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Carrier Phase Modulation

Phase Shift Keying (PSK)

Information is impressed on the phase of the carrier. As data changes from symbol period to symbol period, the phase shifts.

∀m,m∈12…M: s m t=A P T tcos2π f c t+2πm-1M

m m 1 2 … M s m t A P T t 2 f c t 2 m 1 M

(1)

Example 1

Binary s 1 t

s 1 t

or s 2 t

s 2 t

Representing the Signals

An orthonormal basis to represent the signals is

ψ 1 t=1 E s A P T tcos2π f c t

ψ 1 t 1 E s A P T t 2 f c t

(2)

ψ 2 t=-1 E s A P T tsin2π f c t

ψ 2 t -1 E s A P T t 2 f c t

(3)

The signal

S m t=A P T tcos2π f c t+2πm-1M

S m t A P T t 2 f c t 2 m 1 M

(4)

S m t=Acos2πm-1M P T tcos2π f c t-Asin2πm-1M P T tsin2π f c t

S m t A 2 m 1 M P T t 2 f c t A 2 m 1 M P T t 2 f c t

(5)

The signal energy

E s =∫-∞∞A2 P T 2tcos22π f c t+2πm-1Mdt=∫0TA212+12cos4π f c t+4πm-1Mdt

E s t A 2 P T t 2 2 f c t 2 m 1 M 2 t 0 T A 2 1 2 1 2 4 f c t 4 m 1 M

(6)

E s =A2T2+12A2∫0Tcos4π f c t+4πm-1Mdt≈A2T2

E s A 2 T 2 1 2 A 2 t 0 T 4 f c t 4 m 1 M A 2 T 2

(7)

(Note that in the above equation, the integral in the last step before the aproximation is very small.) Therefore,

ψ 1 t=2T P T tcos2π f c t

ψ 1 t 2 T P T t 2 f c t

(8)

ψ 2 t=-2T P T tsin2π f c t

ψ 2 t 2 T P T t 2 f c t

(9)

In general,

∀m,m∈12…M: s m t=A P T tcos2π f c t+2πm-1M

m m 1 2 … M s m t A P T t 2 f c t 2 m 1 M

(10)

and ψ 1 t

ψ 1 t

ψ 1 t=2T P T tcos2π f c t

ψ 1 t 2 T P T t 2 f c t

(11)

ψ 2 t=2T P T tsin2π f c t

ψ 2 t 2 T P T t 2 f c t

(12)

s m = E s cos2πm-1M E s sin2πm-1M

s m E s 2 m 1 M E s 2 m 1 M

(13)

Demodulation and Detection

r t = s m t+ N t , for some m∈12…M

r t s m t N t , for some m 1 2 … M

(14)

We must note that due to phase offset of the oscillator at the transmitter, phase jitter or phase changes occur because of propagation delay.

r t =A P T tcos2π f c t+2πm-1M+φ+ N t

r t A P T t 2 f c t 2 m 1 M φ N t

(15)

For binary PSK, the modulation is antipodal, and the optimum receiver in AWGN has average bit-error probability

P e =Q 2 E s N 0 =QAT N 0

P e Q 2 E s N 0 Q A T N 0

(16)

The receiver where

r t =±A P T tcos2π f c t+φ+ N t

r t ± A P T t 2 f c t φ N t

(17)

The statistics

r 1 =∫0T r t αcos2π f c t+ φ ^ dt=±∫0TαAcos2π f c t+φcos2π f c t+ φ ^ dt+∫0Tαcos2π f c t+ φ ^ N t dt

r 1 t 0 T r t α 2 f c t φ ^ ± t 0 T α A 2 f c t φ 2 f c t φ ^ t 0 T α 2 f c t φ ^ N t

(18)

r 1 =±αA2∫0Tcos4π f c t+φ+ φ ^ +cosφ- φ ^ dt+ η 1

r 1 ± α A 2 t 0 T 4 f c t φ φ ^ φ φ ^ η 1

(19)

r 1 =±αA2Tcosφ- φ ^ +∫0T±αA2cos4π f c t+φ+ φ ^ dt+ η 1 ≈ ±αAT2cosφ- φ ^ + η 1

r 1 ± α A 2 T φ φ ^ t 0 T ± α A 2 4 f c t φ φ ^ η 1 ± α A T 2 φ φ ^ η 1

(20)

where η 1 =α∫0T N t cos ω c t+ φ ^ dt

η 1 α t 0 T N t ω c t φ ^

is zero mean Gaussian with variance≈α2 N 0 T4

variance α 2 N 0 T 4

Therefore,

P e ¯=Q2αAT2cosφ- φ ^ 2α2 N 0 T4=Qcosφ- φ ^ AT N 0

P e Q 2 α A T 2 φ φ ^ 2 α 2 N 0 T 4 Q φ φ ^ A T N 0

(21)

which is not a function of α

and depends strongly on phase accuracy.

P e =Qcosφ- φ ^ 2 E s N 0

P e Q φ φ ^ 2 E s N 0

(22)

The above result implies that the amplitude of the local oscillator in the correlator structure does not play a role in the performance of the correlation receiver. However, the accuracy of the phase does indeed play a major role. This point can be seen in the following example:

Example 2

x t ′ =-1iAcos-2π f c t ′ +2π f c τ

x t ′ -1 i A 2 f c t ′ 2 f c τ θ ′

(23)

x t =-1iAcos2π f c t-2π f c τ ′ -2π f c τ+ θ ′

x t -1 i A 2 f c t 2 f c τ ′ 2 f c τ θ ′

(24)

Local oscillator should match to phase θ

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Last edited by Tuan Do-Hong on Nov 16, 2007 11:25 pm US/Central.