1
\$\begingroup\$

Studying the characteristics of an oscillator with a phase noise analyzer, but I don't quite understand the difference between absolute and residual noise. Any help would be greatly appreciated

\$\endgroup\$
  • \$\begingroup\$ Did you not do any research? community.keysight.com/thread/19932 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 28 at 18:03
  • \$\begingroup\$ What is your oscillator? Does it have an input, or is it just a free-running oscillator? If it has no input, it doesn't make sense to talk about the residual phase noise. \$\endgroup\$ – Justin Jun 28 at 19:09
  • \$\begingroup\$ you don't normally talk about residual noise of an oscillator. Amplifier aka buffer aka isolator, divider, multiplier, mixer, yes, it's the output phase noise less the input phase noise, ie the irreducible phase noise of the component itself. You could regard the tune input of a VCO as a noise input, but it's trivial to hammer that with a big decoupling capacitor, and then all output noise is due to the VCO, and its impossible-to-remove tuning components. \$\endgroup\$ – Neil_UK Jun 28 at 19:51
0
\$\begingroup\$

Absolute phase noise is the phase noise measured on the output signal of a one port device. Generally, when phase noise plots or tables are shown on the data sheet of an oscillator or signal generator, it is the absolute phase noise measurement data. Residual phase noise is a measure of the noise added to an input signal by a 2-port device. Some signal generators specify the residual phase noise in their data sheets. Think of the signal generator as a two port device (in this case, a frequency multiplier) where the “input port” is the external frequency reference (generally 10 MHz) and the “output port” is the RF output.

Having residual phase noise data enables users to calculate the absolute noise of their signal generator at the RF output when using an external 10 MHz signal as the reference input. ref

\$\endgroup\$
  • \$\begingroup\$ Yes I did research and read that article, but it doesn't make sense to me, hence why I asked. Is the difference simply the addition of noise in the residual measurement? Wouldn't subtracting this noise after the measurement just yield the absolute? \$\endgroup\$ – Sal M Jun 28 at 18:13
  • 1
    \$\begingroup\$ Yes but two different absolutes. The signal minus the 10MHz reference \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 28 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.