how to use barcode scanner in asp.net c# 0.5A sin 4H = -0.5H . H f$ff = - sm-1 A in VS .NET

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0.5A sin 4H = -0.5H . H f$ff = - sm-1 A
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The peak-to-peak amplitude of stem position is H less than A. Then the gain due to hysteresis is stem amplitude over signal amplitude:
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Phase and gain are plotted vs. dimensionless amplitude A/H in Fig. 5.6. Because gain and phase both change with the amplitude of the contIroller-output oscillation, gain and period of the closed loop are both variable in the presence of hysteresis. A condition like this is cause for concern, in that stability is also variable. Loop gain should be checked for selected values of amplitude to determine whether stability is conditional. example 5.1 Consider a Ijrocess consisting of dead time, an integrating capacity, and hysteresis being cont,rolled prol)ortionallg. From Fig. 5.6, a high n/ii ratio results in gain al)l)roaching unity and phase lag approaching zero, just as if t,hcre were no hysteresis present. Therefore, proportional band can b e set t o I)roduce ,4-anlplitude damping (i.e., loop gain of 0.5) without J/ hysteresis:
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For the loop to be stable, its gain must be less than 1.0 for all finite values of aml)litude. Table 5.1 summarizes the gain and phase of the various elements in the 1001, for selected values of amplitude. Hecause loop gain is less than 1.0 for all finite values of amplitude, the loop is stable.
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TABLE 5.1 Loop Gain vs. Amplitude for Proportional Control
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1; 5 2
1 0.9 .o 0.8 0.5
6.5 0 11 30
90 83.5 79 60
4.3 4.0 4.55 6.0
0.69 0.64r& 0.71 0.95
0.5 0.48 0.45 0.38
This example was chosen as typical of some closed loops containing hysteresis. Although the change in period with amplitude causes the gain of the integrating element to vary, the effect is more than offset by the change in gain of the hysteresis element. Consequently the loop is stable as long as the band has been set for high-amplitude conditions. If adjusted for a given damping with a disturbance of low amplitude, however, the loop could become undamped in t)he face of a severe upset. But the small variation in loop gain (0.75 to 1.0) corresponding to a wide amplitude range (2 to m) makes this event unlikely. Loops containing two integrations are capable of a limit cycle, however. An example would be a non-self-regulating process such as liquid level, with a proportional-plus-reset controller. The gain product of the two integrating elements will vary as the square of the period, more than can be offset by the gain of hysteresis. Under these conditions, loop gain varies inversely with amplitude.
example 5.2
Consider a 1001, consisting of a dead-time plus integrating process of t)ime constant 7,, hysteresis, and a proportional-plus-reset controller. Let the reset time be set for 30 phase lag and the proportional band for >i-amplitude damping at A/H of 2. Table 5.2 summarizes the effect of hysteresis.
TABLE A/H 5.2 Loop Gain vs. Amplitude for Two-mode Control GH -4Hy de -h deg -+R
7 /7d
0.9576/q 1.91 3.9
GPHGGH
4.0 2.0 1.8
0.75 0.50 0.45
14.5 30.0 33.7
60.0 30.0 14.8
15.5 30.0 41.5
6.0 12.0 24.4
0.34 0.50 1.00
With the existing settings, the control loop limit-cycles at an amplitude of 1.8H and a period of 24.4~~. Reduction of the proportional band by half will change the amplitude only to 2H but will reduce the period to 127d. Minimum period may be attained by reducing the proportional band to the point, where the amplitude becomes objectionable. Increasing
Nonlinear Control Elements
the reset time reduces the amplitude while increasing the period of the cycle. Therefore a long reset time and a fairly narrow band will give the best combination of amplitude and period for this loop. A limit cycle caused by hyst cresis will appear as a nearly square wave on a flow record--this is it s most distinguishing characteristic. And it will not bc possible to damp the wave by adjustment of the proportional band; this is typical of loops containing a nonlinear element. Hysteresis can be minimized by superimposing a high-amplitude, highfrequency signal on the controller output. The process would not respond to it, but the sticking element would. There is sufficient noise in some measurements to create this effect. But the best solution is to close the loop around the hysteresis element, alone, with a proportional controller-this amounts to installing a posit ioner on the valve. 3lore will be said about positioners in Chap. 6.
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