The block diagram of a conventional feedback control system contains the following transfer functions: Gc = Kc(1 + 1/5s) Gv = 1 Gm = 1/s + 1 Gp = Gd = 5e^−2s/10s + 1 (a) Plot the Bode diagram for the open-loop transfer function. (b) For what values of Kc is the system stable? (c) If Kc = 0.2, what

Question

The block diagram of a conventional feedback control system contains the following transfer functions:

[latex]\begin{aligned}G_{c} &=K_{c}\left(1+\frac{1}{5 s} ight) \quad G_{v}=1 \G_{m} &=\frac{1}{s+1} \G_{p} &=G_{d}=\frac{5 e^{-2 s}}{10 s+1}\end{aligned}[/latex]

(a) Plot the Bode diagram for the open-loop transfer function.

(b) For what values of [latex]K_{c}[/latex] is the system stable?

(c) If [latex]K_{c}=0.2[/latex], what is the phase margin?

(d) What value of [latex]K_{c}[/latex] will result in a gain margin of [latex]1.7[/latex] ?

Accepted Answer

a) Using Eqs. 14-56 and 14-57

[latex]\begin{aligned}A R_{O L}\left(\omega_{c} ight) &=\left|G_{O L}\left(j \omega_{c} ight) ight|=1 (14-56)\\phi_{O L}\left(\omega_{c} ight) &=\angle G_{O L}\left(j \omega_{c} ight)=-180^{\circ} (14-57)\end{aligned}[/latex]

[latex]\begin{aligned}& AR _{O L}=\left(K_{c} \sqrt{\frac{1}{25 \omega^{2}}+1} ight)\left(\frac{5}{\sqrt{100 \omega^{2}+1}} ight)\left(\frac{1}{\sqrt{\omega^{2}+1}} ight)(1.0) \&\varphi= an ^{-1}(-1 / 5 \omega)+0+\left(-2 \omega- an ^{-1}(10 \omega) ight)+\left(- an ^{-1}(\omega) ight)\end{aligned}[/latex]

b) Set [latex]\varphi=180^{\circ}[/latex] and solve for [latex]\omega[/latex] to obtain [latex]\omega_{c}=0.4695[/latex]

Then [latex]\left. AR _{o L} ight|_{\omega-\omega_{c}}=1=K_{c u}(1.025)[/latex]

Therefore, [latex]K_{c u}=1 / 1.025=0.976[/latex]

System is stable for [latex]K_{c} \leq 0.976[/latex]

c) For [latex]K_{c}=0.2[/latex], set [latex]AR _{O L}=1[/latex] and solve for [latex]\omega[/latex] to obtain [latex]\omega_{g}=0.1404[/latex]

Then [latex]\varphi_{g}=\left.\varphi ight|_{\omega=\omega_{g}}=-133.6^{\circ}[/latex]

From Eq. 14-61, [latex]P M=180^{\circ}+\varphi_{g}=46.4^{\circ}[/latex]

d) From Eq. 14-60

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

[latex]G M=1.7=\frac{1}{A_{c}}=\frac{1}{\left. AR _{O L} ight|_{\omega=\omega_{c}}}[/latex]

From part b), [latex]\left.\quad A R_{O L} ight|_{\omega=\omega_{c}}=1.025 K_{c}[/latex]

Therefore [latex]\quad 1.025 K_{c}=1 / 1.7 \quad[/latex] or [latex]\quad K_{c}=0.574[/latex]

Question 14.21: The block diagram of a conventional feedback control system ...

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