Derive the characteristic equation for a control system with the following transfer functions: $G_{c}=K_{c}$ ,
$G_{v} G_{p}=\frac{0.5}{(s+1)(0.5 s+1)}, G_{m}=\frac{6}{s+3}$
Is the system stable for (a) $K_{c}=9$ , (b) $K_{c}=11$ , (c) $K_{c}=13$ ?

Check your answers using simulation and by direct substitution.

Question

Derive the characteristic equation for a control system with the following transfer functions: [latex]G_{c}=K_{c}[/latex],

[latex]G_{v} G_{p}=\frac{0.5}{(s+1)(0.5 s+1)}, G_{m}=\frac{6}{s+3}[/latex]

Is the system stable for (a) [latex]K_{c}=9[/latex], (b) [latex]K_{c}=11[/latex], (c) [latex]K_{c}=13[/latex] ?

Check your answers using simulation and by direct substitution.

Accepted Answer

The closed loop transfer function for set point changes is given by

[latex]\frac{Y}{Y_{s p}}=\frac{K_{m} G_{c} G_{v} G_{p}}{1+G_{c} G_{v} G_{p} G_{m}}[/latex]

Substituting the information in the problem gives

[latex]\frac{Y}{Y_{s p}}=\frac{K_{c}(s+3)}{(s+1)(0.5 s+1)(s+3)+3 K_{c}}=\frac{K_{c}(s+3)}{0.5 s^{3}+3 s^{2}+5.5 s+3+3 K_{c}}[/latex]

So the characteristic equation is

[latex]0.5 s^{3}+3 s^{2}+5.5 s+3+3 K_{c}=0[/latex]

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

[latex]-3 \omega^{2}+3+3 K_{c}+\left[5.5 \omega-0.5 \omega^{3} ight] j=0[/latex]

Thus, we have:

[latex]\left\{\begin{array} { c }{ - 3 \omega ^ { 2 } + 3 + 3 \mathrm { K } _ { c } = 0 } \{ 5 . 5 \omega - 0 . 5 \omega ^ { 3 } = 0 }\end{array} \Rightarrow \left\{\begin{array}{c}\omega=\sqrt{11} \K_{c, \max }=10\end{array} ight. ight.[/latex]

So the answers to parts (a)-(c) are:

(a) stable

(b) unstable

(c) unstable

The following plot shows the Simulink responses and confirms the above answers:

Question 11.28: Derive the characteristic equation for a control system with...

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