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Phase difference of two 1 Ghz analog signal measurement using DC12D1800RF

Quang Duc Pham
Prodigy 20 points
Other Parts Discussed in Thread: WAVEVISION5

Hello,

I am planing to buy an  AD converter DC12D1800RF for a project. The purpose of project is to precisely measure the phase difference of two very high frequency signals. The frequency of two input signal is about 1Ghz.

In order to measure the phase difference of two input signals, first, the input signals (analog signals) are connected and sampled by two corresponding ports of the AD converter. Then the sampled signals are loaded to the computer for processing (band-pass filter for instance). The processed signals are integrated with each other for measuring the phase difference. 

Because the sampling rate of the AD converter is 1.8Ghz, this means there will be about 1.8Ghz x 12 bits data read and acquired by the computer in 1 second. The amount of data is too large to process.

So my question is how sampled signals are read to the computer and is it possible for me to measure the phase of two 1Ghz signals using  DC12D1800RF?

Please help,

Sincerely,

Quang.

  • 0
    TI__Mastermind 44680 points

    Hi Quang

    What you plan to do should be possible using the ADC12D1800RFRB evaluation module.

    Data from the ADC is captured by the on-board FPGA and stored internally in buffers that can handle as many as 32k samples per channel per storage event. Then the stored data is transferred over the USB interface to the WaveVision5 GUI. Once the data is available in the GUI it can be analyzed using the GUI tools, or the time domain plot data can be exported into a file for further analysis in your software platform of choice.

    As long as 32k samples on each channel is enough for your needs, everything should work fine.

    Regards,

    Jim B

     

  • 0 in reply to Jim Brinkhurst1
    Prodigy 20 points

    Hello Jim B,

    You've been very helpful

    thank you,

    Quang.