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frequency spectrum cadence

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shshprsd

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I am testing DAC for dynamic performance. since it is 12 bit dac so I choosen 4096 coherent samples for dft so that it could handle all the codes. with 4096 samples I have 1Ghz of scope area to analyse since my sampling frequency is 2GSPS, so my second harmonic component is seen as a foldback component close to fundamental component. but when I increased my samples to 8192 scope area in increased to 2 Ghz so I can see a seperate 2nd harmonic component which is lower than previous foldback component. now I am in doubt to consider 2nd harmonic obtained after 8192 to be correct or not.
 

Depending on how you chose your frequencies, multiple harmonics might be falling at the frequency of the folded second harmonic. It is very easy to verify. If you feel unsure, you need to tell us the exact frequencies.
 

I choosen 666mhz as input frequency and 2 GSPS sampling frequency with 4096 samples scope window is of 1 Ghz so second harmonics which is 1332 Mhz folds back after 1 Ghz and sits at 1G-(1G-1332M) it is 668 Mhz, very close to the fundamental frequency, this component is identified as a small hump in the side lobe of fundamental frequency.
 

Your frequency choice is a little dangerous. As you can see below, many harmonics are falling very close to your fundamental:
sampling rate: 2000MHz
fundamental frequency: 666MHz
2nd harm: 668MHz
3rd harm: 2MHz
4th harm: 664MHz
5th harm: 670MHz
6th harm: 4MHz
7th harm: 662MHz
8th harm: 672MHz
9th harm: 6MHz
10th harm: 660MHz
11th harm: 674MHz
Moreover, are you actually using 666015.625Hz or exactly 666MHz? In the first case, you would be using coherent sampling (the DAC clock and the source clock must be locked) while in the second case you would have your signal energy spread over several bins hiding the higher order harmonics.
 

the choice of frequencies comes from the specification of DAC it is required to give sfdr >=50 dB
upto 33% of sampling frequency.

ok to satisfy the condition of coherent sampling fs/fin = 4096/Mcycle.
I have choosen Mcycle a prime no 1361 which gives fin = 664.55078125Mhz. while fs= 2GSPS.
does this way satisfying coherent sampling is different from changing fin as well as fs to bring Mcycle to close prime no.
the second case you described is not understood by me "signal energy spread over several bins " bins means what?
 

Your frequency choices are correct!

If you did not use coherent sampling, the energy of your signal would not all fall into one bin but would be split into adjacent bins. This is not your case. The humps you see might be due to the harmonics just buried in your noise floor. So, to answer your initial question, the capture with 8192 points might be more accurate (especially if you kept the coherent sampling condition).
 
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