calculate the minimum sampling rate

[mani bindu]

A bandpass signal has lower frequency 1002.5 khz and higher frequency 1046 khz. Determine the minimum sampling rate so as to have a minimum guard band of 4khz between two spectrum replicas.

 

[weetabixharry]

The signal bandwidth is 1046-1002.5 = 43.5kHz. Add a guard band of 4kHz and the required minimum sampling rate is 47.5kHz.

 

[cherryman]

Sampling rate 47.5kHz to catch signal with frequency 1046 kHz?
Shouldn't be (1046 + 4) * 2 = 2100 kHz?

 

[FvM]

Shouldn't be (1046 + 4) * 2 = 2100 kHz?
The keyword is "bandpass signal". You can apply undersampling.

The signal bandwidth is 1046-1002.5 = 43.5kHz. Add a guard band of 4kHz and the required minimum sampling rate is 47.5kHz.

Not quite right. Also in bandpass sampling the sampling frequency must be at least double the signal bandwidth: fs >= 2B

 

[weetabixharry]

Could you clarify this, FvM? From what you're saying, I guess we're not talking about complex baseband signals (which is what I'm familiar with). Was there something in the wording of the original question that made this clear?

 

[FvM]

Without quadrature sampling, the sampling frequency must be high enough to avoid overlapping of positive and negative products. https://en.wikipedia.org/wiki/Undersampling

 

[weetabixharry]

Without quadrature sampling

Is it clear from the original question that this is the case? I'm worried that I've worked too much with complex baseband representation and can't recognise a real signal when I see it...

 

[FvM]

Is it clear from the original question that this is the case? I'm worried that I've worked too much with complex baseband representation and can't recognise a real signal when I see it...

It's not explicitely stated. But obviously the question is just an excercise problem about sampling. Complex signals and quadrature demodulation are advanced stuff beyond basic sampling theory.

I didn't even realize this background of your first post, I suspected a simple calculation error...

 

[u7karsh]

According to me, the frequenc must be (1046 + 2)*2 = 2096Khz.
For details, conside the diagram below...



Here as we can see B represents our upper limit, ie., 1046kHz. now if on one side, we add a ban of 2kHz, we will get this 2k and another 2k from the periodic signal.. thus eventually a 4k guard band. Thus the sampling frequency must be 2*(1046 + 2)

 

[FvM]

Utilizing bandpass sampling, B is 42.5 kHz, not 1046 kHz (plus guard band). See the link quoted above.

Bandpass sampling has been already discussed by Shannon https://en.wikipedia.org/wiki/Nyquist–Shannon_sampling_theorem#Sampling_of_non-baseband_signals