EC312 Lesson 16: Propagation
Objectives:
(a) Define reflection, refraction and diffraction.
(b) Describe the characteristics of ground waves, sky waves, and space waves.
(c) Calculate the radio horizon distance for space waves based on antenna height.
(d) Compute received power level for a communication system using Friis Free Space equation.
I. Wireless Propagation
Propagation is the study of how to get a signal from Point A to Point B. It sounds simple, but it is THE most
fundamental aspect of wireless communications. The wired guys don’t really have to worry about propagation
per se, however, wireless transmission requires a fundamental understanding of how electromagnetic waves
move through the atmosphere.
1. Two categories of propagation
In general, we can think about radio frequency propagation in two broad categories: Large Scale and Small
Scale. Our emphasis in EC312 is on understanding Large Scale propagation, although many engineers have
devoted their entire careers to understanding and modeling Small Scale propagation. In fact, you’ve likely
experienced Small Scale propagation numerous times without ever realizing. The classic example would be
driving down the highway while talking on your cell phone and experiencing a dropped call.
Large Scale Propagation – The behavior of the radio channel over large distances (100’s or 1000’s of λ’s).
Received power is directly related to distance between Tx and Rx, and is stationary with respect
to time.
Small Scale Propagation – The behavior of the radio channel over a small local area (1-10 λ’s) and/or small
time durations. Received power fluctuates rapidly based on position, speed, direction of travel,
etc. of the mobile.
Before we deep dive into all the glories of Large Scale Propagation in real-world environments, it behooves us
to look at the most basic way we can transmit energy from Point A to Point B in an environment devoid of
terrain, mountains, buildings, ground, or atmosphere. Such an environment is known as Free Space, and
conveniently, wireless propagation in such an environment is known as Free Space Propagation.
2. Free Space Propagation
Let’s consider the following scenario. You have a brand-new iPhone (or Samsung phone as the case may be),
have just signed up for a super-fast LTE plan, and would like to upload a photo, surf the web, browse facebook,
or just plain make a phone call. To make that happen, your phone has to transmit that information over the air
to the nearest LTE cell tower (cost: $5 Million, that’s why your phone bill is $100/month), which happens to be
5 miles away.
Question: Will your signal make it to the tower and will it have sufficient power to “close the link” and allow
you communicate? Or will you suffer the fate of a cellular “dead zone”?
Recall from our previous lecture what happens when we transmit energy out an antenna:
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Presume I have a wireless transmitter which is going to output some power: pt. I’m going to connect the output
of the transmitter an antenna:
Now, if I had a perfect isotropic antenna, it would radiate that energy equally in all directions, and create a
spherical wavefront: it is completely unfocused
Now, a perfect isotropic antenna really doesn’t exist. All real antennas will concentrate the energy of the wave
in one or more directions:
Because we’re radiating the same amount of power, but in a smaller area, we say that the antenna has a gain
associated with it, due to the fact that it’s concentrating the energy in a smaller area of space. Another way of
thinking about this is a ratio of energy densities:
gt 
energy density antenna boresight
energy density isotropic sphere
We defined something in the previous lesson called the effective radiated power (ERP or EIRP), which is the
gain of an antenna (with respect to an isotropic radiator) multiplied by its input power. This represents the
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amount of power that the communication system would have produced if the antenna were a perfect isotropic
radiator.
EIRP  input power × antenna gain
EIRP  pt gt
Note That: The Antenna Gain/EIRP equation is only valid when we’re in the far field of the antenna, or a
distance away of:
d ff 
D
λ
2D2

is the largest physical dimension of the antenna [meters]
is the wavelength [meters]
Note Also That: A lot of people believe very strongly that the Far Field starts at a distance of 10 wavelengths
(10λ) away. That is a rough rule-of-thumb that you can use to be safe or as an approximation, but the real farfield distance is based on the above equation.
To propagate the energy away from the transmitter, we need to enlarge our sphere—just like blowing up a
balloon. We’re going to keep blowing it up larger and larger, until it intersects with our receiver.
Now, just like a balloon, as we keep blowing it up, the total energy is going to remain constant, however, the
energy density is going to decrease. Our density here is going to be based on the surface area of a sphere:
Asphere  4 d 2
d
Is the distance between transmitter and receiver [meters]
Thus, our energy density at the receiver is:
pg
EIRP
 t t2
2
4 d
4 d
Is the power density in W/m2
pd 
pd
Now, the last thing we need is to turn that power density into an actual received power. So what we need is a
“catcher’s mitt” to catch all of that energy for us. Luckily, we call that an antenna. The physically larger the
antenna, the more of that energy it can “catch”. That’s why long-range antennas are so large – the energy
they’re collecting is spread out over an incredibly large area.
Thus, the problem of determining the received power is just a geometry problem: how much of the power
density intersects with the effective area of the receive antenna.
The Effective Area for an antenna is conveniently defined as:
gr  2
Ae 
4
3
λ
gr
is the wavelength [meters]
Is the gain of the receive antenna
Finally, we can put all this together and find out what our received power is going to be:
prec 
prec
pt
gt
gr
λ
d
pt gt g r  2
EIRP
A

e
2
4 d 2
 4 d 
Is the received power [Watts]
Is the trasmit power [Watts]
Is the gain of the transmit antenna [linear units]
Is the gain of the receive antenna [linear units]
Is the wavelength [meters]
Is the distance between transmitter and receiver [meters]
This is known as the Friis Free Space Equation. It is fundamental to understanding how power falls of as a
function of distance for wireless communications.
Two most common mistakes: Using dB values instead of linear values, and failing to get the
wavelength/distance units correct.
Let’s go back to our example.
Example
Your cell phone transmits at a power level of 500 mW, and an antenna gain of 2.0 dB. The cell tower has an
antenna gain of 8.0 dB, and is a distance of 5 miles away. For LTE, you’re transmitting at 700 MHz.
Will your signal make it to the tower and will it have sufficient power to “close the link” and allow you
communicate? Or will you suffer the fate of a cellular “dead zone”?
Note that: The Friis Free Space equation is technically only valid for Free Space environments (although many
situations will mimic free space). So the question is: what happens when we add back in the mountains, the
buildings, the earth, and the atmosphere?
II. Large Scale Propagation – Three Modes
Large scale propagation is composed of three modes listed below in order of their dominance:



Reflection – Energy reflects off large (relative to λ) conductive objects.
Diffraction – Bending of energy around objects.
Scattering – Diffuse re-radiation of energy off rough (smaller than λ) objects.
Let’s look at these briefly one at a time.
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1. Reflection The angle of reflection is equal to the angle of incidence. Reflection introduces a 180˚ phase
shift.
Refraction (with reflection): The radio wave is bent due to the changing index of refraction of the material.
2. Diffraction Diffraction can be explained by Huygen’s principle – all points on a wavefront serve as point
sources for secondary wavefronts:
Consider a Transmitter and Receiver where an object is blocking the direct line-of-sight path between Tx and
Rx. The signal can diffract around the object such that energy can get to the Rx even though it is shadowed.
Note: The more deeply the receiver is shadowed, the lower the received power. At some point, the receiver
won’t be able to receive any signal.
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3. Rough Surface Scattering Sometimes called Diffuse Scattering or Diffuse Reflection, scattering happens
when energy impacts a rough surface and is re-radiated in numerous directions.
Image Source:
http://www.ee.washington.edu/research/ersl/ResearchLinks/WavePropRough.htm
So those are the basic modes of propagation. What happens when we add in a real earth and a real atmosphere?
The earth and the earth’s atmosphere have the greatest impact on signals in the VLF – HF range (3 kHz – 30
MHz). It’s not that the earth and atmosphere don’t affect signals at higher frequencies (in fact, they do!) it’s
just at those higher frequencies start to come into play and dominate the effects of the earth/atmosphere.
Let’s look at what happens to these lower frequencies first before moving on to the higher frequencies.
Radio Band
Frequency
Propagation Via
VLF
Very Low Frequency
3 - 30 kHz
- Guided between the earth and the ionosphere
LF
Low Frequency
30 - 300 kHz
- Guided between the earth and the ionosphere
- Ground Waves
MF
Medium Frequency
300 - 3000 kHz
- Ground waves
- E layer ionospheric refraction at night, when D layer absorption
disappears
HF
High Frequency (Short
Wave)
3 - 30 MHz
- E layer ionospheric refraction
- F layer ionospheric refraction
VHF
Very High Frequency
30 - 300 MHz
- Line-of-sight
UHF
Ultra High Frequency
300 - 3000
MHz
- Line-of-sight
SHF
Super High Frequency
3 - 30 GHz
- Line-of-sight
EHF
Extremely High Frequency
30 - 300 GHz
- Line-of-sight limited by absorption
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For VLF-HF communications, there are two basic modes a radio wave travels from the transmitter to a
receiving antenna:



Ground wave
Space wave
Sky wave
The frequency of the radio wave is the most important factor in determining the performance of each type of
propagation.
4. Ground Wave Propagation A ground wave is a radio wave that travels along the earth’s surface (also
referred to as a surface wave). A ground wave must be vertically polarized.
Lower frequencies travel efficiently as ground waves because they are diffracted by the surface of the earth.
Ground waves thus follow the curvature of the earth and can travel beyond the horizon, for hundreds of miles.
Ground wave propagation is strongest in the Low and Medium frequency ranges. Ground wave propagation
constitutes the main signal path for signals in the frequency range from 30 kHz-3 MHz.
5. Sky Wave Propagation Sky waves are radiated by an antenna into the upper atmosphere where they are
reflected back to earth. This bending is caused by refraction in the ionosphere. The air molecules of the
ionosphere are subject to severe radiation from the sun. Ultraviolet radiation causes the molecules to ionize, or
separate into charged particles, positive and negative ions.
The ionosphere is composed of 3 layers, D, E and F (although F is subdivided into F1 and F2).
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The direction of reflection depends on the angle at which the radio wave enters the atmosphere and the different
degrees of ionization of the layers, as well as the frequency of the transmission.
A receiver that lies between the end of the ground wave propagation and the first sky wave reflection will not
receive the transmitted signal. This region is termed the skip zone.
6. Space Wave Propagation A space wave refers to the radio wave that travels directly in a straight line from
the transmitting antenna (Line-of-sight). These waves are not refracted, and do not follow the curvature of the
earth.
The chief limitation of a space wave is that it is limited to line-of-sight distances. The range of space wave
propagation is limited by the curvature of the earth and height of the antennas above the earth’s surface.
If an antenna is at height h, the distance, d, to the radio horizon (which is the maximum range for space wave
communications from that antenna) is given by the formula
d  2h
In this formula, the height of the antenna is in feet, and the distance to the horizon is in miles.
h
h
In the picture below, where we have one antenna of height t and a second antenna of height r , the maximum
separation at which they can still communicate by line-of-sight is
2h t  2h r
D=
Example
What is the longest line-of-sight communication range between a transmitter whose transmitting antenna is 350
feet high and a receiver whose receiving antenna is 25 feet high?
III. Log-Normal Model
Ok, so that covers the VLF – HF range of the frequency spectrum. Those bands are mostly used for
narrowband, long-distance communication. Most terrestrial wireless communications operate in the “sweet
spot” of the wireless spectrum (aka the “beachfront property”) in the VHF and UHF bands. In this range, the
earth and atmosphere play a far smaller role, and propagation becomes dominated by the specific local
environment.
Let’s consider the following scenario. Suppose we convince the ECE Department to build a cell tower on the
top of Rickover Hall, and you’re driving down McNair Road. The signal you receive will be a combination of
Reflection, Diffraction, and Scattering, as shown in the image below. The problem is that we call it “mobile”
radio for a reason: you want to be able to drive, move about the local environment, and communicate on your
cell phone at the same time.
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Diffr
a
Sc
tte
g
rin
R
ec
e fl
actio
n
n
tio
As you move about the environment, the three propagation modes will have an impact on the instantaneous
received signal in different ways. In the example shown above, you receive a nice strong signal reflected from
Mahan Hall, with a little bit of signal energy coming from diffraction off the back corner of Nimitz Library,
along with some energy scattered by the clock tower. As you move towards Alumni hall, the direct line-ofsight signal to the tower will be blocked, as will most of the strong reflected signals; diffraction is now the
dominant mode. Conversely, if you moved towards Rickover, you would receive a nice strong line-of-sight
signal from the tower, along with a strong reflection from the Northeast side of Nimitz as well as scattering
from all the parked cars in the Triangle Lot.
So the question remains: Using your brand-new iPhone (or Samsung phone as the case may be), will your
signal make it to the tower and will it have sufficient power to “close the link” and allow you communicate? Or
will you suffer the fate of a cellular “dead zone”?
What happens when you put all three major modes of propagation together? How do you create a simple
easy-to-utilize model of the resulting signal power?
Clearly, the Friis Free Space equation is out, and the ground wave/sky wave effects are so small that they can be
neglected. Although numerous sophisticated models exist (and are used to varying degrees in both commercial
and military systems), by far the simplest and most common way to describe propagation in such an
environment is the Log-Normal or Log-Distance model. This model is widely used to not only predict
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coverage for a particular mobile user, but also for predicting the interfering signal power that the mobile user
will experience from other RF sources.
We will use the Log-Normal model in today’s lab to illustrate the short-comings of Friis Free Space, but this
material is beyond the scope of the course. You can look at Appendix A at the end of these notes to see an
explanation of the Log-Normal or Log-Distance model or to assist you in your lab.
Chris Anderson, LCDR Jesse Atwood, LCDR Jennie Wood, CDR Hewitt Hymas
Appendix A. Log-Normal Model.
First, a quick definition. We use the term Path Loss to denote the received signal power at a specific
transmitter-receiver distance relative to the transmit power, or mathematically:
PL  d   PT  PR  d 
PT
PR  d 
d
dB
Is the transmitted power [dBm]
Is the received power [dBm]
Is the distance between transmitter and receiver [meters]
Over the years, wireless engineers have observed that Average Path Loss for a particular environment generally
follows a d n relationship, where the variable n is known as the Path Loss Exponent, and is specific to that
environment. Researchers have also observed that when they made numerous measurements at a specific
distance (but in different local environments), the variation in received signal power obeyed a “bell curve”
distribution about the local mean (the “bell curve is formally known as a “Normal” or “Gaussian” distribution).
Plotted on a log scale, the results look something like this:
PL
(dB)
Average
Path Loss
Gaussian distribution of
Path Loss about that
particular distance
Distance Dependent Mean
1m
10m
100m
1km
10
10 km
Dist
(m)
We call this Log-Normal Path Loss. Average Path Loss obeys a linear relationship (straight line) on a log
scale, and the variation in received power at that distance follows a normal distribution. The slope of the line is
the Path Loss Exponent, and is determined experimentally for the particular scenario of interest.
Mathematically Log-Normal Path Loss is given by:
PL  d 
PL  d0 
n
d 
PL  d   PL  d0   10n log10  
 d0 
Is the path loss at a distance d [dB]
Is the path loss at some close-in reference distance d0 (d0 << d).
Can either be measured or calculated with the Friis Free Space equation.
Is the path loss exponent
Note That: Antenna gains, wavelength, etc. are embedded in the model ( PL  d0  , and n) parameters.
Changing the configuration means we will end up with different model parameters and different results.
A couple of Real World Examples:
Path Loss in a medium-dense forest environment at 2.7 GHz using omnidirectional antennas. Path Loss is
referenced to d0  3m .
C. R. Anderson, H. I. Volos, and R. M. Buehrer, “Characterization of low antenna ultra wideband propagation in a forest
environment,” IEEE Transactions on Vehicular Technology, Vol. 62, No. 7, pp. 2878-2895, September 2013.
Path Loss in five German cities for a cellular network at 900 MHz using omnidirectional antennas. Path Loss is
referenced to d0  100m
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S. Y. Seidel, et. al., ”Path loss, scattering and multipath delay statistics in four European cities for digital cellular and
microcellular radiotelephone,” IEEE Transactions on Vehicular Technology, Vol. 40, No. 4, pp. 721-730, November 1991.
Values for Path Loss Exponents have been tabulated for a number of environments, a few representative ones
are given here.
Path Loss Exponents for Different Environments at 900 MHz
Environment
Urban Area
Dense Urban Area
In Building with Line-of-Sight
In Building Obstructed
Factory Floor Obstructed
Retail Stores
Path Loss Exponent
2.7 to 3.5
3 to 5
1.6 to 1.8
4 to 6
2 to 3
1.8 to 2.5
Source: Wireless Communications: Principles and Practice, 2nd Edition by Theodore S. Rappaport.
So after all that discussion, we still haven’t answered the question:
Example
Using your brand-new iPhone (or Samsung phone as the case may be), will your signal make it to the tower and
will it have sufficient power to “close the link” and allow you communicate? Or will you suffer the fate of a
cellular “dead zone”?
Recall: Your cell phone transmits at a power level of 500 mW, and an antenna gain of 2.0 dB. The ECE
Department’s cell tower has an antenna gain of 8.0 dB.
Let’s assume you’re at the entrance to Gate 8, which would put you approximately 1.0 km away from the tower.
From the table above, the USNA campus most closely matches “Urban Area”, so let’s use a Path Loss Exponent
that’s exactly in the middle of 2.7 and 3.5, or 3.1.
First, we need to calculate the path loss at a close in reference distance. The choice is technically arbitrary, but
is typically 1 meter, as it makes the math much easier to work with. If you can believe that 1 meter away from
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your phone you have Free Space conditions (but double-check the Far Field Distance!), we can use the Friis
Free Space equation to calculate the path loss:
 0.5W 1.585  6.310  0.428m   5.80 103 W
prec  d 0  1m  

2
2
 4 d 
 4 1m  
PL  d 0  1m   PT  PR  d 0  1m   27 dBm  7.6 dBm  19.4 dB
2
pt gt g r  2
From there, we can plug and chug into our model:
d 
PL  d   PL  d 0   10n log10  
 d0 
 1000m 
PL  d  1000m   19.4dB  10  3.1 log10 

 1m 
PL  d  1000m   112.4 dB
And the power received at the cell tower is:
PL  d  1000m   PT  PR  d  1000m 
PR  d  1000m   PT  PL  d  1000m 
PR  d  1000m   27 dBm  112.4 dB
PR  d  1000m   85.4 dBm
Note That: This is actually weaker (by 15 dB, or a factor of 30) than the received power at a distance of 5
miles (8 km) that was predicted by the Friis Free Space equation for the exact same scenario. This illustrates
the danger of using the Friis equation in scenarios that are not free-space.
Incidentally, a received signal power of -85.4 dBm is still sufficient to “close the link” and communicate with
the tower.
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