digital data transmission

Digital Data Transmission

ECE 457

Spring 2005

Analog vs. Digital

Analog signals Value varies continuously

Digital signals Value limited to a finite set

Binary signals Has at most 2 values Used to represent bit values Bit time T needed to send 1 bit Data rate R=1/T bits per second

t

x(t)

t

x(t)

t

x(t) 1

0 0 0

1 1

0T

Information Representation

• Communication systems convert information into a form suitable for transmission

• Analog systemsAnalog signals are modulated (AM, FM radio)

• Digital system generate bits and transmit digital signals (Computers)

• Analog signals can be converted to digital signals.

Digital Data System

Principles of Communications, 5/E by Rodger Ziemer and William TranterCopyright © 2002 John Wiley & Sons. Inc. All rights reserved.

Figure 7-1 Block diagram of a digital data system. (a) Transmitter.(b) Receiver.

Components of Digital Communication

• Sampling: If the message is analog, it’s converted to discrete time by sampling. (What should the sampling rate be ?)

• Quantization: Quantized in amplitude. Discrete in time and amplitude

• Encoder: – Convert message or signals in accordance with a set of

rules– Translate the discrete set of sample values to a signal.

• Decoder: Decodes received signals back into original message

Different Codes

0 1 1 0 1 0 0 1

Performance Metrics

• In analog communications we want,• Digital communication systems:

– Data rate (R bps) (Limited) Channel Capacity– Probability of error – Without noise, we don’t make bit errors– Bit Error Rate (BER): Number of bit errors that occur

for a given number of bits transmitted.

• What’s BER if Pe=10-6 and 107 bits are transmitted?

)()(ˆ tmtm ≅

eP

Advantages

• Stability of components: Analog hardware change due to component aging, heat, etc.

• Flexibility: – Perform encryption– Compression– Error correction/detection

• Reliable reproduction

Applications

• Digital Audio Transmission

• Telephone channels

• Lowpass filter,sample,quantize

• 32kbps-64kbps (depending on the encoder)

• Digital Audio Recording

• LP vs. CD

• Improve fidelity (How?)

• More durable and don’t deteriorate with time

Baseband Data Transmission


Figure 7-2 System model and waveforms for synchronous baseband digital data transmission. (a) Baseband digital data communication system. (b) Typical transmitted sequence. (c) Received sequence plus noise.

• Each T-second pulse is a bit.

• Receiver has to decide whether it’s a 1 or 0

( A or –A)

• Integrate-and-dump detector

• Possible different signaling schemes?

Receiver Structure


Figure 7-3 Receiver structure and integrator output. (a) Integrate-and-dump receiver. (b) Output from the integrator.

Receiver Preformance

• The output of the integrator:

• is a random variable.

• N is Gaussian. Why?

−+−+

=

+= ∫+

sentisANAT

sentisANAT

dttntsVTt

t

0

0

)]()([

∫+

=Tt

t

dttnN0

0

)(

Analysis

• Key Point – White noise is uncorrelated

2

)!?()(2

)]()([

)(

?][

][][][

0)]([])([][

0

0

2

2

22

0

0

0

0

0

0

0

0

0

0

0

0

0

0

TN

eduncorrelatisnoiseWhiteWhydtdsstN

dtdssntnE

dttnE

WhyNE

NENENVar

dttnEdttnENE

Tt

t

Tt

t

Tt

t

Tt

t

Tt

t

Tt

t

Tt

t

=

−=

=

=

=−=

===

∫ ∫

∫ ∫

∫

∫ ∫

+ +

+ +

+

+ +

δ

Error Analysis

• Therefore, the pdf of N is:

• In how many different ways, can an error occur?

TN

enf

TNn

N

0

)/( 02

)(π

−

=

Error Analysis

• Two ways in which errors occur:– A is transmitted, AT+N<0 (0 received,1 sent)

– -A is transmitted, -AT+N>0 (1 received,0 sent)


Figure 7-4 Illustration of error probabilities for binary signaling.

•

• Similarly,

• The average probability of error:

== ∫

−

∞−

−

0

2

0

/ 2)|(

02

N

TAQdn

TN

eAErrorP

AT TNn

π

==− ∫

∞ −

0

2

0

/ 2)|(

02

N

TAQdn

TN

eAErrorP

AT

TNn

π

=

−−+=

0

22

)()|()()|(

N

TAQ

APAEPAPAEPPE

• Energy per bit:

• Therefore, the error can be written in terms of the energy.

• Define

TAdtAETt

t

b22

0

0

== ∫+

00

2

N

E

N

TAz b==

• Recall: Rectangular pulse of duration T seconds has magnitude spectrum

• Effective Bandwidth: • Therefore,

• What’s the physical meaning of this quantity?

)(TfsincAT

TBp /1=

pBN

Az

0

2

=

Probability of Error vs. SNR


Figure 7-5PE for antipodal baseband digital signaling.

Error Approximation

• Use the approximation

1,2

2

1,2

)(

0

2

2/2

>>≅

=

>>≅

−

−

zz

e

N

TAQP

uu

euQ

z

E

u

π

π

Example

• Digital data is transmitted through a baseband system with , the received pulse amplitude A=20mV.

a)If 1 kbps is the transmission rate, what is probability of error?

HzWN /10 70

−=

3

237

6

0

2

33

1058.22

4104001010

10400

1010

11

−−

−−

−

−

×=≅

=×=×

×===

===

z

eP

BN

AzSNR

TB

z

E

p

p

π

b) If 10 kbps are transmitted, what must be the value of A to attain the same probability of error?

• Conclusion:

Transmission power vs. Bit rate

mVAAA

BN

Az

p

2.6310441010

3247

2

0

2

=⇒×=⇒=×

== −−

Binary Signaling Techniques


Figure 7-13Waveforms for ASK, PSK, and FSK modulation.

ASK, PSK, and FSK

Amplitude Shift Keying (ASK)

Phase Shift Keying (PSK)

Frequency Shift Keying

==

==0)(0

1)()2cos()2cos()()(

b

bcccc nTm

nTmtfAtfAtmts

ππ

−=+=

==1)()2cos(

1)()2cos()2cos()()(

bcc

bcccc nTmtfA

nTmtfAtftmAts ππ

ππ

−==

=1)()2cos(

1)()2cos()(

2

1

bc

bc

nTmtfA

nTmtfAts π

π

1 0 1 1

1 0 1 1

1 0 1 1

AM Modulation

PM Modulation

FM Modulation

m(t)

m(t)

Amplitude Shift Keying (ASK)

• 00

• 1Acos(wct)

• What is the structure of the optimum receiver?

Receiver for binary signals in noise


Figure 7-6 A possible receiver structure for detecting binary signals in white Gaussian noise.

Error Analysis

• 0s1(t), 1s2(t) in general.• The received signal:

• Noise is white and Gaussian.

• Find PE

• In how many different ways can an error occur?

Tttttntsty

OR

Tttttntsty

+≤≤+=

+≤≤+=

002

001

),()()(

),()()(

Error Analysis (general case)

• Two ways for error: » Receive 1 Send 0

» Receive 0Send 1

• Decision: » The received signal is filtered. (How does this

compare to baseband transmission?)

» Filter output is sampled every T seconds

» Threshold k

» Error occurs when:

kTnTsTv

OR

kTnTsTv

<+=

>+=

)()()(

)()()(

002

001

• are filtered signal and noise terms.• Noise term: is the filtered white Gaussian

noise.• Therefore, it’s Gaussian (why?)• Has PSD:

• Mean zero, variance?• Recall: Variance is equal to average power of the

noise process

00201 ,, nss

)(0 tn

20 )(2

)(0

fHN

fSn =

dffHN 202 )(2∫

∞

∞−

=σ

• The pdf of noise term is:

• Note that we still don’t know what the filter is.• Will any filter work? Or is there an optimal one?

• Recall that in baseband case (no modulation), we had the integrator which is equivalent to filtering with

2

2/

2)(

022

πσ

σn

N

enf

−

=

fjfH

π2

1)( =

• The input to the thresholder is:

• These are also Gaussian random variables; why?

• Mean: • Variance: Same as the variance of N

NTsTvV

OR

NTsTvV

+==

+==

)()(

)()(

02

01

)()( 0201 TsORTs

Distribution of V

• The distribution of V, the input to the threshold device is:


Figure 7-7 Conditional probability density functions of the filter output at time t = T.

Probability of Error

• Two types of errors:

• The average probability of error:

−−==

−==

∫

∫

∞−

−−

∞ −−

σπσ

σπσσ

σ

)(1

2))(|(

)(

2))(|(

02

2

2/)]([

2

01

2

2/)]([

1

2202

2201

TskQdv

etsEP

TskQdv

etsEP

k Tsv

k

Tsv

)](|[2

1)](|[

2

121 tsEPtsEPPE +=

• Goal: Minimize the average probability of errror

• Choose the optimal threshold

• What should the optimal threshold, kopt be?

• Kopt=0.5[s01(T)+s02(T)]

•

−=

σ2

)()( 0102 TsTsQPE

Observations

• PE is a function of the difference between the two signals.

• Recall: Q-function decreases with increasing argument. (Why?)

• Therefore, PE will decrease with increasing distance between the two output signals

• Should choose the filter h(t) such that PE is a minimummaximize the difference between the two signals at the output of the filter

Matched Filter

• Goal: Given , choose H(f) such that is maximized.

• The solution to this problem is known as the matched filter and is given by:

• Therefore, the optimum filter depends on the input signals.

)(),( 21 tsts

σ)()( 0102 TsTs

d−=

)()()( 120 tTstTsth −−−=

Matched filter receiver


Figure 7-9 Matched filter receiver for binary signaling in whiteGaussian noise.

Error Probability for Matched Filter Receiver

• Recall• The maximum value of the distance,

• E1 is the energy of the first signal.

• E2 is the energy of the second signal.

=

2

dQPE

)2(2

1221210

2max ρEEEE

Nd −+=

∫

∫+

+

=

=

Tt

t

Tt

t

dttsE

dttsE

0

0

0

0

)(

)(

222

211

dttstsEE

)()(1

21

21

12 ∫∞

∞−

=ρ

• Therefore,

• Probability of error depends on the signal energies (just as in baseband case), noise power, and the similarity between the signals.

• If we make the transmitted signals as dissimilar as possible, then the probability of error will decrease ( )

−+=

2/1

0

122121

2

2

N

EEEEQPE

ρ

112 −=ρ

ASK

• The matched filter:

• Optimum Threshold:

• Similarity between signals?

• Therefore,

• 3dB worse than baseband.

)2cos()(,0)( 21 tfAtsts cπ==

)2cos( tfA cπ

TA2

4

1

( )zQN

TAQPE =

=

0

2

4

PSK

• Modulation index: m (determines the phase jump)

• Matched Filter:

• Threshold: 0

• Therefore,

• For m=0, 3dB better than ASK.

)cos2sin()(),cos2sin()( 12

11 mtfAtsmtfAts cc

−− −=+= ππ

)2cos(12 2 tfmA cπ−−

))1(2( 2 zmQPE −=

Matched Filter for PSK


Figure 7-14 Correlator realization of optimum receiver for PSK.

FSK

•

•

• Probability of Error:

• Same as ASK

))(2cos()(),2cos()( 21 tffAtstfAts cc ∆+== ππ

T

mf =∆

)( zQ

Applications

• Modems: FSK

• RF based security and access control systems

• Cellular phones

digital data transmission

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