Understand AM signal reception -- No Q component?

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(2pifc*t) and i am receiving
that
signal in antenna.

2. In Daughter board, this signal is multiplied by cos(2pifct) and
sin(2
pifct) in two separate branches (assuming Direct conversion –
Baseband conversion)

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

4. Branch 2: x(t)cos(2pifct) * sin(2pifct) --> 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(2pifct) -
xi(t)sin(2
pifct) – 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

On 03/24/2012 10:41 PM, senthil murugan wrote:

Baseband conversion)

1. Branch 1: x(t)cos(2pifct) * cos(2pifct) → LPF → x(t) → FPGA

2. Branch 2: x(t)cos(2pifct) * sin(2pifct) → LPF → 0 (since cos
and sin are orthogonal) → FPGA

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

There is also mixing in the DSP, see tuning notes:
http://files.ettus.com/uhd_docs/manual/html/general.html#tuning-notes

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

Is my understanding correct?

Not quite. This assumes that x(t) is symmetric, and more so, that the
phase of the carrier is the same phase as the down conversion chain.