calculate the minimum sampling rate

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mani bindu

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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.
 

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

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

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

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 said:
Without quadrature sampling
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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...
 

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...
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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...
 
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)
 

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