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WAVEVISION5: WAVEVISION5:the difference between Int SFDR and Peak SFDR

zhikai zhu
Prodigy 70 points
Part Number: WAVEVISION5

Dear Support,

When using WAVEVISION5, I find the int SFDR and the peak SFDR on the FFT readouts panel. I can't find any reference information about the two kinds of SFDR, so I don't know how to select between them.

Can anyone tell me some information about them? 

Thanks!

  • 0
    TI__Mastermind 27420 points

    Hi Zhikai,

    Can you please send a picture of the FFT readout panel? I don't see these SFDR references on my version of Wavevision5.

    Regards,

    Rob

  • 0 in reply to Rob Reeder
    Prodigy 70 points

    Hi Rob,

    By default the FFT readouts panel shows the int SFDR, you can switch to peak SFDR by selecting the "Peak" in "HD/IMD Calc.Method" option on the "FFT Controls" panel, then you can find the SFDR value on the FFT readouts panel changes. Picture 1 shows that.

    What's more, you can right click on the wave window, and select "Plot readouts",then it shows the Int SFDR and Peak SFDR. Picture 2 shows that.

    You can find the difference between int SFDR and peak SFDR is very big sometimes.

    Picture 1

    Picture 2

  • 0 in reply to zhikai zhu
    TI__Mastermind 27420 points

    Hi Zhikai,

    Okay, I think I understand your question better, thank you for the additional detail.

    You would choose integrated when using non-coherent sampling (analog sign and clock signal not locked) and you would choose the "peak" option when using coherent sampling (locked analog signal and clock).

    Regards,

    Rob

  • 0 in reply to Rob Reeder
    Prodigy 70 points

    Hi Rob,

    Thanks for your reply. But I don't know much about non-coherent sampling and coherent sampling, and can't truly understand what your "locked" mean. Can you tell me how to distinguish between non-coherent sampling and coherent sampling?

    By the way, I refer to the explanation of "coherent sampling" on wikipedia.org, but I don't know whether the explanation is the same with what you mean.

  • 0 in reply to zhikai zhu
    TI__Mastermind 27420 points

    Hi Zhikai,

    Here is the shortened version of the difference between, Coherent sampling vs. non-coherent sampling in terms of your ADC setup....

    Coherent sampling: both the analog input frequency and clock frequency are locked in phase with a reference frequency, typically this mean you use the 10MHz reference output on one signal generator and connect it to the input of the 10MHz reference of the other signal generator. Then the signal generator with the input 10MHz ref will need to show on the display something like "EXTREF". This means the two sig gens are reference together. Next, you need to select no windowing or rectangular windowing in the FFT capture. once you have a set sample frequency and number of FFT point in mind, you need to calculate the exact analog input frequency to apply. That exact frequency needs to be entered in the signal generator and possibly in the data capture software. This ensure all the FFT points are bin centered. For example: Fs: 1000MSPS, FFT points: 32768, Analog Input: 100MHz, this equals and analog input of 100.007486MHz, put this exact frequency in the signal generator. See this link which can help you arrive to this number https://en.wikipedia.org/wiki/Coherent_sampling or you can use the HSDC-Pro SW at this link as well, http://www.ti.com/tool/DATACONVERTERPRO-SW download this SW and the coherent frequency calculate is built into this GUI.

    If you use Non-coherent sampling: then your signal generators do not need to be reference locked together as described. However, your analog input frequency should be "smeared". Meaning if you want to sample a 100MHz analog input signal, use something close to 100MHz that is a prime number...example would be 99.1235MHz. Also in the FFT capture you need to make sure you are using the windowing option, like Blackman-Harris, etc. This will ensure the FFT capture spreads out the samples appropriately and you are not simply capturing the same point along the input sinewave signal.

    Regards,

    Rob

  • 0 in reply to Rob Reeder
    Prodigy 70 points

    Hi Rob,

    Thanks very much.