Consider a third-order process with Gm = Gv = 2 and Gp = Gd = 2/(s + 2)^3 . (a) Find the roots of the characteristic equation for a proportional controller Gc = Kc for Kc = 1, 8, and 27. Classify each case as stable, unstable, or marginally stable. Derive a formula for the offset for a disturbance

Question

Consider a third-order process with [latex]G_{m}=G_{v}=2[/latex] and [latex]G_{p}=G_{d}=\frac{2}{(s+2)^{3}}[/latex]

(a) Find the roots of the characteristic equation for a proportional controller [latex]G_{c}=K_{c}[/latex] for [latex]K_{c}=1,8[/latex], and 27. Classify each case as stable, unstable, or marginally stable. Derive a formula for the offset for a disturbance change as a function of [latex]K_{c}[/latex].

(b) Show that there is no offset for a PI controller using the Final Value Theorem for a disturbance change.

Accepted Answer

[latex]\frac{Y}{D}=\frac{G_{d}}{1+G_{c} G_{v} G_{p} G_{m}}=\frac{\frac{8}{(s+2)^{3}}}{1+K_{c} imes 1 imes \frac{8}{(s+2)^{3}} imes 1}=\frac{8}{s^{3}+6 s^{2}+12 s+8+8 K_{c}}[/latex]

The characteristic equation for above is shown as:

[latex]s^{3}+6 s^{2}+12 s+8+8 K_{c}=0[/latex]

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

[latex]-6 \omega^{2}+8+8 K_{c}+\left[12 \omega-\omega^{3} ight] j=0[/latex]

Thus, we have:

[latex]\left\{\begin{array} { c }{ - 6 \omega ^ { 2 } + 8 + 8 \mathrm { K } _ { c } = 0 } \{ 1 2 \omega - \omega ^ { 3 } = 0 }\end{array} \Rightarrow \left\{\begin{array}{l}\omega=2 \sqrt{3} \K_{c m}=8\end{array} ight. ight.[/latex]

So [latex]K_{c}=1[/latex] is stable; [latex]K_{c}=8[/latex] is marginally stable, and [latex]K_{c}=27[/latex] is unstable

For a step change [latex]D=\frac{1}{s}[/latex], applying the final value theorem:

Offset: [latex]\underset{t ightarrow \infty}{\lim} y(t)=\underset{s ightarrow 0}{\lim}\left[s \frac{8 \frac{1}{s}}{s^{3}+6 s^{2}+12 s+8+8 K_{c}} ight]=\frac{1}{1+K_{c}}[/latex]

(b)

[latex]\frac{Y}{D}=\frac{G_{d}}{1+G_{c} G_{v} G_{p} G_{m}}=\frac{\frac{8}{(s+2)^{3}}}{1+K_{c}\left(1+\frac{1}{ au_{I} s} ight) imes 1 imes \frac{8}{(s+2)^{3}} imes 1}=\frac{8 au_{I} s}{ au_{I} s(s+2)^{3}+K_{c}\left( au_{I} s+1 ight)}[/latex]

For a step change [latex]D=\frac{1}{s}[/latex], applying the final value theorem:

Offset: [latex]\underset{t ightarrow \infty}{\lim} y(t)=\underset{s ightarrow 0}{\lim}\left[s \frac{8 \frac{1}{s} au_{I} s}{ au_{I} s(s+2)^{3}+K_{c}\left( au_{I} s+1 ight)} ight]=0[/latex]

So there is no offset for PI controller.

Question 11.27: Consider a third-order process with Gm = Gv = 2 and Gp = Gd ...

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