Phase Shift Keying (PSK) - Digital To Analog Conversion and Transmission Tutorials Part IV
http://ingenuitydias.blogspot.com/2014/04/phase-shift-keying-psk-digital-to.html
Phase Shift Keying (PSK) - Basic Tutorials
We vary the phase shift of the carrier signal to represent digital data.
The bandwidth requirement, B is:
B = (1+d)xS
In QPSK, we parallelise the bit stream so that every two incoming bits are split up and PSK a carrier frequency. One carrier frequency is phase shifted 90o from the other - in quadrature.
nThe two PSKed signals are then added to produce one of 4 signal elements. L = 4 here.
QPSK and its implementation
Problem Example - 1
Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0.
Solution
For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz.
We vary the phase shift of the carrier signal to represent digital data.
The bandwidth requirement, B is:
B = (1+d)xS
PSK is much more robust than ASK as it is not that vulnerable to noise, which changes amplitude of the signal.
Binary phase shift keying(See Diagram Above)
Implementation of BASK
Quadrature PSK
To increase the bit rate, we can code 2 or more bits onto one signal element.
Binary phase shift keying(See Diagram Above)
Further Reading suggested :
Quadrature PSK
To increase the bit rate, we can code 2 or more bits onto one signal element.
In QPSK, we parallelise the bit stream so that every two incoming bits are split up and PSK a carrier frequency. One carrier frequency is phase shifted 90o from the other - in quadrature.
nThe two PSKed signals are then added to produce one of 4 signal elements. L = 4 here.
QPSK and its implementation
Problem Example - 1
Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0.
Solution
For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz.