bit error rate in digital photoreceivers · the detected signal is above or below the threshold...

5

Γ. Έλληνας, Διάλεξη 13-14, σελ. 29Μάθημα HMY 455: Συστήματα και Δίκτυα Επικοινωνιών με Οπτικές Ίνες

Bit Error Rate in Digital Photoreceivers

In the previous slides, we saw that the photoreceiver makes a decision as to whether the recovered waveform is above (“1”) or below (“0”) the threshold level. When noise is present, a wrong decision can be made, i.e. we have a bit error.

We will now examine techniques for calculating the bit error probability and hence the BER.


Digital Photoreceiver

Recovered pulse train

(output voltage)

© Prentice-Hall


© Wiley

Example of a bit error

• Bit errors are a consequence of the noise present on the received signal. Since the noise is random and probabilistic, it can be described using a random variable.


S0

S1

D0

D1

Only two types of bit can be sent in a binary system: “1”s and “0”s. These events are mutually exclusive, so we have Pr(S0) + Pr(S1) = 1.S0 is the event “0” was sentS1 is the event “1” was sent

Only two types of decision can be made:the detected signal is above or below the threshold level, i.e. either a “1” or a “0” is detected. These events are mutually exclusive, so we have Pr(D0) + Pr(D1) = 1.D0 is the event “0” was detectedD1 is the event “1” was detected

Example of a bit error


S0

S1

D0

D1

Conditional probabilities

D1 .S0

D0 .S0

D0 .S1D1 .S1

A total of four mutually exclusive outcomes are possible in a binary

communications system


Conditional probabilities

D1 .S0

D0 .S0

D0 .S1D1 .S1

The shaded regions represent events thatgive a bit error:

• D1. S0 = a “1” is detected and a “0”was sent

• D0 .S1 = a “0” is detected and a “1”was sent

• These two events are mutually exclusive, hence:).().()errorbit ( 1001 SDPSDPP rrr +=


Baye’s formula( )( )1

1010

.)/(SP

SDPSDPr

rr =

Rearranging gives: ( ) )/()(. 10110 SDPSPSDP rrr =

Similarly, we have: ( ) )/()(. 01001 SDPSPSDP rrr =

Thus the bit error probability can be written as:

)/()()/()()errorbit ( 101010 SDPSPSDPSPP rrrrr +=

Probability of bit error


The previous formula can be used to calculate the bit error probability provided:

we know what the probabilities of sending “0”s and “1”s are (often we have Pr(S0) = Pr(S1) = 0.5)and we can obtain the conditional probabilities Pr(D1/S0) and Pr(D0/S1).

We can obtain Pr(D1/S0) and Pr(D0/S1) if we know what the PDFs associated with reception of the bits “0” and “1” in the presence of noise are.

These processes can be very accurately approximated by gaussian random variables; the gaussian PDF is plotted on the next slide.

Probability of bit error


Gaussian PDF

∞≤−≤∞−=−−

xexpmx2

2

2)(

221)( σ

πσ

m

p(x)


The gaussian PDF occurs very widely in many applications (and for that reason is also called the Normal distribution).

One reason for this is the central limit theorem. This theorem tells us that if we take the sum of a large number of independent variables X1, X2, .... Xn, and if each of these makes a small contribution to the sum X = X1 + X2 + .... + Xn, then the PDF of X will approach a gaussian shape as n →∞.The proof is beyond the scope of this course, but the idea can be illustrated best by an example, e.g. roll n dice and add their values. If this event is repeated enough times, you get a gaussian distribution.www.users.on.net/zhcchz/java/quincunx/quincunx.8.html

Gaussian PDF


Properties of the gaussian PDF

∫∞

∞−

= 1)(xp

5.0)()( =≥=≤ mXPmXP rr by symmetry

mean: mX =

σ is the standard deviation: when p(x) is used to describe the probability of detecting a noise current (or voltage) then σ represents the rms value of the noise current (or voltage).


Obtaining probabilities from the gaussian PDF

When calculating the bit error probability later on, we will have to evaluate probabilities such as:

∫∞

=≥1

)()( 1x

r dxxpxXP

This expression cannot be calculated analytically, we must use numerical techniques. We define:

∫∞

−=k

y dyekQ 22

21)(π

This can be obtained numerically and then plotted:


Q(k)


To calculate:[ ] dxexXP

x

mxr ∫

∞−−=≥

1

22 2)(

212

1)( σ

πσLet: ymx

=−σ

⎟⎠⎞

⎜⎝⎛ −

=≥

=≥ ∫∞

−

−

σ

πσ

mxQxXP

dyexXP

r

mx

yr

11

2/1

)(

21)(

1

2



m

p(x)

⎟⎠⎞

⎜⎝⎛ −

=≥

=≥ ∫∞

σmxQxXP

dxxpxXP

r

xr

11

1

)(

)()(1

x1



In the context of our digital photoreceiver, we can say that output voltage v(t) generated immediately after the amplifier stage in response to the transmission of “0” and “1” will have mean values of Vm0 and Vm1 for these two pulses. The threshold level (Vth) will be set between these two values.

However, noise (due e.g. to thermal and amplifier contributions) will be superimposed on these mean values, and the distributions will follow that of a gaussian PDF. Hence the received voltages for “0” and “1” have PDFs given by p0(v) and p1(v) respectively:

Towards BER .....


Pr(D1/S0)

Pr(D0/S1) Vm0

Vm1

detected voltage, v

Vth

p0(v)

p1(v)

Assume σ0 = σ1 = σ

Towards BER .....


)/()()/()( 101010 SDPSPSDPSPP rrrre +=

We saw earlier that the bit error probability is:

If we assume that “ones” and “zeros” are equally likely to be sent, then Pr(S0) = Pr(S1) = 0.5 and:

[ ])/()/( 100121 SDPSDPP rre +=

By considering an NRZ waveform with Vm0 = 0, and picking a threshold

midway between this and Vm1, i.e. Vth= Vm1 /2, show that:

( )σ21me VQP =

QUESTION


Bit Error Rate in Digital Photoreceivers

In the previous slides, we saw that the photoreceiver makes an error whenever noise “pushes” the waveform to the “wrong side”of the threshold level.

We also saw that we could model this process using the gaussian distribution.

We will now finish our treatment by showing how BER is related to SNR.


Digital Photoreceiver

Recovered pulse train

(output voltage)

© Prentice-Hall

Bit errors can be made here;the number depends on the SNR

of the received signal


)/()()/()( 101010 SDPSPSDPSPP rrrre +=

We saw earlier that the bit error probability is:

If we assume that “ones” and “zeros” are equally likely to be sent, then Pr(S0) = Pr(S1) = 0.5 and:

[ ])/()/( 100121 SDPSDPP rre +=

We will consider a NRZ waveform with Vm0 = 0, and pick a threshold midway between this and Vm1, i.e. Vth= Vm1 /2. We refer to this as a

unipolar waveform.

Towards BER .....


)/( 01 SDPr

Vth Vm10

v

p0(v) p1(v)

2

2

220

21)( σ

πσ

v

evp−

=

∫∞

=≥=thV

thrr dvvpVvPSDP )()()/( 001

Towards BER .....


• Using the relationship:

⎟⎠⎞

⎜⎝⎛ −

=≥σ

mxQxXPr1

1 )(we have:

⎟⎠⎞

⎜⎝⎛=

≥=

σth

thrr

VQ

VvPSDP )()/( 01

Towards BER .....


)/( 10 SDPr( )

2

21

221

21)( σ

πσ

mVv

evp−−

=

∫∞−

=≤=thV

thrr dvvpVvPSDP )()()/( 110

v

0 Vth Vm1

p0(v) p1(v)

Towards BER .....


• By symmetry, we have:

v

0 Vth Vm1

Green area = black area

3Vth

∫∞

=thV

r dvvpSDP3

110 )()/(

p1(v)

Towards BER .....


• Using ⎟⎠⎞

⎜⎝⎛ −

=≥σ

mxQxXPr1

1 )(

we have:

⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛ −

=

≥=

σ

σ

th

mth

thrr

VQ

VVQ

VvPSDP

1

10

3

)3()/(

Towards BER .....


[ ]

⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛=

+=

σ

σ

2

)/()/(

1

100121

m

th

rre

VQ

VQ

SDPSDPP• Hence:

• Now, remember that σ is the rms noise voltage, so:

mean square noise power ∝ σ2

Towards BER .....


• Also, if ones and zeros are equally likely,

• Hence the SNR is:

mean square signal power ∝ [ ] 212

121

202

1mmm VVV =+

2

21

2σmV

• Comparing with the bit error probability,

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛=

221 SNRQVQP m

e σ

Bit Error Probability


From plot of Q function,for Pe = 10-9, need tofind Q(k) = 10-9, whichgives k = 6.0.

plot of Q function


Hence we have for Pe = 10-9:

9102

−=⎟⎟⎠

⎞⎜⎜⎝

⎛=

SNRQPe

From the plot of Q(k) versus k, we have k = 6.0,i.e.:

0.720.62

=⇒= SNRSNR

In dB, we have SNR = 10 log10(72.0) = 18.6 dB

Pe and SNR


- 1 0 -5 0 5 1 0 1 5 2 0 2 51 0

- 2 5

1 0- 2 0

1 0- 1 5

1 0- 1 0

1 0- 5

1 00

-5-10 0 5 10 15 20 25

1

10-5

10-10

10-15

10-25

10-20

SNR (dB)

Bit

erro

r pro

babi

lity

BER versus SNR for unipolar NRZ


Note that we have used Q(k) in these calculations; most textbooks make use of the complementary error function erfc(x) defined as:

duexx

u∫∞

−=22)(erfc

πIt is straightforward to show this is related to Q(k) as follows:

⎟⎠

⎞⎜⎝

⎛=2

erfc21)( kkQ

(MATLAB, for example, uses erfc(x), not Q(x))

Complementary error function

bit error rate in digital photoreceivers · the detected signal is above or below the threshold...

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