8/2/2019 EEE461Pres16

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EEE 461 Communication Systems II

Lecture Presentation 16

Aykut HOCANIN

Dept. of Electrical and Electronic Engineering

Eastern Mediterranean University

c Dr. Aykut HOCANIN

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7.3 (Couch) Coherent Detection of BandpassBinary Signals

On-Off Keying

The OOK signal is represented by

s1(t) = A cos(ct + c), 0 < t T binary 1s2(t) = 0, 0 < t T binary 0

For coherent detection, a product detector is used.

The bandpass noise is represented by

n(t) = x(t)cos(ct + n) y(t)sin(ct + n)

where the psd of n(t) is Pn(f) =N02

and n is uniformly distributed randomvariable which is independent of c.

The noise power in the received signal is

E[x2(t)] = 20

= E[n2(t)] = 2(N0/2)(2B) = 2N0B

The optimum threshold is VT = A/2.

c Dr. Aykut HOCANIN

Eastern Mediterranean University2/8

EEE 461 Communication Systems II

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Transmitter

Low-pass

filter

or

Matched

filterH(f)

Sample

and hold

at t0

Threshold

Device

+

Noise n(t)

Channel

s(t)

Digital

input m

Baseband

analog

output

r0(t) r0(t0)

Clock

Digital

output

m~

Receiver

r(t)=s(t) + n(t)

X

)cos(2cc

t +

Figure 1: Coherent detection of OOK or BPSK signals.

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EEE 461 Communication Systems II

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Pe = Q

A2

8N0B

(narrowband filter) (1)

where B is the bandwidth of the LPF.

The energy in the difference signal is

Ed =

T0

[A cos(ct + c) 0]2dt =

A2T

2

and the BER becomes

Pe = Q

A2T

4N0

= Q

EbN0

(matched f ilter) (2)

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Binary-Phase-Shift Keying

The BPSK signal is represented bys1(t) = A cos(ct + c), 0 < t T binary 1s2(t) = A cos(ct + c), 0 < t T binary 0

The noise power in the received signal is

E[x2

(t)] = 2

0 = E[n2

(t)] = 2(N0/2)(2B) = 2N0B

The optimum threshold is VT = 0.

Pe = QA2

2N0B (narrowband filter) (3)

where B is the bandwidth of the LPF.

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The energy in the difference signal is

Ed = T

0

[2A cos(ct + c)]2dt = 2A2T

and the BER becomes

Pe = Q

A2T

N0

= Q

2

EbN0

(matched f ilter) (4)

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Frequency-Shift Keying

The FSK signal is represented bys1(t) = A cos(1t + c), 0 < t T binary 1s2(t) = A cos(2t + c), 0 < t T binary 0

where the frequency shift is 2F = f1 f2.

The output noise power is

E[n20(t)] = 2

0= E[x21(t)] + E[x

2

2(t)] = E[n21(t)] + E[n

2

2(t)] = 4N0B

The optimum threshold is VT = 0.

Pe = QA2

4N0B

(bandpass f ilters) (5)

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EEE 461 Communication Systems II

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The energy in the difference signal is

Ed = T

0

[A cos(1t + c) A cos(2t + c)]2dt = A2T

when s1(t) is orthogonal to s2(t) or (f1 f2) R.

The BER becomes

Pe = Q

A2T

2N0

= Q

EbN0

(matched f ilter) (6)

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Comparing the Coherent Binary Detectors

0 5 10 1510

-1 0

10-8

10-6

10-4

10-2

100

P

b

Eb/N

0(dB)

B P S KB F S K

10.5 dB 13.5 dB

To achieve Pb = 10-6

BPSK requires Eb/N0 = 10.5 dB

BFSK requires Eb/N0 = 13.5 dB

For the same probabilityof bit error BPSKrequires half the power.

For the same probabilityof bit error BPSKrequires half the power.

BPSK is 3 dB betterthan coherent BFSK

BPSK is 3 dB betterthan coherent BFSK