Reciprocal Mixing: The trouble with oscillators

Transcription

1 Reciprocal Mixing: The trouble with oscillators

2 Tradeoffs in RX Noise Figure(Sensitivity) Distortion (Linearity) Phase Noise (Aliasing) James Buckwalter

3 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"). Note that commercial generators don t do better than around -160 dbc/hz James Buckwalter 3

4 What does phase noise look like? James Buckwalter 4

5 Simple Phase Noise Model Consider an oscillator with some noise lo When that noise is small, V t A t t sin Our power spectrum indicates lo sin cos V t A t t t lo lo lo A S f S S 4 VV lo lo lo lo James Buckwalter 5

6 Phase Noise Definition We can define the phase noise from the ratio of the noisy signal to the ideal carrier component L A S 4 lo S lo lo 10log10 10log 10 A lo lo 4 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 6

7 Characterizing Phase Noise L Psideband flo f,1hz f 10log10 Pcarrier James Buckwalter 7

8 Approximating the Phase Noise k 1 i i i i i1 log log L f a f f b U f f U f f i1 Phase noise is generally broken into well-defined regions. 1/f 3 1/f Constant James Buckwalter 8

9 Phase Noise Definition Let s relate this back to a timing parameter Phase jitter: s q s q This generally looks like ( t) = 4ò S qq ( f )sin ( p ft) df S qq f S qq ( ) ( t) = c( p f osc t) sin p ft ò ( p ft) - s t - æ ( f ) = c f ö osc ç è f ø ( df = c p f t osc ) pt ( t) = ct +kt ò - sin ( x) ( ) =10 L w James Buckwalter 9 x dx = c( p f osc ) t { }/10

10 Phase Noise Definition Cycle-to-cycle Jitter: æ 1 s t ç è f osc ö ø = c f osc James Buckwalter 10

11 Approximating the Phase Noise k 1 i i i i i1 log log L f a f f b U f f U f f i1 By parameterizing the phase noise, we can find the expected phase jitter b i a i is the slope a k 1 ai ai 1 1 bi/10 ai/10 i fi 1 fi 1 fi i1 10 F i James Buckwalter 11

12 Calculating Phase Noise In 1/f region, { } = -0 log 10 f L f ( ) - log 10 ( f i ) ( )+ b i f ò h L{ f } df =10 b l log 10 f l f l ( ) ò f l f h ( ) 10 log 10 f - df =10 b l 10 f l æ f =10 b l f h - f ö l l ç è ø f ò h f - df =10 b l 10-1 f f l l f f l f h If treated as constant, ò f l f h { } L f x æ f ö 10 b l x ç è ø f l df =10 L f x { } 10 f h =10 b l 10 f l { } = -0 log 10 f x L f x æ ( f h - f l ) =10 b f ö l l ç è ø æ f ( f h - f l ) =10 b l f h - f ö l l ç è ø æ f ö l ç è ø f h James Buckwalter f x f x f ò h f - df =10 b l 10-1 f f l l f f l ( ) - log 10 ( f i ) ( )+b i ( f h - f l ) æ = f ö l ç è ø f = f f x h l f h f h 1

13 Oscillator Conditions G( jw o )Z ( jw ) o = 1 ÐG( jw o )Z ( jw ) o = pn James Buckwalter 13

14 1 1 Z R R R o j 1 scr 1Q Q sl j Z R 1Q Look at perturbed behavior near resonance j LC Oscillator Tank 1 j o Q o o Z R R o 1 jq 1 jq 1 1 jq o o o James Buckwalter 14

15 LC Oscillator Now add negative resistance to impedance of tank Z ( w o + Dw )» R 1+ jq Dw w o James Buckwalter 15

16 LC Oscillator Now add negative resistance to impedance of tank 1 R 1 Zo gm 1 1 jq1 jq1 gm 1 jq o o R o R Zo jq o James Buckwalter 16

17 LC Oscillator Phase Noise { } = 10log v noise L Dw æ ç è v s ö ø James Buckwalter 17

18 LC Oscillator Phase Noise (cont) v noise v noise ( ) i n = Z w o + Dw = æ ö ç R çç Q Dw ç è ø w o Df 4kTG R = 4kTGR æ w o ö ç è QDw ø James Buckwalter 18

19 LC Oscillator Phase Noise (cont) v noise = v s æ w 4kTGR o ö ç è QDw ø v s v noise v s = 4kTG P osc æ ç è w o ö QDw ø James Buckwalter 19

20 LC Oscillator Phase Noise (cont) 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 { } = 10log ktg L Dw æ ç è P osc æ w o ö ç è QDw ø ö ø James Buckwalter 0

21 Leeson s Formula This expression can be empirically expanded to describe noise in other regimes. L kt o 1/ f 10 log 1 1 P osc Q At low frequency L At high frequency kt o 10log Posc Q L kt 10log Posc 1/ f James Buckwalter 1

22 Phase Noise Regions L kt o 1/ f 10 log 1 1 P osc Q { } = 10log ktg L Dw æ ç è P osc æ w o ö ç è QDw ø ö ø L kt 10log Posc James Buckwalter

23 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 3

24 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 4

25 Phase Noise Impairment (I) 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. James Buckwalter 5

26 Phase Noise Impairment (II) SNR PN = P int - P sig + SNR min For cell standards, in-band jammer can be 40 db greater than desired signal. SNR min should be > 0 db SNR PN < = 60dB James Buckwalter 6

27 Phase Noise Impairment How do we specify the noise due to the phase noise? 1 SNR PN = f h L Df ddf In 1/f region, ò f l { } { } = -0 log 10 f L f ( ) - log 10 ( f i ) ( )+ b i SNR PN = 1 L{ f h f } l f h - f l ( ) From earlier.. James Buckwalter 7

28 A Cautionary Note Summary L{ f h f } l = P sig - P int - SNR min -10log 10 f h - f l ( )

29 GSM Frequency Plan

30 A Cautionary Note Summary L{ f h f } l = P sig - P int - SNR min -10log 10 f h - f l ( )

31 Phase Noise Impairment (IV) Example: GSM channel spacing (600 khz) L{ f h f } l = L 519kHz L 100kHz L{ f h f } l = P sig - P int - SNR min -10log 10 f h - f l { } = -99dBm- -43dBm 519kHz { } = -113dBc+ 0log 10 ç ( ) -10log ( kHz) = -113dBc æ ö = -99dBc è100khz ø ( ) We specify that the oscillator should have a phase noise of under -99 dbc at a 100 khz offset frequency. James Buckwalter 31

32 Basic FDD System What about our out-of-band Blockers?

33 Reciprocal Mixing Assume RX at 10 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

34 Reciprocal Mixing: LTE Example Calculate the SNR phase noise SNR PN = P int - P sig + SNR min SNR PN =10dBm- -110dBm The phase noise is Translate to 10 MHz L 10MHz ( ) +10dB =130dB L{ f h f } = -SNR -10log l f PN 10 h - f l ( ) L{ 49 MHz} = -130dBc -10log ( 10 60MHz - 40MHz) L{ 49 MHz} = -130dBc -73dB = -03dBc { } = -03dBc+10log ( ) = -197dBc James Buckwalter 34

35 Reciprocal Mixing: LTE Example { } = -197dBc L 10MHz Note the noise floor for the phase noise is around 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

36 Reciprocal Mixing: LTE Example Let s ask the opposite question. Given the noise floor and oscillator power, how much transmit power can we tolerate? L 50MHz The phase noise is { } = -164dBm-10dBm = -174dBc SNR PN = -10log 10 SNR PN = -10log 10 ( f h - f ) l - L{ 50MHz} SNR PN =174dBc - 73dB =101dB ( 60MHz - 40MHz ) - (-174dBc) James Buckwalter 36

37 Reciprocal Mixing: LTE Example The phase noise SNR is related to P int = SNR PN + P sig - SNR min P int =101dB -110dBm-10dB = -19dBm This isn t much blocker power. Consider the out-ofband blocker levels that we have discussed for 800 MHz and.4 GHz. We MUST filter signals to avoid reciprocal mixing! James Buckwalter 37

38 Conclusions Oscillator phase noise is a silent killer in interference limited systems. Requirements for phase noise are dictated by phase noise in 1/f^ and thermal noise floor.