# sampling rate for band pass signal

The sampling frequency for band pass signal which is having frequency from 4 kHz to 6 kHz(rectangular shape from 4 to 6 kHz), can I prefer the sampling frequency of 12 kHz(2*the highest frequency present is message signal) or 6 kHz (2*bandwidth of message signal(6-4)kHz), can u please suggest which one to prefer?

• Welcome to SE.DSP. Could you please rephrase your question in a clearer fashion?You can borrow inspiration from [ Is there a condition for bandpass sampling? ](dsp.stackexchange.com/questions/25400/…). And your 2*bandwidth of message signal(6-4)kHz is probably 4 KHz – Laurent Duval Nov 30 '15 at 20:11
• Normally, one would use a sampling frequency above (not at) twice the highest frequency present in a (mostly) band-limited signal. – hotpaw2 Nov 30 '15 at 20:21
• tkanks to all for your valuable information – user18568 Dec 2 '15 at 18:08
• Sir I just want the minimum nyquist rate for sampling frequency. Consider the sampling is normal not for specific application. In general the sampling frequency(message signal is in frequency domain is rectangular wave from 4 kHz to 6 kHz )? – user18568 Dec 2 '15 at 18:12

Sampling is frequency domain convolution by a frequency domain impulse train with the impulses spaced apart by the sampling frequency. If there is overlap between the images thus created, then there will be information loss. As you can see in the illustration below, all of the sampling frequencies $f_s$ = 12 kHz, $f_s$ = 6 kHz, and $f_s$ = 4 kHz will work fine. I tilted the band tops to indicate positive and negative frequencies. The first curve is the signal's spectrum before sampling. The unit is kHz. Using a small sampling frequency has the benefit that any processing will have less samples that need to be computed per time unit. For example filters designed to a performance specification can be lower-order than with a higher sampling frequency.

Sometimes it is useful if the signal is oversampled though. Things like polynomial interpolation work better because of the more dense sampling. A Hilbert transformer would have more room for the transition band. If processing is non-linear, it may be possible to avoid aliasing of harmonics by filtering after the non-linear processing; with critical sampling the harmonics might alias and contaminate the signal.

Using a low sampling frequency (6 kHz or 4 kHz) that aliases the bands may require more work later if you need to have the band back at the original frequency.

So the optimal sampling frequency depends on what you are going to do with the signal.

You don't specify what you intend to do with the samples, but I woudn't sample at either of the two frequencies you're contemplating.

The well-known (bandpass) sampling formulas indicate that, if your purpose is to reconstruct the original signal from its samples, you may sample at either 4 kHz, anywhere between 6 and 8 kHz, or above 12 kHz.

However, these calculations are only valid if there is no signal at the very ends of the spectrum, which is not true in your case. That means you shouldn't sample at exactly 4, 6, 8 or 12 kHz.

(Recall that a continuous signal sampled at rate $f_s$ will keep its original spectrum, and in addition its spectrum will be replicated and shifted up and down to all integer multiples of $f_s$.)

If you sample at 12 kHz, the rectangular spectrum at 4-6 kHz will be adjacent to a rectangular replica at 6-8 kHz. I would sample higher than 12 kHz to avoid this.

If you sample at 6 kHz, you'll get two adjacent rectangles, one at -2 to 0 kHz and another at 0 to 2 kHz. This is also undesirable.

If you want to demodulate the signal using bandpass sampling, you may sample at 5 kHz. That gives you a spectrum replica centered at 0 Hz, and with some care you may be able to low-pass filter it appropriately.

The lowest acceptable values for the $f_s$ sample rate, to avoid spectral replication overlap (what many people call "aliasing"), are described by: where $f_c$ and $B$ are the center freq and bandwidth of your analog bandpass signal. Variable $m$ is an integer. Set $f_c$ = 5 kHz, $B$ = 2 kHz, and $m$ = 1. The above expression's two ratios will be 8 kHz and 6 kHz. So your $f_s$ can be anywhere in the range of 6 -to- 8 kHz.

If you then set $m$ = 2, then the two ratios will be 4 kHz and 4 kHz. So your $f_s$ can be 4 kHz. If you next set $m$ = 3 the two ratios will be 2.667 kHz and 3 kHz. Values in that freq range are less than $2B$, violate Nyquist, and cannot be used. Stop at this point because there's no reason to set $m$ to a value larger than 3.