isotropic radiation patter equation 1

[systemseng]

I'm trying to plot an ideal isotropic radiation pattern (meters vs dB). So far, I've managed to find some equations in order to do this...

ReceivePower(RxP) = TransmitPower(TxP) - FreeSpaceLoss(FSL)

*Assuming that there is no loss internally within the receiver or transmitter, and that there is no gain between either of the systems.

TxP = InputPower / 4*Pie*Radius^2
FSP = -27.55 dB + 20*log[frequency(MHz)] + 20*log[distance(m)]

Given the transmit frequency, the only variable would be the distance from the signal origin.

The isotropic pattern should be a sphere, but it seems that the 1/R^2 is heavily influencing my graph. So instead of a steady decline in power, I immediately see a steep drop as the power goes to zero as distance increases.

Are these the equations I should be using to calculate the propagation data?

 

[VEBill]

"...4*Pie..." Yummie.

By the way, tomorrow is 'Pi Day' (3/14). Happy Pi Day.

The number I learned was 32.25 (instead of your 27.55), but it depends on your units.

Wiki link:

http://en.wikipedia.org/wiki/Free-space_loss

It's only a sphere in 3-D, and even then only for a given power density value. On a flat graph it is a square-law decline. Compared to short distance (near field), the power does actually approach zero (essentially) fairly quickly depending on your scales. In other words, adjust your X and Y scales.

Hint: try a log scale on the power (amplitude) axis.

 

[biff44]

Some say Pi are squared, but I always say Pie are round!

 

[systemseng]

Sorry my mistake (pi). I'm assuming that when I use the TxP and FSL equations, the radius and distance (for the respective equations) are the same variable. Given these parameters, I am seeing a 1/R^2 decrease in power as the distance increases. So I guess I am plotting the correct 2D graph (distance vs power).

Just so that I have this straight, the isotropic 3d graphs that I'm viewing (that show a sphere) only demonstrate distribution of the signal and not the distribution of power? (

http://www.hp.com/rnd/images/pdf_html/antennas_figure6.jpg)

When I graph my isotropic signal, it in a 3d space, it looks more like rotating a 1/R^2 function around the y-axis. Hopefully I'm seeing this properly.

Lastly, I'm not entirely sure what you meant by trying a log scale on the amplitude axis, but I am unable to take the log of most of those numbers since, most of those consist of negative values. I appreciate all the help.

 

[VEBill]

To make a spherical shape in 3-D, pick a value and note that the locations where that value occurs forms the surface of a sphere (for such a perfectly isotropic source).

To plot all values in 3-D you need to introduce colour as the fourth axis.

If your path loss is giving negative values (which is fine), then assign an absolute power (say 1 watt) to the source and calculate the smaller-but-always-positive power levels all the way out. Then you can convert the scale to log if it helps.

(If you still find negative power, then the Universe will implode shortly anyway... ;-) )

 

[Higgler]

Simple loss formulas are also usually a far field calculation. I realized that one day when I was working at lower frequencies, like 100 Mhz. my answers were all wrong in my antenna chamber.
If you get very close to an antenna, the simple formulas in the antenna near field don't work. Hence, beware if for exampley you want results from 0.1" to 1" from an antenna that's radiating at a wavelength of 4 inches.

kch

 

[logbook]

This is all a bit backwards. If you radiate 1W isotropically you get 1W spread over the sphere at all distances. That's it.

You had it in your first equation for TxP, although that formula gives power flux density at a distance from the antenna (watts per metre squared).