

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) + $10) + 4x(O)

s2 + 4s 4 16

Since x(0) = 6-05 and g(0) = 1, X(s) becanes

The inverse Laplace transform of X ( s ) gives

f- x(t) = 0.05 e-2t m s 2 6 t + 0.3175 e-2t sin 2. 3 t

This equation gives the time xespQme x(t) . The response mme x(t1 versus t can be obtained easily by use of

NATLAE. Noting that X(s) can be written as

X(s) = 0.05~2 + 1.2s 2 s 2 + 4 s + 1 6 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 tit1 e(lriesponse x(t)3 xlabelrt secq> ylabel(k(t)')

B-6-17. In this system F is the input and x2 is the output. From Figure 6-57, we obtain the following equations:

Laplace transforming these tm -atimsr assuming zero i n i t i a l conditions, we obtain

By eliminating XI(s3 f r a these two equtltionsr we get

simplifying this last equation, we get

from which xe obtain

By substituting numerical ~ l u e s for 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. x2(s) can be obtained from

The inverse Laplace transform of X ~ ( S ) gives

The response curve x2(t ) versus t can be obtained with MATLAB as follows: F i r s t note tha 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)')

B-5-18. Referring to the figure showr to the right, we ham

1 = - J - + C 2 5 , Z2(s) = Rl + - I

Zl(d R2 C l S

H e n c e R2 Zr(s1 = Z ~ ( S ) = RrCls + 1

R ~ C ~ S + 1 CIS

Thus

By rmbstituting the given nrrmerical wlues for Rl, R 2 , C1, and C2, we obtain

men ei (t 1 = 5 V ( S*p input )l is applied to the system, we have

The reqmnse curve e,(t) versus t can be obtained by entering the follmj-lng MA'lTAEl program 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 that 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 mtion for the system is

whesex1(0) = 0 and G(OE = O . The laplace transform of this equation is

By substituting the given numefical values of k, P, and c3 into this las t equation, we obtain

Solving for X(s) ,

X ( s > = 10 - 5 - - 2 - - 1 10 (s2 + 100)(s2 + 4) 96 s2 + 4 s2 + loo

The inverse Laplace tsansfarm of X(s) gives the response x ( t ) -

5 ' sin I&) x(t) = (sin 2t - - 5

B-7-2. The equation of motion for the system is -

By substituting the *ven nwrtlerlcal va3ues into this equation, we get

+ 24;; + 2 0 0 ~ = 5 sin 6t or

+ 123; + fOOx = 2.5 s i n 6t

73king the Laplace transform aE th is last equation, we o b i n

2 6 (S t 12s + 100)X(s) = 2.5 --

s - 1 - 6

Solving for ~ ( s ) ,

The i m s e laplace transfarm of X ( s ) gives

3 eat txe 8t + sin 6t - m s 6t x(t) = - e4t s i n 8t + - 1856 464

1 EJS) -

C s - - - 1 1 RCs + 1 E i ( d R + - 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 X (3) and P(s) can thus b obtained as 1

H e n c e

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 + G ( j W ) 1 1 1 [ Ll

NgXt, referring to Equation 131 we have the transfer function G (s) between X ( 8 ) and P ( s l as fallows: 2

2

H e n c e

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 ~ , ( j w ) I P sin kt +&I

B-7-5. 2 2 tension = mu s = 0.1 X 6.28 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 klokr , we obtain

centrifuqal force , nUJ2r - 0.15 gravitational force m9 0.2598 h

Solving for W , we obtain

B-7-7. - The equations of m t i o n for the system are

MZ + & + m = m t d r s i n u t :

Tf 10 % of the excitation form is to 1s transmitted to the foundation, the tranmnissibility must be qua1 to 0.1. Thus

Since 5 is desired to be 0.2, we substitute J = 0.2 into this last equation*

Solving for ' , we find I

86

Nating tha t & > O , velnuat have!* = 22.28. men

3 k = 17.7 x 10 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

m z + & + k z = - m y

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 the amplitude of the relative displacemer~t z.

If w<< W , we have n

So the acceleration of the base is ptopartional. to z.

B-7-9. - Define the displacement of spring k as y. 'Then the eqmations of motion for the system are 2

The f o r e f(t) transmitted to the foundation is

By taking laplace transforrrrs of Equations (1) and (21, assuming zero ink t i a l conditions, WF obtain

By eliminating YCs] fran the 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. Then the equations of mtim for the system ape 2

where p = P s inwt, Rewriting these equatfom,

Laplace transforming these two 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

WE motion transmissibility TR is

?RIP vibration amplitude I X( j W ) I 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

2 By substituting %/m, = U Into this last equation, ve get

Hence

I p ( Y ( J W ) I - =-- p

ma LrlL ka ka "a-

m,

The amplitude of dbration 05 mass m, is P&.

8-7-12. Assuming ma11 angles Q1 and e2 the equations of nation for the system rriay be obtained as fo Llm:

Rewriting these equations, we obtain

which can be simplified to

To find the natural frequencies of the free yibntionr we assume the motion to be harmonic. That is, we assume

= A sin W t , €I2 = B s i n W t

u 0-

1 = -A& s i n t, e2 = -B&* s i n w t

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 the coefficients of Equations (3) and (4 ) must be equal t o zero, or

This determinant equation determines the natural frequencies of the system- Equation (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 A t the first natura l freqwncy W =J=, we obtain from Equation

(3) the following expression:

kit2 -

[ ~ o t e that we 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.

A t the second natural frequency & = LJ2 we obtain from Equation (3)

ka - A - m - - m l l - 2

2 - - -

E3 ka* ka2 - 9 - + k a L L - - - - m 1 I mlt2 P m 1 t 2 m 2 1 2

[we obtain the same result fran Equation (4). 1 At the second mode, the amplitude ratio A D becomes - m /m or A = - (m2/ml)~. mis means that 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 r thesystemare *.

mx = - (x + f l s i n O)kl - (x -I2 sin €Ilk2

For small angle B, we have sin 8 + 8 and cos +3 1 and the preceding two mat ions becnne

Notice that if flkl = [2k2t then the coupling terms become zero and ~I?Q

equations become independent. However, in this problem

Therefore, mupling exists between Equation (11 and Equation (2). To f i n d the natural frequencies for the system, assume the following

harmonic motion:

x = A s h u t r B =: T3 s i n W t

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 na tura l frequencies of the system. Equation (5) can ke rewritten as


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 ol~tain

2 a4 - 1 4 0 W + 3920 = O

Solving this qua t ion for W *, we get

H e n c e

The first natural frequency is Cdl = 6.2205 rad/s and the second natural frequency is u = 10.065 rad/s.

2 To determine the modes of vibration, notice tha 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) the amplitude ratio A h becomes as follows: 1

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 ratio A/B k c m e s as 2

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

k l = k , k2=2k, k 3 = k r m l = m , m 2 = m

Equations (740) and (7-41) become as follows:

Also, that satisfies Equations (7-38) and (7-39) bcmes as follows:

X)ef ine

By substituting Ld12 into Equation (I), we obtain

Similarly, by substituting uZ2 into Equation (13, get

[We get the sarrte result if we substitute dl2 or aZZ i n to Equation ( 2 ) , ] Hence we have

ses m he S a m n ( w i t 1 e same

t ion by ency w . Fig

the s: 2) r tht lures ( i

u n e m 2 msse 3 ) and

which means that: in the first mode of vibration (with frequenq dl), the ma5 E in t ? d i m In the second mod .brat io I E r e q in opposite d i r ; by th amount l u ) s r i m on next page deplct 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 .* .I

x = - A ~ * s i n a I t , y = - B w 2 s i n d t

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


( t~ 2 - 7.2985)C d2 - 13.7016) = 0

Hence r

The frequency of the first mode is 2.7016 rad/s and the frequency of the 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

Hence, at the first m o d e of vibration, two masses rnm in the same direction, while a t the second mode of vibration, two masses move i n opposite d i m - tions.

Next, we shaU obtain the vibrations x ( t ) and y(t) subjected to the given initial conditions. Laplace transforming Equations (1) and ( 2 ) ,

Substituting the given initial condition3 into the preceeding t w o equations we get

~liminating Y(S) frm Equations ( 5 ) and ( 6 ) ,


Similarly, we can obtain Y ( s ) as follows:

To obtain the responses x(t) and y ( t ) to the given initial conditions, we rewrite X(s) and Y ( s ) as follows:

Possible MATLAB programs to plot 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_mrn to obtain vibration x(t) *****

num = [0.05 0 0.5 0 01. den=[ 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)')

- ****#I M 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 s for the system are

Referring to Equation 17-54) we have

Case ( a ) : For the initial condition..;

x(0) = 0.2807, G(o) - O I y ( O ) = I, ~ 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 into the l a s t eya t ion and simplifying, we get

To obtain plots of x ( t ) versus t and y ( S ) versus t, w e may enter the following M4TLAE program into the m p u t e s . The resulting plots are shown in Figure (a) .

Case (b): For the i n i t i a l conditions

x(0) = 1.7808, G(03 = 0 , y ( 0 ) = -1, ;(o) = 0

% ***** MATLAB propo 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)') title(Responses x(t) and y ( t ] ~ due to initial conditions (a)') xlabel (It sec') ylabel (k(t) and y(t33

we obtain the following expressions for ~ ( s ) and ~(s):

I

k MATMI3 pragram for obtaining plots of x( t ) versus t and y(t) v;ersus t is 5- below. The resulting plots 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]: For the i n i t i a l conditions I

~ { o r = 0.5, ;f(o) = O, y ( o ) = - a s , + ( o ) = o

WE obtain the followhq expressions for XIS] and Y(s}:

A MATLAB program to obtain plots of x ( t ) versus t and y(t) versus t is s h u n below. The resulting plots are shown in f iqure (c] .

% ***** 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 - are s h m to the right, The transfer function ~(s)/R(s) is

B-8-2. Simplified block diagrams for the system are shown $elm.

The transfer function ~ ( s ) h ( s ) is

B-8-3, Define the i n p u t impedance and feedback impedance as Z and Z respxtively, as shown in the figure bel,>w. 1 2

Then

Since e l 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. Let us define d i s p l a m n t s e, x , and y as shown in the figure klow .

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:

Since in such a systm 1 ~ a / [ o (a + b) 1 I is designed to be very large -red to I, ~(s)/B1sI may be simplified to

We 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 , there 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.

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 speed. Referring to Figure (a) shorn below, a block diagram for the system 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 R al I + - bs s aI t a2 bs + k

Since such a speed mtruller is usually designed such that

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 cons tan 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

Theref ere,

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 M is specifimf as 0.05, we haw P

or

Rewriting,

Solving Ear the damping ratio 5 w e obtain

The s e t t l i n g time ts is specified as 2 seconds. So we have

or

Iheref ore,

-8-10. The closed-loop transfer f u n e i o n of the system is

c(s1 - - loo R l s ) s3 + 2s2 + 10s + 100

This system is unstable m u s e two m l e x pte closgd-loop poles are in the right half plane. To vi: tstable r e spnse , WE may enter the folrmring MATLAB program il .er . The resulting unstable response olrve is shown in L~IC r LYU e: UIL next page.

To make the system stable, it is necessary to reduce the gain o f the system or add an appropriate -nsator.

malize It0 the .h- .I=<-

-con juq the ur

cDmPut am-- --

num = [0 O 0 1001; den=[ l 2 10 lC101; t - 0:o. 1 5 ; s tv Inun ldqt ) €?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, = 7 t : - 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~tlon for the system is

From this quation, we obts in

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

K

G I s ) = 2s + .L I 3t - = 1 - 1 + 3- S 2 8 + f , + m h S

2s - 1 the s ta t ic velocity error constalt Kv is

This value must be equal to or greater than 50, Hentx,

Thus, the-conditions to be satisfied can be surmaarized as follows:

O < R h < l

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, we determined a set of values of K and Kh as follows:

R = 5000, Kh = 0.0198

With these values of 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 nve r se Laplace transform of C(s) gives

The steady-state error e,, for s unit ramp input is

= l i m -

?he steady-state e r r o r can also 6e obbined by use of the f ina l value theorem. Since t h e error signal ~ ( s ) 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 fo r stability to be

3 O > R > O

8-8-15. Since the system is of higher order (5th order), it is easier to find the range of gain K fo 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 wi l l generate a plot of the root loci for the system. The resulting root-locus plat is s h m also on n e page.

num= [O 0 Q I 2 43; den=[ ! 11.4 39 43.6 24 01; 1 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 sem 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 ld )* + I24 + 2R)jkJ

This equation can be rewritten as

By equating the real part; and imaginary part equal to zero, respectively, we obtain

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 that goes to i n f i n i t y crosses the 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:

-1.2130~ + 39 x 1.2130~ - 2% = 15-61 K = for hJ = 1.2130 2

R = -2.1509~ + 39 X 2.1509~ - 24 = 67-51 for OJ = 2.1509 2

K = -3 . ~ 5 5 3 ~ + 39 x 3 - 7 5 ~ 3 ~ - 24 , 163.56 for W = 3.7553 2

Based or1 the K values above, we obtain t h e range of gain K for s t a b i l i t y as follows: me system is stable if

13-8-16. A MATLPlB program to plot the root loci and asymptotes for the following system

is given blow and the result5ng root-locus plot is shorn on next page-

N o t e that 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]; I

dena = [ I 2.1 0.4538 0.083 19 0.00571 93; 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')

B-8-17. The open-loop transfer function G(s) is

The fallawing M A T L W pragram will generate a rmt-lms plot. The resulting plot is shown on n e page.

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 the crossing pint of t;le rwt I m s branch that goes to infinity and the imaginary axis. Hence, we first f ind the crossing frquency and then find the mrrespding gain mlue .

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


( d 4 - 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

u4 - L O ~ J ~ + Z K = 0 C I. )

Equation ( 2 ) can be rewritten as

By suketitutimg Equation ( 2 ) in to Equation (11, we find

t ~ 4 - lo& + 7 k 1 2 = 0

which yields

Since W = 5 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 stability .

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 f ram t h e nlagnitude condition.

A m the lead rompensator beromes as f=rllows:

The f ed forward transfer function is

?"he following MATLAB program w i l l generate a root-locus plot. The 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

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 5 and s = -2.

B-8-19. The MATUlB program a i m below generates a root-locus p lo 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 j W axis as shown in the figure k l a w , From the figure we obtain . -

sin 8 = - Q n

Note also that = 0.6 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 sim- taneaus equations far a.

By substituting Equa t ion (1) into Equation (21,


~y equating t he real part and imaginary part of t h i s l a s t equation to zero, respctively, we obtain

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 for a = 0.5623

K = -1.8281a3 + 2.1875a2 + 3a = -1759.74 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 the 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 2 c as + bs + ab)

s3 + s + ~ ( s 2 + as 4 bs + abl

Since the dominant closed-1- pales are located at s = -1 f jl, the 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 4 abK = (s2 4 2s + 2 ) l s + K - 2 )

+ (aK + bK - 2R + 3)s + Kab - Z ( K - 2 )

?he remainder of divis ion must be zero. Hence we set

Kab - 2 ( K - 23 = 0

Since a is specified as 0 = 5 r hy s u k t i t ~ t i n g a = 0.5 into these two equations, we obtain

By substituting Equation (1) in to Equation (21, we haw

Then, by substituting K = 2 i n t o Eguation I l ) r we get

2 b = 1 . 5 X 2 - 3 - 0

The PFD controller with K = 2 and b = 0 kcoms

Thus, the controller beccxnes a P'D controller. The open-loop transfer funct ion 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 is shown in Figure (a ) . -

A Bode diagram of the ml controller is shown in Figure (b).

F i g u r e (b)

5 9 - 5 ,

Lead ne.t:wor-k

Lag network

B-9-6, The equation of motion for the q s k m is

'Ihe A_ transf o m of this equation, using zero i n i t i a l conditions, givfs

Hence

Notice that th 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

1 ,-(k/bSt e ( t ) = - 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

we o m i n ""'I - 1

/ G ( j d ) = 9 0 O - tan-' Wb k

Figure (a )

The 6teady-state output es,(t) is therefore given by

1 e It] = - d X -1 W b SS

sin( Crl t + 90'- tan -) k

Next, substituting [ = 0.1 m, k = 2 ~ / m , and b = 0-2 N-si'm into G( jLJ ) gives

A Bode diagram of G ( j t d ) is shown below.

Bode Dlaqmn (Problem 0.96)

B-9-7. Noting that

we have

B-9-8. A possible MATLAB progmn for obtaining a Bode diagram of the given G C S 1 is s h m on next page. The resul t ing Bode diagram is ahown also art next page.

- num=[O 0 0 320 6401; den=[l 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 ~ ( s ) is sham below. The resul t ing Bode diagram is shown on next page.

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)')

'3-9-10. A pssible WfTI.AB program for obtaining a Nyquist plot of the giwn ~ ( s ) is shown bzlow. Note Chat to plat G ( j W ) locus only fox w 0, we use the following command:

The resulting Nyquist plot is shown an n.ext gage.

C

num = [O 0 0 1 J; d e n = [ l 0.8 1 01; w = O.I:O.1:100; [re,im,w] = nyquist(num,dqw); plot(re,im) Y = 1-3 3 -4 23; ~ s ( v ) ; axis('squate? ~d titlewyquist Plot (Problem B-9-10$':1 XI abeI('Rea1 Axis'] ylabelC1Lmag h 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 for obtaining a Nyquist plot of the given ~ ( s ) is sham below- The resulting N y q u i s t plot is shown on next page. Since none of the open-loop ples l i e in the right-haif s plane and frm the Nyquist plot it can be seen that the G { j W ) lorus daes not encircle the -1 + jO point, the system is stable.

num- [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-9-1 I)') d a b e l ( R d Axis') ylabel(7mag h i s ' )

B-9-12. A possible EaTLRB prqram for obtaining a Nyquist plot of the given GIs) is s h m below. me resulting N y q u i s t plot is shown un next 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 the -1 + jO mint t w i c e a 8 a 1 f s =rid f r u n W = - w to W = 0 to W = & . Refer- ring to the Nyquist stability criterion (see page 497). we hare

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 hn open-loop poles in the right-half s plane, because

s3. 0 . 2 ~ 2 + s + 1

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. A closed-loop systm with the following open-loop transfer function

K E ( S ) H ( S ) = ( T ~ > O )

s2(~is * 1) 11 3

is tmskable, w h i l e a closed-loop system with the follbwing open-laap transfer function is stable.

Nyquist plots of tkse two systems are shgvnm next page. Ndte that G(jw 1 H l j L . d ) laci s ta 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

! r T - - e 1 1 (Stable ) I

B-9-14. For t h i s system

~y setting K = 1, we draw a Nyquist diagram as shown below. Note that

The ~ ~ q i i s t locus crosses the negative real axis at CT = -0.442. Hence for stability, w e require

The same result can also €x obtained 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 in to G ( j w ) yields

K ~(j2-029) = ( m s 2.029 - 2.029 sin 2.029)

1 + 2.029~

The critical value of K for stability ran k obtained by letting G(jZ.029) = -I, or

0.4421 K = L

Thus, the 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 the damping ratio of 0.25. Define the frequency corresponding to the angle OF -130 to be

Solving this last equation for dl, we f ind W 1 = 1.491. nus , 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'- 90.. - 1 8 0 ~

The magnitude IG(j2)1 with R = 3.46 becomes

Thus, the gain margin is 1.26 dB. The Bode diagram of G ( j W ) with R = 3.46 is shown below.

B-9-16. Note that

j Cc' + 0.1 TO - - ZK(10j 4 + f 3 K j ~ + o . 5 j W ( ~ j J + l l J j L J + l )

W e shall plot the W e diagram when 2R = 1. That is, w e plot the Bade

diagram o f ~ o j d + I

'(jd' = j w ( 2 j LO c l)(jd+ 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 rums lies above the -18~' line for all Ld , the gain margin is + ~ s dB.

B-9-17. Let us use the following lead compensator:

%?.+I I s + -

G,(s) = k O ( = R, T

d ~ s + 1 1 S f - e T

Since K, is specified as 4.0 s-1, we have

Ts + 1 IC, = l i m sK,d K

- = K , d K = 4 s+o, U T s + 1 s ( O . l s + l ) ( s + I )

n Le t us set K = 1 and define R c q = K. Then

N e x t , plot a Elode diagram of

The following MATLAB pryram prduces the W e diagram s h m klm.

num = I0 0 0 41: d e n = I O . f 1.1 1 01; bodelnum,den) 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' (& = 40 O)

1 -FN

d i s d e l ; e n ~ i r ~ c d as 0.2174- f ~ t us choose, inskead o f 0.2174,M t o b e 0.21, or

Next s k p is to dctcrmi rkc t l ~ e mrncr rrcquencies @ = TJT and d = l/(d T) or tlie f pad mrp?tisalor. Notc? tlraI; the miximutn phase-lead angle &, occurs at ttie qemtr ic man or thc t w o carnor frequencies, or W = ~ / ( G T ) . 'I'l~e arwunt or tile nlorl l i icaLior~ i n tile rnagrlitude curve at tu'= l / ( & T ) due to t h e inclusion or the term (Ts + l}/(d Ts + 1) is

we need to find the frequency p i n t where, when the lead cmpnsator is a d d d , the total magnitude kmnes 0 dB. The magnitude I ~ ( j w ) I is -6.7778 dB corresponds to &J = 2.81 rad/s. we select this frequency to be the new gain crosswer f rquency 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 4.1064 41; step(nu m, d en) grid titIe('Ur Jt-Step Response of Compensated System (Problem B-9- 17)') KEabel('1 wc'] ylabeI('0utpu t 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. To satisfy the xequirments, try a lead compensator Gc(s) of the form

D e f i n e

where K = Kc&. Since the 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 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 angle nr lin requirement is 42 .ZO. We may as: 'ed to h 48'. This means tha t 5.8 ' Llt13 - t he shift in the gain crosswer frequency.

s i n &, = . - 1 +oC

3tisfy *

num pha: :d to a

se lead se m r g

I r q i ~ te for

Since

l - o l

F$, = 48 mrreqmnds to a = 0.114735. Note that ol = 0.15 -corresponds t o & = 4 7 . 6 5 J b . ) Whetherwe chmse8,= 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.

The nc& step is to determine the corner frequencies W = 1SS and ti, = l/(o( 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

Note that

We n e d to f ind the Frequency point where , when the lead rompensator is added, the to ta l magnitude beccmes O dB- The frequency at which

the magnitude of G ~ ( ~ L L ' S is 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 the frequency p i n t where I G l ( j ~ ) I = -8.239 dB m s a t k 3 = 11..4 rad/s. Not ing tha t this frequency corresponds to 1 / ( K T ) , or

we obtain

The lead c a n p n s a t o r thus determined is

- "

T s -F 4.4152 cc(s) = yC = Kc

1 S f - s + 29.4347

where Kc is determined as

The folLowhg MATTAB program produces the W e diagram of the l e a d ccan- pensator 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 1 which is shown on next page.

From this Bode diagram, it is clearly seen that the phase margin is approximately 52", the gain margin is + w dB, and ??, = 50 s-1; a13 specifications am met. ~ ~ I I S , t h e designed s v s t m is satisfactory.

e unit- and th

Next, we shall obtain thl ~d unit-- responses of the original uncompensated systm 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 ples of the campensated system are as follows:

The MATLkB proqsam given belw produces the 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 S e c ' ) 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("t Sec') 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 ?at the rompensated systm f o l l m the input ramp very closely. ;he compensaW system the error in fellowing t h e i npu t can be s : 0 < t < 0.5, but it is almost zero for 0 . 5 X t . (The steady state u r v r in themit-rampresponse is 0.C2. )

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. Thus,

Y = 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 (U - Z) = bl(k - 1

bl(i - 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(s1 from the l a s t two equations,

which can k simplified to

J - kAS 5

Yo- - - - k2

Using 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 a - i x 2 = - -

X2 aT U

aT

This is the state q u a t i o n .

F m Equation (6) we have

Frm Equation ( 7 ) we get

Hence

This is the output equation.

B-10-4. Note t ha t

Hence


which is w i n l e n t to

This equation can be wri t t en as

where

13ef ine

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 state space representation for the givlen system.

Method 1: The systm 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 ~ r the 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

Ikf ine

Then

Hence, the sta te space representation for the system by this approaeh fs

B-10-6. The equations for the circuit are

~ubstitution of ql = xlt $ - = x i n t o the last t w u equations yields - X ~ ' 9 2 3

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. The equations for the system are

Nuti- tht from Equations (1 1 and ( 2 )

Also, frm mat ion (3)

a .. = X I - 3(x1 - U) + 3(x1 - i - U) + 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 - 3 i ; + 3 j r - y = u - 2 G + u

['She same result can be obtained by use of Equation (10-51). ]

B-10-9. The eigenvector zi associated with an eigenvalue Xi is a -or

that sat is f ies the following equation :


example of eigenvectors corresponding to eigerlvalue

[I:] - -

For Xi = A *

An example

B-10-lo. Wf ine

The eigenvedors for this matrix can be determined by solving the following equation for x. -


which, i n turn, gives x = arbitrary constant 1

X* = 0

x3 = arbitrary ~ n s t a n t

H e n c e ,

where a and b are arbitrary nonzero constant. Notice that

Note that

a m linearly independent vectors. The eigerneors of matrix B involve two M l inearly independent eigenveetsrs.

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


x = arbitrary canstant 1

3 = arbitrary constant

x3 = arbitrary mnstant

Hence,

where a, b, and c are arbitrary nonzero m s t a n t , Notie that

Thus, the eigenvectors of matrix C involve three l inearly indepndent eigen- L

vectors.

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

Noking that

B-10-15. The solution of the state quat ion is

Since x(0) = 0 and u(t) - l(t). we obtain CH -

x ( t ~ = I f e ~ , ' ~ B d T

Note that Yb

where

Thus

At: R. = & -l[(s1 - A ) - ~ ]

Y b.

Hence

-TI B d T = a. b.m

Alternate solution Referring to Example 10-9, we have for the unit-step input the following solution:

Since x(O) = 0 in the present case, we have w. w

x ( t l = jI-'(ett - I J B W m .*I-

where

and

and x ( t ) is given by -

3-10-16. The solution to the state 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 taking 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 = . The 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 0-1 7)') xlabell't Sec') ylabel('y(t)")

Note that t h e respnse reaches zero as t increases. This is because the 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 -25.1026 -5.008 1 0 0 0 E 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("Output y(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. The following M A W pregsam produces the unit-ramp response. The resulting response curve is shown on next page.

A = [-5.03247 -25.1626 -5.008 1 0 0 0 1 01 ;

B = [1;0;0]1; C = 10 25.04 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- 10-1 8)') xlabelrt Set') ylabel('lnput and Output')

CHAPTER 11

8-11-1. Since

sin(tclRT + 0) = sinlrlm m s 0 + cosd sin 8 we obtain

~ [ x ( r c r ) l = cm e [ s i n ~ ~ l + sin o 7, ~ Q ) S Y J ~ I

-1 + s in 8 1 - z mslFT

-1 1 - 22 msWT + z - ~

- - (ms O ) Z - ' I S ~ ~ W T ) + (sin 0 ) (1 - z'l msdT) -2

1 - 22-= cosLdT * z

- - sin 8 + 2-I s i n ( W T - 8)

1 - 2 f 1 m s W T + 2-2

By defining m = k - h, we obtain

Since x (KT - hT) = 0 for k < h, the limit of sumation in can be 1 00 k = h

&zinged to X . n u s k = O

ad d

!Zince X(z) = X l ( z ) X 2 ( z ) t by mmparing Eqwtims (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= (1 - e'aT 1 e -aT

3-~-

z -aT ( z - 1 ) I z - e s - 1 z - e -aT

Hence

The inverse z transform of X ( z ) gives

-aT -aw ~ ( k ) = L - Q- e

B-11-7. X(Z) can be rewritten as

Then

Hence

B-11-8. Since -T

x ( z > = II - e ]z- ( z - 1)(2 - eWT}

we have -T

X(z) = 1 - e - - - I I z - I - 2 - 1 z - e -T

Thus

B-11-9. Referring to Table 11-1, we f ind

By writing 6' = a, we obtain

H e n c e we have

%king the inverse z transform of each tern of this last equatizm, we obtain

3 2 1:-1 + .3*.k-1 a ( k ) = a k a

H e n c e

Since a = e'l, we ktave

-k-2 ~ ( k ) = k(k * 1) k = 0, I f 2, ...

B-11-10. 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

xl l ) = x(0) + 1 = 1

x(2) = x(1) + a = 1 + a

x/31 = x(2) t a2 = 1 + a + a 2

+ ,k-1 = 1 - a k x ( k ) = l + a + a ' + - . - l - a

l a f 1, a #0)

= k ( a = 1)

If a = 0, then

~ ( k ) = 0 k = 0, I, Z r ... (a = 0 )

- 2 The pulse transfer function for G I s ) including the z e m r d e r hold is obtained as follaws?

P

- 3 . The pul se trans fer function forthe systemcanberemittenas

The inverse z transform of this l a s t equation gives

Clamparing this d i f f esence 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 0

h = 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 + l) - u(L) 1.

Referring to Section 11-6, the state equation for the system can k given by

The output equation is

N o t e that a numkr of different state space represenbtions are possible far the 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 e is y . ) Then we have t k following continuous-time state equation and a u t p u t equation:

x = - a x + u

The discretized state 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 (11-58) as follows:

8-11-15. Referring to Equation (11-501, the purse transfer functfm F(z ) is given by

- 6 The discretized state equation is

where

GCT) and HIT) can be obtained as follows: C .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

- 7 The discretized s-te equation is

GITP and HIT) m be obtained as fellows: Since L. LI*

we have

Hence

Thus, the discretized state space representation of the system becames as f ollaws :

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 space representation of the system is

w h e r e ~ ( k ) = m(k) + m(k)

Notice that the cneffirients bg, bl, and b2 affect only matrices C and D.

Next, consider the case where the system has no input (u = 0). For such a case, MATLAX? prduces a pulse transfer funckian of the fcllawing form :

If the original system bad an input function u such that

MATLAB prcduces the pulse transfer function

resenta 3 y ield

tion wl- t h e pc

The following FlA?ZAB program w i l l yield a discrete-time state-space rep1 Len the sampling period T is 0.1 s. This p r q m w i l l alsc ~Lse transfer fmction,

Eased an the MkTLAB output , the discrete-time state equation is

num = 10 0 03; den=[] 2 101; [AB,C,D J = 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)

n u m =

0 0 0

denz =

1.0000 -1.7288 0.8 I87

I

I

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 iwn by

Note that the original differential equation system given by n y + 2; + 10y = 0, y(0) = I, ;(o) = o

has the response curve (free vibration curve) as s h m in Figure ( a ) ,

Figure (a) F Sec

m e MRTLAB program 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') I

ylabel('0utput y(t) of Original Continuous-Time System')

The response curve (free vibration curve) of the discreti zed 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 blm.

rmml = [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) u i d 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')

r -

The pulse transfer function f o r this system m y be obtained by use of t he MATUB prqram s h m on n& page.

-

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 is

n m - 0.00912-1 + 0,0082~'2 G(2) = -

denz 1 - 1.72362-1 + 0.74082-2

This result is identical w i t h the pulse transfer function obtained in P K Q ~ ~ B I I A-11-12.

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 the ICro- necker 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 75025 121393 19H18 317811 514229

Note t ha t column 1 corresponds to k = 0 and c o l m 31 corresponds to k = 30. Thus, the Fibonacci series is givw by

The following MATLMl program will generate a plot of the mit-step response up to k = 50. The resulting unit-step response is s h m b l o w .

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

- 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)> xtabel('k') ylabel('y(k)')

B-11-23. A MATLAB prQgram to discretize the given equation is presented below. (The sampling period T is 0.05 s. )

I

~ l i s MAT& program produces the output as shown on next p a p .

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 1

grid titIcfUnit-Ramp Response (Ftoblem B- 1 1-22)') xlabel('kl) ylabel("y(k)3 I

From the MATLRB output, the discrete-time state equation is given &bow.

1.0259 0.0504 0

1.0389 k .0259 0

-0.0006 6.0000 1.10000 0 - 0560 q ( k + 13 6.0247 -0.0006 0 1.0000 0.0250

The END

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