where λ = 300/fMHz
π = Pi or 3.14159
d = Distance or range in meters
If you want to know the range,
you can rearrange the equation:
d2 = (PtGtGrλ2) / (16π2Pr)
d = √ (PtGtGrλ2) / (16π2Pr)
Don’t let this seemingly complex
equation bother you. It is just a bunch
of values that you multiply together and
divide by. You do have to square a couple values and use pi and maybe take a
square root, but that is easy if you have
a basic scientific calculator. The harder
part is finding out what the various
individual values are to work with.
The first thing to note is that in the
RF and wireless world, we work with the
power of a signal instead of voltage. As
you can see, the received power is directly proportional to the transmit power.
Next, consider the antenna gain G.
This is the numerical power gain, NOT
expressed in decibels (dB). All practical
antennas have some gain because they
exhibit directional characteristics. The
ideal theoretical antenna is called an
isotropic radiator and is an RF point
source that radiates equally well in all
directions creating a spherical antenna
radiation pattern. Real antennas don’t
have that kind of radiation pattern as
they tend to radiate better in some
directions rather than others.
For most applications, using a plain
old half wave dipole or quarter wave
ground plane antenna, the power gain
is 1.64. Other antennas (like a Yagi or
collinear) have much greater gains. And
remember that the gain is not really an
actual multiplication in power level.
Instead, the gain comes strictly by way
of focusing or concentrating the available power into a narrower beam. This
gives you what is called the effective
radiated power (ERP). In wireless applications, such gain is just as effective as
increasing the actual transmit power.
Another important thing to note is
that the wavelength of the signal is of
major importance. The longer the wavelength, the greater the range. Recall that
wavelength is related to frequency by
the simple expression given earlier: λ =
300/fMHz. What this tells you is that
signals at the lower frequencies travel
farther than at the higher frequency
signals. Over the years, we have pushed
upward into the UHF and microwave
spectrum to get more space for new
radio signals and services. There is lots
of room in the microwave bands, but as
the frequency gets higher, the range
becomes much less by the square of
the wavelength. That is a major factor.
A good example is that cell phone
sites or base stations operating in the
800 to 900 MHz range have a much
greater area of coverage that those operating in the 1800 to 1900 MHz range.
You have to use more cell sites at the
higher frequencies to get the same
coverage. It is much more expensive. Of
course, you can overcome that limitation
by increasing transmit power or increasing either or both antenna gains and
heights. But there are regulations and
other limits that may thwart those plans.
One final thing. This formula is only
good for ground and direct or space
wave propagation. Ground waves are
radio signals that hug the earth and are
the main wave used in low frequency
operation (< 3 MHz). Think AM radio.
Space waves are the direct line-of-sight
(LOS) waves characteristic of VHF, UHF,
and microwaves. Think TV or two-way
radio. The formula does not work accurately for sky waves that are refracted
from the ionosphere for long distances.
These waves are the so-called short
waves in the 3 to 30 MHz range.
As you can see from the basic
formula, the signal loss is inversely
proportional to the square of the distance. Doubling the distance quadruples the loss. And there is a way to
find what that path loss is in dB.
dB loss = 37 dB + 20 log(f) + 20 log(d)
In this formula, f is the frequency in
MHz and d is the range or distance in
miles. Just use your scientific calculator
to find the logarithms. And don’t forget,
we are dealing with power loss here.
The formula assumes isotropic
antennas so the result is usually the
worst case. An example calculation
PATH LOSS CALCULATION
What is the path loss for a 900 MHz
signal over a range of four miles?
dB loss = 37 dB + 20 log(f) + 20 log(d)
dB loss = 37 dB + 20 log(900) + 20 log( 4)
dB loss = 37 + 59 + 12 = 108
This would be expressed as -108 dB.
Call it Ploss.
is given in the Path Loss Calculation
THE IMPORTANCE OF
One thing not shown in any of the
formulas is the receiver sensitivity.
Sensitivity is a measure of how much
gain the receiver has ... that is how
much it amplifies small signals. It is a
common specification in most wireless
transceiver chips, boards, and modules.
And wow, is it important. Like the transmit power, it is the key to establishing a
solid wireless link between two points.
The combination of transmit power,
path loss, and receiver sensitivity are
related as this formula shows.
Pr = Pt - Ploss
In this formula, Pr is the received
power and Pt is the transmit power
both given in dBm, or decibels referenced to one millliwatt. You calculate
Pt with the expression:
Pt (dBm) = 10 log (Pt/0.001)
For example, assume a transmit
power of 600 milliwatts (0.6 watt). In
dBm, this is:
Pt (dBm) = 10 log (600) = 27. 8 dBm
Now you can figure the received
power. Assume the path loss of 108
dB as calculated earlier. The received
power then is:
Pr = 27. 8 - 108 = - 80. 2 dBm
What this tells you is with the
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