# [Solved] Simulation of received signal strength for Half Wave dipole

Discussion in 'Wireless & RF Design' started by abhaymv, Mar 27, 2016.

1. ### abhaymv Thread Starter Active Member

Aug 6, 2011
105
5
Hi,

This question has to do with antenna theory, particularly that of a half wave dipole. I'm coding this with MATLAB as I'm comfortable with this (no toolboxes used) but this is not a programming question.
In my simulation, I think the received signal strength I get is way weaker than what I'm supposed to get. I want to check if this error has to do with the way I have interpreted certain formulae used for half wave dipoles.
My reference is "Antenna Theory" by C. A. Balanis.

I'm interested in simulating the signal strength (in dBm) for a half wave dipole antenna. I am considering in particular the antenna used in MICAz wireless measurement system, whose datasheet can be found here.
Some relevant parameters from the datasheet I've used in the simulation:
Maximum outdoor range is given as 75 m to 100 m
Maximum transmit power: 0 dBm. (Datasheet says RF power).
Frequency of operation: 2.48 GHz

I'm simulating the received signal strength in an area around the antenna (A sixty meter by sixty meter area with the antenna placed in the center, so its 30 meters to each extremes). I am not plotting a radiation pattern. The received signal strength in the plane with the maximum directivity is plotted as a surface plot.

My approach is to first find average power density (time averaged Poynting vector pg 36, eqn 2-8) as a function of distance from the antenna using the equation:
$W_{av} = \eta \times \frac{|I_{0}|^2}{8 \pi^2 r^{2}}sin^3(\theta)$.
(Ref: Section 4.6, pg 162 Half Wavelength dipole)
Here angle theta is 90° in the direction of maximum directivity.
I found I0 using the equation for radiated power, which is known as 0 dBm or 1 mW.
$P_{rad} = \eta \frac{|I_{0}|^2}{8 \pi} C_{in}(2 \pi)$

The text gives the value of Cin(2pi) as 2.435. Substituting for Prad, I get I0 = 5.2 mA. This can be substituted in equation for $W_{av}$ and we can find average power density for all desired values of r. I used a distance matrix R that calculates r for all (x,y) in the plane within -30 m <x < 30 m and -30m < y < 30m with a resolution of about 3 cm. So its a 2000 X 2000 matrix. Now I have Wav for each of these points.

According to my understanding, Received signal strength in watts will be Wav multiplied by maximum effective area of the half wave dipole, given as $0.13 \lambda^2$. The device is operated under frequency of operation 2.48 GHz, so lambda is around 0.1209 m. Thus I get the received signal strength.
In dBm, Pr (dBm) = 10*log10(Pr).

The problem: According to my simulation, received signal strength falls below -90 dBm within 30 meters distance from the transmit antenna. But, the datasheet clearly specifies an outdoor range of 75 to 100m, thus I should only get -90 dBm at those distances.

Also: In a paper based on this device, it says that the received power at 30 m should be -55 dBm, but I'm getting -95 dBm. However, I don't know if this is 100% correct.

Have I gone wrong anywhere in the calculations?

2. ### abhaymv Thread Starter Active Member

Aug 6, 2011
105
5
Problem solved. Was a silly mistake as I had calculated dBW rather than dBm, leading to the difference. Thread may be closed/deleted. i don't see an option for doing this myself.