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?
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?
