Question

Problem 2: In this problem we explore the power penalty involved in going to higher level signal modulations, i.e. from QPSK to 16PSK. (a) Find the minimum distance between constellation points in 16PSK modulation as a function of signal energy Es. (b) Find aM and BM such that the symbol error probability of 16PSK in AWGN is approximately (c) Using your expression in part (b), find an approximation for the average symbol error probability of 16PSK in Rayleigh fading in terms of (d) Convert the expressions for average symbol error probability of 16PSK in Rayleigh fading to expressions for average bit error probability assuming Gray coding. (e) Find the approximate value of required to obtain a BER of 10A-3 in Rayleigh fading for QPSK and 16PSK (assuming Gray coding). What is the power penalty in going to the higher level signal constellation at this BER?

Answer 1

Let us start by defining what is PSK. PSK(Phase Shift Keying) is a digital modulation process which transfers data by modulating the phase of the carrier wave. A two dimensional representation of the signal produced by the PSK technique is called as a constellation diagram and individual points represented in the diagram are called as constellation points. These points are basically a combination of the amplitude and phase of the carrier signal wave and represent a symbol. The distance of each point from the origin represents the power or amplitude of the signal.

a) In a constellation diagram, the constellation points are located in approximately a circle around the origin and the distance between two points can be represented by the formula:

$d_{16psk}=2\sqrt{E_{s}}\sin (\frac{\pi }{M})$

where M=16 for 16 PSK signal, and E_s is the energy of the signal

substituting the value in the above equation,we get:

l6psk 16

=>

d16psk = 2V E, * 0.195

=>

d16psk = 0.390 VE,