could anyone tell me the process of the demodulation of I\Q signal in the Rx part of AD9361?
To demodulate the RF modulated signal, it is split two ways and one side is multiplied by LO (and it becomes I) and the other side is multiplied by LO shifted by 90deg (and it becomes Q).
You can consult resources on the internet regarding this subject. There are many available.
Signal is not demodulated. Rx digital output is digital samples of I and Q. Demodulation would have to be done off chip in a BBP.
thank you for your reply. I am still confused. Rx part receives signal from RF port and the signal is analog. the analog signal needs be demodulated into I\Q signal before the ADC. Am I right? could you explain the process between RF port and ADC in the Rx part.
Please see UG-570 starting from page 35 (other sections of the same document will explain the rest of the blocks).
AD9361 and AD9364 Integrated RF Agile Transceiver Design Resources | Analog Devices
You are correct, received signal is an RF signal that is amplified, filtered (analog, digital), downconverted, digitized and demodulated into I and Q digital outputs. See documentation for more detail.
thank you for reply. I have already seen the document you refer to. I still have a question, in the process of modulation, modulator take the :
1. I multiply it by a LO.
2. Q multiply it by a LO.
3. add the resulting waveforms together.
corresponding to modulation, what will be taken to get the I\O signal from RF signal ? I can't find the the answser in the document,Did I miss something?
thank you for your reply. Do you mean that the RF signal is split two signals (I signal and Q signal) after demodulated,then I signal and Q signal are processed ,respectively? But in the block diagram of AD9361,
there is just only one channel after the mixer, could you tell me where is my problem?
The diagram is a simplified block diagram. AD9361 uses IQ modulators and IQ demodulators.
Have a look at:
AD9361 and AD9364 [Analog Devices Wiki]
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