Suppose a control system is modeled by GvGp = 0.25/(s + 1)^3 , Gm = 4, and Gc = Kc. Find the maximum value of Kc for a proportional controller for which the system is stable, using direct substitution, and verify your result using simulation. Replace the controller with a PD controller (Kc = 10).

Question

Suppose a control system is modeled by [latex]G_{v} G_{p}=[/latex] [latex]\frac{0.25}{(s+1)^{3}}, G_{m}=4[/latex], and [latex]G_{c}=K_{c} .[/latex] Find the maximum value of [latex]K_{c}[/latex] for a proportional controller for which the system is stable, using direct substitution, and verify your result using simulation. Replace the controller with a PD controller [latex]\left(K_{c}=10 ight) .[/latex] Determine the range of [latex] au_{D}[/latex] for which the system is stable. Check your results using simulation.

Accepted Answer

a) Proportional controller:

We derive the transfer function as follows:

[latex]\begin{gathered}\frac{Y}{Y_{S P}}=\frac{K_{m} G_{c} G_{v} G_{p}}{1+G_{m} G_{c} G_{v} G_{p}} \\frac{Y}{Y_{S P}}=\frac{K_{c} \frac{1}{(s+1)^{3}}}{1+K_{c} \frac{1}{(s+1)^{3}}}=\frac{K_{c}}{(s+1)^{3}+K_{c}}=\frac{K_{c}}{s^{3}+3 s^{2}+3 s+1+K_{c}} (1)\end{gathered}[/latex]

The characteristic equation of [latex](1)[/latex] is the following:

[latex]s^{3}+3 s^{2}+3 s+1+K_{c}=0[/latex] (2)

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

[latex]-3 \omega^{2}+1+K_{c}+\left[3 \omega-\omega^{3} ight] j=0[/latex]

Thus, we have:

[latex]\left\{\begin{array} { c }{ - 3 \omega ^ { 2 } + 1 + K _ { c } = 0 } \{ 3 \omega - \omega ^ { 3 } = 0 }\end{array} \Rightarrow \left\{\begin{array}{c}\omega=\sqrt{3} \K_{c, \max }=8\end{array} ight. ight.[/latex]

We conclude that the system will be stable if

[latex]K_{c}<8[/latex] (3)

Simulation results are in Figure S11.29a.

b) PD controller:

We derive the transfer function as follows:

[latex]\begin{gathered}G_{c}=K_{c}\left(1+ au_{D} s ight) \\frac{Y}{Y_{S P}}=\frac{K_{m} G_{c} G_{v} G_{p}}{1+G_{m} G_{c} G_{v} G_{p}} \\frac{Y}{Y_{S P}}=\frac{K_{c}\left(1+ au_{D} s ight) \frac{1}{(s+1)^{3}}}{1+K_{c}\left(1+ au_{D} s ight) \frac{1}{(s+1)^{3}}} \=\frac{K_{c}\left( au_{D} s+1 ight)}{s^{3}+3 s^{2}+3 s+1+K_{c}+ au_{D} K_{c} s} (4)\end{gathered}[/latex]

where [latex]K_{c}=10[/latex].

The characteristic equation of (4) is the following:

[latex]s^{3}+3 s^{2}+\left(10 au_{D}+3 ight) s+11=0[/latex] (5)

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

[latex]-3 \omega^{2}+11+\left[\left(10 au_{D}+3 ight) \omega-\omega^{3} ight] j=0[/latex]

Thus, we have:

[latex]\left\{\begin{array}{c}-3 \omega^{2}+11=0 \\left(10 au_{D}+3 ight) \omega-\omega^{3}=0\end{array} \Rightarrow au_{D, \min }=\frac{1}{15}=0.0667 ight.[/latex]

Question 11.29: Suppose a control system is modeled by GvGp = 0.25/(s + 1)^3...

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