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RMS Jitter From Phase Noise

strigois

I am trying to convert Phase Noise to RMS Jitter(radians), but I'm having trouble following the units through the process.

To get RMS Jitter, in radians, from Phase Noise you must integrate the Phase Noise.  What are the units of integrated Phase Noise and how do they cancel.  The equation I am currently using for this is A = Phase Noise (L(f)) + 10*log10(frequency2- frequency1) and to generate the RMS Jitter value in radians I am using sqrt(2*10^(A/10)).

Additionally, what are the units for 10*log10(frequency2- frequency1) and if L(f) is the ratio of Pcarrier and Poffset in dBm, and that has units of dBc/Hz, what would be the units if it were converted to a linear value.

Thank you for any insight you might have on these questions.

Parents

  • Dear Grant,

    > ...question about integrating L(f) over a bandwidth is more of a process based one.  The equation I'm using to get "area" is:

    > A = L(f) + 10*log(frequncy2-frequncy1)

    > Will this give me the value I'm looking for?

    Unless I am not understanding your question, Grant, I do not believe this will (in the general case) provide the area representative of L(f) between frequency2 and frequency1. Why do I suggest this?

    L(f), in general, is not a constant value with frequency. It is expressed in units of dBc/Hz due to its rather large dynamic range. Its definition relates the noise power in a 1 Hz bandwidth to the entire carrier noise power. Hence, to estimate the rms value over an arbitrary frequency interval,  you are interested in the power, expressed in radian^2, represented by the area under the curve between frequency2 and frequency1. If L(f) happens to be a constant value between L(frequency2) and L(frequency1), say, Lo dBc/Hz,  then the area under the curve between frequency2 and frequency1 is:

    (frequency2 - frequency1)*[10^(Lo/10)]  and is multiplied by 2 to get the total spectral power between frequency2 and frequency1 by the definition of L(f). 

    This is similar to your expression but the area, A in your expression, must be expressed in radian^2 - not in logarithmic units.

    For example, suppose for a very good 100 MHz clock the phase noise is constant at -150 dBc/Hz between 100 kHz and 1 MHz, then the area under the curve is 1.80e-09 radian^2 or the rms jitter is 0.0675 ps rms.

    In general, L(f) is not constant with frequency and hence the area computation must be performed in the linear frequency domain.

    I hope I  understood your question Grant!

    Shawn

Reply

  • Dear Grant,

    > ...question about integrating L(f) over a bandwidth is more of a process based one.  The equation I'm using to get "area" is:

    > A = L(f) + 10*log(frequncy2-frequncy1)

    > Will this give me the value I'm looking for?

    Unless I am not understanding your question, Grant, I do not believe this will (in the general case) provide the area representative of L(f) between frequency2 and frequency1. Why do I suggest this?

    L(f), in general, is not a constant value with frequency. It is expressed in units of dBc/Hz due to its rather large dynamic range. Its definition relates the noise power in a 1 Hz bandwidth to the entire carrier noise power. Hence, to estimate the rms value over an arbitrary frequency interval,  you are interested in the power, expressed in radian^2, represented by the area under the curve between frequency2 and frequency1. If L(f) happens to be a constant value between L(frequency2) and L(frequency1), say, Lo dBc/Hz,  then the area under the curve between frequency2 and frequency1 is:

    (frequency2 - frequency1)*[10^(Lo/10)]  and is multiplied by 2 to get the total spectral power between frequency2 and frequency1 by the definition of L(f). 

    This is similar to your expression but the area, A in your expression, must be expressed in radian^2 - not in logarithmic units.

    For example, suppose for a very good 100 MHz clock the phase noise is constant at -150 dBc/Hz between 100 kHz and 1 MHz, then the area under the curve is 1.80e-09 radian^2 or the rms jitter is 0.0675 ps rms.

    In general, L(f) is not constant with frequency and hence the area computation must be performed in the linear frequency domain.

    I hope I  understood your question Grant!

    Shawn

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