AD6688 datasheet的SFDR问题

Question

1.图17中文字表述Ain=15dBFs，但我看图像中幅度大概在-20dBFs，这是什么原因呢，求解答 
 2.图21和图17的两个输入频率是一致的，但我看图17中文字表示Ain=-15dBFs时，SFDR等于96dBFs，而对图21中，当输入幅度在-15dB时，图中绿线SFDR（dBFs)大概在-70多，为什么差这么多呢，这是图错了吗

Sirgeo · Accepted Answer

Hi, 
 1- The loss is due to the balun and the matching network. Both will introduce loss. 
 2.- Figure 17 SFRD measurement is when is in DCC0 mode, that example has 8 decimation (bandwidth of 375 MHz). While in the figure 21 the system is in full bandwidth mode (1.2 GHz bandwidth). 
 Bigger the bandwidth, SFDR will be larger, but if we reduce bandwidth to 375 MHz will go up because of the filtering done in DCC0 mode. 
 
 Thanks 
 Sergio

bulling · Answer

感谢您的回答，我想再请教一下DCC0 模式是什么模式，这方面我不太清楚，麻烦您了

Sirgeo · Answer

Sorry, made a mistake. Is not DCC is DDC (Digital Down Converter). This is a feature embedded in the AD6688 chip. That allows to de-mix the signal digitally by using an NCO (numerical controller oscillator). On DDC mode you can select the bandwidth by choosing the filtration, and after applying filter the signal will be decimating depending on the filter factor. If you want to read more about it, please referrer to pages 42 to 62 on the AD6688 spec sheet. 
 
 Thanks

AD6688 datasheet的SFDR问题

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