The total distance between the transmitting and receiving antenna of a
microwave link at 10GHz, is 30 Km. the height of the Tx antenna is
above ground level is 20 m. the maximum acceptable total path loss is
169 dB. Furthermore there is hill located 10 km away from the
transmitter antenna with a height of 80m.

calculate the height of the receiver antenna for the path loss to be
just equal to the maximum acceptable value?

wrote in message
oups.com...
The total distance between the transmitting and receiving antenna of a
microwave link at 10GHz, is 30 Km. the height of the Tx antenna is
above ground level is 20 m. the maximum acceptable total path loss is
169 dB. Furthermore there is hill located 10 km away from the
transmitter antenna with a height of 80m.

calculate the height of the receiver antenna for the path loss to be
just equal to the maximum acceptable value?

1. Antenna essentially operates in free space with no near field ground
losses because the wavelength is extremely small (3.3cm) compared to 20m
antenna height at the transmitter.

2. The effect of the 80m hill 10Km away is negligible. The arc tan is only
..008 degrees, thus the transmitter hardly "sees" it.

3. The path loss seen by an atenna at 0 feet is then 131.8 dB (Path Loss =
20log(4*pi*d/lambda)), which is much less than the desired 169dB maximum.

Answer: 0 Feet.

"Stefan Wolfe" wrote in message
...

wrote in message
oups.com...
The total distance between the transmitting and receiving antenna of a
microwave link at 10GHz, is 30 Km. the height of the Tx antenna is
above ground level is 20 m. the maximum acceptable total path loss is
169 dB. Furthermore there is hill located 10 km away from the
transmitter antenna with a height of 80m.

calculate the height of the receiver antenna for the path loss to be
just equal to the maximum acceptable value?

1. Antenna essentially operates in free space with no near field ground
losses because the wavelength is extremely small (3.3cm) compared to 20m
antenna height at the transmitter.

2. The effect of the 80m hill 10Km away is negligible. The arc tan is only
.008 degrees, thus the transmitter hardly "sees" it.

3. The path loss seen by an atenna at 0 feet is then 131.8 dB (Path Loss =
20log(4*pi*d/lambda)), which is much less than the desired 169dB maximum.

Answer: 0 Feet.

Oops, that's a path loss of 141.16dB. Same answer, however.

On Tue, 6 Feb 2007 22:35:04 -0500, "Stefan Wolfe"
wrote:

2. The effect of the 80m hill 10Km away is negligible. The arc tan is only
.008 degrees, thus the transmitter hardly "sees" it.

Actually, Stefan, the transmitter cannot see through it at all.

Answer: 0 Feet.

Our Lincolnshire student needs to think back to simple trigonometry to
answer this. The path loss, once the height of the hill is made
irrelevant, is sufficient as Stefan points out.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Tue, 6 Feb 2007 22:35:04 -0500, "Stefan Wolfe"
wrote:

2. The effect of the 80m hill 10Km away is negligible. The arc tan is only
.008 degrees, thus the transmitter hardly "sees" it.

Actually, Stefan, the transmitter cannot see through it at all.

Answer: 0 Feet.

Our Lincolnshire student needs to think back to simple trigonometry to
answer this. The path loss, once the height of the hill is made
irrelevant, is sufficient as Stefan points out.

73's
Richard Clark, KB7QHC
The only way I could figure the hill in is if one is trying to Knife-edge
over the hill. Otherwise the hill is totally irrelevant to the problem.

"Richard Clark" wrote in message
...
On Tue, 6 Feb 2007 22:35:04 -0500, "Stefan Wolfe"
wrote:

2. The effect of the 80m hill 10Km away is negligible. The arc tan is only
.008 degrees, thus the transmitter hardly "sees" it.

Actually, Stefan, the transmitter cannot see through it at all.

Exactly . And taking into account the 15 degree bend of the radio horizon
(even at 10 GHz) vs the arctan of .008 degrees for the the hill, reduced
somewhat by the 20 foot transmitter tower to .006 degrees, the hill itself
is invisible at a 30km far field. The bend of the propagating waves which
extends the radio horizon clearly mitigates any possible effects of the 80m
hill.

On Wed, 7 Feb 2007 00:08:19 -0500, "Stefan Wolfe"
wrote:


"Richard Clark" wrote in message
.. .
On Tue, 6 Feb 2007 22:35:04 -0500, "Stefan Wolfe"
wrote:

2. The effect of the 80m hill 10Km away is negligible. The arc tan is only
.008 degrees, thus the transmitter hardly "sees" it.

Actually, Stefan, the transmitter cannot see through it at all.

Exactly . And taking into account the 15 degree bend of the radio horizon
(even at 10 GHz) vs the arctan of .008 degrees for the the hill, reduced
somewhat by the 20 foot transmitter tower to .006 degrees, the hill itself
is invisible at a 30km far field. The bend of the propagating waves which
extends the radio horizon clearly mitigates any possible effects of the 80m
hill.


Hi Stefan,

You seem to be simultaneously agreeing and disagreeing. The hill is
either in the way, or it is not. It is in the way. To count on an
intermediary, such as suggested by Jimmie, knife-edge propagation or
bend of the waves, is probably not in our student's syllabus. Besides,
I have seen neither you nor Jimmie offer the attenuation presented by
such refractions (and the attenuation is not marginal). Without
quantifiables, the path budget cannot be calculated.

The problem, as stated, has a clear answer in looking over the hill by
raising one antenna, the question informs us that is the answer and
that is simply resolved with trig (albeit, including the radius of
earth and accounting for its curvature).

73's
Richard Clark, KB7QHC

Richard Clark wrote:

Hi Stefan,

You seem to be simultaneously agreeing and disagreeing. The hill is
either in the way, or it is not. It is in the way. To count on an
intermediary, such as suggested by Jimmie, knife-edge propagation or
bend of the waves, is probably not in our student's syllabus. Besides,
I have seen neither you nor Jimmie offer the attenuation presented by
such refractions (and the attenuation is not marginal). Without
quantifiables, the path budget cannot be calculated.

The problem, as stated, has a clear answer in looking over the hill by
raising one antenna, the question informs us that is the answer and
that is simply resolved with trig (albeit, including the radius of
earth and accounting for its curvature).

73's
Richard Clark, KB7QHC

Might I recommend a program misleadlingly called Radio Mobile. This
piece of software will all let you check many real world situations.
This program is not real easy to use, it's a lot worse than Windows or
Office (ok, not worse than Office), but it is worth learning unlike the
previous 2 mentioned.

The site you need for the software is

http://www.cplus.org/rmw/english1.html

and then you need to download a large amount of terrain data, which is
freely available from NASA. It looks like the way this is handled has
changed since I did it last, so I can't comment on how it is done now.

tom
K0TAR

"Richard Clark" wrote in message
...
On Wed, 7 Feb 2007 00:08:19 -0500, "Stefan Wolfe"
wrote:


"Richard Clark" wrote in message
. ..
On Tue, 6 Feb 2007 22:35:04 -0500, "Stefan Wolfe"
wrote:

2. The effect of the 80m hill 10Km away is negligible. The arc tan is
only
.008 degrees, thus the transmitter hardly "sees" it.

Actually, Stefan, the transmitter cannot see through it at all.

Exactly . And taking into account the 15 degree bend of the radio horizon
(even at 10 GHz) vs the arctan of .008 degrees for the the hill, reduced
somewhat by the 20 foot transmitter tower to .006 degrees, the hill
itself
is invisible at a 30km far field. The bend of the propagating waves which
extends the radio horizon clearly mitigates any possible effects of the
80m
hill.


Hi Stefan,

You seem to be simultaneously agreeing and disagreeing. The hill is
either in the way, or it is not. It is in the way. To count on an
intermediary, such as suggested by Jimmie, knife-edge propagation or
bend of the waves, is probably not in our student's syllabus. Besides,
I have seen neither you nor Jimmie offer the attenuation presented by
such refractions (and the attenuation is not marginal). Without
quantifiables, the path budget cannot be calculated.

The problem, as stated, has a clear answer in looking over the hill by
raising one antenna, the question informs us that is the answer and
that is simply resolved with trig (albeit, including the radius of
earth and accounting for its curvature).

73's
Richard Clark, KB7QHC

There are lots of programs out here for calculating path loss for LOS
situations but figuring in the path via knife edge defraction is something I
have always tried to avoid.. In real life there are too many variables that
effect this and you could have a signal that would tend to fade. Over the
years I have forgotten or lost interest in figuring the impractical.

Jimmie

The total distance between the transmitting and receiving antenna of a
microwave link at 10GHz, is 30 Km. the height of the Tx antenna is
above ground level is 20 m. the maximum acceptable total path loss is
169 dB. Furthermore there is hill located 10 km away from the
transmitter antenna with a height of 80m.

calculate the height of the receiver antenna for the path loss to be
just equal to the maximum acceptable value?

1. Antenna essentially operates in free space with no near field ground
losses because the wavelength is extremely small (3.3cm) compared to 20m
antenna height at the transmitter.

2. The effect of the 80m hill 10Km away is negligible. The arc tan is only
.008 degrees, thus the transmitter hardly "sees" it.

3. The path loss seen by an atenna at 0 feet is then 131.8 dB (Path Loss =
20log(4*pi*d/lambda)), which is much less than the desired 169dB maximum.

Answer: 0 Feet.

Definitely NOT 0 feet.

Even without the hill in the way the curvature of the earth means that the
radio horizon is at about 20km with the tx on a 20m mast. 0.6 Fresnel
clearance occurs at about 6km. and the path loss exceeds 180dB.!!
The Rx mast needs to be at about 25m to obtain a 0.6 fresnel zone clearance,
WITHOUT a hill.This would give about 150dB path loss.

Adding the 80m hill at 10km gives a single knife edge diffraction, that
increases the path loss enormously to about 200dB!!

This path loss does not change significantly until the Rx antenna height
gets so large that near line of sight is achieved. That is over 100m!! To
obtain the 169dB required figure the mast height would have to be about
250m!!!!!!!

Regards
Jeff