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antenna hight
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? |
antenna hight
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. |
antenna hight
"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. |
antenna hight
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 |
antenna hight
"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. |
antenna hight
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antenna hight
"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. |
antenna hight
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 |
antenna hight
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 |
antenna hight
hello
Jeff i am very thank full to your help, i have got Rx height 101.2 Regards naqvi |
antenna hight
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 (Tx) 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? Whats the height of the receive antenna? (required) Is the hill directly in the line of sight of the receive antenna? yes it is b/w the LOS |
antenna hight
wrote
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? _____________ The height above mean sea level of the tx and rx sites, and the terrain profile for the path would be necessary to answer this ~ accurately. But for a smooth earth model, the graphic at the link below will give some insight. It shows that a height of around 270 meters would be needed for the receive antenna, using a K-factor of 1.33 and 0.6 fresnel clearance for an 80 m hill 10 km downrange. The path loss then would be about 142 dB. http://i62.photobucket.com/albums/h8.../10GigPath.gif RF |
antenna hight
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antenna hight
wrote in message oups.com... hello Jeff i am very thank full to your help, i have got Rx height 101.2 Regards naqvi Who is right, you or Jeff? You are more than 100% off from each other. In truth, i could receive that signal holding a hand held 10GHz receiver while sitting on the ground. The 80m hill is nothing from an observer 10KM away...only .006 degree from the top of the transmitter tower. It is part of the horizon. I love it when you guys talk like you are sol knowledgeable yet lack the common sense to conceptualize the problem as it really exists. |
antenna hight
wrote in message oups.com... hello Jeff i am very thank full to your help, i have got Rx height 101.2 Regards naqvi Please show your work. |
antenna hight
"Stefan Wolfe" wrote in message ... wrote in message oups.com... hello Jeff i am very thank full to your help, i have got Rx height 101.2 Regards naqvi Please show your work. Hi Stephan Your posts seem to inply that the receive antenna will "see" the 20 meter high transmitting antenna when the receiver antenna is in the shadow of the 80 meter hill. It seems that the receiver needs to be out of the shadow of the hill unless you are able to estimate refraction from the hill. But, your aparent confidance in the statement "0 feet" makes me wonder if I have this problem wrongly analyzed. I have so much confidance in Richard Fry's data that I had accepted his estimation of 270 meters to be as close as you can estimate. Do I misunderstand your post about what minimum height is needed? Jerry |
antenna hight
"Jerry Martes" wrote in message news:iLuyh.5336$384.156@trnddc05... "Stefan Wolfe" wrote in message ... wrote in message oups.com... hello Jeff i am very thank full to your help, i have got Rx height 101.2 Regards naqvi Please show your work. Hi Stephan Your posts seem to inply that the receive antenna will "see" the 20 meter high transmitting antenna when the receiver antenna is in the shadow of the 80 meter hill. It seems that the receiver needs to be out of the shadow of the hill unless you are able to estimate refraction from the hill. But, your aparent confidance in the statement "0 feet" makes me wonder if I have this problem wrongly analyzed. I have so much confidance in Richard Fry's data that I had accepted his estimation of 270 meters to be as close as you can estimate. Do I misunderstand your post about what minimum height is needed? Jerry Hi Jerry, how many shadows have you seen that are 20Km long? |
antenna hight
From: "Stefan Wolfe"
how many shadows have you seen that are 20Km long? ______________ Can we count the shadow of the moon on the earth during a fully-eclipsed sun? If so, that's a bit more than 20 km. RF |
antenna hight
"Stefan Wolfe" wrote in message ... "Jerry Martes" wrote in message news:iLuyh.5336$384.156@trnddc05... "Stefan Wolfe" wrote in message ... wrote in message oups.com... hello Jeff i am very thank full to your help, i have got Rx height 101.2 Regards naqvi Please show your work. Hi Stephan Your posts seem to inply that the receive antenna will "see" the 20 meter high transmitting antenna when the receiver antenna is in the shadow of the 80 meter hill. It seems that the receiver needs to be out of the shadow of the hill unless you are able to estimate refraction from the hill. But, your aparent confidance in the statement "0 feet" makes me wonder if I have this problem wrongly analyzed. I have so much confidance in Richard Fry's data that I had accepted his estimation of 270 meters to be as close as you can estimate. Do I misunderstand your post about what minimum height is needed? Jerry Hi Jerry, how many shadows have you seen that are 20Km long? Hi Stephan Tell me where I have misunderstood the problem. I assumed the transmitting antenna was Lower than the top of the hill. But, you seem to imply that the transmitter can be seen even when the hill is blocking the "view" to it. I have actually never measured a shadow longer that a few feet, but I assumed they continued to exist to infinity when an object blocks them from view. Jerry |
antenna hight
"Jerry Martes" wrote in message news:2Yvyh.37369$5U4.35764@trnddc07... "Stefan Wolfe" wrote in message ... "Jerry Martes" wrote in message news:iLuyh.5336$384.156@trnddc05... "Stefan Wolfe" wrote in message ... wrote in message oups.com... hello Jeff i am very thank full to your help, i have got Rx height 101.2 Regards naqvi Please show your work. Hi Stephan Your posts seem to inply that the receive antenna will "see" the 20 meter high transmitting antenna when the receiver antenna is in the shadow of the 80 meter hill. It seems that the receiver needs to be out of the shadow of the hill unless you are able to estimate refraction from the hill. But, your aparent confidance in the statement "0 feet" makes me wonder if I have this problem wrongly analyzed. I have so much confidance in Richard Fry's data that I had accepted his estimation of 270 meters to be as close as you can estimate. Do I misunderstand your post about what minimum height is needed? Jerry Hi Jerry, how many shadows have you seen that are 20Km long? Hi Stephan Tell me where I have misunderstood the problem. I assumed the transmitting antenna was Lower than the top of the hill. But, you seem to imply that the transmitter can be seen even when the hill is blocking the "view" to it. I have actually never measured a shadow longer that a few feet, but I assumed they continued to exist to infinity when an object blocks them from view. Well you see Jerry, the reason you only see it for a few feet is because the attenuation of the light varies inversely with the distance from the object that blocks the light. I think you have done a good job in making my point. Thanks, |
antenna hight
"Stefan Wolfe" wrote in message ... "Jerry Martes" wrote in message news:2Yvyh.37369$5U4.35764@trnddc07... "Stefan Wolfe" wrote in message ... "Jerry Martes" wrote in message news:iLuyh.5336$384.156@trnddc05... "Stefan Wolfe" wrote in message ... wrote in message oups.com... hello Jeff i am very thank full to your help, i have got Rx height 101.2 Regards naqvi Please show your work. Hi Stephan Your posts seem to inply that the receive antenna will "see" the 20 meter high transmitting antenna when the receiver antenna is in the shadow of the 80 meter hill. It seems that the receiver needs to be out of the shadow of the hill unless you are able to estimate refraction from the hill. But, your aparent confidance in the statement "0 feet" makes me wonder if I have this problem wrongly analyzed. I have so much confidance in Richard Fry's data that I had accepted his estimation of 270 meters to be as close as you can estimate. Do I misunderstand your post about what minimum height is needed? Jerry Hi Jerry, how many shadows have you seen that are 20Km long? Hi Stephan Tell me where I have misunderstood the problem. I assumed the transmitting antenna was Lower than the top of the hill. But, you seem to imply that the transmitter can be seen even when the hill is blocking the "view" to it. I have actually never measured a shadow longer that a few feet, but I assumed they continued to exist to infinity when an object blocks them from view. Well you see Jerry, the reason you only see it for a few feet is because the attenuation of the light varies inversely with the distance from the object that blocks the light. I think you have done a good job in making my point. Thanks, Hi Stephan I now understand now that you do think you can receive 10 GHz signals while the receiver is in the shadow caused by the mountain between you and the transmitter. Do I understand you correctly? Jerry |
antenna hight
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 |
antenna hight
On Wed, 07 Feb 2007 22:24:26 -0600, Tom Ring
wrote: 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. Hi tom, I was involved in Beta testing this. The current version allows dynamic map loading over the Internet, and overlays of Mapquest, Tiger or many other mapping programs available. I've been using it heavily for the last few months and its world of variables allows for finely grained analysis. However, it also allows for massive headaches if all you are looking for is a simple solution. In short, no quantifiables are going to follow from your recommendation - unless I do it. This lil Red Hen isn't interested in cooking that bread. 73's Richard Clark, KB7QHC |
antenna hight
" In truth, i could receive that signal holding a hand held 10GHz receiver while sitting on the ground. The 80m hill is nothing from an observer 10KM away...only .006 degree from the top of the transmitter tower. It is part of the horizon. I love it when you guys talk like you are sol knowledgeable yet lack the common sense to conceptualize the problem as it really exists. So you are saying that you can achieve 30km at 10GHz to a hand held receiver at ground level with a 300 foot hill in the way!!! I am sure that you could not do this with any sensible power even at 2m let alone 10GHz. Perhaps it is your concept of what is going on that is wrong. Have you heard of Fresnel Zones?? When obstructions come within the first Fresnel zone significant attenuation occurs. With the situation that you are describing the path is totally obstructed, with the path only possible due to diffraction from the hill top. The hill top impinges to at least the top of the 5th Fresnel Zone, hence the attenuation is very high. As the height of the Rx antenna increases the attenuation is still very high until the hill top only start to intersect with of the first zone (antenna height~150m). It then drops quite rapidly until there is true line of sight and bottoms out when the hill top is clear of the second zone. You make a great deal of the hill only being 0.006 degree at the horizon. If you plot it accurately and with reference to the Fresnel zones, it does make a big difference. With the Rx antenna at ground level the top of the Fresnel zones are never below the horizon, which is completely different to the situation when the hill is there (-5th Zone obstructed). Also, without the hill you only have to raise the Rx antenna to about 15m to achieve line of sight compared to 200m with the hill there!!!!!! Quite a big difference I think you will agree, and one that your 'conceptualisation doesn't seem to allow for!! 73 Jeff |
antenna hight
"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 |
antenna hight
On 7 Feb, 15:43, "Richard Fry" wrote:
wrote The total distance between the transmitting and receivingantennaof a microwave link at 10GHz, is 30 Km. the height of the Txantennais 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 transmitterantennawith a height of 80m. calculate the height of the receiverantennafor the path loss to be just equal to the maximum acceptable value? _____________ The height above mean sea level of the tx and rx sites, and the terrain profile for the path would be necessary to answer this ~ accurately. But for a smooth earth model, the graphic at the link below will give some insight. It shows that a height of around 270 meters would be needed for the receiveantenna, using a K-factor of 1.33 and 0.6 fresnel clearance for an 80 m hill 10 km downrange. The path loss then would be about 142 dB. http://i62.photobucket.com/albums/h8.../10GigPath.gif RF hi Richard i have read ur response very carefully and its perfect 142dB Path loss but problem is that what's the Rx height if i suppose Tx and Rx install on the same height then my Rx height after the earth buldge and knife e calculation i have got 100m Rx its totally wrong because if i increase that (100m) Rx height its mean get the problem at fresnel Zone need help thanks regard naqvi |
antenna hight
naqvi wrote
i have read ur response very carefully and its perfect 142dB Path loss but problem is that what's the Rx height if i suppose Tx and Rx install on the same height then my Rx height after the earth buldge and knife e calculation i have got 100m Rx its totally wrong because if i increase that (100m) Rx height its mean get the problem at fresnel Zone _________________ Suggest that you print my graphic, and look for different heights of the transmit and receive antennas that still clear the 80 meter hill by using a straight edge that always crosses the location of the hill at the elevations shown in my plot. It is best to provide more path clearance at the 80 meter hill than I showed, because some K-factor variations could steer that 10 GHz beam into that hill, and cause loss of a usable signal for the receiver. Also remember that this plot was done over a smooth earth. The true elevations at the endpoints of the path and at the hill could be considerably different than I showed. RF |
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