Chapter 9 Counters
80 MHz Timebase. Your measurement may return 1600 ± 1 cycles depending on the phase of the signal with respect to the timebase. As your frequency becomes larger, this error of ±1 cycle becomes more significant; Table
Table
Task | Equation | Example 1 | Example 2 |
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Actual Frequency to Measure | F1 | 50 kHz | 5 MHz |
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Timebase Frequency | Ft | 80 MHz | 80 MHz |
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Actual Number of Timebase | Ft/F1 | 1600 | 16 |
Periods |
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Worst Case Measured Number of | (Ft/F1) – 1 | 1599 | 15 |
Timebase Periods |
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Measured Frequency | Ft F1/(Ft – F1) | 50.125 kHz | 5.33 MHz |
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Error | [Ft F1/(Ft – F1)] – F1 | 125 kHz | 333 kHz |
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Error % | [Ft/(Ft – F1)] – 1 | 0.06% | 6.67% |
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•Method 1b (measuring K periods of F1) improves the accuracy of the measurement. A disadvantage of Method 1b is that you have to take K + 1 measurements. These measurements take more time and consume some of the available USB bandwidth.
•Method 2 is accurate for high frequency signals. However, the accuracy decreases as the frequency of the signal to measure decreases. At very low frequencies, Method 2 may be too inaccurate for your application. Another disadvantage of Method 2 is that it requires two counters (if you cannot provide an external signal of known width). An advantage of Method 2 is that the measurement completes in a known amount of time.
•Method 3 measures high and low frequency signals accurately. However, it requires two counters.
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