A process (including valve and sensor-transmitter) has the approximate transfer function,G(s) = 2e^−0.2s/(s + 1) with time constant and time delay in minutes. Determine PI controller settings and the corresponding gain margins by two methods: (a) Direct synthesis (τc = 0.3 min). (b) Phase margin

Question

A process (including valve and sensor-transmitter) has the approximate transfer function, [latex]G(s)=2 e^{-0.2 s} /(s+1)[/latex] with time constant and time delay in minutes. Determine PI controller settings and the corresponding gain margins by two methods:

(a) Direct synthesis [latex]\left( au_{c}=0.3 min ight)[/latex].

(b) Phase margin [latex]=40^{\circ}[/latex] (assume [latex] au_{I}=0.5 min[/latex] ).

(c) Simulate these two control systems for a unit step change in set point. Which controller provides the better performance?

Accepted Answer

a) [latex]K=2, au=1, heta=0.2, au_{c}=0.3[/latex]

Using Eq. 12-11, the PI settings are

[latex]K_{c}=\frac{1}{K} \frac{ au}{ au_{c}+ heta}, \quad au_{I}= au[/latex] (12-11)

[latex]K_{c}=\frac{1}{K} \frac{ au}{ heta+ au_{c}}=1 \quad, \quad au_{I}= au=1 \min[/latex]

Using Eq. 14-58

[latex]A R_{O L}(\omega)=K_{c} A R_{G}(\omega)[/latex] (14-58)

[latex]\begin{aligned}&-180^{\circ}= an ^{-1}\left(\frac{-1}{\omega_{c}} ight)-0.2 \omega_{ c }- an ^{-1}\left(\omega_{c} ight)=-90^{\circ}-0.2 \omega_{c} \& ext { or } \quad \omega_{c}=\frac{\pi / 2}{0.2}=7.85 rad / min\end{aligned}[/latex]

Using Eq. 14-57,

[latex]\phi_{O L}\left(\omega_{c} ight)=\angle G_{O L}\left(j \omega_{c} ight)=-180^{\circ}[/latex] (14-57)

[latex]A_{c}=\left. AR _{O L} ight|_{\omega=\omega_{c}}=\sqrt{\frac{1}{\omega_{c}{ }^{2}}+1}\left(\frac{2}{\sqrt{\omega_{c}{ }^{2}+1}} ight)=\frac{2}{\omega_{c}}=0.255[/latex]

From Eq. 14-60, [latex]G M=1 / A_{c}=3.93[/latex]

[latex]G M riangleq \frac{1}{A R_{c}}[/latex] (14-60)

b) Using Eq. 14-61,

[latex]P M riangleq 180+\phi_{g}[/latex] (14-61)

[latex]\varphi_{g}=P M-180^{\circ}=-140^{\circ}= an ^{-1}\left(-1 / 0.5 \omega_{g} ight)-0.2 \omega_{g}- an ^{-1}\left(\omega_{g} ight)[/latex]

Solving, [latex]\omega_{g}=3.04 rad / min[/latex]

[latex]\left. AR _{O L} ight|_{\omega=\omega_{g}}=1=K_{c} \sqrt{\left(\frac{1}{0.5 \omega_{g}} ight)^{2}+1}\left(\frac{2}{\sqrt{\omega_{g}{ }^{2}+1}} ight)[/latex]

Substituting for [latex]\omega_{g}[/latex] gives [latex]K_{c}=1.34[/latex]. Then from Eq. 14-8

[latex]\phi=\angle G= an ^{-1}(I / R)[/latex] (14-8)

[latex]-180^{\circ}= an ^{-1}\left(\frac{-1}{0.5 \omega_{c}} ight)-0.2 \omega_{c}- an ^{-1}\left(\omega_{c} ight)[/latex]

Solving, [latex]\omega_{c}=7.19 rad / min[/latex]

From Eq. 14-56,

[latex]A R_{O L}\left(\omega_{c} ight)=\left|G_{O L}\left(j \omega_{c} ight) ight|=1[/latex] (14-56)

[latex]A_{c}=\left. AR _{o L} ight|_{\omega=\omega_{c}}=1.34 \sqrt{\left(\frac{1}{0.5 \omega_{c}} ight)^{2}+1}\left(\frac{2}{\sqrt{\omega_{c}{ }^{2}+1}} ight)=0.383[/latex]

From Eq. 14-60, [latex]G M=1 / A_{c}=2.61[/latex]

c) By using Simulink-MATLAB, these two control systems are compared for a unit step change in the set point.

The controller designed in part a) (Direct Synthesis) provides better performance giving a first-order response. Part b) controller yields a large overshoot.

Question 14.23: A process (including valve and sensor-transmitter) has the a...

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