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## Q. 6.EX.25

Effect of Sampling on Stability
Determine the additional phase lag due to the digital sampling in Example 6.15 and reconcile that difference with the observed performance of the continuous and digital implementations shown in the example. How slowly could you sample if it was necessary to limit the decrease in the $PM$ to less than $20°$ ?

## Verified Solution

The sample rate in Example 6.15 was selected to be $T_s = 0.05 \text{ sec}.$ We can see from Fig. 4.22 that the effect of the sampling is to hold the application of the control over one sample period, thus the actual delay varies between zero and one full sample period. Therefore, on the average, the effect of the sampling is to inject a time delay of $T_s/2 = 0.05/2 = 0.025 = T_d \text{ sec}.$ From Eq. (6.71), we see that the phase lag due to this sampling at the crossover frequency of $5 \text{ rad/sec },$ where we measure the $PM,$ is $∠G_D = −ωT_d = −(5)(0.025) = −0.125 \text{ rad } = −7°.$ Therefore, the $PM$ will decrease from $45°$ for the continuous implementation to $38°$ for the digital implementation. Fig. 6.59(a) shows that the overshoot, $M_p,$ degraded from $1.2$ for the continuous case to $≈1.27$ for the digital case, which is predicted by Eq. (6.32) and Fig. 6.38.

$∠G_D(jω) = −ωT_d$          (6.71)

$ζ ≅ \frac{PM}{100} .$          (6.32)

In order to limit the phase lag to $20°$ at $ω = 5 \text{ rad/sec},$ we see from Eq.(6.71) that the maximum tolerable $T_d = 20/(5 ∗ 57.3) = 0.07 \text{ sec},$ so that the slowest sampling acceptable would be $T_s = 0.14 \text{ sec}.$ Note, however, that this large decrease in the $PM$ would result in the overshoot increasing from $≈20%$ to $≈40%.$