?s about propogation/losses/noise floor on some bands...

[pianoplayer88key]

What's a typical noise floor range (average rural, suburban, urban, whatever)? for:
13.553-13.567MHz (assuming a 5kHz wide SSB AM mono transmission) (i.e. what field strength averages 1-2dB above noise, assuming a decent receiving antenna?)
902-928MHz (whose fringe would go farther, an FM transmission or an AM transmission? (assume mono, frequency response cut off at 5 kHz))
24-24.25GHz (same ? as 902-928MHz)

Also, what are the through-the-air transmission losses for those frequencies?

For example, assuming I...

transmit 15,848uV/m @ 30m on 13.553MHz in USB 5kHz BW AM, how far might my signal go before it sinks down to < 2dB S/N ratio?
transmit 50mV/m @ 3m on 902MHz (or a little higher to keep emissions in the band), or 250mV/m @ 3m on 24GHz (a bit higher to keep emissions in-band)?

How far might this go in Southern California (assuming TX location might be Whittier, Jurupa, Moreno Valley, Colton, San Gabriel, Long Beach, Bellflower, Ocean Beach, Poway, etc.) until the signal gets down to about what you get on an AM broadcast station between the purple and blue lines on radio-locator (or outside the blue line with a decent receive antenna)?
Assume the antenna is a small whip at 900MHz or 24GHz, or a loop the size of a select-a-tenna at 13MHz.

 

[Andy_Brown]

> What's a typical noise floor range (average rural, suburban,
> urban, whatever)?

the daytime ambient noise floor on the HF
band is generally below 0dBuV/m, quasi peak, in a 9kHz bandwidth.

http://www.rsgb.org/emc/emcwebsite/pdf/rsgbmeasurements_1b.pdf


> 13.553-13.567MHz (assuming a 5kHz wide SSB AM mono
> transmission) (i.e. what field strength averages 1-2dB above
> noise, assuming a decent receiving antenna?)
> 902-928MHz (whose fringe would go farther, an FM
> transmission or an AM transmission? (assume mono, frequency
> response cut off at 5 kHz))
> 24-24.25GHz (same ? as 902-928MHz)
>
> Also, what are the through-the-air transmission losses for
> those frequencies?

Free space loss = 32.4 + 20xLog F(MHz) + 20xLog R(Km)
F is the RF frequency expressed in MHz.
R is the distance between the transmitting and receiving antennas.

You must also consider path loss contributing to the overall attenuation
depending on line of sight obstructions and antenna height(s) transmit and
receive.

http://didier.quartier-rural.org/implic/ran/sat_wifi/sigprop.pdf


> For example, assuming I...
>
> transmit 15,848uV/m @ 30m on 13.553MHz in USB 5kHz BW AM,
> how far might my signal go before it sinks down to < 2dB S/N
> ratio?
> transmit 50mV/m @ 3m on 902MHz (or a little higher to keep
> emissions in the band), or 250mV/m @ 3m on 24GHz (a bit
> higher to keep emissions in-band)?
>
> How far might this go in Southern California (assuming TX
> location might be Whittier, Jurupa, Moreno Valley, Colton,
> San Gabriel, Long Beach, Bellflower, Ocean Beach, Poway,
> etc.) until the signal gets down to about what you get on an
> AM broadcast station between the purple and blue lines on
> radio-locator (or outside the blue line with a decent
> receive antenna)?
> Assume the antenna is a small whip at 900MHz or 24GHz, or a
> loop the size of a select-a-tenna at 13MHz.
>

I'll let you do the math.

Normally more than a 2db margin is desirable due to fluctuation in received power as a result of signal fading caused by multipath, bad line of sight and weather conditions.

Hope this is helpful.

Andy<P ID="signature">______________
Electricity is really just organized lightning.
~George Carlin</P>

 

[rfry]

Line-of-sight analysis is appropriate for point-point applications such as in microwave systems. In general, the further apart the end terminals are, the higher above local ground one or both of the antennas must be, to maintain a LOS path over terrain obstructions and earth curvature.

But if you are hoping for "broadcast" type service, probably you should be analyzing groundwave propagation -- which for frequencies 2 MHz and above is progressively more lossy, and provides rather small coverage areas for the power radiated.
//

 

[pianoplayer88key]

> Line-of-sight analysis is appropriate for point-point
> applications such as in microwave systems. In general, the
> futher apart the end terminals are, the higher above local
> ground one or both of the antennas must be, to maintain a
> LOS path over terrain obstructions and earth curvature.
>
> But if you are hoping for "broadcast" type service, probably
> you should be analyzing groundwave propagation -- which for
> frequencies 2 MHz and above is progressively more lossy, and
> provides rather small coverage areas for the power radiated.
>
> //
>

I was thinking point-to-point applications. For example, running the TX at a church service when for some reason I'm unable to be there, or am late getting there.

 

[pianoplayer88key]

> the daytime ambient noise floor on the HF
> band is generally below 0dBuV/m, quasi peak, in a 9kHz
> bandwidth.
>
ht> tp://www.rsgb.org/emc/emcwebsite/pdf/rsgbmeasurements_1b.pdf
>
>

> Free space loss = 32.4 + 20xLog F(MHz) + 20xLog R(Km)
> F is the RF frequency expressed in MHz.
> R is the distance between the transmitting and receiving
> antennas.


Ok, I did that calculation for 13.56MHz, and got the following several #s:
21km: 81.4895796853
29km: 84.2931537486
127km: 97.1212682097
141km: 98.0295760437
156km: 98.9076857577
163km: 99.2889458787
178km: 100.053593837

So, what do I do with those figures? I know that the field strength at 30m is 15848uV/m. How (using those #s) do I figure out the field strenghts at the specified distances?

>
> You must also consider path loss contributing to the overall
> attenuation
> depending on line of sight obstructions and antenna
> height(s) transmit and
> receive.
>
http> ://didier.quartier-rural.org/implic/ran/sat_wifi/sigprop.pdf
>
>
> I'll let you do the math.
>
> Normally more than a 2db margin is desirable due to
> fluctuation in received power as a result of signal fading
> caused by multipath, bad line of sight and weather
> conditions.
>
> Hope this is helpful.
>
> Andy
>

 

[Andy_Brown]

> > the daytime ambient noise floor on the HF
> > band is generally below 0dBuV/m, quasi peak, in a 9kHz
> > bandwidth.
> >
> ht>
> tp://www.rsgb.org/emc/emcwebsite/pdf/rsgbmeasurements_1b.pdf
>
> >
> >
>
> > Free space loss = 32.4 + 20xLog F(MHz) + 20xLog R(Km)
> > F is the RF frequency expressed in MHz.
> > R is the distance between the transmitting and receiving
> > antennas.
>
>
> Ok, I did that calculation for 13.56MHz, and got the
> following several #s:
> 21km: 81.4895796853
> 29km: 84.2931537486
> 127km: 97.1212682097
> 141km: 98.0295760437
> 156km: 98.9076857577
> 163km: 99.2889458787
> 178km: 100.053593837
>
> So, what do I do with those figures? I know that the field
> strength at 30m is 15848uV/m. How (using those #s) do I
> figure out the field strenghts at the specified distances?

Convert uV/m to dBu
e.g. 15848uV/m = 83.9995 dBu
subtract the loss you calculated
e.g. at 21km you get 83.9995-81.4896 = 2.5099 dBu
convert back to uV/m 2.5099 dBu -> 1.33504e-06 V (1.33 uV)
so at 21.03 km you have 2.51 dB above the noise floor
The starting measurement should be at the transmitting antenna, not 30m away.
This may be throwing things off ... I have to think about that.

Remember, dBu references 1 uV or uV/m.

There are many calculators out on the net you can use.
Here's one:
http://www.mogami.com/e/cad/db.html


<P ID="signature">______________
Electricity is really just organized lightning.
~George Carlin</P>

 

[pianoplayer88key]

> Convert uV/m to dBu
> e.g. 15848uV/m = 83.9995 dBu
> subtract the loss you calculated
> e.g. at 21km you get 83.9995-81.4896 = 2.5099 dBu
> convert back to uV/m 2.5099 dBu -> 1.33504e-06 V (1.33 uV)
> so at 21.03 km you have 2.51 dB above the noise floor
> The starting measurement should be at the transmitting
> antenna, not 30m away.
> This may be throwing things off ... I have to think about
> that.
>
> Remember, dBu references 1 uV or uV/m.
>
> There are many calculators out on the net you can use.
> Here's one:
> http://www.mogami.com/e/cad/db.html
>

Hmm... I'm trying to get from, say, Whittier, or Jurupa, CA, to El Cajon. I only need 4.5kHz frequency response, and about 2-3dB SNR, although more would be nice.

 

[rfry]

> Hmm... I'm trying to get from, say, Whittier, or Jurupa, CA,
> to El Cajon. I only need 4.5kHz frequency response, and
> about 2-3dB SNR, although more would be nice.
______________

These are 90+ mile path lengths!

It isn't possible to do what you want to with the power/radiated field restrictions applying to unlicensed transmitters -- sorry.

The surface wave at your proposed 13.56 MHz frequency is strongly attenuated by ground losses, and wouldn't provide a useful signal beyond a mile or two.

The antenna heights needed for line-of-sight paths of those lengths would be impractical (100s, or possibly 1000s of feet).

The only real chance you would have to link those two endpoints would be by HF skywave propagation, but that would take considerably more radiated power than you are legally allowed.

 

[rfry]

> > So, what do I do with those figures? I know that the
> >field strength at 30m is 15848uV/m. How (using those #s) do I
> > figure out the field strenghts at the specified distances?
>
>
> Convert uV/m to dBu
> e.g. 15848uV/m = 83.9995 dBu
> subtract the loss you calculated
> e.g. at 21km you get 83.9995-81.4896 = 2.5099 dBu
> convert back to uV/m 2.5099 dBu -> 1.33504e-06 V (1.33 uV)
> so at 21.03 km you have 2.51 dB above the noise floor
____________

The formula posted to calculate free-space path loss is valid for power, but not for field strength, sorry.

The simplest approach in this case with a known field strength at a known distance (15,848 µV/m at 30 meters from the transmit antenna) is to multiply the initial field by the ratio of the distances. For example, the field strength in this scenario at a distance of 21 km over a line-of-sight path with no reflections will be 15,848 x (30/21,000) = 22.6 µV/m. This approach is based on the fact that radiated field strength is inversely proportional to distance. IOW when distance doubles, field strength drops to one-half its starting value.

So for the posted distance of 178 km, field strength would be about 2.7 µV/m -- again, for a line-of-sight path with no reflections. The problem being, though, that a line-of-sight path between two endpoints 178 km apart on the surface of the earth is virtually impossible to achieve unless shooting between mountaintops or very tall towers. And without a line-of-sight path, the additional losses due to earth curvature and terrain obstructions would make this path unusable.

As I wrote earlier, the best chance of linking these distant endpoints on HF will be by skywave propagation, and that will take a lot more power than is legally authorized for unlicensed transmitters in the US.
//

 

[Andy_Brown]

> The formula posted to calculate free-space path loss is
> valid for power, but not for field strength, sorry.

I stand corrected, one needs to convert from field strength to power (dBm not dBu).

>This approach is based on the fact
> that radiated field strength is inversely proportional to
> distance. IOW when distance doubles, field strength drops to
> one-half its starting value.

That is TOTALLY incorrect. It's the inverse SQUARE law!!

When distance doubles, field strength drops to one-fourth its starting value.

http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html#c4<P ID="signature">______________
Electricity is really just organized lightning.
~George Carlin</P>

 

[rfry]

> > (Richard Fry quote): This approach is based on the fact
> > that radiated field strength is inversely proportional to
> > distance. IOW when distance doubles, field strength drops
> > to one-half its starting value.

Andy_Brown response:
> That is TOTALLY incorrect. It's the inverse SQUARE law!!
> When distance doubles, field strength drops to one-fourth
> its starting value.
________________

Sorry, but no -- POWER drops to 1/4, voltage (field strength) drops to 1/2.

Here are the appropriate equations (for a 1/2-wave dipole in the direction of maximum radiation):*

P = (1.64*P)/(4*Pi*R^2)
E = [49.2*P^(1/2)]/R

Where
P = transmitted power, watts
E = field strength in V/m
R = distance in meters

Note in the equations above that the reduction in received power is related to the square of the distance, but the reduction in received field strength is a simple, inverse (linear) relationship.

The web page you linked to does not appear to recognize this relationship. Every college and professional engineering reference text on the principles of electromagnetic propagation does so.

___* Equations are quoted from "Reference Data for Radio Engineers," 6th Edition, page 27-7, equations 14 and 15.

//

 

[Andy_Brown]

> The web page you linked to does not appear to recognize this
> relationship. Every college and professional engineering
> reference text on the principles of electromagnetic
> propagation does so.

Uh, that page was at Georgia State, but I digress.

> > radiated field strength is inversely proportional to
> > distance. IOW when distance doubles, field strength drops
> > to one-half its starting value.

You keep pushing that a theoretical question should be answered by the use
of field strength (voltage) measurements. Two points:

One, it is Newton's inverse square law that is the foundation for propagation
loss, unless you want to go one step further and discuss Maxwell's equations.
The law starts with the power relationship, not otherwise.

Two, according to "Reference Data for Radio Engineers" 28-17 "...to compute the field accurately [under real world conditions]it is necessary to calculate the two components separately [free space wave and reflected wave]and to add them in correct phase ..." "Measured field strengths usually show large deviations from point to point because of reflections ... etc."

That is why I answered from a power perspective, which is definitely how to approach a theoretical transmitter-antenna-propagation question without getting involved with a bunch of icky sky wave equations I doubted the poster had any interest in.

Practically speaking, your rough approximation method ignores too many variables most of which I mentioned in my initial answer to the post.

<P ID="signature">______________
Electricity is really just organized lightning.
~George Carlin</P>

 

[rfry]

> You keep pushing that a theoretical question should be
> answered by the use of field strength (voltage) measurements.
> Two points:
> One, it is Newton's inverse square law that is the
> foundation for propagation loss, unless you want to go
> one step further and discuss Maxwell's equations.
> The law starts with the power relationship, not otherwise.

Radiated power falls to 1/4 when distance doubles (the inverse square law). And as a result of that reduction, field strength at that location falls by the square root of 1/4 -- which is 1/2 (as I wrote). Note that you denied this engineering reality in capital letters in a post of yours earlier in this thread.

It is equally accurate to describe radiated fields in terms of voltage as in terms of power. Actually, voltage is far more common in practice. But regardless of the method used, the answers should agree if the right equations and analysis steps are used. And if the approach you used in your post http://www.radio-info.com/mods/board?Post=664660&Board=engineering is valid, why did you not arrive at the same answer for field strength at 21 km as I did? You calculated 1.33 µV/m, I calculated 22.6 µV/m.

> Two, according to "Reference Data for Radio Engineers" 28-17
> "...to compute the field accurately [under real world
> conditions]it is necessary to calculate the two components
> separately [free space wave and reflected wave]and to add
> them in correct phase ..." "Measured field strengths
> usually show large deviations from point to point because of
> reflections ... etc."

I analyzed a free-space path with no reflections, and I said so in my post. You also analyzed a free-space path with no reflections.

My comments went on to state that a 90+ mile free-space path between two end terminals on the surface of the earth was not very likely, and that earth curvature and terrain obstructions would give additional loss, making the path unusable.

I have a decent background in this area. You might want to visit my website http://rfry.org for more information.
//

 

[pianoplayer88key]

Ok, so HF might not be useful for crossing multiple county lines in southern California. (and I'm also doubtful about forward scatter / tropo / e-skip / is there something more reliable?) working on 900MHz (50mV/m @ 3m) or 24GHz (250mV/m @ 3m).

Now, several questions:
What is the average noise level below 9kHz?
Groundwave losses are mimimal, right?
how do I calculate antenna (in)efficiency at those frequencies? (transmitting antenna for example would probably be several meters long laid on the floor indoors (or it might be outdoors.))
How much power is safe for near-field human exposure? Do I correctly understand that extremely long wavelengths are not as hazardous as short wavelengths? (like would 250 watts to 1kW be safe?)
How, based on transmitter power & antenna inefficiency, do I calculate the field strength at various locations?
Is it possible to transmit a lower sideband amplitude-modulated mono signal with a frequency response of 4.5kHz with a carrier frequency of 8.5kHz?

Or, is there some propogation that would be safe for near-field human exposure, and reliably cover up to 120 miles above 300 GHz, assuming transmitting antenna is indoors, or outdoors not mounted more than 8 feet above ground level, and the receive antenna is indoors?

 

[rfry]

> ... Or, is there some propogation that would be safe for
> near-field human exposure, and reliably cover up to 120
> miles above 300 GHz, assuming transmitting antenna is
> indoors, or outdoors not mounted more than 8 feet above
> ground level, and the receive antenna is indoors?
__________

Unfortunately for your project -- physics, together with the FCC's constraints on unlicensed transmitters dictate that there is no practical, legal means available to you to accomplish what you wish to do.
//