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Re: [Discuss-gnuradio] Understand AM signal reception -- No Q component
From: |
Josh Blum |
Subject: |
Re: [Discuss-gnuradio] Understand AM signal reception -- No Q component ?? |
Date: |
Sun, 25 Mar 2012 01:09:59 -0700 |
User-agent: |
Mozilla/5.0 (X11; Linux x86_64; rv:10.0.2) Gecko/20120216 Thunderbird/10.0.2 |
On 03/24/2012 10:41 PM, senthil murugan wrote:
> 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.
>
There is also mixing in the DSP, see tuning notes:
http://files.ettus.com/uhd_docs/manual/html/general.html#tuning-notes
> 6. 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.
Because of the phase difference, your received signal is better
described as x(t) * e^(j*phi)
So, if x(t) = Re{x(t)} and phi = 0, then you get your answer.
> 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?
>
Using this above logic, you should get this baseband signal:
[xr(t) + j*xi(t)] * e^(j*phi)
-Josh