why signal are observed at IQ path ?

[HustLiliAn]

Hi, everyone :

I am confusing about a phenomenon I observed from a simple demo. The demo is that a sine signal is transmitted by the USRP through I path.  I observed sine signal at both I path and Q path of the received signal for the same USRP. The Q path signal is not smooth as the I path and the amplitude is smaller.  However, I think the Q path should not exsist the signal.    I understand this problem as the figure shows and the attachment is the demo

[ErikL · Accepted solution]

The USRP allows Tx and Rx LO are independent with a common reference allowing you to tune to two frequencies at once. This also means phase of the two LO's are different. This means that the acquired signal will disperse energy among I and Q but will have a common complex magnitude. If plotted on a constellation diagram it would look like a rotated constellation.

[HustLiliAn]

Ok , Erik , thanks , I got it , and I should revise my  theoretic analysis by adding the influence of the LO's radom initial phase.

However , another problem is that if I transmit a constant array such as [1 1 1 1 1.......] through I path,  why I can not observed the DC signal at the receive path?  Whether will it be filtered out ?