Conversion of Additive Noise to Phase Noise
RF Microelectronics by Razavi contains the following snippet in section 8.7.3 concerning the analysis of phase noise in oscillators:
We write x(t)=Acos(ω0t)+n(t)
where n(t)
denotes the narrowband additive noise (voltage or current). It can be proved that narrowband noise in the vicinity of ω0
can be expressed in terms of its quadrature components:
n(t)=nI(t)cos(ω0t)−nQ(t)sin(ωot)
where nI(t)
and nQ(t)
have the same spectrum of n(t)
but translated down by ω0
(Fig. below) and doubled in spectral density.
I don't see how the math adds up though. Taking the Fourier transform of n(t)
,
Sn(ω)=
1
2
[SnI(ω−ω0)+SnI(ω+ω0)]+
j
2
[SnQ(ω+ω0)−SnQ(ω−ω0)]
If the quadrature components are the same as mentioned so SnQ(ω)=SnI(ω)
,
Sn(ω)=
1−j
2
SnI(ω−ω0)+
1+j
2
SnI(ω+ω0)
Doesn't this show that the spectral density of SnI
and SnQ
is
2
√
2
that of Sn
rather than double, in order for the magnitude to equal?
We write x(t)=Acos(ω0t)+n(t)
where n(t)
denotes the narrowband additive noise (voltage or current). It can be proved that narrowband noise in the vicinity of ω0
can be expressed in terms of its quadrature components:
n(t)=nI(t)cos(ω0t)−nQ(t)sin(ωot)
where nI(t)
and nQ(t)
have the same spectrum of n(t)
but translated down by ω0
(Fig. below) and doubled in spectral density.
I don't see how the math adds up though. Taking the Fourier transform of n(t)
,
Sn(ω)=
1
2
[SnI(ω−ω0)+SnI(ω+ω0)]+
j
2
[SnQ(ω+ω0)−SnQ(ω−ω0)]
If the quadrature components are the same as mentioned so SnQ(ω)=SnI(ω)
,
Sn(ω)=
1−j
2
SnI(ω−ω0)+
1+j
2
SnI(ω+ω0)
Doesn't this show that the spectral density of SnI
and SnQ
is
2
√
2
that of Sn
rather than double, in order for the magnitude to equal?
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