A feedback control system has the open-loop transfer function, GOL(s) = 0.5Kc^e−3s/(10s + 1). Determine the values of Kc for which the closed-loop system is stable using two approaches: (a) An approximate analysis using a 1/1 Padé approximation for e^−3s and the direct substitution method. b) An

Question

A feedback control system has the open-loop transfer function, [latex]G_{O L}(s)=0.5 K_{c} e^{-3 s} /(10 s+1) .[/latex] Determine the values of [latex]K_{c}[/latex] for which the closed-loop system is stable using two approaches:

(a) An approximate analysis using a [latex]1 / 1[/latex] Padé approximation for [latex]e^{-3 s}[/latex] and the direct substitution method.

(b) An exact analysis based on substitution of [latex]s=j \omega[/latex]. (Hint: Recall Euler's identity.)

(c) Compare your answers for (a) and (b) using computer simulation.

Accepted Answer

The characteristic equation is

[latex]1+\frac{0.5 K_{c} e^{-3 s}}{10 s+1}=0[/latex] (1)

a) Using the Pade approximation

[latex]e^{-3 s} \approx \frac{1-(3 / 2) s}{1+(3 / 2) s}[/latex]

in (1) gives

[latex]15 s^{2}+\left(11.5-0.75 K_{c} ight) s+\left(1+0.5 K_{c} ight)=0[/latex]

Substituting [latex]s= j \omega[/latex] in above equation, we have:

[latex]-15 \omega^{2}+1+0.5 K_{c}+\left[11.5 \omega-0.75 K_{c} \omega ight] j=0[/latex]

Thus, we have:

[latex]\left\{\begin{array} { c }{ - 1 5 \omega ^ { 2 } + 1 + 0 . 5 \mathrm { K } _ { c } = 0 } \{ 1 1 . 5 \omega - 0 . 7 5 K _ { c } \omega = 0 }\end{array} \Rightarrow \left\{\begin{array}{c}\omega=0.760 \K_{c m}=15.33\end{array} ight. ight.[/latex]

b)

Substituting [latex]s=j \omega[/latex] in (1) and using Euler's identity.

[latex]e^{-3 j \omega}=\cos (3 \omega)-j \sin (3 \omega)[/latex]

gives

[latex]10 j \omega+1+0.5 K_{c}[\cos (3 \omega)-j \sin (3 \omega)]=0[/latex]

Then,

[latex]1+0.5 K_{c} \cos (3 \omega)=0[/latex] (2)

and

[latex]10 \omega-0.5 K_{c} \sin (3 \omega)=0[/latex] (3)

From (2) and (3)

[latex] an (3 \omega)=-10 \omega[/latex] (4)

Eq. 4 has infinite number of solutions. The solution for the range [latex]\pi / 2<3 \omega<3 \pi / 2[/latex] is found by trial and error to be [latex]\omega=0.5805[/latex].

Then from Eq. [latex]2, K_{c}=11.78[/latex]

The other solutions for the range [latex]3 \omega>3 \pi / 2[/latex] occur at values of [latex]\omega[/latex] for which [latex]\cos (3 \omega)[/latex] is smaller than [latex]\cos (3 imes 5.805)[/latex]. Thus, for all other solutions of [latex]\omega[/latex], Eq. 2 gives values of [latex]K_{c}[/latex] that are larger than [latex]11.78[/latex]. Hence, stability is ensured when

[latex]0

To solve Eqs. 2 and 3 , another way is to use Newton's method. With initial guess [latex]K_{c}=5, \omega=0[/latex] (steady state), the solution to Eqs. 2 and 3 is:

[latex]K_{c}=-2, \omega=0[/latex]

With a different initial guess ( e.g., [latex]K_{c}=5, \omega=5[/latex] ), the solution is:

[latex]K_{c}=11.78, \omega=0.5805[/latex]

Again, [latex]\omega_{ c }=0.5805[/latex] and the stability is ensured when

[latex]0<\dot{K}_{c}<11.78[/latex]

c)

Question 11.20: A feedback control system has the open-loop transfer functio...

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