This example refers to the system of automatically adjusting the opening and closing of the pupil of the human eye. A simplified diagram of this system is given in Figure 8.41. Here, the constant a, which is the inverse of the time constant of the pupil of the eye, is usually 0.5 sec; the constant

Question

This example refers to the system of automatically adjusting the opening and closing
of the pupil of the human eye. A simplified diagram of this system is given in Figure 8.41. Here, the constant a, which is the inverse of the time constant of the pupil of the
eye, is usually 0.5 sec; the constant K is the gain constant of the pupil; and F(s) is the
feedback transfer function of the signal from the optic nerve, where T is the time
constant, which is usually 0.2 sec. For simplicity, assume that T is zero. Using the
Bode diagrams, determine the range of values of K for which the closed-loop system
is stable.

Accepted Answer

The open-loop transfer function has the form

[latex]G(jω)F(jω)=\frac{K}{(jω+0.5)³}=\frac{8K}{(2jω+1)³}, K>0[/latex]

To plot the diagram of the amplitude A, define [latex]A_1=20 log(8K)[/latex] and [latex]A_2=-60 log|2jω+1|=-60 log \sqrt{4ω²+1}[/latex]. The plot of [latex]A_2[/latex] has a break point at ω = 1/2 = 0.5 rad/sec, and is shown in Figure 8.42. The plot of [latex]A_1[/latex] is a horizontal line which passes through the point 20 log(8K). The diagram of the amplitude A is the sum of [latex]A_1[/latex] and [latex]A_2[/latex], which means that the diagram A essentially is the plot [latex]A_2[/latex] moving upwards or downwards depending on the sign of [latex]A_1[/latex]. To plot the diagram of the phase φ, define [latex]φ_1=/\underline{8K=}0°[/latex] and [latex]φ_2=-3 an^{-1}(2ω)[/latex]. The diagram of φ is the sum of [latex]φ_1[/latex] and [latex]φ_2[/latex], shown in Figure 8.43. The critical frequency [latex]ω_c[/latex] for which the total phase becomes -180° satisfies the following equation

[latex]φ(ω_c)=-180° or -3 an^{-1}(2ω_c)=-180° or [/latex]

[latex]2ω_c= an 60° =\sqrt{3}[/latex]

Hence, the critical frequency is [latex]ω_c=0.87[/latex]rad/sec. The gain margin [latex]K_g[/latex] is given by

[latex]K_g=-20 log|G(jω_c)F(jω_c)|=-20 log(8K)+60 log\sqrt{4ω²_c+1}[/latex]

[latex]=-20 log(8K)+18[/latex]

For the closed-loop system to be stable, the gain margin [latex]K_g[/latex] and the phase margin [latex]φ_p[/latex]
must both be positive. For [latex]K_g[/latex] to be positive, we must have 20 log(8K) < 18 or K < 1. For values of K less than the total amplitude diagram of G( jω)F( jω) will cross over the 0 dB axis at a frequency whose value is less than [latex]ω_c[/latex]. From Figure 8.43 we conclude that for[latex]ω<ω_c[/latex] the phase φ is less (in absolute value) than -180°, and hence [latex]φ_p>0[/latex]. Therefore, for the closed-loop system to be stable it must hold that 0 < K < 1.

Question 8.5.2: This example refers to the system of automatically adjusting...

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