# How does the PA4000 Measure a 1MHz waveform?

### Question:

How does the PA4000 Measure a 1MHz waveform?

The topic often comes up of how you measure a 1MHz waveform with a sample rate of 1MHz and bandwidth of 1MHz.

Firstly, let’s look at the 1MHz bandwidth. This refers to the analog performance of the measurement channel. With a 1MHz bandwidth, the signal arriving at the Analog to Digital Convertor (ADC) has minimal distortion. Since both the voltage and the current channel have the same bandwidth, the phase shift seen by the ADC is the same.

The second issue is making a measurement of a 1MHz waveform with a 1MHz sample rate. If you sample at any frequency below two times the frequency of the measured signal you cannot determine the frequency of the signal. Also, if you sample at rate that returns the same points (same phase angle) from the waveform each time, you will not have enough data to calculate an accurate rms value. Imagine sampling a 1MHz waveform at 1 Msps. You would end up with a DC signal.

The problem extends to signals that have a frequency below half the sample rate. A good example of this problem would be sampling a 250kHz signal at 1Msps. You will only ever end up with 4 different samples taken on the waveform. This will not give a true representation of the waveform for the purpose of rms calculations.

The PA4000 deals with this problem by not relying on the samples to determine the frequency of the waveform. The PA4000 has a tuned analog circuit that is designed to measure the frequencies that, if measured by sampling alone, would cause problems. Once the frequency of the signal has been established, the PA4000 adjusts the sample rate to avoid the problem of taking samples that would not represent the waveform accurately.

The goal of the PA4000 is to select a sample rate that, during the sample period of the measurement (between 0.2 seconds and 2 seconds), is to "cover" as much of the waveform as possible at a variety of phase angles. Going back to the example of sampling a 250kHz waveform at 1Msps. In this case we would only ever get 4 different points. However, if we sampled 990ksps we would end up covering 99 different points on the waveform.

The calculation for the example above is as follows:

If the sample rate is 990ksps, then one sample is taken every 1.0101uS. On a 250kHz waveform, there would be 3 samples per cycle. The 4th sample, at 4.0404uS, would be displaced from the 1st sample by 404nS. With a 404nS displacement, and a 4uS cycle, there will be 4uS / 404nS samples taken before they start to repeat. This will give a much more accurate representation of the waveform.

This technique, known as under sampling is very commonly used in instrumentation. For many years, high-end multi-meters have been using this technique; long before high speed ADCs were available. This method can result in measurements that are extremely accurate. In fact, the results can be much more accurate than using a high speed ADC as opposed to a low speed ADC with good analog bandwidth since these often have lower non-linearity errors than high speed ADCs.

Another article that explains under sampling, from the perspective of an oscilloscope can be found at:

Real-Time Versus Equivalent-Time Sampling

This FAQ Applies to:

Applications:

Product: PA4000

FAQ ID : 69196

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