receiver noise and sensitivity
DESCRIPTION
Receiver Noise and Sensitivity. Abdul Rehman. Receiver Noise. • Optical receivers convert incident optical power. into electric current through photodiode. • The relation I p = R ⋅ P in assumes that such a. conversion is noise free but this is not true even - PowerPoint PPT PresentationTRANSCRIPT
Receiver Noise
• Optical receivers convert incident optical powerinto electric current through photodiode
• The relation I p = R ⋅ Pin assumes that such aconversion is noise free but this is not true evenfor a perfect receiver
• Two fundamental mechanisms are– Shot noise– Thermal noise
• This causes fluctuations in the current evenwhen the incident optical power has a constantpower. Additional noise is generated if inputpower is fluctating
PPhotocurrentt
average
I p = R ⋅ Pin
• This relation still holds good if we consider photocurrentas the average current
• Electrical noise induced by current fluctuations affectsthe receiver performance
4.4 RECEIVER NOISE
Shot noise
Thermal noise
I p = R ⋅ Pin
SNR4.4.1 Noise mechanisms4.4.1.1 Shot noise
Spectral Density: It is often called simply the spectrum of the signal. Intuitively, the spectral density captures the frequency content of a stochastic process and helps identify periodicities
Average current
−∞
∞
∫ )
Ip
Constant optical power
I (t ) = I p + i s (t )
I p = RPin
is(t)
stationary random process
Poisson statistics
Gaussian
Wiener-Khinchin theorem
i s (t )i s (t + τ) =
Autocorrelation function
S S (f e i 2 πfτdf
Spectral density
is(t)
t
Theory Of StochasticsProcess
In a stochastic or random process there is some indeterminacy in its future evolution described by probability distributions. This means that even if the initial condition (or starting point) is known, there are many possibilities the process might go to, but some paths may be more probable and others less so.
THEORY OF POISSONSTATISTICS
• A sequence of independent random events is onein which the occurrence of any event has no effecton the occurrence of any other. One example issimple radioac-tive decay such as the emission of
663 KeV photons by a sample of 137Cs. In contrast,the fissions of nuclei in a critical mass of 235U arecorrelated events in a ‘chain re- action" in whichthe outcome of each event, the number of neutronsreleased, affects the outcome of subsequent
Negative Frequency
The concept of negative frequency is purely mathematical. This can be easily proven by some simple trigonometric math. However in comms systems the 'negative' frequencies is simply copies of you signal symetric about the sampling frequency. Thus if you sample (for the carrier) at 433MHz and your data sits at 433.1MHz then you see an image at 432.9Mhz as well. If you move this to the 0Hz point negative frequencies needs to be used.
However, as mentioned negative frequency is not a physical phenomenan. Imagine a sime wave of -100Hz - I can't.
s s
White noise SS(f)Two-sided
qIp
f
One-sided
2qIp
fNoise variance τ=0
δ2 = i 2 (t ) =∞
∫ S s (f )df−∞
= 2qI p ∆f
HT (f )2
2 2
SS(f)
qIp
SS(f)
2qIp
−∆f ∆ff
0 ∆ff
effective noise bandwidth
At output of photodiode: bandwidth of photodiodeInput to decision circuit: bandwidth of receiver
Preamp Amp
One − sided
Photodiode Filter
S s (f ) = 2qI p HT (f )
σs
∞
= ∫ 2qI p HT (f )0
df∞
= 2qI p ∫ HT0
(f ) 2 df
⇒ ⎬ dark current
s
∆fd ∞
= ∫ HT0
(f ) 2 df
Stray light ⎫
Thermal generation ⎭(also shot noise )
σ 2 = 2q (I p + I d )∆fcan be reduced bylow pass filter
4.4.1.2 Thermal noiseRandom thermal motion of electrons in RL
IT(t)
Ip is iT CT RL
2
2
∆
2
Thermal noiseI (t ) = I p + I s (t ) + I t (t )
thermal
iT(t): stationary Gaussian random processflat up to ∼ 1THz
ST (f ) =2k BT
RL
(Two − sided ) kB=1,38⋅10-23 J/K
σT = i T2 (t ) =∞
∫ S T (f )df−∞
=∆f
−∫f
2k BTR L
df =4k BT ⋅ ∆fR L
σT independent of I p
∆f =∞
∫ HT (f )0
df
σT =
i s (t ) ⎫
iT (t )⎭
2 2 2
Also thermal noise from pre-ampfield effectbipolar
Amplifier noise figure2 4k BTFn ∆f
R L
⎬ : independent random processes (Gaussian )
Total variance
σ = σ s + σT = 2q (I p + I d )∆f +4k BTFn ∆f
R L
2I p
σ
=
4.4.2 p-i-n ReceiversSNR
Neglect CT:
Ip is iT RL
SNR = 2 =
Thermal limit
R 2Pin2
2q (RPin + I d ) ∆ f +4 k BTFn ∆fRL
SNR ≈ R 2Pin2
4k BTFn ∆ fRL
R 2Pin2R L
4k BTFn ∆ fSNR ∝ Piin2
SNR ∝ R L (C L neglected )
=
2
NEP: Noise Equivalent Power
RLR 2Pin2
4k BTFn ∆f= 1
Pin2 4k BTFn
∆f RLR 2
NEP =Pin
∆f=4k BTFn
RLR 2
=hν
ηq4k BTFn
RL
Pin = NEP ⋅ ∆f
Shot noise limit σs>>σT
σ s = 2q (I p + I d )∆f = 2q (RPin + I d ) ∆ f
Increases linearly with Pin
Obtained for large Pin
Id neglected
ηq
=
thttdw'(==)1p
d
h'(t=)1
∞
hdTT==
−∫
−
−
SNR ≅ R 2Pin2
2qRPin ∆ f= RPin
2q ∆ f= hν Pin
2q ∆ f
ηPin
2hν∆f Shot noise limit SNR ∝ Pin
E p = Pin TB
E p = Area in a bit
E p
E p
= PinT B =
= N p hν
Pin
B
Pin = BE p = BN p hν
∆f =B
2(Nyquist )
Piin= 2∆f ⋅ N p hν
SNR ≅η ⋅ 2∆f ⋅ N p hν2hν∆f
= ηN p Shot noise limit
η 1≅
Np=100SNR=100
SNRdB=10log10SNR=10⋅log100 dB=20dB
Thermal noise dominates Np≈1000
2
⎛⎟
s
4.4.3 APD Receivers
Higher SNR (for same Pin)
Ip=MRPin=RAPD ⋅Pin
Also multiplication noiseThermal noise the sameShot noise changes due to multiplicationM is a random variable
p − i − n diode
APDExcess noise factor
F A = k A M + (1 − k A ) ⎜ 2 −⎝
σ 2 = 2q (RPin + I d )∆f
σ s = 2qM 2F A (RPin + I d )∆f
1 ⎞
M ⎠
Electrons initiate and
Holes initiate and
αe > αh
αh > αe
k A =
k A =
αh
αe
αe
αh
2
2 2
2
2
k A = 0
k A = 1
F A = 2 −
F A = M
1M
Fig. 4.14
Small kA best
SNR = I p
σ s + σT
= (MRPiin )2qM F A (RPin + I d )∆f+ 4k BTFn ∆f
RL
=
2 =
ηq
Thermal noise limit
SNR ≅M 2R 2Pin2
4 k BTFn ∆fRL
RLR 2
4k BTFn ∆fM 2Pin2
improved by M 2
Shot noise limit
SNR ≅ M 2R 2Pin2
2qM F ARPin ∆fRPin
2qF A ∆f
= hν Pin
2qF A ∆f= ηPin
2hνF A ∆f
reduced by F A
There is optimum Mopt that maximizes SNR
22
⎡ ⎛2
⎝⎣
⎞⎤
⎠⎦
3 2 2 2
B [k M + (1 − k )(2M − M )]+ C
SNR =
dSNRdM
A
2
SNR = M (RPin )
2q (RPin + I d )∆f ⋅ M ⎢k A M + (1 − k A ) ⎜ 2 −1
M ⎥ +4 k BTFn ∆f
RL
AM 2
B [k AM 3 + (1 − k A )(2M 2 − M )]+ C= B [k AM + (1 − k A )(2M − M )]+ C 2AM − AM B [3k AM2 + (1 − k A )(4M − 1)] = 0
3 2A A
B [kM 3 + (1 − k A )(2M 2 − M )] + C 2 − MB [3k A M 2 + (1 − k A )(4M − 1)] = 0
M 3 (Bk A − B 3k A ) + M 2 [B (1 − k A )4 − B (1 − k A )4] + M [− B (1 − k A )2 + B (1 − k A )] + 2C = 0
Bk A M 3 + B (1 − k A )M − 2C = 0
3 RL=k A M + (1 − k A )M =2C
B
2 4k BTFn ∆f
2q (RPin + I d )∆f
= 4k BTFn
qRL (RPin + I d)
1.55 µm
APD receiver:
RL=1kΩ
Fn=2
R=1A/W
Id=2nA
Fig. 4.15 Pin (dBm)
⎛ 4k BTFn ⎞= ⎜
13
⎠⎝
Neglect (1-kA)M:
M opt ⎜ k AqRL (RPin + I d ) ⎟
critical
Si APD:GaInAs APD:
kA<<1kA ≈0,7
Mopt≈100Mopt ≈10
4.5 RECEIVER SENSITIVITY
BERBER ≤ 10−9
Sensitivity : Prec ⇒ BER = 10− 9
4.5.1 Bit-Error Rate
PDF ! !
12
12
I>IDI<ID
⇒⇒
bit 1bit 0
I<IDI>ID
for bit 1⇒for bit 0⇒
errorerror
p(1):p(0):
probability of receiving 1probability of receiving 0
P(0/1): probability of deciding 0 when 1 is receivedP(1/0): probability of deciding 1 when 0 is received
BER = p (1) ⋅ P (0 / 1) + p (0) ⋅ P (1 / 0)
p (1) = p (0) =
BER = [P (0 / 1) + P (1 / 0)]Thermal noise: Gaussian, zero mean, variance σT2
2APD :s
2s
2
0
2
∫
D
0
Shot noise:
p − i − n : Gaussian , zero mean ,
Gaussian ?, zero mean ,
σ2 = 2q (I p + I d )∆f
σs = 2qM 2F A (RPin + I d )∆f
σ2 = σT + σ2
bit 1 and 0 have different average value and variance
bit − 1 :
bit − 0 :
σ1
σ2
P (0 / 1) =I D
−∞ σ1
1
2πe
−(I −I 1 )2
2σ1dI
P (1 / 0) =∞
I∫ σ0
1
2πe
−(I −I 0 )2
2 σ2dI
= ⋅1 22
1
P (0 / 1) =
I D − I12σ1
∫−∞
1
σ1 2πe−y 2
2σ1dy
1 2
2 π
I D − I12σ1
∫ e−∞
−y dy = ⋅2 π
∞
∫ eI1 − I D
2σ1
−y 2dy
= erfc2
I 1 −I D
2σ1
y =I − I 0
2σ0
dy =dI
2σ0
∞
∫
[1 I 1 −I 0
4]
P (1 / 0) =∞
∫I D − I 0
2σ 0
1
σ 0 2πe−y 2
2σ 0dy
1 2 −y 2 1= ⋅ e dy = erfc2 π I D − I 0 2
2σ 0
BER = erfc + erfc2σ1
I D −I 0
2σ 0
I D −I 0
2σ0
12
20
σ0
σ1
P(0/1)I0 ID I1
P(1/0)
minimum for:
BER = [P (0 / 1) + P (1 / 0)]
1
σ0 2πe
−(I D −I 0 )2
2 σ2=1
σ1 2πe
−(I D −I 1 )2
2 σ1
2 2
=2 2
=2 2 ≈ 2
+ ln σ0 +(I D − I 0 )2
2σ0= + ln σ1 +
(I D − I 1 )2
2σ1
(I D − I 0 )2 (I 1 − I D )2
2σ0 2σ1+ ln σ1 − ln σ0
(I D − I 0 )2 (I 1 − I D )2
σ0 σ1
+ 2 lnσ1 (I 1 − I D )2
σ0 σ1
I 1 − I D
σ1
=I D − I 0
σ0
= Q
σ0I 1 − σ0I D = σ1I D − σ1I 0
σ0I 1 + σ1I 0 = I D (σ0 + σ1 )
I D =
=
1 ⎡
4 ⎣ ⎦
σ 0I 1 + σ 1I 0
σ 0 + σ 1
σ1 =σ0 ⇒ I D =σ0 I 1 +σ0 I 0 I 1 + I 0
2σ0 2
For I 1 − I D
σ1
=I D − I 0
σ0
= Q
Holds for p-i-n receivers, where σT2>>σS2
BER = ⎢erfcQ
2+ erfc
Q ⎤ 1
2 = ⎥ 2 erfcQ
2
Q =I 1 − I D
σ1
=I 1 −σ0I 1 + σ1I 0
σ0 + σ1
σ1
=σ0I 1 + σ1I 1 − σ0I 1 − σ1I 0
σ1 (σ0 + σ1 )
= I 1 − I 0
σ1 + σ0
1 Q 1
erfc ≅1
x πe− x 2 x >> 1
BER = erfc2
≈ ⋅2 2 Q
2
1
πe
−Q 2
2
=1
2πQe
−Q 2
2
Reasonable for Q>3
Q = 6 ⇒ BER ≈ 10−9
1 1
2 2 2
2 2
2
2
2 2
4.5.2 Minimum Average Received Power
Assume: I0=0 (P0=0)
I1 = MRP1 = 2MRPrec
Prec = (P1 + P0 ) = P12 2
σ1 = σS + σT
σ0 = σT , no short noise variance -- '0' bit
σS = 2qM 2 FA (RP1 + Id )∆f ≅ 2qM 2 FA R2Prec ∆f
σT =4k B TFn ∆f
R L
Q = I 1 − I 0
σ1 + σ0
= I 1
σ1 + σ0
= 2MRPrec
σS + σT + σT
1
2
2
2
⎛ 2MRPrec ⎞2 2
⎜ Q2
2
22
2
BER = erfc2
Q2
BER ⇒ Q ⇒ Prec
Q σ S + σT2 + Q ⋅ σT = 2MRPrec
Q σ S + σT2 = 2MRPrec − Q ⋅ σT
σ S + σT2 =2MRPrec
Q− σT
σ S + σT = ⎜ ⎝ ⎠
+ σT −4MRPrec σT
Q
2qM F AR 2Prec ∆f =4M 2R 2
QPrec −
4MRPrec σT
Q
2
⋅
Q ⎛R ⎝
σ ⎞
QR
MRQ
Prec = qMF A ∆f +σT
Q
Prec =Q 2
MRqMF A ∆f +
Q 2 σT
MR Q
Prec= ⎜QqFA ∆f+ T
M ⎠
p-i-n receiverM=1, σT dominates
Prec ≈ σT =Q
R4k BTFn ∆f
RL
∝ B
16
dBm
Q ⎛R ⎝
⎟
⎛ 1 ⎞
M ⎠
Numerical example R = 1 A /W
σT = 0,1 µAQ = 6 (BER = 10− 9 )
Prec ≈⋅ 0,1 ⋅ 10−6W = 0,6 µW
Prec = 10 log0,6 ⋅ 10−6
10−3dBm = 10 log 6 ⋅ 10−4 dBm
= −32,2 dBm
APD receiver:
Prec= ⎜QqF A ∆f +σT ⎞
M ⎠
If σT dominant ⇒ Prec reduced by factor MBut also shot noise contributes
F A = k AM + (1 − k A ) ⎜ 2 − ⎟⎝
⎟⎥ + ⎬⎝⎣
dPrec Q ⎧ ⎡ ⎛ 1 ⎞⎤ σT ⎫dM R ⎩ ⎝ M ⎠⎦ M ⎭⎣
2
2
2
⎣ ⎦
Prec =Q
R⎧⎨Qq∆f⎩
⎡ ⎛ 1⎢k AM + (1 − k A ) ⎜ 2 − M
⎞⎤ σT ⎫
⎠⎦ M ⎭
Optimum M
= ⎨Qq∆f ⎢k A + (1 − k A ) + ⎜ 2 − ⎟⎥ 2 = ⎬ 0
Qq∆f [k AM 2 + (1 − k A )] − σT = 0
k AM+ (1 − k A ) = σT
Qq∆f
k AM = σT
Qq∆f− (1 − k A )
M = 1 σ⎡ T ⎤k A ⎢Qq∆f + k A − 1⎥
2
⎡ ⎤⎥⎥⎦⎣
M opt =1
k A
σT
Qq∆f+ k A − 1
≈1
k A
σT
Qq∆f= σT
k AQq∆f
Qq∆f [k AM opt + (1 − k A )]− σT = 0
σT
M opt
= Qq∆f⎢k AM opt + (1 − k A )⎢
1
M opt
Q ⎧⎜
1 ⎞ ⎤ ⎡
M opt ⎟ ⎥Prec
⎩ ⎣ ⎢
⎪
⎥ ⎪
σM
Q ⎣ ⎦
= ⎣ ⎦
⎢⎣
=
Q
= ⎨Qq∆fR ⎪
⎡ ⎛ ⎢ k A M opt + (1 − k A ) ⎜ 2 − ⎢ ⎝
⎟ ⎥ + Qq∆f ⎢ k A M opt + (1 − k A ) ⎠ ⎦ ⎣
1 ⎤ ⎫ ⎥ ⎬
M opt ⎦ ⎭
by replacing back the value of T
= Qq∆f ⎡ 2k A M opt + 2 (1 − k A )⎤R
2Q 2 q∆f
R ⎡ k A M opt + 1 − k A ⎤
2Q 2 q∆f ⎡ k A σ T
R Qq∆f
Ideal receiver
⎤+ 1 − k A ⎥
⎦
σT = 0, M = 1, F A = 1
Prec= Qq∆f =R
Q 2
Rq∆f
Np: number of photons in a 1-bit
=
2
Thermal noise limitσ0 = σ1
I 0 = 0
Q = I1 − I0 I1
σ1 + σ0 2σ1
SNR =I 12
σ1
= 4Q 2 BER = 10−9 ⇒ Q = 6
SNR = 4 ⋅ 36 = 144
SNRdB = 10 log144dB = 21,6dB
Shot noise limit
σ0 = 0
σ1 = σS
I 0 = 0
=
I 12
σ 1
Q = I 1 − I 0 I 1
σ 1 + σ 0 σ 1
SNR = 2 = Q 2 = 36
SNRdB = 10 log 36 = 15,6dB
(peak power basis )
SNR = ηN p (shot noise limit)
Q = SNR = ηN p
1 Q 1BER = erfc
2= erfc
2 2η = 1
ηN p
2
BER = 10− 9 or Q = 6
N p = Q 2 = 36
In practice Np≈1000, limited by thermal noise
1 ηN p
2144424443
m
12
4.5.3 Quantum Limit of Photodetection
BER = erfc
assumes Gaussian statistics
Shot noise limit
Small number of photons in a bitPoisson statisticsNp.: average number photons in a bit 1Probability of generating m electron-hole pairs
Pm =N p
m!e−N p
BER = [P (0 / 1) + P (1 / 0)]
0
1 −N p
1 −N p
P (1 / 0) = 0because N p = 0 for bit 0
P (0 / 1) =N p
0!e−N p= e−N p for bit 1
BER = e2
BER = e2
= 10−9
− ln 2 − N p = −9 ln10
N p = 9 ln10 − ln 2 = 20,03 ≈ 20
12
=
c 3 ⋅ 108 m s 3
Quantum limit: Each 1-bit must contain 20 photons:
P1 =N p hν
TB= N p hνB
Prec =P1 + P0
2= P1 =
N p
2hνB = N p hνB
N p =N p 20
2 2= 10 average number of photons per bit
Numerical example
λ = 1,55 µm
ν = = =λ 1,55 ⋅ 10−6 m 1,55
⋅ 1014 Hz
= 1,94 ⋅ 1014 Hz ≈ 2 ⋅ 1014 Hz = 200.000GHz= 200THz
hν = 6,626 ⋅ 10−34 Js ⋅ 1,94 ⋅ 1014 s −1 = 12,85 ⋅ 10−20 J
= 1,285 ⋅ 10−19 J
−3
B = 1Gbit / s
Prec = 10 ⋅ 1,285 ⋅ 10−19 ⋅ 109W = 1,285 ⋅ 10−9W
Prec ,dBm= 10 log1,285 ⋅ 10−9
10dBm = −58,9dBm
In practice:20dB more power is required10⋅100=1000 photons per bit
4.6 RECEIVER DEGRADATION
So far: Noise only degrading mechanismIdeal sensitivityPractice: More power needed. Power penalty!Several sources:Extinction ratio
Intensity noise (in transmitter)Timing jitter
4.6.1 Extinction Ratio
Energy carried by 0-bitsLaser biased above threshold P0>0
rex =P0
P1
P1
P0
1 0 1 1
Prec =
Q = =
⋅=
Find power penalty
Q = I 1 − I 0
σ 1 + σ 0
I 1 = RP1
I 0 = RP0
P1 + P0
2
R (P1 − P0 ) (P1 − P0 ) R (P1 + P0 )σ1 + σ0 P1 + P0 σ1 + σ0
1 − rex R ⋅ 2Prec
1 + rex σ1 + σ0
⋅ = ⋅
Prec (rex ) = ⋅
σT QR
δex = =
p-i-n receiver: σ1 ≈ σ0 = σT
Q =1 − rex R ⋅ 2 ⋅ Prec 1 − rex RPrec
1 + rex 2σT 1 + rex σT
1 + rex σT Q1 − rex R
Prec (0) =
Prec (rex ) 1 + rex
Prec (0) 1 − rex
increases with rex
δex ,dB = 10 log1 + rex
1 − rex
dB Power penalty
Fig. 4.18
rex
Laser biased below threshold: rex≈0,05
δex ,dB= 10 log1 + 0,05
0,95dB = 0,43dB
Laser biased above threshold:δex,dB significant
APD receiver: Power penalty larger by factor of 2 for same rex
2s I
2
2
2
∞
rI2 ∫
4.6.2 Intensity noise
Assumption - optical power incident on the receiver does not fluctuate !
So far:Practice:
No intensity noiseIntensity noise causes current fluctuation extra noise
Simple approach:
σ 2 = σ 2 + σ T + σ 2
σI = R
d
rI =
∆Pin =
∆Pin
Pin
∆Pin
Pin⋅ RPin = rI RPin
∞
= ∫ RIN (f )df =−∞
1
2π −∞RIN (ω)dωRelative Intensity Noise
σ1 ⎫
σ0 ⎭
rI =1
SNR
Typically SNRdB>20dB SNR>100 rI<0,01
Q = I 1 − I 0
σ1 + σ0
reduced because ⎬increased because of σI
Q should be the sameincident power must be increased (i.e. power penalty)
Assume : I 0 = 0 ⇒ σ0 = σT
I 1 = RP1 = 2RPrec
2s I
s
2
4k BTFn ∆f 2 2 2
RL
Q = I 1 − I 0
σ1 + σ 0= 2RPrec
σ 2 + σT + σ 2 + σT
σ2 = 2qRP1 ∆f = 2qR 2Prec ∆f
σs = 2 qRPrec ∆f
σI = rI RPin = rI R 2Prec
σT =4k BTFn ∆f
RL
σT =
Q =
4k BTFn ∆fR L
4qRPrec ∆f +
2RPrec
+ rI R 4Prec +4k BTFn ∆fRL
2
2 2 ⎤2 2
2 2 2
2 2
2 2 2
[
Q 4qRPrec ∆f +4k BTFn ∆f
RL+ rI2R 2 4Prec
= 2RPrec − Q4k BTFn ∆f
RL
⎡Q ⎢4qRPrec ∆f +
⎣
4k BTFn ∆fRL
+ rI R 4Prec ⎥⎦
= 4R Prec + Q4k BTFn ∆f
RL− 4RPrec Q
Rk BTFn ∆fRL
Q q∆f + rI RPrec ] = RPrec − Q
Prec [R − Q rI R ] = Q q∆f + Q
4k BTFn ∆fRL
4k BTFn ∆fRL
+ Q 2q∆fQ
Q
=
Prec (rI ) =4k B Fn ∆f
RL
R (1 − rI2Q 2 )
Prec (0) =4k BTFn ∆f
RL
R
+ Q 2q∆f
Prec (rI ) 1Prec (0) 1 − rI2Q 2
δI ,dB= 10 logPrec (rI )Prec (0)= −10 log(1 − rI2Q 2 )
1 244 3for
Fig. 4.19
Intensity noise parameter rex
δI ,dB < 0,02dB rI < 0,01normal for transmitters
Intensity noise negligible for most digital receiversCan become limiting factor for analog systems
4.6.3 Timing jitter
Fig. 4.10
Timing jitter Fluctuation in signalTiming jitter as random variableSNR reduced Power penaltyAssume:
p-i-n receiverdominated by thermal noisezero extinction ratio
2j
∆ij:σj:
(t)
Q = I 1 − I 0
σ1 + σ0
=(I 1 − ∆i j ) − 0
σT + σ2 + σT
current fluctuation induced by timing jitterrms value of current fluctuation ∆ij
I1hout()
I1hout(0)
0 ∆t
I1hout(∆t)∆ij
4.7 RECEIVER PERFORMANCE
Receiver sensitivity , Prec : Power needed forBER = 10− 9
Fig. 4.21
Bit Rate (Gbit/s)Long sequence of pseudo-random bits (215-1)
Quantum limit
Prec = Ν p hνB = 10hνB
Practical receivers: 20dB worse than quantum limit
Most degradation due to thermal noise
Some degradation due to dispersion
degradation∝BL
Biterrate
rror
InGaAs APD1,3 µm8 Gbit/s
Fig. 4.22
1,5dB dispersion penalty
Received optical power (dBm)
5-6dB improvement with APD over p-i-n
InGaAs APD≈
Quantum limit ≈
Si APD ≈
1000 photons/bit
10 photons/bit
400 photons/bit, 0,8-0,9 µm
So far: direct detection receiversCoherent detection (or heterodyne) receiverswithin 5dB of quantum limit
Received signal
LO laser
But optical preamplifiers also give good sensitivity
PDFA probability density function is any function f(x) thatdescribes the probability density in terms of the inputvariable x in a manner described below.• f(x) is greater than or equal to zero for all values of x• The total area under the graph is 1:
The actual probability can then be calculated by takingthe integral of the function f(x) by the integration intervaof the input variable x.For example: the variable x being within the interval 4.3< x < 7.8 would have the actual probability of