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User Manual | PE11S100X Series Synthesizer
With this simplification the single sideband integrated VCO phase noise, Ф
is given by
where
2
(f
)
is the single sideband phase noise in rads
o
SSB
frequency of the closed loop PLL, and N is the division ratio of the prescaler.
The rms phase jitter of the VCO in rads, Ф, results from the power sum of the two sidebands:
Since the simple integral of (EQ 3) is just a product of constants, we can easily do the integral in the log
domain.
For example if the VCO phase noise inside the loop is -100 dBc/Hz at 10 kHz offset and the loop bandwidth
is 100 kHz, and the division ratio N=100, then the integrated single sideband phase noise at the phase
detector in dB is given by
equivalently Ф = 10
-82/20
While the phase noise reduces by a factor of 20logN after division to the reference, the jitter is a constant.
The rms jitter from the phase noise is then given by T
In this example if the reference was 50 MHz, T
A normal 3 sigma peak-to-peak variation in the arrival time therefore would be ±3
If the synthesizer was in fractional mode, the fractional modulation of the VCO divider will dominate the
jitter. The exact standard deviation of the divided VCO signal will vary based upon the modulator chosen,
however a typical modulator will vary by about ±3 VCO periods, ±4 VCO periods, worst case.
If, for example, a nominal VCO at 5 GHz is divided by 100 to equal the reference at 50 MHz, then the worst
case division ratios will vary by 100±4. Hence the peak variation in the arrival times caused by ΔΣ
modulation of the fractional synthesizer at the reference will be:
T
T
In this example,
j ∑pk
modulation is approximately gaussian, we could approximate T
estimate the rms jitter of the ΔΣ modulator as about 1/3 of T
Hence the total rms jitter Tj, expected from the delta sigma modulation plus the phase noise of the VCO
would be given by the rms sum, where:
18
π
2
2
= (
( f
) B
) / N
o
SSB
2
√
2
=
2
SSB
2
2
= 10log (
(f
dB
= 56 urads rms or 3.2 milli-degrees rms.
= ±T
· (N
j ∑pk
VCO
max
= ±200 ps(104-96)/2 = ±800 ps.
2
/Hz inside the loop bandwidth, B is the 3-dB corner
2
2
)Bπ ⁄ N
) = -100 + 50 + 5 - 40 = -85 dBrads, or
o
Ф / 2π
= T
jpn
ref
= 20 nsec, and hence T
ref
- N
) / 2
min
. If we note that the distribution of the delta sigma
as a 3 sigma jitter, and hence we could
jΔΣpk
or about 266 ps in this example.
jΔΣpk
, in rads
at the phase detector
2
2
(EQ 2)
(EQ 3)
= 178 femtoseconds.
jpn
√
2 T
= 0.756 ps
jpn
(EQ 4)
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