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Edited by amy at 2021-8-26 16:49
Please refer to these links:
https://edadocs.software.keysigh ... need-580279264.html
https://edadocs.software.keysigh ... F%3F-620138640.html
https://edadocs.software.keysigh ... ters-583423317.html
Because the square waveform contains many high frequency harmonics. For example, a 50MHz square waveform, according to the Fourier expansion, contains 50MHz, 150MHz, 250MHz, 350MHz, 450MHz, ... 50×(n+1)MHz harmonics (all sine waveform) components. If you use a low-bandwidth oscilloscope, the high-frequency harmonic components will be attenuated when passing through the oscilloscope. For example, a 50MHz oscilloscope can only leave a 50MHz fundamental wave, so the result displayed by the oscilloscope is a sine waveform. Therefore, it is generally believed that to observe a square waveform, you need to see the fifth harmonic of the square waveform, that is, the 9th harmonic. Therefore, an oscilloscope with a bandwidth of more than 9 times the square waveform frequency is needed.
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