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Automating Manufacturing Systems - Process Control and ...

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plc analog - 22.6<br />

Given,<br />

N = 10, R min<br />

= 0<br />

V max<br />

= 10V<br />

V min<br />

= – 10V<br />

V in<br />

= 4.564V<br />

Calculate,<br />

R = R max<br />

= 2 N = 1024<br />

⎛ ⎞ 0.0098V<br />

V ERROR<br />

=<br />

V max<br />

– V<br />

---------------------------- min<br />

2R<br />

=<br />

⎝<br />

⎠<br />

V I<br />

V in<br />

⎛ ⎞<br />

⎠ R 1<br />

= INT<br />

– V<br />

---------------------------- min<br />

– ( – ) + 0 = 745<br />

V max<br />

V min<br />

⎝<br />

V<br />

V I<br />

– 0<br />

⎞ ⎛<br />

C<br />

= ------------- ( Vmax – V<br />

R – 1<br />

min<br />

) + V min<br />

= 4.565V<br />

⎝<br />

⎠<br />

Figure 22.5<br />

Sample Calculation of A/D Values<br />

If the voltage being sampled is changing too fast we may get false readings, as<br />

shown in Figure 22.6. In the upper graph the waveform completes seven cycles, <strong>and</strong> 9<br />

samples are taken. The bottom graph plots out the values read. The sampling frequency<br />

was too low, so the signal read appears to be different that it actually is, this is called aliasing.

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