[Discuss-gnuradio] Understand AM signal reception -- No Q component ??

[senthil murugan]

Hi all,

I have some doubt in understanding working principle of usrp. I just want to confirm my understanding:

1. I have AM modulated singnal x(t)cos(2*pi*fc*t) and i am receiving that signal in antenna.

2. In Daughter board, this signal is multiplied by cos(2*pi*fc*t) and 
sin(2*pi*fc*t) in two separate branches (assuming Direct conversion -- 
Baseband conversion)

3. Branch 1: x(t)cos(2*pi*fc*t) * cos(2*pi*fc*t) --> LPF --> x(t) --> FPGA

4. Branch 2: x(t)cos(2*pi*fc*t) * sin(2*pi*fc*t) --> LPF --> 0 (since cos and sin are orthogonal) --> FPGA

5. Since Daughter board is doing direct conversion, no mixing in FPGA (i.e) in DDC.

6. In PC, I will have I component as x(t) and Q component as 0 for AM modulated signal.

Is my understanding correct? 

Similarly if I take OFDM modulated signal -- xr(t)cos(2*pi*fc*t) - 
xi(t)sin(2*pi*fc*t) -- output at Daughter board will be xr(t) at I 
component and xi(t) at Q component. This will be send to PC. Is it 
correct?

Thanks
Senthil