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ADS5402EVM: Accurate data conversion methods

Takaki Aoyama
Prodigy 20 points
Part Number: ADS5402EVM
Other Parts Discussed in Thread: ADS5402,

Hello, I am trying to get a sinusoidal signal into a PC using an AD converter.
I am trying to get a sinusoidal signal into a PC using an AD converter.
I am using ADS5402 / ADS5402EVM / TSW1405. The frequency of the input signal is 155 MHz, and when the sampling frequency is set to 800 MHz and a capture is performed, only noise appears, and the FFT result also shows only noise.
Is this because the clock signal is not input?
I thought it was operating with an internal clock, but am I wrong?

こんにちは．
私は，AD変換器を用いて正弦波信号をPCに取り込もうとしています．
使用しているデバイスは，ADS5402 / ADS5402EVM / TSW1405 です．入力信号の周波数は155MHzであり，サンプリング周波数を800MHzと設定しキャプチャーを行ったところ，ノイズしか現れずうまく波形を取得できませんでした．FFT結果についても同様にノイズしか確認できませんでした．
この原因は，クロック信号を入力していないからなのでしょうか．
内部クロックで動作していると考えていましたが，間違っていますでしょうか．

  • 0
    TI__Guru 68094 points

    Hi Takaki-san

     The ADD5402 requires an external clock. Please refer to the EVM users guide for detail.

  • 0 in reply to Kang Hsia
    Prodigy 20 points

    Hi, Kang-san

    Thanks for your answer.

    Can the clock signal be the same as the input signal?

    That is, whether the clock signal can be a sinusoidal shape, and whether the frequency of the clock signal can be the same as that of the input signal?

  • 0 in reply to Takaki Aoyama
    TI__Guru* 91947 points

    Takaki-san,

    The evaluation setup should use a clock from a high-quality signal generator (normally a sine wave around 10dBm) and a sine wave for the analog input from a high-quality signal generator. High-order, narrow-bandpass filters are usually required on clock and input frequency for removing phase noise and harmonic content from the input sine waves. If the two signal generators are not synchronized by an external reference signal to make the clock and input frequency coherent, then the resulting FFT first needs to have a windowing function such as Hanning or Blackman-Harris applied to the data.

    If the clock signal is the same frequency as the input, your output will be a DC level signal as the samples will all be the same value.

    Regards,

    Jim