I posted the same question on the math forum as well.

I was reading this paper on direct conversion receiver, and there's a part on the math of calculating the DC offset due to self mixing that I am a little confused about.

http://www.ee.ucla.edu/~brweb/papers/Journals/RTCAS97.pdf

Refer to 4th paragraph in page 3 of the paper.

1) It says the LO has a Vpp of 0.63v (0dBm in a 50 ohm system)... how does a 0.63v translate to a 0 dBm in a 50 ohm system?

2) following the rest of the paragraph. Suppose I start off with 0 dBm at the mixer, then it goes down to -60dBm at LNA due to the attenuation, then add the LNA/mixer gain of 30dB, that means it would put the self-mixing signal at -30dBm at the output of the mixer? how does -30dBm at the output translate to 10mV DC offset?

3) with a 10mV dc component at the input into an amplifier, how easily would that drive the amplifier into saturation? wouldn't it depends on how much rail to rail voltage is there and how much the peak to peak signal swing is?

 

Not sure my question pertain to just math. Think there's some analog question as well.

I was reading this paper on direct conversion receiver, and there's a part on the math of calculating the DC offset due to self mixing that I am a little confused about.

http://www.ee.ucla.edu/~brweb/papers/Journals/RTCAS97.pdf

Refer to 4th paragraph in page 3 of the paper.

1) It says the LO has a Vpp of 0.63v (0dBm in a 50 ohm system)... how does a 0.63v translate to a 0 dBm in a 50 ohm system?

2) following the rest of the paragraph. Suppose I start off with 0 dBm, then it goes down to -60dBm due to the attenuation, then the LNA/mixer gain of 30dB put it at -30dBm at the output of the mixer? how does -30dBm at the output translate to 10mV?

3) with a 10mV dc component at the input into an amplifier, how would that drive the amplifier into saturation?

For Q1, do you know how to calculate decibels? For power, it is 10xLOG(P/Po). Use as a reference for Po = 1 milliwatt. A dbm is decibels referenced to 1 mW. As for the power P, find the RMS power for 0.63 volts peak to peak across 50 ohms.

 

(1)
1mW in 50 Ω would be 0dBm.
So
P = V^2/R =V^2/50=1E-3
giving
V=√(50E-3)=0.2236 V [rms] = 0.632 V [pk-pk]

(2)
Something looks strange there.

The DC/mean mixer output due to self-mixing of the two LO frequency signals Va & Vb at the mixer input would be

(Va[pk]*Vb[pk])/2

Where Va is the LO input and Vb is feedback component.

A LNA/Mixer gain of 30dB has to fit into this picture as well. The text doesn't elaborate as to how the 30dB gain is apportioned between the LNA and the mixer. Is it all in the LNA? I don't think it matters in the end.

The feedback component to the LNA input is stated as -60dB down on the LO signal. If the LNA/Mixer gain is all in the LNA then the effective value of the feedback signal appearing at the mixer input would be -30dB down on the LO signal.

Suppose this was the case then the effective feedback signal is -30dB down on 0.63V pk-pk or 0.63/31.62=20mV pk-pk. So the mixer DC output due to self mixing would then be 0.315*10mV/2=1.58mV. This doesn't tie up with the 10mV DC offset mentioned in the text.

If the mixer is considered to have unity gain, then the self mixing feedback component would have to be about 63.5mV peak or 127mV pk-pk with the LO input at 0.63 V pk-pk to give a mean mixer output of 10mV. This would imply a LO signal feedback level of about -14dB down on the LO level. Again this doesn't tie in with the attenuation factors stated in the text. With -60dB spurious LO signal feedback to the LNA input the LNA/mixer gain would need to be more like 46dB to give a 10mV self mixing offset at the mixer output.

Perhaps I'm missing something obvious.

(3)
If the down-line amplifier will amplify unwanted DC offset on a very low level useful signal (~30uV) then a 10mV offset would cause problems in a high gain [70 dB] amplifier - notably producing saturation or at least clipping distortion.

 

Hello,

I have moved the answers from the other thread over here and closed the other thread.

Bertus

 

Ok, i think I get 1) and 2) now, but still not so sure about 3)

For 1), A Vpp means peak to peak voltage. Vpp of 0.63V is a peak voltage of half that, or 0.31V.

Recalling that (RMS voltage)(Rms current)= Power, and (RMS voltage)/(RMS current) = characteristic impedance, you can solve for Power...and sure enough you get 1mW, or 0dBm.

Alternatively you can just look up peak voltage on a chart http://www.rfcafe.com/references/electrical/pwr2volts.htm
and get the power.

For 2), Using those same equations backwards, you can go from power being carried on a T-line to peak voltage on the T-line. Or you can just look it up on the table above.

-30dBm = 10 log (w/1mW) => w = 1e-6w

power = (V)^2/100 => V = 10mV

Either way, you get -30dBm is the same as a peak voltage of 10mV on a 50 ohm transmission line.

Still not sure about 3). Is there a good link that talks about it?