For the process described by the transfer function G(s) = 12/(8s + 1)(2s + 1)(0.4s + 1)(0.1s + 1) (a) Find second-order-plus-time-delay models that approximate G(s) and are of the form Ĝ(s) = Ke^−θs/(τ1s + 1)(τ2s + 1) One of the approximate models can be found by using the method discussed in

Question

For the process described by the transfer function

[latex]G(s)=\frac{12}{(8 s+1)(2 s+1)(0.4 s+1)(0.1 s+1)}[/latex]

(a) Find second-order-plus-time-delay models that approximate [latex]G(s)[/latex] and are of the form

[latex]\hat{G}(s)=\frac{K e^{- heta s}}{\left( au_{1} s+1 ight)\left( au_{2} s+1 ight)}[/latex]

One of the approximate models can be found by using the method discussed in Section [latex]6.3[/latex]; the other, by using a method from Chapter [latex]7 .[/latex]

(b) Compare all three models (exact and approximate) in the frequency domain and a FOPTD model.

Accepted Answer

a) Method discussed in Section 6.3:

[latex]\hat{G}_{1}(s)=\frac{12 e^{-0.3 s}}{(8 s+1)(2.2 s+1)}[/latex]

Method discussed in Section 7.2.1:

Based on Figure S14.17, we can obtain the time stamps corresponding to [latex]20 \%[/latex] and [latex]60 \%[/latex] response: [latex]t_{20}=4.034 ; t_{60}=10.09 ; t_{20} / t_{60}=0.4[/latex]. Based on Figure [latex]7.7[/latex], we have [latex]\frac{t_{60}}{ au}=2.0 ; \zeta=1.15[/latex], so we have [latex] au=5.045 .[/latex] Using the slope of the inflection point we can estimate the time delay to be [latex]0.8[/latex]. So we have:

[latex]\hat{G}_{2}(s)=\frac{12 e^{-0.8 s}}{25.45 s^{2}+11.60 s+1}[/latex]

b) Based on Figure S14.17a, we can obtain [latex] heta=0.8 ; au=15-0.8=14.2[/latex]

[latex]\hat{G}_{3}(s)=\frac{12 e^{-0.8 s}}{14.2 s+1}[/latex]

Comparison of three estimated models and the exact model in the frequency domain using Bode plots:

Question 14.17: For the process described by the transfer function G(s) = 12...

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