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[SOLVED] Bandwidth and Sampling Time

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Full Member level 1
Aug 8, 2008
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This probably a basic question about the relation between bandwidth and sampling time. I understood the following terms.

Sampling time: Time between two consecutive samples while sampling a signal periodically
Sampling frequency: Reciprocal of sampling time

I am using a magnetic field instrument in which I select f_low = 100 Hz and f_high = 500 Hz.

a) If I select bandwidth 10 Hz, I get data points at frequency 100 Hz, 105 Hz, 110 Hz and so on.
b) If I select bandwidth 30 Hz, I get data points at frequency 100 Hz, 115 Hz, 130 Hz and so on.

Text book says use narrow band filter to avoid aliasing but if the data points are close like in case (a) then how narrow band filter or bandwidth can cope with aliasing ?

I am also wondering the relation between sampling time and bandwidth. Please help me to understand this concept.

Could you explain your question better? What do you mean by "select bandwidth"? Or "f_low/f_high"? What is your sampling freq?

Nyquist theorem states 2 samples per Hz for a brick wall filter. Typically 2.5 samples per Hz for resolution bandwidth (RBW) is the minimum to prevent modulation aliasing. But often it is much higher sampling rate with enough samples to ensure peak envelope has settled usually >5xT cycles of where T=1/f_RBW

Does it sweep or step the frequency?
Kindly let me know how much time it should take to complete one sweep in case (a) and case (b). In software, the bandwidth is RBW which is 10 Hz in case (a) and 30 Hz in case (b). Practically it takes around 3 to 4 min to complete one sweep. I am wondering calculation wise how it relates to the signal theory and sampling.

It's completely unclear what kind of sampling you're performing and what the said frequency sweep means.

At first sight, it sounds like a case of undersampling, fsample < fsignal/2. If an input signal can be acquired unequivocally depends on the input signal characteristic.

Where ST is sweep time in seconds, k is proportionality constant, Span is the frequency range under consideration in hertz, and RBW is the resolution bandwidth in hertz.
Sweeping too fast, however, causes a drop in displayed amplitude and a shift in the displayed frequency.

Thus if VBW=RBW=10Hz and Span = 400 Hz and k=5 for near full (99%?) using quasi-peak detection...

ST=5*400/100=20 sec

For 30Hz
ST=2000/900=2.22 sec

But why you need minutes?

Thanks Sunny for your reply. Now I tried another measurement with following settings.

Start frequency = 5 kHz
Stop frequency = 250 kHz
Span = 245 kHz

RBW = 100 Hz
Sample time = 30 sec

I observe the time of one complete sweep from start frequency to strop frequency. The whole sweep take around 10 min. I guess k is constant number (constant of proportionality) but I am not sure if it's same for one instrument or not. Because the software provide all the three parameters span, RBW, and sample time or sweep time to set by user then the value of k would be different in different settings because user can change all the three parameters. I could not compare the calculations of sample time or sweep time with actual time it take to complete the sweep on spectrum analyzer.

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