With PXF the frequency range that you specify on the analysis is always the frequency range of the observed signal. Thus, since you specified 1Hz to 10MHz absolute, that is what you get. So you are measuring the transfer function from VDD from baseband to baseband.

For PSR for an oscillator, you probably want the transfer function from baseband to the carrier. So will want to use a relative sweep. That would cause the output frequency of your transfer function to range between f₀ + 1Hz to f₀ + 10MHz. When plotting VDD, you will need to choose the +1 sideband (or perhaps it is the -1 sideband, I always forget, use 'spectre -h pxf' for find out for sure). This will result in you computing the transfer function from 1Hz --> 10MHz to f₀ + 1Hz --> f₀ + 10MHz.

If you chose the output as a pair of node and the input as a voltage source (VDD), then the transfer function will be the voltage gain from the source to the output nodes. So a result of 40dB @ 1Hz would imply that a 1Hz signal at VDD is amplified by a factor of 100 before appearing at the output at f₀ + 1Hz.

You do not have to set mag or pacmag on the source.

-Ken

You can use small-signal FM modulation theory to map the transfer function in voltage to a transfer function in frequency.

-Ken

Any book on communication theory, such as Ziemer and Trantor.

-Ken

Ken,

I did read that section and equation 64. I am a bit confused as to what is Ac, the denominator of eqn 64 ?. Also i sL and U the results from PXF analysis.

Is it possible to show an examle using equation 64. That will be great help.

Thx

ic_engr

As for an example, take a look at http://www.designers-guide.com/Analysis/am-pm-conv.pdf.

-Ken