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关注：1
2013-05-23 12:21
求翻译：有源晶振20MHz，如果用40MHz或60MHz的示波器测量，显示的是正弦波，这是由于方波的傅里叶分解为基频和奇次谐波的叠加，带宽不够的话，是什么意思？ 待解决
 悬赏分：1
- 离问题结束还有
有源晶振20MHz，如果用40MHz或60MHz的示波器测量，显示的是正弦波，这是由于方波的傅里叶分解为基频和奇次谐波的叠加，带宽不够的话，
问题补充： |
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2013-05-23 12:21:38
Active crystal 20mhz, 40mhz or 60MHz oscilloscope measurements, shows a sine wave, which is due to the Fourier decomposition of the square-wave superposition of odd harmonics of the fundamental frequency, bandwidth is not enough
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2013-05-23 12:23:18
Active crystal, if you use 20 MHz MHz 40 or 60 MHz oscilloscope measurements, show the sine wave, square wave because it is a Fourier decomposition as the base frequency and odd harmonics of the overlay, the bandwidth is not sufficient.
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2013-05-23 12:24:58
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2013-05-23 12:26:38
Crystal oscillator with source 20MHz, if by 40MHz or 60MHz oscilloscope measurement, display is a sine wave, this is due to the square wave Fourier decomposition and superposition of odd harmonics of fundamental frequency, bandwidth is not enough,
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2013-05-23 12:28:18
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