Process.Dynamics.and.Control.Seborg.2nd.Ch03.pdf

12345678983.1 a) [ 1e − bt ∞ ] sin ωt = ∫ e − bt 0 ∞ sin ωt e dt = ∫ sin ωt e − ( s + b )t dt − st 0  [− (s + b) sin ωt − ω cos ωt ] = e − ( s + b ) t  ( s + b ) 2 + ω2  0 ω = ( s + b) 2 + ω2 ∞ b) [ 1 e − bt ∞ ] cos ωt = ∫ e 0 − bt ∞ cos ωt e dt = ∫ cos ωt e − ( s + b )t dt − st 0  [− (s + b) cos ωt + ω sin ωt ] = e − ( s + b ) t  ( s + b) 2 + ω2  0 s+b = ( s + b) 2 + ω2 ∞ 3.2 a) The Laplace transform provided is Y ( s) = 4 s + 3s + 4 s 2 + 6 s + 4 4 3 We also know that only sin ωt is an input, where ω = X ( s) = ω 2 = 2 s +ω s2 + 2 2 ( ) 2 = 2 . Then 2 s +2 2 Since Y(s) = D-1(s) X(s) where D(s) is the characteristic polynomial (when all initial conditions are zero), Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp. 3-1 Y (s) = 2 2 2 2 ( s + 3s + 2) ( s + 2) 2 and the original ode was d2y dy + 3 + 2 y = 2 2 sin 2t 2 dt dt with y ′(0) = y (0) = 0 b) This is a unique result. c) The solution arguments can be found from Y (s) = 2 2 2 ( s + 1)( s + 2) + ( s 2 + 2) which in partial fraction form is Y (s) = α1 α a s + a2 + 2 + 12 s +1 s + 2 s +2 Thus the solution will contain four functions of time e-t , e-2t , sin 2 t , cos 2 t 3.3 a) Pulse width is obtained when x(t) = 0 Since x(t) = h – at tω : h − atω = 0 or tω = h/a b) h slope = -a slope = a x(t) x(t) slope = -a x(t) = hS(t) – atS(t) + a(t -tω) S(t-tω) 3-2 3-104. 3-3 .4e -2s .e -6s ) s 3.6) 1 F (s ) = = ( 5 . Thus X(s) = [ a 1 − e −tr s − e − 2tr s + e −3tr s 2 s ] by utilizing the Real Translation Theorem Eq.c) h a ae − stω h e − stω − 1 X ( s) = − 2 + 2 = + s s s s s2 d) Area under pulse = h tω/2 a) f(t) = 5 S(t) – 4 S(t-2) – S(t.4 b) x(t) x1 a a x4 tr 2tr 3tr -a -a x2 x3 x(t) = x1(t) + x2(t) + x3(t) + x4(t) = at – a(t − tr)S(t − tr) − a(t −2tr)S(t − 2tr) + a(t − 3tr)S(t − 3tr) following Eq. 3-101. 3.6 a) X ( s) = α1 = α2 = α3 = b) α α α s ( s + 1) = 1 + 2 + 3 ( s + 2)( s + 3)( s + 4) s + 2 s + 3 s + 4 s ( s + 1) ( s + 3)( s + 4) s ( s + 1) ( s + 2)( s + 4) s ( s + 1) ( s + 2)( s + 3) =1 s = −2 = −6 s = −3 =6 s = −4 X (s) = 1 6 6 − + s+2 s+3 s+4 X (s) = s +1 s +1 = 2 ( s + 2)( s + 3)( s + 4) ( s + 2)( s + 3)( s + 2 j )( s − 2 j ) X ( s) = α + jβ 3 α 3 − j β 3 α1 α + 2 + 3 + s+2 s+3 s+2j s+2j α1 = α2 = s +1 ( s + 3)( s 2 + 4) s +1 ( s + 2)( s 2 + 4) α 3 + jβ 3 = =− s = −2 = s = −3 x(t ) = e−2t − 6e −3t + 6e −4t and 1 8 2 13 s +1 ( s + 2)( s + 3)( s − 2 j ) 3-4 = s = −2 j 1− 2 j − 3 + 11 j = − 40 − 8 j 208 .5 55 55 t S(t) – (t-30) S(t-30) 30 30 20 55 1 55 1 −30 s 20 55 1 T (s) = + − e = + 1 − e −30 s 2 2 2 s 30 s 30 s s 30 s T(t) = 20 S(t) + ( ) 3. we first ignore the time delay term. Xˆ ( s ) = s +1 A B C = + + s ( s + 2)( s + 3) s s + 2 s + 3 Multiply by s and let s → 0 3-5 . Using the Heaviside expansion with the partial fraction expansion. Eq.5 s s ( s + 2)( s + 3) To invert. substitute any s≠-1 to determine α1.1 2  −3   11  x(t ) = − e − 2t + e −3t + 2  cos 2t + 2  sin 2t 8 13  208   208  1 2 3 11 = − e − 2t + e −3t − cos 2t + sin 2t 8 13 104 104 c) X ( s) = α α2 s+4 = 1 + 2 s + 1 ( s + 1) 2 ( s + 1) (1) α 2 = ( s + 4) s =−1 = 3 In Eq. Arbitrarily using s=0. 1. 1 gives 4 α1 3 = + 1 12 12 or α1 = 1 1 3 + and x( t ) = e− t + 3te − t 2 s + 1 ( s + 1) 1 1 1 X (s) = 2 = = 2 2 2 s + s +1  1  3 (s + b ) + ω s +  + 2 4  1 3 where b = and ω = 2 2 X (s) = d) t 1 2 −2 3 x(t ) = e −bt sin ω t = e sin t ω 2 3 e) X(s) = s +1 e −0. 5) = 1 1 − 2 (t −0.5) + e − e 6 2 3 for t ≥ 0.5 3.7 a) Y (s) = α 6( s + 1) 6 α = 2 = 1 + 22 2 s s ( s + 1) s s 6 s2 6 Y (s) = 2 s α2 = s2 b) Y (s) = =6 α1 = 0 s =0 12( s + 2) α1 α 2 s + α 3 = + 2 s s ( s 2 + 9) s +9 3-6 .1 1 = (2)(3) 6 Multiply by (s+2) and let s→ −2 A= B= − 2 +1 −1 1 = = (−2)(−2 + 3) (−2)(1) 2 Multiply by (s+3) and let s→-3 C= − 3 +1 −2 2 = =− (−3)(−3 + 2) (−3)(−1) 3 Then 1 6 1 2 −2 3 Xˆ ( s ) = + + s s+2 s+3 xˆ (t ) = 1 1 − 2 t 2 −3t + e − e 6 2 3 Imposing shift theorem x(t ) = xˆ (t − 0.5) 2 −3( t − 0. s2: α1 + α2 = 0 s1: α3 = 12 0 s : 9α1 = 24 Solving simultaneously. ( s + 2)( s + 3) ( s + 5)( s + 6) ( s + 2)( s + 3) ( s + 4)( s + 6) ( s + 2)( s + 3) ( s + 4)( s + 5) =1 s = −4 = −6 s = −5 =6 s = −6 Y (s) = 1 6 6 − + s+4 s+5 s+6 Y (s) = [(s + 1) = α 3 = 12 α α α ( s + 2)( s + 3) = 1 + 2 + 3 ( s + 4)( s + 5)( s + 6) s + 4 s + 5 s + 6 Y (s) = α1 = d) 8 3 1 2 ] 2 + 1 ( s + 2) = 1 ( s + 2 s + 2) 2 ( s + 2) 2 α s + α4 α α1 s + α 2 + 2 3 + 5 2 2 s+2 s + 2s + 2 ( s + 2s + 2) 3-7 . . −8 3  8   − s + 12  8 1  3  Y (s) = + 3s s2 + 9 α1 = c) α2 = α3 = α2 = .Multiplying both sides by s(s2+9) 12( s + 2) = α 1 ( s 2 + 9) + (α 2 s + α 3 )( s ) or 12 s + 24 = (α 1 + α 2 ) s + α 3 s + 9α 1 2 Equating coefficients of like powers of s. 3-100 t  1 1  ∫ f (t * )dt *  = F ( s ) 0  s  t −τ  1 1 we know that 1  ∫ e dτ = 21 e − τ = s ( s + 1) 0  s [ ] ∴ Laplace transforming yields 2 s ( s + 1) 2 or (s2 + 3s + 1) X(s) = s ( s + 1) s2X(s) + 3X(s) + 2X(s) = 3-8 α5 = ¼ .8 a) From Eq.Multiplying both sides by ( s 2 + 2s + 2) 2 ( s + 2) gives 1 = α1s4 + 4α1s3 + 6α1s2 +4α1s + α2s3 +4α2s2 +6α2s +4α2 + α3s2 +2α3s + α4s + 2α4 + α5s4 + 4α5s3 + 8α5s2 + 8α5s + 4α5 Equating coefficients of like power of s. s4 : α1 + α5 = 0 s3 : 4α1 + α2 + 4α5 = 0 s2 : 6α1 + 4α2 + α3 + 8α5 = 0 s1 : 4α1 + 6α2 + 2α3 + α4 + 8α5 = 0 s0 : 4α2 + 2α4 + 4α5 = 1 Solving simultaneously: α1 = -1/4 Y (s) = α2 = 0 α3=-1/2 α4=0 − 1 / 4s − 1 / 2s 1/ 4 + 2 + 2 s+2 s + 2 s + 2 ( s + 2 s + 2) 2 3. 9 a) X (s) = 6( s + 2) 6( s + 2) = ( s + 9s + 20)( s + 4) ( s + 4)( s + 5)( s + 4) 2  6s ( s + 2)  =0 x(0) = lim  s →∞ ( s + 5)( s + 4) 2     6s ( s + 2)  x(∞) = lim =0 s →0 ( s + 5)( s + 4) 2    x(t) is converging (or bounded) because [sX(s)] does not have a limit at s = −4. it has a limit for all real values of s ≥ 0. See Fig. See Fig. x(t) is diverging (unbounded). S3.9b. and s = −5 only.. x(t) is smooth because the denominator of [sX(s)] is a product of real factors only. S3. 3-9 .e. For the same reason..X(s) = x(t) = 1 − 2te-t − e-2t and b) 2 s ( s + 1) 2 ( s + 2) Applying the final Value Theorem lim x(t ) = lim sX ( s ) = lim t →∞ s →0 s →0 2 =2 ( s + 1) ( s + 2) 2 [ Note that Final Value Theorem is applicable here] 3. i. at s = 3±2j. i. b) 10 s 2 − 3 10 s 2 − 3 X (s) = 2 = ( s − 6 s + 10)( s + 2) ( s − 3 + 2 j ) ( s − 3 − 2 j )( s + 2)   10s 3 − 3s x(0) = lim  2  = 10 s →∞ ( s − 6 s + 10)( s + 2)   Application of final value theorem is not valid because [sX(s)] does not have a limit for some real s ≥ 0.e. x(t) is oscillatory because the denominator of [sX(s)] includes complex factors.9a. S3.3 0.2 0.5 Time Figure S3. since divergence occurs only if [sX(s)] does not have a limit for some real value of s>0.9a. See Fig. This implies that x(t) is not diverging.5 2 2.9b.5 4 4.16 s + 5 16s + 5 = 2 ( s + 9) ( s + 3 j ) ( s − 3 j ) X (s) = 16s 2 + 5s  x(0) = lim  2  = 16 s →∞  ( s + 9)  Application of final value theorem is not valid because [sX(s)] does not have a limit for real s = 0.15 0. Since x(t) is oscillatory.25 x(t) 0.5 Figure S3. x(t) is oscillatory because the denominator of [sX(s)] is a product of complex factors.9c 0.1 0. it is not converging either.5 Time 3 3.5 1 1.5 2 2. Simulation of X(s) for case b) 3-10 5 . Simulation of X(s) for case a) 8000 6000 4000 2000 x(t) c) 0 -2000 -4000 -6000 -8000 -10000 -12000 0 0.05 0 0.05 0 -0.5 1 1.35 0.4 0. 3-11 . (At time t = 0 and t =0.5 Time 3 3.20 15 10 x(t) 5 0 -5 -10 -15 -20 0 0.001 with changes of magnitude 1000 and –1000 respectively).5 1 1. b) and c). Simulink block diagram for cases a). An impulse input should be used to obtain the function’s behavior.9c.5 2 2. The MATLAB command impulse might also be used.9d.5 4 4. Simulation of X(s) for case c) The Simulink block diagram is shown below.5 5 Figure S3. Figure S3. In this case note that the impulse input is simulated by a rectangular pulse input of very short duration. 3. e-2t sin2t. t. e-4t Y(s) = 2 2 A B C = = + + s ( s + 4 s + 3) s ( s + 1)( s + 3) s s + 1 s + 3 ∴ y(t) will contain terms of form: constant. e-2t . e-t. e-2tcos2t A Bs C 2( s + 1) 2( s + 1) = = + 2 + 2 2 2 2 2 s ( s + 4) s ( s + 2 ) s s + 2 s + 22 2( s + 1) 1 A = lim 2 = s → 0 ( s + 4) 2 Y (s) = 2(s+1) = A(s2+4) + Bs(s) + Cs 2s+2 = As2 + 4A + Bs2 + Cs Equating coefficients on like powers of s s2 : 0=A+B → B = −A = − s1 : 2= C → s0 : 2 = 4A → C=2 1 A= 2 3-12 1 2 . e-3t Y (s) = 2 2 A B C = = + + 2 2 s ( s + 2) s+2 s ( s + 4 s + 4) s ( s + 2) ∴ y(t) will contain terms of form: constant.10 a) i) ii) iii) iv) Y(s) = 2 2 A B C = 2 = 2+ + s s+4 s ( s + 4 s ) s ( s + 4) s ∴ y(t) will contain terms of form: constant. te-2t Y (s) = 2 s ( s + 4 s + 8) 2 2 2 2 s 2 + 4 s + 8 = ( s 2 + 4 s + 4) + (8 − 4) = ( s + 2) 2 + 2 2 2 Y (s) = s[(s + 2) 2 + 2 2 ] ∴ b) y(t) will contain terms of form: constant. 11 a) Since convergent and oscillatory behavior does not depend on initial dx 2 (0) dx(0) conditions. x(t) is not oscillatory. See Fig.11c. x(t) is divergent. The denominator vanishes at s=1 ≥0. c) s 3 X ( s) + X (s) = X (s) = 1 s +1 2 1 = ( s + 1)( s 3 + 1) 2 1 1 3 1 3 ( s + j )( s − j )( s + 1)( s − + j )( s − − j) 2 2 2 2 The denominator contains complex factors. S3.11b. assume = = x ( 0) = 0 dt dt 2 Laplace transform of the equation gives s 3 X ( s ) + 2 s 2 X ( s ) + 2 sX ( s ) + X ( s ) = 3 1 3 1 3 s ( s + 1)( s + + j )( s + − j) 2 2 2 2 Denominator of [sX(s)] contains complex factors so that x(t) is oscillatory. x(t) is oscillatory. S3. S3.11a. See Fig. The denominator vanishes at real s = 0.∴ Y(s) = 1 2 − (1 2) s 2 + 2 + 2 2 s s +2 s + 22 y(t) = 1 1 2 − cos 2t + sin 2t 2 2 2 y(t) = 1 (1 − cos 2t ) + sin 2t 2 3. ½. 3-13 . x(t) is not convergent. and denominator vanishes at real values of s= −1 and -½ which are all <0 so that x(t) is convergent. See Fig. 2 s 2 X ( s) − X (s) = s −1 2 2 X (s) = = 2 ( s − 1)( s − 1) ( s − 1) 2 ( s + 1) X (s) = b) 3 s 3 = s ( s + 2 s 2 + 2 s + 1) 3 The denominator contains no complex factors. 5 2 2.11a. The denominator of [sX(s)] vanishes at s = 0.5 1 1.5 0 1 2 3 4 5 time 6 7 8 9 10 Figure S3.5 time 3 3.11b.5 3 2. Simulation of X(s) for case b) 3-14 .5 x(t) 2 1.5 0 -0.11d. x(t) is not convergent. S3.5 1 0. x(t) is not oscillatory.5 5 Figure S3. 3.5 4 4.s 2 X ( s ) + sX ( s ) = 4 s 4 4 = 2 s ( s + s ) s ( s + 1) X (s) = 2 The denominator of [sX(s)] contains no complex factors. Simulation of X(s) for case a) 700 600 500 400 x(t) d) 300 200 100 0 0 0. See Fig. 11d.5 3 3.5 1 1. we Laplace Transform with x ( 0) = dx(0) =0 dt τ1τ2s2X(s) + (τ1+τ2)sX(s) + X(s) = KU(s) 3-15 .80 60 x(t) 40 20 0 -20 -40 0 1 2 3 4 5 time 6 7 8 9 10 Figure S3.11c. Simulation of X(s) for case d) 3.12 Since the time function in the solution is not a function of initial conditions. Simulation of X(s) for case c) 18 16 14 12 x(t) 10 8 6 4 2 0 0 0.5 5 time Figure S3.5 2 2.5 4 4. e– t/τ2. e-t/τ1. t e– t/τ1. e– t/τ2) c) If u(t) =ce-t/τ where τ = τ1 .X ( s) = K U (s) τ1 τ 2 s + ( τ1 + τ 2 ) s + 1 2 Factoring denominator X ( s) = a) K U (s) (τ1 s + 1)(τ 2 s + 1) If u(t) = a S(t) then U(s)= X a (s) = a s Ka s (τ1 s + 1)(τ 2 s + 1) τ1 ≠ τ 2 xa(t) = fa( S(t). e– t/τ2) b) If u(t) = be-t/τ then U(s) = X b ( s) = bτ τs +1 Kbτ (τs + 1)(τ1 s + 1)(τ 2 s + 1) τ ≠ τ1 ≠ τ 2 xb(t) = fb(e-t/τ . e-t/τ1. then U(s) = X c ( s) = τc τ1 s + 1 Kcτ (τ1 s + 1) 2 (τ 2 s + 1) xc(t) = fc(e– t/τ1. cos ωt) 3-16 . sin ωt. e– t/τ2) d) If u(t) = d sin ωt then U(s) = X d (s) = dω s + ω2 2 Kd ( s + ω )(τ1 s + 1)(τ 2 s + 1) 2 2 xd(t) = fd(e– t/τ1. 074 cos 1.37 j s − 0.79 −1.059 j − 0.59 j s = 0.059 j X(s) = 5 + 19. s 3 X(s) + 4 X(s) = X (s) = = 1 s −1 1 1 = 3 ( s − 1)( s + 4) ( s − 1)( s + 1.6 dx − 12 x = sin 3t with x(0) = 0 dt b) sX (s) − 12 X(s) = X (s) = = 3 s +9 2 3 3 = ( s + 9)( s − 12) ( s + 3 j )( s − 3 j )( s − 12) 2 α3 α 1 + j β1 α 1 − j β 1 + + s+3j s−3j s − 12 3-17 .37t + 0.37 j ) =− s = −1.37 j )( s − 0.3.59 s − 0.79 − 1.79 − 1.13 dx3 + 4 x = et 3 dt a) d 2 x(0) dx(0) = = x(0) = 0 dt 2 dt with Laplace transform of the equation.79 − 1.37 j ) α 3 + jβ 3 α 3 − jβ 3 α1 α2 + + + s − 1 s + 1.79 + 1.79 + 1.074 − 0.37t ) 5 19.59 1 19.37 j ) α 3 + jβ 3 = 1 ( s − 1)( s + 1.79t (0.059 sin 1.79 − 1.79 + 1.6 = −0.074 + 0.37 j )( s − 0.59t x(t ) = e t − e − 2e 0.74 − 0.59)( s − 0.59)( s − 0.79 + 1.37 j α1 = α2 = 1 ( s + 4) = 3 s =1 1 5 1 ( s − 1)( s − 0.37 j 1 1 −1.6 + + s − 1 s + 1.37 j s − 0.37 j 1 1 − − 0.59 s − 0.79 − 1. α 1 + j β1 = α3 = = s = −3 j 3 1 4 =− − j − 18 + 72 j 102 102 3 1 = ( s + 9 ) s =12 51 2 − X (s) = x(t ) = − c) 3 ( s − 3 j )( s − 12) 1 4 1 4 1 − j − + j 102 102 + 102 102 + 51 s+3j s −3j s − 12 1 1 (cos 3t + 4 sin 3t ) + e12 t 51 51 d2x dx + 6 + 25 x = e −t 2 dt dt dx(0) = x(0) = 0 dt with s 2 X ( s ) + 6 sX ( s ) + 25X ( s ) + X ( s ) = X ( s) = α1 = 1 or s +1 X( s ) = 1 ( s + 1 )( s + 6 s + 25 ) 2 α α + β2 j α 2 − β2 j 1 = 1 + 2 + ( s + 1)( s + 3 + 4 j )( s + 3 − 4 j ) s + 1 s + 3 + 4 j s + 3 − 4 j 1 ( s + 6 s + 25) = 2 α 2 + jβ 2 = s = −1 1 20 1 ( s + 1)( s + 3 − 4 j ) =− s = −3− 4 j 1 1 − j 40 80 1 1 1 1 1 − − j − − j X ( s ) = 20 + 40 80 + 40 80 s +1 s + 3 + 4 j s + 3− 4 j x(t ) = d) 1 −t 1 1 e − e −3t ( cos 4t + sin 4t ) 20 20 40 Laplace transforming (assuming initial conditions = 0. since they do not affect results) sY1(s) + Y2(s) = X1(s) (1) sY2(s) – 2Y1(s) + 3 Y2(s) = X2(s) (2) 3-18 . For Y2(s) Y2 ( s ) = 2 A B C = + + 2 s+2 ( s + 1) ( s + 2) ( s + 1) ( s + 1) 2 so that y2(t) will contain the same functions of time as y1(t) (although different coefficients). 3-19 .1 sY1(s) + 1 2 X 2 (s) + Y1 ( s ) = X1(s) s+3 s+3 We neglect X2(s) since it is equal to zero. [s(s + 3) + 2]Y1 (s) = ( s + 3) X 1 (s) ( s 2 + 3s + 2)Y1 ( s ) = ( s + 3) X 1 ( s ) Y1 ( s ) = s+3 s+3 X 1 (s) = X 1 (s) ( s + 1)( s + 2) s + 3s + 2 2 1 s +1 s+3 A B C Y1 ( s ) = = + + 2 2 ( s + 1) ( s + 2) ( s + 1) ( s + 1) s+2 Now if x1(t) = e-t then X1(s) = ∴ so that y1(t) will contain e-t/τ . (s+3) Y2(s) = X2(s) + 2Y1(s) Y2 ( s ) = 1 2 X 2 (s) + Y1 ( s ) s+3 s+3 Substitute in Eq.From (2). te-t/τ. e–2t functions of time. 38e-1. ζ = 1.5e-1. y(t)= yˆ (t − 2) is shown in Fig. Y (s) − (1 − 4s )e −2 s = G(s) = 2 X ( s) s + 3s + 1 a) (1) The standard form of the denominator is : τ2s2 + 2ζτs + 1 From (1) .5 0 -0.5 1 0. S3.11t) Using MATLAB-Simulink. s2Y(s) + 3sY(s) + Y(s) = 4 e-2ssX(s) −e-2sX(s) Rearranging.5 3-20 .5 Thus the system will exhibit overdamped and non-oscillatory response. Output variable for a step change in x of magnitude 1.3.5t cosh(1.5 and Y(s) = 2 X(s) = s ( s + 3s + 1) s Therefore yˆ (t ) = −1.5t sinh(1.5 0 5 10 15 20 25 30 Figure S3.5 + 1.5 − (1 − 4 s )e −2 s 1. b) Steady-state gain K = lim G ( s ) = −1 (from (1)) s →0 c) For a step change in x 1.5 -1 -1.14 1.11t) + 7. τ = 1 .14.14 d 2 y (t ) dy (t ) d ( x − 2) +3 + y (t ) = 4 − x(t − 2) 2 dt dt dt Taking the Laplace transform and assuming zero initial conditions. 4 0.1 0 0 0.8 1 time 1.2 0. 10 and 100 3-21 .2 0.8 0.5 0.9 0.15 f (t ) = hS (t ) − hS (t − 1 / h) dx + 4 x = h[S (t ) − S (t − 1 / h)] dt x(0)=0 .2 1.8 2 Figure S3.3 0.3.6 0.15.7 x(t) 0.  1 e−s / h   sX ( s ) + 4 X ( s ) = h − s  s α  1 α X ( s ) = h(1 − e − s / h ) = h(1 − e − s / h )  1 + 2  s ( s + 4)  s s + 4 1 1 1 1 α1 = = . Take Laplace transform.6 0.4 0. α2 = =− s + 4 s =0 4 s s =−4 4 h 1  1 (1 − e − s / h )  −  4 s s + 4 X ( s) = = h 1 e −s / h 1 e −s / h  − − +   4 s s s + 4 s + 4 0 h (1 − e − 4t ) 4 h − 4( t −1 / h ) e − e −4t 4 x(t ) = [ t <0 0 < t < 1/h ] t > 1/h 1 h=1 h=10 h=100 0.6 1.4 1. Solution for values h= 1. ω=2} .3.16 a) Laplace transforming [s Y (s) − sy(0) − y′(0)]+ 6[sY (s) − y(0)] + 9Y (s) = s s+ 1 2 2 (s2 + 6s + 9)Y(s) − s(1) − 2 –(6)(1)= (s2 + 6s + 9)Y(s) = s s +1 2 s +s+8 s +1 2 s + s 3 + s + 8s 2 + 8 (s + 6s + 9)Y(s) = s2 +1 2 Y(s) = s 3 + 8s 2 + 2 s + 8 ( s + 3) 2 ( s 2 + 1) To find y(t) we have to expand Y(s) into its partial fractions Y (s) = A B Cs D + + 2 + 2 2 ( s + 3) s + 3 s +1 s +1 y(t) = Ate-3t + Be-3t + C cost + D sint b) Y(s)= s +1 s ( s + 4 s + 8) Since 42 < 8 we know we will have complex factors. 4 2 ∴ complete square in denominator s2 + 4s + 8 = s2 + 4s + 4 + 8−4 = s2 + 4s + 4 + 4 = (s+2)2 + (2)2 ∴ Partial fraction expansion gives 3-22 { b = 2 . t ≤0 t>0 3-23 . we can Laplace Transform sVC(s) + qC(s) = q Ci(s) Note that c(t = 0) = 0 Also. ci(t) = 0 ci(t) = ci .17 V dC + qC = qCi dt Since V and q are constant.Y (s) = A B ( s + 2) C s +1 + 2 + 2 = 2 s s + 4 s + 8 s + 4 s + 8 s ( s + 4 s + 8) Multiply by s and let s→0 A=1/8 Multiply by s(s2+4s+8) A(s2+4s+8) + B(s+2)s + Cs = s + 1 As2 + 4As + 8A + Bs2 + 2Bs + Cs = s + 1 Y (s) = → B = −A = − s2 : A+B=0 s1 : 4A + 2B + C = 1 s0 : 8A = 1 →A= 1 1 3 C = 1 + 2  − 4  = 8 8 4 → 1 8 1 8 (This checks with above result) 1 / 8 (− 1 / 8)( s + 2) 3/ 4 + + 2 2 s ( s + 2) + 2 ( s + 2) 2 + 2 2 1 1  3 y(t) =   −   e-2t cos 2t +   e-2t sin 2t 8 8 8 3. . Concentration response of the reactor effluent stream.17. Ci=50 Kg/m3 and q = 0. (Consider V = 2 m3. 3-24 .1 to invert (τ =V/q) V − t    c(t) = ci 1 − e q      Using MATLAB. the concentration response is shown in Fig.4 m3/min) 50 45 40 35 c(t) 30 25 20 15 10 5 0 0 5 10 15 20 25 30 Time Figure S3.17. S3. Ci ( s ) = ci s so that sVC(s) + qC(s) = q ci s or C(s) = qci ( sV + q ) s Dividing numerator and denominator by q C(s) = ci V   s +1 s q  Use Transform pair #3 in Table 3. a constant.Laplace transforming the input function. 3.18 a) If Y(s) = KAω s( s 2 + ω2 ) and input U(s) = Aω = 1 {A sin ωt} (s + ω2 ) 2 then the differential equation had to be dy = Ku (t ) dt b) Y(s) = α1 = with y(0) = 0 α ω α α s KAω = 1+ 2 2 2 + 2 3 2 2 2 s( s + ω ) s s + ω s +ω KAω s + ω2 2 = s →0 KA ω Find α2 and α3 by equating coefficients KAω= α1(s2+ω2) + α2s2+α3ωs KAω = α1s2 + α1ω2 + α2s2 + α3ωs s2 : 0 = α1 + α2 s: 0 = α3 ω → α2 = −α1 = → α3 = 0 KAω KA / ω ( KA / ω) s = − 2 2 2 s s( s + ω ) s + ω2 KA (1 − cos ωt ) y(t) = ω ∴ Y(s) = 3-25 − KA ω . 3-26 . which is a function of frequency. which is constant for all ω ii) The amplitudes of the two sinusoidal quantities are: y : KA/ω u: A Thus their ratio is K/ω.c) A u(t) -A 2KA/ω y(t) 0 Time i) We see that y(t) follows behind u(t) by 1/4 cycle = 2π/4= π/2 rad.