# ch06 - 6-1 6-2 b)Process output will be unbounded because...

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Unformatted text preview: 6-1 6-2 b)Process output will be unbounded because some poles lie in the right half plane. c)By using Simulink-MATLAB 51015202530-16-14-12-10-8-6-4-22x 108TimeOutputFigure E6.1b. Response of the output of this process to a unit step input. As shown in Fig. S6.1b, the right half plane pole pair makes the process unstable. 6.2a)Standard form = )1)(1()1(21+τ+τ+τsssKab)Hence )12)(15.()12(5.)(5+++=-ssessGsApplying zero-pole cancellation: )15.(5.)(5+=-sesGsc)Gain = 0.5 Pole = -2 Zeros = No zeros due to the zero-pole cancellation. 6-3 d)1/1 Pade approximation: )2/51()2/51(5sses+-=-The transfer function is now )2/51()2/51(15.5.)(ssssG+-×+=Gain = 0.5 Poles = -2, -2/5 Zeros = + 2/5 6.3)1()1()()(1+τ+τ=ssKsXsYa, sMsX=)(From Eq. 6-13 y(t)= KMττ-τ+=ττ--τ-τ-11/11/1111tataeKMea)KMKMyaa1111)(ττ=ττ-τ+=+b)Overshoot →y(t) > KMKMeKMtaττ-τ+τ-1/111or τa-τ1> 0 , that is, τa> τ1)(1/211<ττ-τ-=τ-KMforeKMyta6-4 6.4)1)(1()1()()(21+τ+τ+τ=sssKsXsYa, τ1>τ2, X(s) = M/s From Eq. 6-15 τ-ττ-τ-τ-ττ-τ+=τ-τ-21/212/2111)(tataeeKMtya)Extremum →)(=ty6-5 Since τ1> τ2, τa< 0. d) If an extremum in y exists, then from (a) =τ-τ-2111teττ-ττ-1211aaττ-ττ-τ-τττ=12212111lnaat6.5Substituting the numerical values into Eq. 6-15 Case (i) : y(t) = 1 (1 + 1.25e-t/10-2.25e-t/2) Case (ii(a)) : y(t) = 1 (1 -0.75e-t/10-0.25e-t/2) Case (ii(b)) : y(t) = 1 (1 -1.125e-t/10+ 0.125e-t/2) Case (iii) : y(t) = 1 (1 -1.5e-t/10+ 0.5e-t/2) 5101520253035404550-0.20.20.40.60.811.21.41.6Timey(t)case(i)case(ii)acase(ii)bcase(iii)Figure S6.5. Step response of a second-order system with a single zero. 6-6 Conclusions: τa> τ1gives overshoot. 0 < τa< τ1gives response similar to ordinary first-order process response. τa< 0 gives inverse response. 6.6)(1)(1)()(2121sUsKsKsUsKsUsKsY+τ+=+τ+=)1()()1()()(121211+τ++τ=+τ++τ=ssKsKKsssKKsKsUsYPut in standard K/τform for analysis: )1(1)()()(121+τ++τ==sssKKKsUsYsGa)Order of G(s) is 2 (maximum exponent on s in denominator is 2) b)Gain of G(s) is K1. Gain is negative if K1< 0. c)Poles of G(s) are: s1= 0 and s2= –1/τs1is on imaginary axis; s2is in LHP. d)Zero of G(s) is: 211121KKKKKsa+τ-=+τ-=If 211<+τKKK, the zero is in RHP. 6-7 Two possibilities: 1. K1<0 and K1τ+ K2>0 2. K1> 0 and K1τ+ K2< 0 e)Gain is negative if K1< 0 Then zero is RHP if K1τ+ K2> 0 This is the only possibility. f)Constant term and e-t/τterm. g) If input is M/s, the output will contain a tterm, that is, it is not bounded....
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## This note was uploaded on 04/10/2008 for the course CHE 242 taught by Professor Cummings during the Spring '08 term at Vanderbilt.

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ch06 - 6-1 6-2 b)Process output will be unbounded because...

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