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Solution Manual For System Dynamics, 4th Edition

Solution Manual For System Dynamics, 4th Edition offers textbook solutions that are easy to follow, helping you ace your assignments.

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    Contents Preface v Chapter 2 The Laplace Transform 1 Chapter 3 Mechanical Systems 12 Chapter 4 Transfer-Function Approach to Modeling Dynamic Systems ~~ 24 Chapter 5 State-Space Approach to Modeling Dynamic Systems 43 Chapter 6 Electrical Systems and Electromechanical Systems 89 Chapter 7 Fluid Systems and Thermal Systems 111 Chapter 8 Time-Domain Analysis of Dynamic Systems 125 Chapter 9 Frequency-Domain Analysis of Dynamic Systems 152 Chapter 10 Time-Domaim Analysis and Design of Control Systems 176 Chapter 11 Frequency-Domain Analysis and Design of Control Systems 198 iii

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    Solutions to B Problems B-2-1. £(t) =0 t <0 = te 2t t>0 Note that _1 PA = Referring to Equation (2-2), we obtain Fs) = J [£(6)] = of, [bet] = —1 (s +2)? B-2-2. (a) £,(6) =0 t <0 = 3 sin(5t + 45°) tt =0 Note that 3 sin (5t + 45%) = 3 sin 5t cos 45° + 3 cos 5t sin 45° =F sin 5t +f cos 5t So we have F(s) = JIE (6)]=->—_-5 3 _s 1 & 1 Z2+52 2 s2 + 52 = 3 s+5 JZ 524 25 (b) f(t) = 0 t<0 = 0.03(1 - cos 2t) t>0 Fo(s) = J, [£5(6)) = 0.03 2 - 0.03 5 ___0.12 s s2 +22 g(s2 +4) 1 |

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    B-2-3. £(t) = 0 t<0 =t%@ go Note that Lez) = 2 s3 Referring to Equation (2-2), we obtain Fs) = of) [£(t)] = J) [b%eat] = —2— (s + a)3 B-2-4. £(t) = 0 t <0 =cos 2¢)t cos 3 t t >0 Noting that cos 2c t cos 3t = (cos 5wt + coset) we have F(s) = LG [£(6)] = J lslcos Sut + cos a/t)] _ 1 s . s N (s? +13 w?)s 2\2 vw? 2+ w? (2+ 25whE? + wh) B-2-5. The function f(t) can be written as £(t) = (¢ - a) 1(t - a) The Laplace transform of £(t) is "as F(s) = § [£(6)] =f [(6 -a) L(t -a)] = &— s B-2-6. £(t) =c 1(t - a) - c 1(t - b) The Laplace transform of f(t) is F(s) =c £95 _ oc eS _ c (gas _ obs) s s s 2 + StudyXY

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    B-2-7. The function f(t) can be written as £6) = 29-223 ac - 2) + 2:5 1c - a) a a? 5 a So the Laplace transform of f(t) becomes F(s) = 0 [e(e)) = 40-1 _ 125 L ~(a/5)s , 2.5 1 as a: Ss a s a2 s =-L (10 - 12.5 e7(a/5)s 4 3 5 o-as) als As a approaches zero, the limiting value of F(s) becomes as follows: —(a/5 _, lim F(s) = lim -10-12.5¢ (a/5)s 4 2.5 eas a=0 a=>0 as 2 (10 - 12.5 e"(a/5)s 4 2.5 ~25) =m SS a—+0 2 als = lim 2:5se(@/b)s _ 356 cas a0 2as d — — yn (2.5 e7(a/5)s _ 5 5 cas) =1ip RO a—0 da ER 2a : 0.5 s e~(a/5)s 4 2.5 g e=@s = lim > a=0 --=0.5s+2.5s _ 28 _g 2 2 B-2-8. The function f(t) can be written as fe) =-24 ¢ - 24 1(¢ Ep: Np (t - a) 1(t - a) a3 a2 2 a3 So the Laplace transform of f(t) becomes .24 124 1 Yas _ 24 ee Fs) =—3 "7-7 7s © a3 §2 2a (1 a gas =) =T3 \ 2 Ts 2 3 | + StudyXY

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    The limiting value of F(s) as a approaches zero is - -kas _ g-as Lin F(s) = lin 2A =ase #8 - e8) B € a—0 a0 ads = 24(1 - as e~'#as — e=as) = 14mg —Q - - a—>0 2 ads? -%as as? — 24(-s e + === e~%3s + g eS) =1im ———— a>0 3a2s? _d go e735 + 2% eas 4 gas) = lim 92 a->0 3 als S 4s , 8 _-kas , 85(-S)~%as _ ge-as a] >. +e + 565) e se =p ——— a->0 2as d -%as -as -as da (de -as e -4e 7) = lim ——————— a0 dd, da 2 I _ “25 eS _ 5 eT yas 5-¢ BS, gg e708 =i ——— a—=+0 1 =-25 -s +4s =s B-2-9. f(ee) = 1im f(t) = lim sF(s) t->00 s->0 = lim s 5(s +2 - 5x2 _ jp 0 s(s +1) 1 2-10. : : ( ) B-2-10 £(0+) = 1im £(t) = lim s 2s +2 =0 t—0+ S »>o0 s(s + 1)(s +3) 4 + StudyXY

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    B-2-11. Define , y =x Then . y(0+) = x(0+) The initial value of y can be obtained by use of the initial value theorem as follows: y(0+) = lim sY(s) S200 Since . ¥(s) = J Iy(£)] = Jf, [X(t)] = sX(s) ~ x(O+) we obtain y(0+) = lim s¥(s) = lim s[sX(s) ~- x(0+)] s>0® S-> 00 = lim [s2X(s) - sx(0+)] SU cd B-2-12. Note that pd [+ 0] = sF(s) ~ £(0) dt PAE ww) = s28(s) - s£(0) - £(0) dt? Define a(t) = 2 ze) dt? Then Bey =p = ~ g(0 5 Es XL | 90) = ss) - a0) = s[s2r(s) - s£(0) - £(0)] - £(0) = s38(s) - s2£(0) - s£(0) — £(0) B-2-13. a a. = s+5 1 2 (a) Fs) “gins +3 sF1 573 where +5 =_4 a EH, LE =5*5 =-2 -_ az x =) 1 5

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    Fy (s) can thus be written as 2 __1 Ble) =g3T 7543 and the inverse Laplace transform of Fy (s) is f(t) =2et et (b) P(e) mera) of, "2 fs 2 s(s+1)(s+2) 's s+1 s+2 where = 3(s + 4) =3x4_5 UT +2) 2 s=0 a =3(s +4) =_3xX3 _ 9 2" s(s +2) (-1) x1 s=-1 a =3(s +4) = 3x2 _g4 3 s(s +1) (-2)(-1) s = -2 F (s) can thus be written as F(s)=-6__-9 ,_3 2 s s+1 s+2 and the inverse Laplace transform of F (8) is £,(t) =6-9et+3elt B-2-14. 6s + 3 6 3 (a) F(s) = ——5= = 2+ —5— 1 <2 s 52 The inverse Laplace transform of F (s) is f(t) =6 + 3t A ) 6 + StudyXY

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    (b) a b b SSPE SEL EEL. TH 2 (s+1)(s +2) 3 (s +2) where 55 + 2 =5+2_ 4 a =—07 12 +2 (s ) s = -1 _5s +2 _-10 +2 _ by = | 2+1 ° s=-2 b= (2H) _5(s +1) - (5s +2 “3 \ s +1 2 1 ds s =-2 (s +1) s=-2 _5(-1) - (-10 +2) _ 5 12 F (s) can thus be written as -. 8 3 F (s) ==, 8 , 2 2 s+1l (s+ 2)2 s+2 and the inverse Laplace transform of F,(s) is £,(t) =-3et spre? +32 B-2-15. 252 +45 +5 _ 2 5 F =E& 22 720 =2 + <= ©) =e D s+17s(s +1) —2+-2 +5 5 _,__3 ,5 s +1 s s +1 s +1 s The inverse Laplace transform of F(s) is fe) =28()-3eT +5 7 | T= StudyXY

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    B-2-16. 2 F(s) _-St+25+d4 2 4 s2 s s2 The inverse Laplace transform of F(s) is £6) = &(t) +2 +4t B-2-17. ro) . CC ev1-a s2 + 2s + 10 (s+ 1)2 + 32 =..s8*1 3 1 (s+ 1)2 + 32 (s+1)2+32 3 Hence £(t) = et cos 3t - + e-t sin 3t B-2-18. Fs) =-82*t2s+5 _.a , b,c s2(s + 1) 2s s+1 where a=_82+25+5 -5 s+1 s=0 p={25+2)(s+1)-(s2+ 25 +5) 22-5. 3 (s +1)2 1 s=0 2 — c=-_S2+25+5 = 1-2+5 _ s2 1 s =-1 Hence 5 =3 4 F(s) = —&5 + — + ——— (=) s2 s s+ 1 The inverse Laplace transform of F(s) is £(t) =5t -3+4et 8 + StudyXY

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    B-2-19. 2 10 b c Ss + a Fs) =— =" — = — 2 — 4 ——— ——— (s + 1)2(s + 4) (s +1)2 s+ 1 s + 4 where a= _25+10 --2+10__8 s +4 3 3 s=-1 p=-2(s+4) - (25 +10) -6-8_-2 = = > = (s + 4)2 3 9 s=-1 c=_25+10 _=8+10 _ 2 (s + 1)2 9 9 s=-4 Hence 8 rr 2 r ,2_ 1 F(s) 3 (s+ 1)2 9 s+1 9 s+4 The inverse Laplace transform of F(s) is F(t) = 8 tet - 2 ot 4 2 gat 3 9 9 B-2-20. 1 1 1 1 Fe) = t= (hz) s2(s? + w?) s2 52 +w?2 Jw? The inverse Laplace transform of F(s) is f(t) = 1 (t - 1 sin ct) w? w B-2-21. — = Cc _ e-as) _ -B_ o-as F(s) 52 (1 -e7%) Ss © a>o0 The inverse Laplace transform of F(s) is f(t) = ct - c(t - a)l(t - a) -b 1(t - a) 9 I~ StudyXY

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    B-2-22. ¥+ax=o0, x(0) = 5, %(0) = 0 The Laplace transform of the given differential equation is [s2ZxX(s) - sx(0) ~ %(0)] + 4X(s) = 0 Substitution of the initial conditions into this last equation gives (s2 + 4)X(s) = 5s Solving for X(s), we obtain 5s X(s) = —==2— s2 +4 The inverse Laplace transform of X(s) is x(t) = 5 cos 2t This is the solution of the given differential equation. B-2-23. ¥+w Zx=t, x(0) = 0, x(0) = 0 The Laplace transform of this differential equation is $2x(s) + ew 2x(s) = —L- n 2 s Solving this equation for X(s), we obtain Xs) = —L of 1 1 1 s2(s2 + w? 2 2+ w 2) W2 The inverse Laplace transform of X(s) is x(t) = —1— (¢ - 2 sing t) wy? Wn n This is the solution of the given differential equation. B-2-24. X+2%+x=1, x(0) = 0, x(0) = 2 The Laplace transform of this differential equation is 2[s%K(s) - sx(0) - %(0)] + 2[sX(s) - x(0)] + X(s) = 1 10 | T= StudyXY

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    Substitution of the initial conditions into this equation gives 2[s2(s) - 21 + 2[sK(s)] + X(s) = L- or : (252 + 2s + 1)X(s) = 4 + Solving this last equation for X(s), we get X(s) = —& XL s(2s? + 2s + 1) [ENS EEO — 262 +25 +1 (2s? + 25 + 1) _ 2 0.5 = (s +0.5)2 + 0.25 sls + 0.5)% + 0.25] _ 4x0.5 41 _ _(s+0.5) +0.5 (s +0.52+052 5 (s+0.52+0.5 The inverse Laplace transform of X(s) gives x(t) = 4 e705 gin 0.5t +1 - e705 cos 0.5 - e705 sin 0.5¢ =1+3¢e05 gin 0.5t - 0-5 cos 0.5 B-2-25. X + x = sin 3t, x(0) = 0, *(0) = 0 The Laplace transform of this differential equation is 3 s2(s) + X(s) = =—>— s? + 32 Solving this equation for X(s), we get Xs) = —u3 3 1 1 3 (2+1)(s2+9) 8 s2+1 8 &240 The inverse Laplace transform of X(s) gives x(t) = 2 sin t - + sin 3t 11 i + StudyXY

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    B-3-1. J=%mR? = % x 100 x 0.52 = 12.5 kgm? B-3-2. Assume that the body of known moment of inertia J is turned through a small angle 6 about the vertical axis and then released. The equation of motion for the oscillation is § = - J k8 where k is the torsional spring constant of the string. This equation can be written as +X 0-0 0 or - 2 + = 8 ow, 8=0 where k Ww = [= n Jo The period Tg of this oscillation is ny = 2K. 2 () [oN V % Next, we attach a rotating body of unknown moment of inertia J and measure the period T of oscillation. The equation for the period T is T = 2m (2) [x _ J By eliminating the unknown torsional spring constant k from Equations (1) and (2), we obtain 23, 2m fF To T Hence 2 T J = — 0, The unknown moment of inertia J can therefore be determined by measuring the period of oscillation T and substituting it into Equation (3). 12 | T= StudyXY

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    B-3-3. Define the vertical displacement of the ball as x(t) with x(0) = 0. The positive direction is downward. The equation of motion for the system is . mx = mg with initial conditions x(0) = O m and x(0) = 0 m/s. So we have X=g x =gt + x(0) = gt x=%gt? + x(0) = % gt? xX Assume that at t = t the ball reaches the ground. Then 100m 100 = % x 9.807 t;2 from which we obtain = 4.516 s 5 The ball reaches the ground in 4.516 s. The velocity of the ball when it hits the ground is %(4.516) = 9.807 x 4.516 = 44.288 m/s B-3-4. Define the torque applied to the flywheel as T. The equation of motion for the system is gB=1, 80)=0 6(0)=0 from which we obtain 6 = -L [:] 7 t By substituting numerical values into this equation, we have =_T 20 x 6.28 = 5p X 5 Thus T = 1256 N-m B-3-5. Jo =-1 (T = braking torque) Integrating this equation, 6=- Le + 8(0), 6(0) = 100 rad/s Substituting the given numerical values, 20 = - -L x 15 + 100 J 13 v + StudyXY

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    Solving for T/J, we obtain T _ T= 5.33 Hence, the deceleration given by the brake is 5.33 rad/s2. The total angle rotated in 15-second period is obtained from Tt? ,4 5 a(t) = - - + 6(0)t + 6(0), 68(0) = 0, 8(0) = 100 as follows; 2 8(15) = - 5.33 x £- +100 x 15 = 900 rad B-3-6. The equations for the system are Fk (x-y) # £; WF k(x -y) = ky } ¥ l x Eliminating y from the two equations gives k. Kk, k. Fek, (2 = 12 p11 x Kk tk, | ky tk, EE 1 2 The equivalent spring constant eq is then obtained as = eq TI, LT k, k 2 Next, consider the figure shown below. Note that A ABD and ACBE are similar. So we have Em AD BD which can be rewritten as A keg. = H_axl 2 OB - OC —5— xy PTT us = G+ Lo on 22 OB + 0A — or TE D 0 Ek B 0C(0B + % OA) = OA(OB - % OC) Solving for oc, we obtain 14 Vv + StudyXY

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