Average

Relationship Between Peak Power and
Average Power in the xMax System

Chao-Hwa Chen
xG Technology, Inc.
Ft. Lauderdale, FL USA

August 14, 2007

1. Problem Formulation and AssumptionThe current version of the xMax signal is a pulsed signal with the central frequency at 915 MHz. When bit “1” is transmitted, we send a pulse to represent “1.” When “0” is transmitted, we do not senda pulse at all.

Since the xMax signal is a pulsed signal, the peak power is not the same as the average power. In the continuous wave (CW) system, the average power is equal to the peak power.The purpose of this document is to find the relationship between peak power and average power. For simplicity, we assume that the pulse model is a sine pulse model, which is defined as

(1)where fc is the central frequency and Tp is the pulse duration. and the multiplication between fc and (2Tp) is an integer. Figure 1 shows the example of the sine pulse model with A = 1, fc = 50and Tp = 1.

Since the peak amplitude is A, the peak power Ppeak = A2/2.


Figure 1: the sine pulse model, where A = 1 and Tp = 1 sec.

2. The Average Power Derivation

Before we derivethe average power, we derive the energy of the sine pulse p(t) first.

(2)

Therefore, the average bit energy Eb is derived as

(3)

Then the average power Pavg can be derived as

,(4)

where Rb is the bit rate, Tb is the bit duration and  is the duty cycle which is equal to .

Therefore, the ratio between peak power over average power can be described as

. (5)Assume that the xMax signal’s pulse width Tp is 40 ns, Table 1 shows different duty cycles and ratios between peak power and average power with different bit rates based on Equation (5).

Rb (inMbps)  Ppeak/Pavg (in dB)
1.65 0.066 17.8
2.88 0.1152 15.41
3.5 0.14 14.56
5 0.2 13.01
7.5 0.3 11.25
10 0.4 10.0
15 0.6 8.24
17 0.68 7.70
Table 1: duty cycles and the ratios between peak...