reciprocal mixing: the trouble with oscillators · reciprocal mixing: lte example •note the noise...
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Reciprocal Mixing:The trouble with oscillators
Tradeoffs in RX
• Noise Figure(Sensitivity)
• Distortion (Linearity)
• Phase Noise (Aliasing)
©James Buckwalter 2
Phase Noise
• Phase noise is the frequency domain representation of rapid, short-term, random fluctuations in the phase of a waveform, caused by time domain instabilities ("jitter").
©James Buckwalter 3
Note that commercial generators don’t do better than around -160 dBc/Hz
What does phase noise look like?
©James Buckwalter 4
Simple Phase Noise Model
• Consider an oscillator with some noise
• When that noise is small,
• Our power spectrum indicates
©James Buckwalter
sinlo loV t A t t
sin coslo lo loV t A t t t
2
4VV lo lo lo lo
AS f S S
5
Phase Noise Definition
• We can define the phase noise from the ratio of the noisy signal to the ideal carrier component
• Independent of amplitude. Note that the units of phase noise are dBc per Hz. dBc means a relative level with respect to the carrier.
©James Buckwalter
2
10 102
410log 10log
4
lolo
lo
lolo
AS S
LA
6
Characterizing Phase Noise
©James Buckwalter
10
,110log
sideband LO
carrier
P f f HzL f
P
7
Approximating the Phase Noise
• Phase noise is generally broken into well-defined regions.
©James Buckwalter
1
1
1
log logk
i i i i i
i
L f a f f b U f f U f f
1/f3 1/f2 Constant
8
Phase Noise Definition
• Let’s relate this back to a timing parameter
• Phase jitter:
• This generally looks like
©James Buckwalter 9
sq
2 t( ) = 4 Sqq
f( )sin2 p ft( )-¥
¥
ò df S
qqf( ) =10
L w{ }/10
s
t
2 t( ) = ct +kt2
Sqq
f( ) = cf
osc
f
æ
èç
ö
ø÷
2
sq
2 t( ) = c 2p fosc
t( )2 sin2 p ft( )
p ft( )2
-¥
¥
ò df = c2p f
osct( )
2
pt
sin2 x( )x2
-¥
¥
ò dx = c 2p fosc( )
2
t
Phase Noise Definition
• Cycle-to-cycle Jitter:
©James Buckwalter 10
st
2 1
fosc
æ
èçç
ö
ø÷÷ =
c
fosc
Approximating the Phase Noise
• By parameterizing the phase noise, we can find the expected phase jitter
©James Buckwalter
1
1
1
log logk
i i i i i
i
L f a f f b U f f U f f
1 1 1/10 /102 10 10
1
1
2 10 110
i i
i i
a akb a i
i i i
i
af f f
ai is the slope
Fi
bi
11
Calculating Phase Noise
• In 1/f2 region,
• If treated as constant,
©James Buckwalter
L f{ } = -20 log
10f( )- log
10f
i( )( )+ bi
12
L f{ }fl
fh
ò df =10b
l10
102log
10fl( )
10log
10f -2( )
fl
fh
ò df
=10b
l10
fl
2 f -2
fl
fh
ò df =10b
l10
fl
2 -1
ffl
fh
=10b
l10
fl
2 f -2
fl
fh
ò df =10b
l10
fl
2 -1
ffl
fh
=10b
l fl
fh- f
l
fh
æ
èçç
ö
ø÷÷
L f
x{ } = -20 log10
fx( )- log
10f
i( )( )+bi
L fx{ }
fl
fh
ò df =10L f
x{ } 10f
h- f
l( ) =10b
lf
l
fx
æ
èçç
ö
ø÷÷
2
fh- f
l( )
10b
lf
x
fl
æ
èçç
ö
ø÷÷
2
fh- f
l( ) =10b
l fl
fh- f
l
fh
æ
èçç
ö
ø÷÷®
fl
fx
æ
èçç
ö
ø÷÷
2
=f
l
fh
æ
èçç
ö
ø÷÷® f
x= f
hf
l
Oscillator Conditions
©James Buckwalter
G jw o( )Z jwo( ) = 1
ÐG jwo( )Z jwo( ) = 2pn
13
LC Oscillator Tank
• Look at perturbed behavior near resonance
©James Buckwalter
1 1
1 1
1
1
o
o
o
o
Z R RR j
sCR Q QsL j
Z Rj
Q Qj
1 21 1 1
o
oo o
R RZ
jQjQ jQ
14
LC Oscillator
• Now add negative resistance to impedance of tank…
©James Buckwalter
Z wo+ Dw( ) »
R
1+ 2 jQDw
wo
15
LC Oscillator
• Now add negative resistance to impedance of tank…
©James Buckwalter
1 1
11 1 1 1 2
2
o
m
m
o o o
o
o
RZ
gjQ jQ g jQ
R
RZ
jQ
16
LC Oscillator Phase Noise
©James Buckwalter
L Dw{ } = 10logv
noise
2
vs
2
æ
èçç
ö
ø÷÷
17
LC Oscillator Phase Noise (cont)
©James Buckwalter
vnoise
2 = Z wo+ Dw( )
2 in
2
Df
vnoise
2 =R
2QDw
wo
æ
è
çççç
ö
ø
÷÷÷÷
2
4kTG
R= 4kTGR
wo
2QDw
æ
èçö
ø÷
2
18
LC Oscillator Phase Noise (cont)
©James Buckwalter 19
vnoise
2
vs
2=
4kTGRw
o
2QDw
æ
èçö
ø÷
2
vs
2
vnoise
2
vs
2=
4kTG
Posc
wo
2QDw
æ
èçö
ø÷
2
LC Oscillator Phase Noise (cont)
©James Buckwalter
• A factor of ½ is included in noise expression to indicate AM-PM / AM-AM distinction.
• Phase noise is related to
– Oscillator Q
– Oscillator power
– Noise in tank circuit
L Dw{ } = 10log2kTG
Posc
wo
2QDw
æ
èçö
ø÷
2æ
èçç
ö
ø÷÷
20
Leeson’s Formula
• This expression can be empirically expanded to describe noise in other regimes.
• At low frequency
• At high frequency
©James Buckwalter
2
1/210log 1 1
2
fo
osc
kTL
P Q
2
1/210log
2
fo
osc
kTL
P Q
2
10logosc
kTL
P
21
Phase Noise Regions
©James Buckwalter
2
1/210log 1 1
2
fo
osc
kTL
P Q
L Dw{ } = 10log2kTG
Posc
wo
2QDw
æ
èçö
ø÷
2æ
èçç
ö
ø÷÷
2
10logosc
kTL
P
22
Classical Phase Noise Theory
• Use a linear time invariant model
• Assume all noise sources are white
• In reality, the oscillator is a large-signal circuit
• Transconductor might be on only part of the time
• Oscillator noise is cyclostationary
• Leeson’s formula while intuitive is empirical!
©James Buckwalter 23
Why is Phase Noise Important?
• Channels are spaced tightly to provide a large number of users.
• Reciprocal Mixing
• Adjacent channel power might be much higher than desired signal. Mixing smears the strong signal into the desired channel.
©James Buckwalter 24
Phase Noise Impairment (I)
©James Buckwalter 25
P
int- P
sig+ SNR
min= SNR
PN
• Compare the adjacent channel power to the desired power to determine the signal level that will fall into the desired band.
Phase Noise Impairment (II)
• For cell standards, in-band jammer can be 40 dB greater than desired signal.
• SNRmin should be > 20 dB
©James Buckwalter
SNRPN
= Pint
- Psig
+ SNRmin
SNR
PN< 40 + 20 = 60dB
26
Phase Noise Impairment
• How do we specify the noise due to the phase noise?
• In 1/f2 region,
• From earlier..
©James Buckwalter
SNRPN
=1
L Df{ }fl
fh
ò dDf
SNRPN
=1
L fh
fl{ } f
h- f
l( )
L f{ } = -20 log
10f( )- log
10f
i( )( )+ bi
27
A Cautionary Note
• Summary
L f
hf
l{ } = Psig
- Pint
- SNRmin
-10log10
fh- f
l( )
GSM Frequency Plan
A Cautionary Note
• Summary
L f
hf
l{ } = Psig
- Pint
- SNRmin
-10log10
fh- f
l( )
Phase Noise Impairment (IV)
• Example: GSM channel spacing (600 kHz)
• We specify that the oscillator should have a phase noise of under -99 dBc at a 100 kHz offset frequency.
©James Buckwalter
L fh
fl{ } = L 519kHz{ } = -99dBm- -43dBm( ) -10log
10600kHz( ) = -113dBc
L 100kHz{ } = -113dBc+ 20log10
519kHz
100kHz
æ
èç
ö
ø÷ = -99dBc
31
L f
hf
l{ } = Psig
- Pint
- SNRmin
-10log10
fh- f
l( )
Basic FDD SystemWhat about our out-of-band Blockers?
Reciprocal Mixing
• Assume RX at 2120 MHz and TX at 1970 MHz.
• Therefore, frequency offset is 50 MHz.
• TX signal is 10 dBm, RX sensitivity is -110 dBm.
• What is the required phase noise if the reciprocal mixing is 10 dB less that RX sensitivity?
©James Buckwalter 33
Reciprocal Mixing: LTE Example• Calculate the SNR phase noise
• The phase noise is
• Translate to 10 MHz
SNRPN
= Pint
- Psig
+ SNRmin
SNRPN
=10dBm- -110dBm( )+10dB =130dB
©James Buckwalter 34
L fh
fl{ } = -SNR
PN-10log
10f
h- f
l( )
L 49MHz{ } = -130dBc -10log10
60MHz - 40MHz( )L 49MHz{ } = -130dBc -73dB = -203dBc
L 10MHz{ } = -203dBc+10log
104.9( ) = -197dBc
Reciprocal Mixing: LTE Example
• Note the noise floor for the phase noise is around -174 dBm.
• If the noise floor is -164 dBm, this suggests that the oscillator power must be 33 dBm. This is very high oscillator power (and dc power consumption).
©James Buckwalter 35
L 10MHz{ } = -197dBc
Reciprocal Mixing: LTE Example
• Let’s ask the opposite question. Given the noise floor and oscillator power, how much transmit power can we tolerate?
• The phase noise is
©James Buckwalter 36
SNRPN
= -10log10
fh- f
l( ) - L 50MHz{ }
SNRPN
= -10log10
60MHz - 40MHz( ) - -174dBc( )SNR
PN=174dBc -73dB =101dB
L 50MHz{ } = -164dBm-10dBm = -174dBc
Reciprocal Mixing: LTE Example
• The phase noise SNR is related to
• This isn’t much blocker power. Consider the out-of-band blocker levels that we have discussed for 800 MHz and 2.4 GHz. We MUST filter signals to avoid reciprocal mixing!
Pint
= SNRPN
+ Psig
- SNRmin
Pint
=101dB -110dBm-10dB = -19dBm
©James Buckwalter 37
Conclusions
• Oscillator phase noise is a silent killer in interference limited systems.
• Requirements for phase noise are dictated by phase noise in 1/f^2 and thermal noise floor.