Lecture 20

Coherent Detection of MPSK For MPSK, signal space can be divided into M regions, having angular width 2 / M. Below is the signal space shown for QPSK, with possible phases 0, /2, , 3 /2 radians

COMMUNICATION SYSTEMS Lecture # 20 21st Apr 2007 Instructor

WASEEM KHAN

Centre for Advanced Studies in Engineering

The received signal can be expressed as At the transmitter, binary digits are collected two at a time for each symbol interval Two sequential digits instruct the modulator as to which of the four waveforms to produce si(t) can be expressed as:

si (t )

2E cos( T

0t

2 i ) M

where: E: signal energy over each symbol duration T 0: carrier frequency

0

t

i

(t )

2 cos T

0

t

2

(t )

The upper integrator calculates X The lower integrator calculates Y

2E cos T

or r (t )

T

1,... M

where

i

i

cos

0

2 i ) M

t

0

t

sin

i

n (t )

sin

0

t

0 i

n (t )

t

T

1 ,... M

0 i

t

T

1,... M

2 i M

The detector has to calculate the transmitted phase i, if it lies within region 1, the received symbol is s1, if within region 2, then s2, if within region M, then sM.

At the receiver, we define two reference signals

1

2E cos( T

si (t )

2 sin T

T 0

T 0

0

t

r (t )

r (t )

1

2

( t ) dt

( t ) dt

The computation of the received phase angle can be accomplished by computing the arctan of Y/X Where: X: is the inphase component of the received signal Y: is the quadrature component of the received signal : is the noisy estimate of the transmitted i

The demodulator selects the that is closest to the angle

i

Or it computes | i - | for each i prototypes and chooses i yielding smallest output

1

PSK Receiver Example: QPSK

Probability of Bit-error for PSK For binary signalling, with equal a piriori probabilities, PB

Q

a1 a 0 2 0

For BPSK (antipodal signalling),

a1

Eb

a0

Eb

This is true when matched filter is employed in place of integrator. The variance of noise at the output of the matched filter N0 is 2 2

Hence for BPSK PB

2

Q 2

Eb N

0

Q

2Eb N0

Bit error probability for coherently detected MPSK

2

PB for BPSK and QPSK* are same because QPSK is a combination of two orthogonal BPSK channels. For MPSK, with equally likely symbols,

PE ( M )

2Q

2 Es sin N0

M

where PE(M) is the probability of symbol error, Es is symbol energy. * When bits are gray-coded

Gray coding

Problem

A possible set of QPSK symbols with corresponding group of bits is represented in this figure. If the transmitted symbol S1, due to noise is detected as S4, two bits will be in error; i.e. one symbol error means two bit-errors.

00

11

10

S4

S1

S3

S2 01

Consider another possible set of symbols, shown below. 10

11

00 S4

S1

S3

S2

For a BPSK system, what is the minimum Eb/N0, in dB, required to get BER not more than 2 x 10-3. If N0 = 1uW/Hz, signal power is 1 mW, how much maximum bit-rate is possible?

This is gray-coding, in which, any two neighbouring symbols differ from one another in only one bit position.

01

2

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