eecs 270c / winter 2013prof. m. green / u.c. irvine 1 random processes (1) random variable: a...

EECS 270C / Winter 2013 Prof. M. Green / U.C. Irvine 1

Random Processes (1)

Random variable: A quantity X whose value is not exactly known.

Probability distribution function PX(x): The probability that a random variable X is less than or equal to a value x.

0.5

1

x

PX(x)

Example 1:

€

X ∈ [−∞,+∞]Random variable


0.5

1

x

PX(x)

x1 x2

€

P X ∈ [x1,x2 ]( ) =P (x2 )−P (x1)

Probability of X within a range is straightforward:

If we let x2-x1 become very small …



Probability density function pX(x): Probability that random variable X lies within the range of x and x+dx.

€

pX (x) ⋅dx=PX (x+dx)−PX (x)

€

⇒ pX (x) =dPX (x)dx

0.5

1

x

PX(x)

x

pX(x)

dx

€

P X ∈ x1,x2[ ]( ) = pX (x) dxx1

x2∫



Expectation value E[X]: Expected (mean) value of random variable X over a large number of samples.

€

E [X ] ≡X = x⋅pX (x)dx−∞

+∞

∫Mean square value E[X2]: Mean value of the square of a random variable X2 over a large number of samples.

€

E [X 2 ] = x2 ⋅pX (x)dx−∞

+∞

∫

Variance:

€

E (X −X )2[ ] ≡σ2 = x−X( )

2

pX (x)dx−∞

+∞

∫

Standard deviation:

€

σ = E (X −X )2[ ]



Gaussian Function

€

f(x) =1

σ 2πexp

−(x−X )2

2σ 2

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

x

€

f(x)

€

X

€

1

σ 2π

€

0.607

σ 2π

€

X −σ

€

X +σ

2σ

€

f(x)dx=1−∞

+∞

∫

1. Provides a good model for the probability density functions of many random phenomena.

2. Can be easily characterized mathematically .

3. Combinations of Gaussian random variables are themselves Gaussian.

€

σ, X( )


Joint Probability (1)

€

P X ∈ x,x+dx[ ] andY ∈ y,y+dy[ ]( ) =pX (x) ⋅pY (y) ⋅dxdy

€

P (x,y) ≡P X ≤x andY ≤y( )

If X and Y are statistically independent (i.e., uncorrelated):

Consider 2 random variables:


Consider sum of 2 random variables:

€

Z=X +Y

x

y

€

x+y=z0

€

x+y=z0 +dz

dx

dy = dz

€

P Z∈ z0, z0 +dz[ ]( ) = pX (x)pY (y) dxdystrip

∫∫

€

= pX (x)pY (z0 −x) dx−∞

∞∫ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥dz

€

pZ(z0 )

determined by convolutionof pX and pY.



*

Example: Consider the sum of 2 non-Gaussian random processes:



3 sources combined:

*



4 sources combined:

*



Central Limit Theorem:Superposition of random variables tends toward normality.

Noise sources



Fourier transform of Gaussians:

€

pX (x) =1

σ X 2πexp

−(x−X )2

2σ X2

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

€

P X (ω) =exp−12σ X

2ω2 ⎛

⎝ ⎜

⎞

⎠ ⎟F

€

P Z∈ z0,z0 +dz[ ]( ) = pX (x)pY (z0 −x) dx−∞

∞∫ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥dz

Recall:

€

pZ(z0 ) = pX (x)pY (z0 −x) dx−∞

∞∫ F

€

P Z(ω) =P X (ω) ⋅PY (ω)

€

=exp−12σ X

2 ω2 ⎛

⎝ ⎜

⎞

⎠ ⎟⋅exp−

12σY

2ω2 ⎛

⎝ ⎜

⎞

⎠ ⎟

€

=exp−12(σX

2 +σy2 )ω 2

⎛

⎝ ⎜

⎞

⎠ ⎟F -1

€

pZ(z) =1

2π σ X2 +σY

2( )

exp−(z−Z)2

2 σ X2 +σY

2( )

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

Variances of sum of random normal processes add.


Autocorrelation function RX(t1,t2): Expected value of the product of 2 samples of a random variable at times t1 & t2.

€

RX (t1,t2 ) =E X (t1) ⋅X (t2 )[ ]

For a stationary random process, RX depends only on the time difference

€

RX (τ ) =E X (t) ⋅X (t+τ )[ ] for any t

€

RX (0) =σ2

Note€

τ =t1 − t2

Power spectral density SX(ω):

€

SX (ω) =E X (t) ⋅e−jωtdt−∞

+∞

∫2 ⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

SX(ω) given in units of [dBm/Hz]


€

RX (τ ) =12π

SX (ω) ⋅ejωτdω

−∞

∞∫

Relationship between spectral density & autocorrelation function:

€

⇒ RX (0) =σ2 =

12π

SX (ω)dω−∞

∞∫

Example 1: white noise

€

SX (ω)

€

RX (τ )

infinite variance(non-physical)

€

SX ω( ) =K

€

RX (τ ) =K2π

⋅δ t( )


Example 2: band-limited white noise

€

SX (ω)

ω

€

ωp

€

−ωp

€

RX (τ )

€

σ 2 =12Kωp

€

RX (τ ) =σ2e

−ωp τ τ

€

K

€

SX ω( ) =K

1+ω2

ωp2

x

€

pX (x)

€

−σ

€

+σ

For parallel RC circuitcapacitor voltage noise:

€

K =in2

Δf⋅R2 =2kBTR

€

ωp =1RC

€

σVC2 =

kBTC


Random Jitter (Time Domain)

Experiment:

datasource

CDR(DUT) analyzer

CLK

DATA RCK


Jitter Accumulation (1)

€

Tosc =1fosc

Free-runningoscillator output

Histogram plots

Experiment:Observe N cycles of a free-running VCO on an oscilloscope over a long measurement interval using infinite persistence.

NT

τ1 τ2 τ3 τ4

trigger €

σ1

€

σ 2

€

σ 3

€

σ 4


Observation:As τ increases, rms jitter increases.

€

τ

€

στ2

proportionalto τ2

proportional to τ

€



Noise Spectral Density (Frequency Domain)

osc osc+

Sv( )

€

Ltotal(Δω) =10 logP1Hz ωosc +Δω( )

Ptotal

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

Power spectral densityof oscillation waveform:

€

dBmHz[ ]

Ltotal includes both amplitude and phase noise

Ltotal(Δω) given in units of [dBc/Hz]

€

dBcHz[ ]

Δω (log scale)

€

Ltotal Δω( )

1/ 2 region (-20dBc/Hz/decade)

Single-sideband spectral density:

1/ 3 region (-30dBc/Hz/decade)


Noise Analysis of LC VCO (1)

active circuitry

C L R -R

€

ωr =1

LC

€

Z j ωr +Δω( )[ ] =j ωr +Δω( )L

1−ωr

2 +2ωrΔω + Δω( )2

ωr2

≈ jL⋅ωr

2

2Δω

Consider frequencies near resonance:

€

Q =RωrL

C L

+

_

vcinR

noise from resistor

€

ωrL=RQ

⇒ Z j ωr +Δω( )[ ] ≈ jR2Q

⋅ωr

Δω

€

ωr +Δω

€

ωr

€

Z( jω) = jωL

1−ωωr

⎛

⎝ ⎜

⎞

⎠ ⎟2


Noise current from resistor:

€

inR2 =

4kTR

⋅ΔfC L

+

_

vcinR

€

vc2 =inR

2 ⋅|Z( jω) |2

=4kTR

Δf ⋅Rωr Δω2Q

⎡

⎣ ⎢

⎤

⎦ ⎥2

=4kTR⋅ωr Δω2Q

⎡

⎣ ⎢

⎤

⎦ ⎥2

⋅Δf

Noise Analysis of LC VCO (2)

€

L Δω{ } =10 ⋅logF ⋅kTPsig

1+ωr

2Q ⋅Δω

⎛

⎝ ⎜

⎞

⎠ ⎟2 ⎧

⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪1+

ω1/ f 3

Δω

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

Leeson’s formula (taken from measurements):

Where F and ω1/f3 are empirical parameters.

dBc/Hz

spot noise relative to carrier power


Oscillator Phase Disturbance

Current impulse Δq/Δt

_+Vosc

t t

ip(t)

Vosc(t) Vosc(t)

Vosc jumps by Δq/C

• Effect of electrical noise on oscillator phase noise is time-variant.• Current impulse results in step phase change (i.e., an integration).

current-to-phase transfer function is proportional to 1/s

ip(t)

€

τ 1

€

τ 2

€

Δφ=0

€

Δφ < 0

ip(t)


Impulse Sensitivity Function (1)The phase response for a particular noise source can be determined at each point τ over the oscillation waveform.

€

Γ(τ ) ≡Δφ(τ )Δq

⋅qmaxImpulse sensitivity function (ISF):

€

=C ⋅Vmax(normalized to signal amplitude)

change in phasecharge in impulse

t

τ

€

Vosc(t)

€

Γ(τ )

€

Vmax

Example 1: sine wave

t

τ

€

Vosc(t)

€

Γ(τ )

Example 2: square wave

Note Γ has same period as Vosc.


Impulse Sensitivity Function (2)

€

H(s)h(t)

€

iin

€

φout

Recall from network theory:

€

Φout(s)Iin(s)

=H(s)LaPlace transform:

€

φout(t) = h(t,τ ) ⋅iin(τ ) dτ0

t

∫Impulse response:

time-variant impulse response

€

Γ(τ ) ≡Δφ(τ )Δq

⋅qmax ⇒ Δφ(τ ) =Γ(τ )qmax

⋅ΔqRecall:

ISF convolution integral:

€

φ(t) = Γ(τ )qmax0

t

∫ ⋅u(t−τ) ⋅ i(τ ) ⋅dτ[ ] =Γ(τ )qmax0

t

∫ ⋅i(τ ) ⋅dτ

from q

€

=1 forτ ∈ (0,t)

€

Γ(τ ) = ck cos kωoscτ +θk( )k=0

∞

∑

Γ can be expressed in terms of Fourier coefficients:


Case 1: Disturbance is sinusoidal:

€

i(t) =I0 cos mωosc +Δω( )t[ ] , m = 0, 1, 2, …

€

=I0

2qmax

cksin (k+m)ωosc +Δω[ ]t+θk{ }

(k+m)ωosc +Δω+sin (k−m)ωosc +Δω[ ]t+θk{ }

(k−m)ωosc +Δω

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪k=0

∞

∑

negligible significant only form = k

(Any frequency can be expressed in terms of m and Δω.)

€

φ(t) = I0qmax

ck coskωoscτ +θk( ) ⋅cos mωosc +Δω( )t[ ]{ } dτ0

t∫k=0

∞

∑

€

Γ(τ )

€

≈I0

2qmax

⋅cm⋅sin Δω t+θm( )

Δω




€

φ(t) ≈I0

2qmax

⋅cm⋅sin Δω t+θm( )

Δω⇒ φ2 =

I02

8qmax2 ⋅

cm2

Δω( )2

For

€

i(t) =I0 cos mωosc +Δω( )t[ ]

ω

I

2 osc

€

×I0

2qmax

c0ω1

€

×I0

2qmax

c1ω1

€

×I0

2qmax

c2ω1

Current-to-phase frequency response:

oscωoscω1

ω1

ω1 ωosc ω1 2ωoscω1 2ωosc ω1


ω

Case 2: Disturbance is stochastic:


MOSFET current noise:

thermalnoise

1/fnoise

€

in2 (f)Δf

=4kTγgm +gm2 Kf

CgfA2/Hz

€

in

€

φ2 Δf ≈in2 Δf8qmax

2⋅cm2

Δω( )2

€

Sφ Δω( )

Δω

€

×c0

€

×c1

€

in2

Δf

ωosc 2ωosc

€

×c2

thermal noise

€

4kTγgm

€

in2

Δf

ωωosc 2ωosc

€

×c0

€

gm2 2π ⋅Kf

Cgω

1/f noise



€

Sφ Δω( )

Δω

€

×c0

€

×c1

€

in2

Δf

ωωosc 2ωosc

€

×c2

€

Sφ (Δω) =1

8qmax2

4kTγgm ⋅

ck2

0

∞

∑Δω( )

2+2π gm

2 Kf

Cg

⋅c02

Δω( )3

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥

due to 1/f noise

due to thermal noise

Total phase noise:

€

c02 = Γ ( )

2

ck2

k=0

∞

∑ = Γrms( )2

ωn



€

Sφ (Δω) =1

8qmax2

4kTγgm ⋅Γrms( )

2

Δω( )2+2π gm

2 Kf

Cg

⋅Γ ( )

2

Δω( )3

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

€

4kTγgm ⋅Γrms( )

2

Δω( )2=2π gm

2 Kf

Cg

⋅Γ ( )

2

Δω( )3

€

⇒

€

Δωn,phase=π

2kT⋅gmγCg

⋅Γ

Γrms

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2

noise corner frequency ωn

Δω (log scale)

€

Sφ Δω( ) (dBc/Hz)

€

Δωn,phase

1/(Δω3 region: −30 dBc/Hz/decade



t

τ

€

Vosc(t)

€

Γ(τ )

t

τ

€

Vosc(t)

€

Γ(τ )

Example 1: sine wave Example 2: square wave


Example 3: asymmetric square wave

t

τ

€

Vosc(t)

€

Γ(τ )

€

Γ > 0 will generate more 1/(Δω3 phase noise

€

Γrms is higher will generate more 1/(Δω2 phase noise



Effect of current source in LC VCO:

Vosc+ _

Due to symmetry, ISF of this noise source contains only even-order coefficients − c0 and c2 are dominant.

Noise from current source will contribute to phase noise of differential waveform.



ID varies over oscillation waveform

€

in2

Δf=4kTγgm(t)

=(4kTγ) ⋅μCoxWL

⋅VGS (t)−Vt( ) ⎡

⎣ ⎢

⎤

⎦ ⎥

Same period as oscillation

€

in02

Δf=(4kTγ) ⋅μCox

WL

⋅VGS(DC ) −Vt( ) ⎡

⎣ ⎢

⎤

⎦ ⎥Let

Then

€

in2

Δf=in02

Δf⋅α(t)

€

α(t) =VGS (t)−VtVGS(DC ) −Vt

where

€

Γeff(τ ) =Γ(τ ) ⋅α(τ )We can use


ISF Example: 3-Stage Ring Oscillator

M1A M1B M2A M2B M3A M3B

MS1 MS2 MS3

R1A R1B R2A R2B R3A R3B+

Vout

−

fosc = 1.08 GHzPD = 11 mW

Red curve:Unperturbed oscillation waveform

Blue curve:Oscillation waveform perturbed by impulse


ISF of Diff. Pairs

ISF by tx6 for differential ring osc

-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF by tx6


-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF by tx5


-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF by tx4


-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF by tx3


-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF by tx2

ISF by tx1 for 3stage differential ring osc

-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

Radian

ISF by tx1

€

ΓM1A

€

ΓM1B

€

ΓM2A

€

ΓM2B

€

ΓM3A

€

ΓM3B

€

Γrms =1.86Γ =−0.26

for each diff. pair transistor


ISF of Resistors

€

ΓR1A

€

ΓR2A

€

ΓR3A

€

Γrms =1.72Γ =−0.16

for each resistor


ISF of Current Sources

ISF by tail tx3 for differential ring osc

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 1 2 3 4 5 6 7

Radian

ISF by tail tx3


-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 1 2 3 4 5 6 7

Radian

ISF by tail tx2


-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 1 2 3 4 5 6 7

Radian

ISF by tail tx1

€

ΓMS1

€

ΓMS2

€

ΓMS3

ISF shows double frequency due to source-coupled node connection.

€

Γrms =1.00Γ =−0.12

for each current source transistor


Phase Noise Calculation (Thermal noise)

Using: Cout = 1.13 pF

Vout = 601 mV p-p

qmax = 679 fC

€

L Δf{ } =6⋅Γrms(dp)

2

8π 2Δf 2⋅4kTγgm(dp)

qmax2

+ 6⋅Γrms(res)

2

8π 2Δf 2⋅4kT Rqmax2

+ 3⋅Γrms(cs)

2

8π 2Δf 2⋅4kTγgm(cs)

qmax2

€

322Δf 2

€

122Δf 2

€

70Δf 2

€

⇒ L Δf{ } =514Δf 2

= −112 dBc/Hz @ Δf = 10 MHz


Phase Noise vs. Amplitude Noise (1)

€

Vosc(t) =Vc +v(t)[ ] ⋅exp j ωosct+φ(t)( )[ ]

ωosct

v

v Spectrum of Vosc would include effects of both amplitude noise v(t) and phase noise (t).

How are the single-sideband noise spectrum Ltotal(Δω) and phase spectral density S(ω) related?



t t

i(t) i(t)

Vc(t) Vc(t)

0=Δtosc

qt

ωΔ

=Δ

Recall that an input current impulse causes an enduring phase perturbation and a momentary change in amplitude:

Amplitude impulse response exhibits an exponential decay due to the natural amplitude limiting of an oscillator ...


Δω

€

Lamp Δω( )

€

ωcQ

Δω

+

Δω

€

Ltotal Δω( )

Phase noise dominates at low offset frequencies.


€

Lφ Δω( )

Δω


€

Vosc(t) = Vc +v(t)( ) ⋅cosωosct+φ(t)( )

≈ Vc +v(t)( ) ⋅ (cosωosct)−φ(t) ⋅ (sinωosct)[ ]

=Vc (cos ωosct)−φ(t) ⋅Vc (sinωosct) +v(t) ⋅ (cosωosct)

Phase & amplitude noise can’t be distinguished in a signal.


noiseless oscillation waveform

phase noise

component

amplitude noise

component

Amplitude limiting will decrease amplitude noisebut will not affect phase noise.

ωosc

Sv(ω)

phase noise

amplitude noise

ω


Sideband Noise/Phase Spectral Density

€

Vosc(t) =Vc ⋅cosωosct+φ(t)( )

≈Vc ⋅ (cos ωosct)−φ(t) ⋅ (sinωosct)[ ]

€

Vc ⋅ (cosωosct)−Vc ⋅φ(t) ⋅ (sinωosct)

€

PphasenoisePsignal

=

12Vc

2 ⋅φ2

12Vc

2

=φ2

€

Lphase Δω( ) =12⋅Sφ Δω( )

noiseless oscillation waveform

phase noise

component


Jitter/Phase Noise Relationship (1)

€

στ2 ≡

1ωosc

2⋅E φ(t+τ )−φ(t)[ ]

2 ⎧ ⎨ ⎩

⎫ ⎬ ⎭

€

=1

ωosc2

⋅ E φ2 (t+τ )[ ] +E φ2 (t)[ ] −2E φ(t) ⋅φ(t+τ )[ ]{ }

€

Rφ (τ ) =12π

S (Δω) ⋅ej(Δω )τd(Δω)

−∞

∞∫Recall R and S(Δω) are a Fourier transform pair:

€

⇒ στ2 =

2ωosc

2⋅Rφ (0)−Rφ (τ )[ ]

€

Tosc =1fosc

NT

τ1 τ2 τ3 τ4

€

σ 1

€

Rφ (0)

€

Rφ (0)

€

2Rφ (τ )autocorrelation functions

€

σ 2

€

σ 3

€

σ 4


€

Rφ (0) =12π

Sφ (Δω)d(Δω)−∞

∞

∫

Rφ (τ ) =12π

Sφ (Δω) ⋅ej (Δω )τd(Δω)

−∞

∞

∫


€

στ2 =

1πωosc

2⋅ Sφ (Δω) 1−e j(Δω )τ

( )−∞

∞

∫ d(Δω)

=1

πωosc2

⋅ Sφ (Δω) 1− (cos Δωτ )−j (sin Δωτ )[ ]−∞

∞

∫ d(Δω)

=4

πωosc2

⋅ Sφ (Δω) ⋅sin2 Δωτ

2

⎛

⎝ ⎜

⎞

⎠ ⎟

0

∞

∫ d(Δω)


Let

€

Sφ (Δω) =a

(Δω)2

€

στ2 =

4πωosc

2⋅

a(Δω)2

⋅sin2(Δω)τ

2

⎛

⎝ ⎜

⎞

⎠ ⎟

0

∞

∫ d(Δω)

€

=4

πωosc2

⋅aπτ4

€

=a

ωosc2

⋅τ

Consistent with jitter accumulation measurements!


Jitter from 1/(Δω noise:2

Let

€

Sφ (Δω) =b

(Δω)3

€

στ2 =

4πωosc

2⋅

b(Δω)3

⋅sin2(Δω)τ

2

⎛

⎝ ⎜

⎞

⎠ ⎟

ε

∞

∫ d(Δω)

€

=ζ ⋅τ 2

Jitter from 1/(Δω noise:3

^

^

^

^

€

=afosc2

⋅τ wherea≡(2π )2 ⋅a^



Δf

€

Sφ Δf( ) (dBc/Hz)

-100

-20dBc/Hzper decade

• Let fosc = 10 GHz• Assume phase noise dominated by 1/(Δω)2

€

Sφ Δf( ) =a

(Δf)2

€

Sφ 2 ⋅10 6( ) =

a

2 ⋅10 6( )

2=10−10 ⇒ a=400

Setting Δf = 2 X 106 and S =10-10:

Let τ = 100 ps (cycle-to-cycle jitter):

σ = 0.02ps rms (0.2 mUI rms)

€

στ2 =

afc2⋅τ =

400

10 ⋅10 9( )

2⋅τ = 4 ⋅10−18

[ ] ⋅τ

Accumulated jitter:

€

στ = 2 ⋅10−9[ ] ⋅ τ

2 MHz


More generally:

f

€

Sφ Δf( ) (dBc/Hz)

Δfm

Nm

-20 dBc/Hzper decade

€

στ =Δfmfosc

⎛

⎝ ⎜

⎞

⎠ ⎟⋅10Nm 20⋅ τ ps[ ]

€

στ

Tosc=Δfm⋅10

Nm 20⋅ τ UI[ ]

€

στ2 =

afosc2 ⋅τ =

Δfmfosc

⎛

⎝ ⎜

⎞

⎠ ⎟2

⋅10Nm 10⋅τ

€

Sφ Δf( ) =a

(Δf)2=(Δfm )

2 ⋅10Nm 10

(Δf)2


€

στ

Tosc→ Δfm⋅10

Nm+10( ) 20⋅ τ = Δfm⋅10Nm 20⋅ τ( )⋅10

0.5

rms jitter increases by a factor of 3.2

Let phase noise increase by 10 dBc/Hz:



Kpd

phasedetector

loopfilter

€

F (s) Kvco

VCO

€

÷N€

+in out

vco

fb

€

φoutφε

=G(s) =Kpd ⋅F (s) ⋅Kvco

2πs⋅1N

Open-loop characteristic:

€

φout =NG(s)1+G(s)

⋅φin +1

1+G(s)⋅φvcoClosed-loop characteristic:



€

G(s) =IchKvco

N⋅

1s2 (C +Cp)

⋅1+sCR1+sCeqR

Recall from Type-2 PLL:

Δω

|G|

z pω

|1 + G|

-40 dB/decade

Δω

€

Sφ Δω( ) (dBc/Hz)

€

Δωn,phase



Δω

€

φoutφvco

jΔω( )

2

1

80 dB/decade

ω

As a result, the phase noise at low offset frequencies is determined by input noise...


• fosc = 10 GHz• Assume 1-pole closed-loop PLL characteristic


f

€

Sφ Δf( ) (dBc/Hz)

Δf0 = 2 MHz

-100-20dBc/Hzper decade

€

Sφ Δf( ) =

a

Δf0( )2

1+ΔfΔf0

⎛

⎝ ⎜

⎞

⎠ ⎟2≈

a

Δf0( )2, Δf <<Δf0

a

Δf( )2, Δf >> Δf0

⎧

⎨

⎪ ⎪

⎩

⎪ ⎪

€

⇒ στ2 =

22π ⋅fosc

2⋅Rφ (0)−Rφ (τ )[ ]

=afosc2

⋅1−e−2π⋅f0 τ

2π ⋅Δf0

€

Rφ (τ ) = Sφ (Δf) ⋅ej(2 πΔf )τ ⋅d(Δf) =

a2π ⋅Δf0( )

⋅e−2π⋅f0τ

−∞

∞

∫



€

στ2 =

afosc2

⋅1−e−2π⋅f0 τ

2π ⋅Δf0≈

afosc2

⋅τ

afosc2

⋅1

2π(Δf0 )

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

€

, τ <<1

2π(Δf0 )

, τ >>1

2π(Δf0 )

€

στ2

(log scale)

τ

€

1(2π) ⋅(2 )MHz

€

slope=afosc2

€

a=4×10 2

For large τ:

στ = 0.02 ps rms cycle-to-cycle jitterΔf0 = 2 MHz

fosc = 10 GHz

For small τ:

στ = 1.4 ps rms Total accumulated jitter


The primary function of a PLL is to place a bound on cumulative jitter:

τ

€

στ2

(log scale)

€

στ2

(log scale)

proportional to (due to thermal noise)

proportional to 2(due to 1/f noise)

τ


Free-running VCO

PLL


L( ) for OC-192 SONET transmitter

Closed-Loop PLL Phase Noise Measurement


Other Sources of Jitter in PLL

• Clock divider

• Phase detectorRipple on phase detector output can cause high-frequency jitter. This affects primarily the jitter tolerance of CDR.


Jitter/Bit Error Rate (1)

Histogram showing Gaussian distribution

near sampling point

1UI

Bit error rate (BER) determined by and UI …

L R€

2σ R

€

2σ L

Eye diagram fromsampling oscilloscope


R

€

2σ

0 T

€

T2

€

t0

€

T −t0

€

PL =1

σ 2π⋅ exp−

x2

2σ 2

⎡

⎣ ⎢

⎤

⎦ ⎥

t0

∞∫ dx

€

PR =1

σ 2π⋅ exp−

T −x( )2

2σ 2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

t0

∞∫ dx

€

pL (t) =1

σ 2π⋅exp−

t2

2σ 2

⎡

⎣ ⎢

⎤

⎦ ⎥

€

pR (t) =1

σ 2π⋅exp−

T −t( )2

2σ 2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

Probability of sample at t > t0 from left-hand transition:

Probability of sample at t < t0 from right-hand transition:

€

2σ



Total Bit Error Rate (BER) given by:

€

BER =PL +PU =1

σ 2π⋅ exp−

x2

2σ 2

⎡

⎣ ⎢

⎤

⎦ ⎥

t0

∞∫ dx+1

σ 2π⋅ exp−

x2

2σ 2

⎡

⎣ ⎢

⎤

⎦ ⎥

T−t0

∞∫ dx

€

=12erfc

t02σ

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟+erfc

T −t02σ

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

€

where (erfct) ≡2

π⋅ exp

t

∞∫ −x2( )dx

€

PL =1

σ 2π⋅ exp−

x2

2σ 2

⎡

⎣ ⎢

⎤

⎦ ⎥

t0

∞∫ dx

€

PR =1

σ 2π⋅ exp−

T −x( )2

2σ 2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

t0

∞∫ dx=1

σ 2π⋅ exp−

x2

2σ 2

⎡

⎣ ⎢

⎤

⎦ ⎥

T−t0

∞∫



€

•

€

•

€

•

€

•

t0 (ps)

log BER

€

σ =5 ps

€

σ =2.5 ps

€

σ =2.5 ps:

€

BER ≤10−12 fort0 ∈ 18ps, 82ps[ ]

€

σ =5 ps:

€

BER ≤10−12 fort0 ∈ 36ps, 74ps[ ]

Example: T = 100ps

(64 ps eye opening)

(38 ps eye opening)

log(0.5)



Bathtub Curves (1)

The bit error-rate vs. sampling time can be measured directly using a bit error-rate tester (BERT) at various sampling points.

Note: The inherent jitter of the analyzer trigger should be considered.

€

J rmsRJ

( )measured

2= J rms

RJ( )

actual

2+ J rms

RJ( )

trigger

2


Bathtub Curves (2)

Bathtub curve can easily be numerically extrapolated to very low BERs (corresponding to random jitter), allowing much lower measurement times.

Example: 10-12 BER with T = 100ps is equivalent to an average of 1 error per 100s. To verify this over a sample of 100 errors would require almost 3 hours!

€

•

€

•

€

•

€

•

t0 (ps)


Equivalent Peak-to-Peak Total Jitter

BER

10-10

10-11

10-12

10-13

10-14

€

J PPRJ

σ, T determine BERBER determines effectiveTotal jitter given by:

€

J PPRJ

€

J TJ = n⋅σ( ) + J PPDJ

€

12nσ

€

p(t)

€

12nσ

Areas sumto BER

€

12.7 ⋅σ

€

13.4 ⋅σ

€

14.1⋅σ

€

14.7 ⋅σ

€

15.3 ⋅σ

eecs 270c / winter 2013prof. m. green / u.c. irvine 1 random processes (1) random variable: a...

Documents

c winter

probability of x

joint probability

random variable x

random variableprof

gaussian random processes

random normal processes

small random processes