Low-pass and band-pass sampling theorem

low pass sampling theorem

A band is limited to (0, f H f_HfH) in the low-pass signal m(t), if fs f_sfs≥2 f H f_H fHuniform sampling at a rate of , then the sampling sequence {m(n T s T_sTs)} can restore m(t) without distortion. where f H f_HfHis the upper limit frequency, fs f_sfsis the sampling rate, T s T_sTsis the sampling interval.
Uniform sampling means sampling at equal intervals.
fs f_sfs=2 f H f_H fHis the minimum sampling rate, also known as the Nyquist sampling rate.
T s T_sTs= 1 f s \frac{1}{f_s} fs1, it can be seen that the smaller the sampling rate, the larger the sampling interval. T s T_sTs= 1 2 f H \frac{1}{2f_H} 2 fH1called the Nyquist sampling interval.
When the sampling rate fs f_sfs<2 f H f_H fH, spectral aliasing occurs.

bandpass sampling theorem

Band limited to ( f L f_LfL f H f_H fH) within the band-pass signal m(t), bandwidth B= f H f_HfH- f L f_L fL, if f H f_HfH=nB+kB (where n∈N, 0≤k<1), then the minimum sampling rate fs f_s of m(t) can be recovered without distortionfs=2B(1+ k n \frac{k}{n} nk).
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If the bandpass signal is sampled at the minimum sampling rate above, the sampled signal spectrum will have neither gaps nor aliasing . It should be noted that if the sampling rate is greater than this value, the requirements may not always be met.
Example: f L f_LfL=100.5MHz, f H f_H fH=100.9MHz, f H f_H fH=252B+0.25B
Solution: Bandwidth: B = f H f_HfH- f L f_L fL=0.4MHz=400kHz
bandpass sampling rate: fs f_sfs=2B(1+ k n \frac{k}{n} nk)=2×400kHz×(1+ 0.25 252 \frac{0.25}{252} 2520.25)=800.8kHz
If the above example uses the low-pass sampling theorem, the sampling rate is greater than or equal to 2 times f H f_HfH, at least 201.8MHz, which is much larger than 800.8kHz, so it is more difficult to realize, so the band-pass signal is solved by the band-pass sampling theorem.
In practice, the lower limit frequency f L f_LfLmuch larger than the bandwidth.
Bandpass sampling rate fs f_sfsand the lower limit frequency f L f_LfLThe relationship curve is shown in the figure below.
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The above figure is based on the formula fs f_sfs=2B(1+ k n \frac{k}{n} nk) drawn, when k=0, fs f_sfsTake the minimum value 2B, when k=1, fs f_sfsTake the maximum value under different n.

Reference Video:
Sampling Theorem

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Origin blog.csdn.net/weixin_42570192/article/details/128773462