Modeling in the Frequency Domain

Transcription

1 T W O Modeling in the Frequency Domain SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Tranfer Function Finding each tranfer function: Pot: V i θ i 0 π ; Pre-Amp: V p V i K; Power Amp: E a V p Motor: J m ( ) 0.5 D m ( ) 0.3 K t R a 5 K t K b R a 5 Therefore: θ m E a And: θ o E a 5 θ m E a K t R a J m (+ J m (D m + K tk b R a )) 0.6 (+.3) 0.8 (+.3) Tranfer Function of a Nonlinear Electrical Network Writing the differential equation, d(i 0 + δi) dt (i +δi) -i i δii δi.. Thu, (i +δi) i +i0 δi ii 0 + (i 0 +δi) 5 v(t). Linearizing i about i 0,

2 Solution to Problem 3 Subtituting into the differential equation yield, dδi dt + i 0 + 4i 0 δi - 5 v(t). But, the reitor voltage equal the battery voltage at equilibrium when the upply voltage i zero ince the voltage acro the inductor i zero at dc. Hence, i 0 5, or i Subtituting into the linearized differential equation, dδi dt + 6.3δi v(t). Converting to a tranfer function, δi V Uing the linearized i about i 0, and the fact that v r (t) i 5 volt at equilibrium, the linearized v r (t) i v r (t) i (i 0 +δi) (i 0 +i 0 δi) 5+6.3δi. For excurion away from equilibrium, v r (t) δi δv r (t). Therefore, multiplying the tranfer function by 6.3, yield, δv r V about v(t) a the tranfer function ANSWERS TO REVIEW QUESTIONS. Tranfer function. Linear time-invariant 3. Laplace 4. G C/R, where c(t) i the output and r(t) i the input. 5. Initial condition are zero 6. Equation of motion 7. Free body diagram 8. There are direct analogie between the electrical variable and component and the mechanical variable and component. 9. Mechanical advantage for rotating ytem 0. Armature inertia, armature damping, load inertia, load damping. Multiply the tranfer function by the gear ratio relating armature poition to load poition.. () Recognize the nonlinear component, () Write the nonlinear differential equation, (3) Select the equilibrium olution, (4) Linearize the nonlinear differential equation, (5) Take the Laplace tranform of the linearized differential equation, (6) Find the tranfer function. SOLUTIONS TO PROBLEMS. a. F e t dt e t 0 0 b. F te t dt 0 e t ( t ) 0 (t +) e t 0

3 4 Chapter : Modeling in the Frequency Domain Uing L'Hopital' Rule. F t 0. Therefore, F 3 e t t. c. F inωt e t dt 0 0 e t + ω ( inωt ω coωt) e t d. F coωt e t dt + ω ( coωt + ω inωt) 0 0 ω +ω +ω ω a. Uing the frequency hift theorem and the Laplace tranform of in ωt, F (+a) +ω. b. Uing the frequency hift theorem and the Laplace tranform of co ωt, F (+a) (+a) +ω. c. Uing the integration theorem, and ucceively integrating u(t) three time, dt t; tdt t ; 3. t dt t3 6, the Laplace tranform of t3 u(t), F 6 4. a. The Laplace tranform of the differential equation, auming zero initial condition, i, (+7)X 5 +. Solving for X and expanding by partial fraction, Or, Taking the invere Laplace tranform, x(t) e-7t + ( co t in t). b. The Laplace tranform of the differential equation, auming zero initial condition, i, ( +6+8)X Solving for X and expanding by partial fraction, X 5 ( + 9)( ) X

4 Solution to Problem 5 Taking the invere Laplace tranform, x(t) 8 65 co(3t) 65 in(3t) 3 0 e 4t e t c. The Laplace tranform of the differential equation i, auming zero initial condition, ( +8+5)x 0. Solving for X and expanding by partial fraction, X X 5 Taking the invere Laplace tranform, x(t) 5 8 e 4t 5 in(3t) + 5 co(3t ) a. Taking the Laplace tranform with initial condition, X-+3+X-4+X Solving for X, Expanding by partial fraction X ( + 4)( + + ). +. X ( + ) 3 5 ( +) + Therefore, x(t) -0. cot - 0. int +e -t (. cot int). b. Taking the Laplace tranform with initial condition, X--+X-4+X Solving for X, Therefore, x(t) 5e -t - e -t + 9te -t - + t. c. Taking the Laplace tranform with initial condition, X--+4X 3. Solving for X,

5 6 Chapter : Modeling in the Frequency Domain 5. Therefore, x(t) 9 8 cot + int t. Program: ym t f5*t^*co(3*t+45); pretty(f) Flaplace(f); Fimple(F); pretty(f) 'b' f5*t*exp(-*t)*in(4*t+60); pretty(f) Flaplace(f); Fimple(F); pretty(f) Computer repone: an a 5 t co(3 t + 45) 3 co(45) - 7 co(45) - 9 in(45) + 7 in(45) ( + 9) an b 5 t exp(- t) in(4 t + 60) in(60) (( + ) in(60) + 4 co(60)) ( + ) ( + ) + 6 (( + ) + 6) 6. Program: ym 'a' G(^+3*+7)*(+)/[(+3)*(+4)*(^+*+00)]; pretty(g) gilaplace(g); pretty(g) 'b' G(^3+4*^+6*+5)/[(+8)*(^+8*+3)*(^+5*+7)]; pretty(g) gilaplace(g); pretty(g) Computer repone: an a

6 Solution to Problem 7 ( ) ( + ) ( + 3) ( + 4) ( ) 468 / / - 7/03 exp(-3 t) + -- exp(-4 t) exp(-t) in(3 t) an 4807 / exp(-t) co(3 t) 556 b ( + 8) ( ) ( ) / exp(-8 t) exp(-4 t) coh(3 t) / / exp(-4 t) 3 inh(3 t) 54 3 / / exp(- 5/ t) 3 in(/ 3 t) 97 7 / exp(- 5/ t) co(/ 3 t) The Laplace tranform of the differential equation, auming zero initial condition, i, ( )Y ( )X. Y Solving for the tranfer function, X a. Cro multiplying, ( ++7)X F. Taking the invere Laplace tranform, d x dt + dx + 7x f(t). dt b. Cro multiplying after expanding the denominator, ( +5+56)X 0F. Taking the invere Laplace tranform, d x dx x 0f(t). dt dt c. Cro multiplying, ( )X (+)F. Taking the invere Laplace tranform, d 3 x dt d x dt + 9 dx df (t) + 5x +f(t). dt dt C The tranfer function i R

7 8 Chapter : Modeling in the Frequency Domain Cro multiplying, ( )C ( )R. 0. Taking the invere Laplace tranform auming zero initial condition, d 6 c dt d 5 c dt d 4 c dt 4 + d 3 c dt 3 + d c dt + 3c d 5 r dt 5 + d 4 r dt d 3 r dt 3 + d r dt + 3r. C The tranfer function i R Cro multiplying, ( )C ( )R. Taking the invere Laplace tranform auming zero initial condition, d 5 c dc d 4 c dt 4 + d 3 c dt d c dt + 5 dc dt + c d 4 r dt 4 + d 3 r dt d r dt + dr dt + r. Subtituting r(t) t 3, d 5 c dc d 4 c dt 4 + d 3 c dt d c dt + 5 dc dt + c. 8δ(t) + ( t + 9t + 3t 3 ) u(t). Taking the Laplace tranform of the differential equation, X-++X-+3xR. Collecting term, ( ++3)X R++. R Solving for X, X The block diagram i then,. Program: 'Factored' Gzpkzpk([ ],[0-55 root([ 5 30])' root([ 7 5])'],5) 'Polynomial' Gptf(Gzpk) Computer repone: an Factored Zero/pole/gain: 5 (+5) (+6) (+7) (+55) (+4.9) (+.087) (^ ) an

8 Solution to Problem 9 Polynomial Tranfer function: 5 ^ ^ ^ ^ ^4 +.30e004 ^ e004 ^ e Program: 'Polynomial' Gtftf([ ],[ ]) 'Factored' Gzpkzpk(Gtf) Computer repone: an Polynomial Tranfer function: ^4 + 5 ^3 + 0 ^ ^5 + 3 ^4 + 9 ^ ^ an Factored 4. Zero/pole/gain: (+4.) (+.35) (^ ) (+.5) (^ ) (^ ) Program: numg[-0-60]; deng[ (root([ 7 00]))' (root([ 6 90]))']; [numg,deng]zptf(numg',deng',e4); Gtftf(numg,deng) Gzpk(Gtf) [r,p,k]reidue(numg,deng) Computer repone: Tranfer function: 0000 ^ e ^ ^ ^ ^ e005 ^3 +.06e006 ^ Zero/pole/gain: 0000 (+60) (+0) (+40) (+30) (^ ) (^ ) r p i i i i e007

9 0 Chapter : Modeling in the Frequency Domain i i i i 0 k [] 5. Program: ym '(a)' Ga45*[(^+37*+74)*(^3+8*^+3*+6)]... /[(+39)*(+47)*(^+*+00)*(^3+7*^+8*+5)]; 'Ga ymbolic' pretty(ga) [numga,denga]numden(ga); numgaympoly(numga); dengaympoly(denga); 'Ga polynimial' Gatf(numga,denga) 'Ga factored' Gazpk(Ga) '(b)' Ga56*[(+4)*(^3+49*^+6*+53)]... /[(^+88*+33)*(^+56*+77)*(^3+8*^+76*+65)]; 'Ga ymbolic' pretty(ga) [numga,denga]numden(ga); numgaympoly(numga); dengaympoly(denga); 'Ga polynimial' Gatf(numga,denga) 'Ga factored' Gazpk(Ga) an Computer repone: (a) an Ga ymbolic an Ga polynimial 3 ( ) ( ) ( + 39) ( + 47) ( ) ( ) Tranfer function: 45 ^ ^ ^ ^

10 Solution to Problem ^7 + 5 ^ ^ ^ ^ e006 ^ e006 an Ga factored Zero/pole/gain: 45 (+34.88) (+6.83) (+.) (^ ) (+47) (+39) (+6.34) (^ ) (^ ) +.75e006 an (b) an Ga ymbolic an Ga polynimial 3 ( + 4) ( ) ( ) ( ) ( ) Tranfer function: 56 ^ ^ ^ ^7 + 5 ^ ^ ^ e006 ^3 +.89e006 ^ an Ga factored Zero/pole/gain: 56 (+47.7) (+4) (^ ) (+87.6) (+80.06) (+54.59) (+.4) ( ) (^ ) 6. a. Writing the node equation, V o V i + V o + V o 0. Solve for V o V i +. b. Thevenizing,

11 Chapter : Modeling in the Frequency Domain 7. Uing voltage diviion, V o V i a Thu, V o V i + + Writing meh equation (+)I I V i -I + (+)I 0 But, I (+)I. Subtituting thi in the firt equation yield, or (+)(+)I I V i I /V i /( ) But, V L I. Therefore, V L /V i /( ). b.

12 Solution to Problem 3 i (t) i (t) ( + )I (+ )I V (+ )I +(+ + )I 0 Solving for I : I ( +) + ( +) + V V Therefore, V L V I V a. Writing meh equation, ( + )I I V i -I + (3 + + /)I 0 Solving for I,

13 4 Chapter : Modeling in the Frequency Domain I + V i Solving for I /V i, I V i But V o I 3. Therefore, G 3 /( ). b. Tranforming the network yield, Writing the loop equation, Solving for I, But, V o I ( + + )I + I I 3 V i + I + ( )I I 0 3 I I +( +)I 3 0 I ( + + ) V i ( + + ) V i. Therefore, V o V i a. Writing the nodal equation yield,

14 Solution to Problem 5 V R V i + V R + V R V C 3 3 V R V C 0 Rewriting and implifying, V R 6 3 V C V i 3 V R V C 0 Solving for V R and V C, V R V i ; V C V i Solving for V o /V i V o V i V R V C V i b. Writing the nodal equation yield, (V V i ) + ( +) V +(V V o ) 0 (V o V ) + V o + (V o V i ) Rewriting and implifying, 0 ( + +)V V o V i V + ( + + )V o V i

15 6 Chapter : Modeling in the Frequency Domain Solving for V o V o Hence, ( + + ) V i. V o V i ( + + ) a. Meh: (+)I - (+)I - I 3 V - (+)I + (7+5)I - (+3)I 3 0 -I - (+3)I + ( )I 3 0 Nodal: V - V + V (+) (V o - V ) +3 + V o 4 or + (V - V o ) +3 + (V o - V) V V o V 3 + V V o V b. Program: ym V %Contruct ymbolic object for frequency

16 Solution to Problem 7 %variable '' and V. 'Meh Equation' A[(+*) V - -(+*) 0 -(+3*) - 0 (3+3*+(5/))] %Form Ak A. A[(+*) -(+*) - -(+*) (7+5*) -(+3*) - -(+3*) (3+3*+(5/))] %Form A. Idet(A)/det(A); %Ue Cramer' Rule to olve for I. GI/V; %Form tranfer function, G I/V. G4*G; %Form tranfer function, G V4/V. 'G via Meh Equation' %Diplay label. pretty(g) %Pretty print G 'Nodal Equation' A[(6*^+*+5)/(6*^+7*+) V -/(3*+) *(V/5)] %Form Ak A. A[(6*^+*+5)/(6*^+7*+) -/(3*+) -/(3*+) (/0)*(*^+3*+30)/(3*+)] %Form A. Iimple(det(A))/imple(det(A)); %Ue Cramer' Rule to olve for I. GI/V; %Form tranfer function, G I/V. 'G via Nodal Equation' %Diplay label. pretty(g) %Pretty print G Computer repone: an Meh Equation A [ +*, V, -] [ --*, 0, --3*] [ -, 0, 3+3*+5/] A [ +*, --*, -] [ --*, 7+5*, --3*] [ -, --3*, 3+3*+5/] an G via Meh Equation an Nodal Equation A [ (6*^+*+5)/(+7*+6*^), V] [ -/(+3*), /5**V] A

17 8 Chapter : Modeling in the Frequency Domain [ (6*^+*+5)/(+7*+6*^), -/(+3*)] [ -/(+3*), (3/5*^+3/0*+3/)/(+3*)] an G via Nodal Equation a. Z 5x Z Therefore, - Z Z b. Z Z Therefore,. G Z + 0 Z + 0 a. Z Z Therefore, Z G + Z 3 Z

18 Solution to Problem 9 b. 5x0 Z x x Z 5x0 5 + Therefore, Z + Z Z 7 ( + 3.8)( +.68) ( + 7)( + 0) 3. Writing the equation of motion, where x (t) i the diplacement of the right member of pringr, ( ++)X -X 0 -X +X F Adding the equation, ( +)X F 4. From which, X F ( +). Writing the equation of motion, ( + +)X ( +)X F ( +)X + ( + +) X 0 Solving for X, X ( + + ) F ( +) 0 ( + +) ( +) ( +) ( + +) ( +)F ( + + ) 5. From which, X F ( +) ( + + ). Let X be the diplacement of the left member of the pring and X 3 be the diplacement of the ma. Writing the equation of motion

19 30 Chapter : Modeling in the Frequency Domain x x F X + (5 + )X 5X 3 0 5X + (0 + 7)X 3 0 Solving for X, 6. Thu, -0 0 X X F (0 + 7) 0 (5 +) F F ( +50+) + 3+ X + X 0 + X X F 0 + F + + Solving for X ; X Writing the equation of motion, F Thu, X F ( + +)X X 0 X + ( + +)X X 3 F X + ( + +)X 3 0 Solving for X 3, X 3 ( + + ) 0 ( + +) F 0 0 ( + +) 0 ( + +) 0 ( + + ) F ( )

20 Solution to Problem 3 8. From which, X 3 F ( ). a. ( + +)X X X 3 F X + ( + 4)X X 3 0 X X + ( +)X 3 0 Solving for X, ( + +) F or, X F + X F + 3 ( ) b. ( ) X (+ ) X 3X 3 0 (+) X + ( + 5+) X 4X 3 F 3X 4X + (4+3) X 3 0 Solving for X 3, (4 ++ 4) (+) 0 X 3 (+) ( + 5+) F F (4 ++ 4) 3 (+) 4 or 9. Writing the equation of motion, X 3 F ( + + )X X X 3 0 X + ( + +)X X 3 F X X + ( + +) 0

21 3 Chapter : Modeling in the Frequency Domain 30. a. Writing the equation of motion, ( )θ ( + 8)θ 0 ( + 8)θ + ( + + )θ T b. Defining θ rotation of J θ rotation between K and D θ 3 rotation of J 3 θ 4 rotation of right - hand ide of K the equation of motion are (J + K )θ K θ T K θ + (D + K )θ D θ 3 0 D θ + (J + D + K )θ 3 K θ 4 0 K θ 3 + (D + (K + K 3 ))θ Writing the equation of motion, ( + + )θ ( + )θ T ( + )θ + ( + )θ 0 Solving for θ Hence, θ ( + +) T ( +) 0 ( + +) ( +) ( +) ( +) T ( +) 3. θ T ( + ) Reflecting impedance to θ 3,

22 Solution to Problem 33 (J eq +D eq )θ 3 T ( N 4 N N 3 N ) Thu, where J eq J 4 +J 5 +(J +J 3 ) N 4 N 3 θ 3 T N + 4 N J N 3 N, and N 4 N N 3 N J eq + D eq D eq (D 4 + D 5 ) + (D + D 3 )( N 4 N 3 ) + D ( N 4N N 3 N ) 33. Reflecting all impedance to θ, {[J +J ( N N ) +J 3 ( N3 N4 ) ] + [f +f ( N N ) +f 3 ( N 3 N 4 ) ] + [K( N 3 N 4 ) ]}θ T N N Subtituting value, {[+(3) +6( 4 ) ] + [+(3) +3( 4 ) ] + 64( 4 ) }θ T(3) Thu, 34. Reflecting impedance to θ, θ T x x T 5 Thu, 35. θ T Reflecting impedance and applied torque to repective ide of the pring yield the following equivalent circuit:

23 34 Chapter : Modeling in the Frequency Domain Writing the equation of motion, θ - θ 3 4T -θ + (+)θ 3 0 Solving for θ 3, θ 3 4T (+ ) 4T 36. Hence, θ 3 T 4. But, θ 4 5 θ 3. Thu, θ 4 T 4/5 Reflecting impedance and applied torque to repective ide of the vicou damper yield the following equivalent circuit:. Writing the equation of motion, ( + )θ θ 3 0T θ + ( +)θ 3 θ 4 0 θ 3 + ( +)θ 4 0 Solving for θ 4,

24 Solution to Problem 35 θ 4 ( +) 0T ( +) ( +) 0 ( +) 0 ( +) 0T ( +) 0 ( +) 0 ( +) Thu, θ 4 T 0 ( +) But, θ L 5θ 4. Hence, 37. θ 4 T 50 ( +) Reflect all impedance on the right to the vicou damper and reflect all impedance and torque on the left to the pring and obtain the following equivalent circuit: Writing the equation of motion, (J eq +K)θ -Kθ 3 T eq -Kθ +(D+K)θ 3 -Dθ 4 0 -Dθ 3 +[J eq +(D+D eq )]θ 4 0 where: J eq J +(J a +J )( N N ) ; J eq J 3 +(J L +J 4 )( N 3 N 4 ) ; D eq D L ( N 3 N 4 ) ; θ θ N N.

25 36 Chapter : Modeling in the Frequency Domain 38. Reflect impedance to the left of J 5 to J 5 and obtain the following equivalent circuit: Writing the equation of motion, [J eq +(D eq +D)+(K +K eq )]θ 5 -[D+K ]θ 6 0 -[K +D]θ 5 + [J 6 +D+K ]θ 6 T θ 6 From the firt equation, θ 5 J eq +(D eq +D)+ (K +K eq ) D+K. But, θ 5 θ N N 3 N N 4. Therefore, θ 6 θ N N 3 N N 4 J eq +(D eq +D)+ (K +K eq ) D+K, where J eq [J ( N 4N N 3 N ) + (J +J 3 )( N 4 N 3 ) + (J 4 +J 5 )], K eq K ( N 4 N 3 ), and D eq D[( N 4N N 3 N ) + ( N 4 N 3 ) + ]. 39. Draw the freebody diagram,

26 Solution to Problem 37 Write the equation of motion from the tranlational and rotational freebody diagram, (M +f v +K )X -f v rθ F -f v rx +(J +f v r )θ 0 Solve for θ, M +f v +K F θ -f v r 0 M +f v +K -f v r f v rf JM 3 +(Jf v +Mf v r ) +(JK +f v r )+K f v r -f v r J +f v r 40. From which, θ F f v r JM 3 +(Jf v +Mf v r ) +(JK +f v r )+K f v r. Draw a freebody diagram of the tranlational ytem and the rotating member connected to the tranlational ytem. From the freebody diagram of the ma, F ( ++)X. Summing torque on the rotating member, (J eq +D eq )θ + F T eq. Subtituting F above, (J eq +D eq )θ + ( ++)X T eq. However, θ X. Subtituting and implifying, T eq [( J eq +) +( D eq +)+]X

27 38 Chapter : Modeling in the Frequency Domain 4. But, J eq +(4) 7, D eq () + 5, and T eq 4T. Therefore, [ + 9 +]X 8 4T. Finally, X T Writing the equation of motion,. (J +K )θ - K θ T -K θ + (J +D 3 +K )θ +Fr -D 3 θ 3 0 -D 3 θ + (J +D 3 )θ 3 0 where F i the oppoing force on J due to the tranlational member and r i the radiu of J. But, for the tranlational member, F (M +f v +K )X (M +f v +K )rθ Subtituting F back into the econd equation of motion, (J +K )θ - K θ T -K θ + [(J + Mr ) +(D 3 + f v r )+(K + K r )]θ -D 3 θ 3 0 -D 3 θ + (J +D 3 )θ 3 0 Notice that the tranlational component were reflected a equivalent rotational component by the quare of the radiu. Solving for θ, θ K (J 3 + D 3 )T, where i the determinant formed from the coefficient of the three equation of motion. Hence, 4. Since Alo, Thu, θ T K (J 3 + D 3 ) X rθ, X T rk (J 3 + D 3 ) K t R a T tall E a ; K b E a ω no load J m +8( 3 ) 4; D m +36( 3 ) 6. θ m E a / 4 ( + 4 (6 + 3 )) / ( )

28 Solution to Problem 39 Since θ L 3 θ m, θ L E a 6 ( ). 43. The parameter are: K t T 5 R a E a 5 ; K b E a ω ; J m π π ; D m 3 4 Thu, θ m E a 3 ( + 3 ( + ()( 4 ))) 3 ( ) 44. Since θ 4 θ m, θ E a ( ). The following torque-peed curve can be drawn from the data given: T v Therefore, K t R a T tall E a ; K b D m. Thu, E a ω no load Alo, J m 5+00( 5 ) 9;

29 40 Chapter : Modeling in the Frequency Domain θ m E a 0 9 ( + 9 ( + 0.)) 0 9 ( + 0.). Since θ L 5 θ m, θ L E a 0 45 ( + 0.) ( + 0.). From Eq. (.45) and (.46), R a I a + K b θ E a () Alo, T m K t I a (J m +D m )θ. Solving for θ and ubtituting into Eq. (), and implifying yield I a E a R a ( + D m J m ) + R a D m + K b K t R a J m () Uing T m K t I a in Eq. (), 46. T m E a K t R a ( + D m J m ) + R a D m + K b K t R a J m For the rotating load, auming all inertia and damping ha been reflected to the load, (J eql +D eql )θ L + Fr T eq, where F i the force from the tranlational ytem, r i the radiu of the rotational member, J eql i the equivalent inertia at the load of the rotational load and the armature, and D eql i the equivalent damping at the load of the rotational load and the armature. Since J eql () + 5, and D eql () + 5, the equation of motion become, (5 +5)θ L + Fr T eq. For the tranlational ytem, ( +)X F. Since X θ L, F ( +)θ L. Subtituting F into the rotational equation, (9 +9) θ L T eq. Thu, the equivalent inertia at the load i 9, and the equivalent damping at the load i 9. Reflecting thee back to the armature, yield an equivalent inertia of 9 4 and an equivalent damping of 9 4. Finally, K t R a ;

30 Solution to Problem K b. Hence, θ m E a 9 (+ 4 9 (9 4 +)) X rθ L θ L. therefore, X E a 4 9 (+ 3 9 ). Since θ L 4 9 ( ) The equation of motion in term of velocity are: θ m, θ L E a 9 (+ 3 9 ). But [M +( f v + f v 3 ) + K + K ]V K V f v 3 V 3 0 K V + [M + ( f v + f v4 ) + K ]V f v4 V 3 F f v3 V f v4 V +[M 3 + f V3 + f v4 ]V 3 (S) 0 For the erie analogy, treating the equation of motion a meh equation yield In the circuit, reitor are in ohm, capacitor are in farad, and inductor are in henrie. For the parallel analogy, treating the equation of motion a nodal equation yield In the circuit, reitor are in ohm, capacitor are in farad, and inductor are in henrie.

31 4 Chapter : Modeling in the Frequency Domain 48. Writing the equation of motion in term of angular velocity, Ω yield (J + D + K )Ω (D + K )Ω T (D + K )Ω + (J + D + (K + K ) )Ω 0 K Ω D Ω 3 + (D + K )Ω 4 0 (J 3 + D + K 3 )Ω 3 D Ω 4 0 For the erie analogy, treating the equation of motion a meh equation yield In the circuit, reitor are in ohm, capacitor are in farad, and inductor are in henrie. For the parallel analogy, treating the equation of motion a nodal equation yield In the circuit, reitor are in ohm, capacitor are in farad, and inductor are in henrie. 49. An input r yield c 5r +7. An input r yield c 5r +7. An input r +r yield, 5(r +r )+7 5r +7+5r c +c -7. Therefore, not additive. What about homogeneity? An input of Kr yield c 50. 5Kr +7 Kc. Therefore, not homogeneou. The ytem i not linear. a. Let x δx+0. Therefore,

32 Solution to Problem 43 δx.. +3δx. +δx in (0+δx) But, in (0+δx) in 0 + d inx dx x0 δx 0+cox δx δx x0 Therefore, δx.. +3δx. +δx δx. Collecting term, δx.. +3δx. +δx 0. b. Let x δx+π. Therefore, δx.. +3δx. +δx in (π+δx) But, in (π+δx) in π + d inx dx δx 0+cox xπ xπ δx δx Therefore, δx+3δx+δx -δx. Collecting term, δx+3δx+3δx 0. If x 0 + δx,... δx + 0δx.. + 3δx. + 30δx e -(δx) But e -(δx) e -0 + de-x dx... Therefore, δx.. + 0δx δx - e -x δx - δx x0 x δx + 30δx - δx, or, δx The given curve can be decribed a follow:.. + 0δx. + 3δx + 3δx. f(x) -4 ; - <x<-; f(x) x; -<x<; f(x) 4; <x<+ Thu, a. b... x + 5ẋ +50x x + 5ẋ + 50x x, or ẋ + 5ẋ +48x c... x + 5ẋ + 50x 4 The relationhip between the nonlinear pring diplacement, x (t) and it force, f (t) i x (t) e f (t) Solving for the force, Writing the differential equation for the ytem by umming force, d x(t) dt f (t) ln( x (t)) () + dx(t) dt ln( x(t)) f (t) ()

33 44 Chapter : Modeling in the Frequency Domain Letting x(t) x 0 + δx and f(t) + δf, linearize ln( x(t)). ln( x) ln( x 0 ) Solving for ln( x), d ln( x) dx x x 0 δx ln( x) ln( x 0 ) δx ln( x 0 ) δx (3) x x x0 x 0 When f, δx 0. Thu from Eq. (), -ln( x 0 ). Solving for x 0, 54. x 0 e -, or x Subtituting x into Eq. (3), ln(- x) ln( 0.63) δx δx Placing thi value into Eq. () along with x(t) x 0 + δx and f(t) + δf, yield the linearized differential equation, or d δx dt + dδx dt δx +δf d δx + dδx +.78δx δf dt dt Taking the Laplace tranform and rearranging yield the tranfer function, δx δf Firt aume there are n plate without the top plate poitioned at a diplacement of y (t) where y (t) 0 i the poition of the untretched pring. Aume the ytem conit of ma M, where M i the ma of the dipening ytem and the n plate, vicou friction, f v, where the vicou friction originate where the piton meet the ide of the cylinder, and of coure the pring with pring contant, K. Now, draw the freebody diagram hown in Figure (b) where W n i the total weight of the n dihe and the piton. If we now conider the current poition, y (0),

34 Solution to Problem 45 Retaurant Plate Dipener the equilibrium point and define a new diplacement, y (t), which i meaured from equilibrium, we can write the force in the pring a Ky (t) Ky (0) + Ky (t). Changing variable from y (t) to y (t), we um force and get, M d y dt + f dy v dt + Ky + Ky (0) + W n 0 () where d y dt d y dt and dy dt dy dt. But, Ky (0) -W n, ince it i the component of the pring force that balance the weight at equilibrium when y 0. Thu, the differential equation become, M d y dt + f dy v dt + Ky 0 () When the top plate i added, the pring i further compreed by an amount, W D K, where W D i the weight of the ingle dih, and K i the pring contant. We can think of thi diplacement a an initial condition. Thu, y (0-) - W D K and dy dt (0-) 0, and y (t) 0 i the equilibrium poition of the pring with n plate rather than the untretched poition. Taking the Laplace tranform of equation (), uing the initial condition,

35 46 Chapter : Modeling in the Frequency Domain M( Y + W D K ) + f v(y + W D K ) + KY 0 (3) or (M + f v + K)Y - W D K (M + f v ) (4) Now define a new poition reference, Y, which i zero when the pring i compreed with the initial condition, or Y Y + W D K Y Y - W D K (5) (6) Subtituting Y in Equation (4), we obtain, (M + f v + K)Y W D F (7) 55. a differential equation that ha an input and zero initial condition. The chematic i hown in Figure (c). Forming the tranfer function, Y F, we how the final reult in Figure (d), where for the removal of the top plate, F i alway a tep, F W D Writing the equation of motion,. ( )Y f ( )Y h 0Y cat F up - ( )Y f + ( )Y h Y cat 0-0Y f Y h x0 6 Y cat 0 Thee equation are in the form AYF, where det(a).534 x 0 8 ( ) ( ) Uing Cramer rule: Y cat F up 0.047( ) ( )( ) Y h F up ( ) ( )( ) Y h Y cat F up ( ) ( )( )