I'm trying to clear up some conceptual confusion I have about digital signal processing.

Say I have a 1000 point digital signal (stored in the memory of my arbitrary waveform generator "AWG"), and I set the AWG to read the signal out at 50 MHz. So every sample is getting played uniformly at 50MSa/s.

How does the rate that I set my AWG to play this signal at (50 MSa/s) determine the bandwith of my signal? I am familiar with the Nyquist sampling theorem as it applies to sampling an analog signal, but not how it applies to reconstructing a signal from digital samples.

Does the actual data (raw amplitude values) in the signal affect this signal bandwidth at all? I'm imagining the AWG playing a sequence of values [0 0 0 0 0 0 0 A 0 0 0 0 ...], and if A is really large, wouldn't the DAC need a faster rise time to reach a higher amplitude value compared to if A was small (less energy it would need to dissipate)? This makes it seem to me like the amplitude values contained in the waveform would affect the rate the AWG is playing them out at, but this has nothing to do with the AWGs sample rate??

Thanks in advance for the help


Capturing digital samples is a form of recording. Playing them back at a rate faster than the recording rate has the effect of frequency shifting the entire spectrum toward higher frequencies.

If you record a 100 Hz audio tone and play it back at 2x the original sample rate, it will become a 200 Hz audio tone. Etc. Hopefully this is enough information for you to understand the concept.

There are techniques which would allow you to playback audio faster, but with the same perceived pitch.

And it is also possible to "re-sample" a data series at a higher frequency, so that it can be played back at a different sample rate, and sound basically identical. But the re-sampling process cannot add new information. It is just a form of interpolation.


If X=1000 points and f= 50 Msps,

  • A random arbitrary signal will span from 1 cycle per 1000 points and 1/2 of the sampling rate (Nyquist Theorem) or 50Khz to 25Mhz.

BW max = f/X to f/2 for random signal of n points sampled at f rate.

Ideally if you redigitize your arbitrary wave sampled at 50MHz with a brick wall LPF at 25MHz it would have infinite group delay at -xx dB attenuation at 25Mhz and so the practical solution is 30% of the sampling rate.

This 30% is apractical max BW is a tradeoff between; harmonic distortion, BW efficiency and phase distortion, group delay at maximum desired f.

Some may use 35 to 40% while others use 50% of sample rate using max theoretical singal BW with no regard to errors above. ( no free lunch)

For example Arbitrary waveform javascript with log option and X =259 terms of Fourier result with 3 repeating wavforms in < 750 points total and X=250 max terms or harmonics of fundamental including DC content at f=0.

All amplitudes and phase and also time sampled arbitrary wave x3, can be modified with mouse . (use log display box [x])

This arbitrary or standard waveform analyzer Fourier can display 250 harmonics including DC (f=0). An impulse of 1 point spans with flat harmonics then quickly rolls down to 0 or -inf dB. where the wavelngth equals 1 point and then generates harmonics above this. But this Fourier Transform is limited to 60dB and 249 harmonics. Try to make a single point pulse... and locate the null frequency and harmonics.

Other info:

Harmonics must be suppressed by n-th order Bessel or maximally flat group delay filter for good results at 30% of sampling rate rather than 50% for best reduction of harmonic distortion and aliasing from subsequent resampling. e.g. 2.4kHz telphony BW using a 8Ksps logarithmic 8 bit ADC gives the equivalent resolution of 14 bits..


The bandwidth of the signal being created by your AWG is still limited by the Nyquist Theorem. The DAC will have its own bandwidth limitation that is not dependent on the AWG. If you were to create 25MHz signal, you would need a DAC with at least that much bandwidth.


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