equation: dBm to mW/cm^2

Fred Abse wrote:
> On Sat, 05 May 2012 11:42:57 -0700, Winston wrote:
>
>> This'll be a 1/4 wave ground plane so I assume it'll receive equally well
>> or poorly in all directions.
>
> Unfortunately no. There's no such thing in reality as the theoretical
> isotropic radiator. The E-plane polar pattern will be similar to half a
> half-wave dipole.
>
> End-on, the gain/sensitivity will be zero, with maximum at right angles to
> the antenna axis. Imagine, if you will, a half bagel, lying
> cream-cheese-side-down on the ground plane:-)

Ah. I stand corrected. (Orthopedic sneakers).

For my purpose, that 'half bagel' pattern will work just fine.
(I assume that the 'reception pattern' for a receiver
  exhibits the same geometry as does the 'radiation pattern'
  for a transmitter for a given antenna, yes?)


>> Five wavelengths at 1000 MHz is about 1.5 m. I'm interested in field
>> intensity at that radius and somewhat beyond, so the far-field answer
>> will work for me.
>
> Five wavelengths is certainly getting there. The trick is to do the math
> for a uniform isotropic, then adjust for the antenna gain in the direction
> of interest.

So, the on-axis 'near field' of an antenna with a gain of 10 is actually
more like 50 wavelengths - rather than say 5 wavelengths - away?

I conjecture that I *should* be calculating watts / cm to fall inversely
with distance (rather than as the inverse cube...) if my 1 GHz receiver is
within say 45 meters of the (gain of 10) antenna?  Holy Moley!

> It's easier with receiving antennas to use volts/meter, rather than
> watts/square meter.

My application is EMC / agency compliance related, where W/cm^2 is the
lingua franca. So I want to be able to express field intensity using
that radix, even if it means doing the involved math to convert from
W/m^2   :)

> For example, the EMF induced in a half-wave dipole is usually taken to be:
>
> e = E * lambda / pi, where e is the induced emf and E is the field
> strength in volts per meter.
>
> Using your antenna's stated isotropic gain of 1dB, and the theoretical
> gain of a half wave dipole (2.15dB), that means that your antenna will be
> 1.15dB down on E lambda / pi.
>
> This is (open circuit) EMF. For a perfectly matched antenna, the voltage
> at the receiver terminals will be half that.

'Sounds as if I need to build a prototype and calibrate
  it against a leveled 1 GHz source of known wattage (and known
  radiation pattern).

> To obtain watts per square meter from volts per meter, square, and divide
> by 377.

Excellent!  That is another good piece of info.

> Forget centimeters, they're not used in engineering anymore.

Thanks for your help, Fred.

I sincerely appreciate it.


--Winston

Winston wrote:

(...)

> So, the on-axis 'near field' of an antenna with a gain of 10 is actually
> more like 50 wavelengths - rather than say 5 wavelengths - away?
>
> I conjecture that I *should* be calculating watts / cm to fall inversely
> with distance (rather than as the inverse cube...)

Er. make that '(rather than as the inverse square...)'


--Winston<--Magnets on the brain.

On Sat, 05 May 2012 14:21:58 -0700, Winston <Winston@Bigbrother.net> wrote:

>Jim Thompson wrote:
>> On Sat, 05 May 2012 11:42:57 -0700, Winston<Winston@Bigbrother.net>  wrote:
>>
>> [snip]
>>>>
>>>>
>>>> The biggie here is that most simple formulas assume plane wave conditions.
>>>> ie. far-field. Near-field conditions are more complicated. I suspect that
>>>> the OP is trying to measure near fields.
>>>
>>> Five wavelengths at 1000 MHz is about 1.5 m.
>>> I'm interested in field intensity at that radius
>>> and somewhat beyond, so the far-field answer will
>>> work for me.
>>>
>> [snip]
>>>
>>> --Winston
>>
>> So mW/cm^2 (power-density) is your absolute Watts divided by the surface area
>> of a sphere with radius equal to the distance of interest (uniform radiator
>> assumed).
>
>That sounds very reasonable isotropically speaking,
>though I recall that the website example was for a
>directional radiator with a gain of 10.
>
>The example implied that my receiver is *always*
>on-axis with the main lobe on both planes, so I
>understand that the field strength would measure
>the same as if the transmitter were 1000 W with
>an isotropic radiator, all else being equal.
>
>I'm here to learn though, so I *am* willing to
>listen to reason.  :)
>
>--Winston

As soon as you have an anisotropic radiator, all "nice" equations are
meaningless.

Your roll-off with distance will be somewhere between 1/r and 1/r^2 depending
on the lobe pattern, and how much you are off axis.

So measure it.
 
                                       ...Jim Thompson
-- 

| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.

Jim Thompson wrote:

> As soon as you have an anisotropic radiator, all "nice" equations are
> meaningless.
>
> Your roll-off with distance will be somewhere between 1/r and 1/r^2 depending
> on the lobe pattern, and how much you are off axis.
>
> So measure it.

Aye aye, Sir.

--Winston

On Sat, 05 May 2012 15:34:56 -0700, Jim Thompson wrote:

> As soon as you have an anisotropic radiator, all "nice" equations are
> meaningless.

Some of them are.

> 
> Your roll-off with distance will be somewhere between 1/r and 1/r^2
> depending on the lobe pattern, and how much you are off axis.

I disagree. At any given angle, the radiated watts per steradian are
constant, irrespective of distance. What changes with radial distance is
the surface subtended by one steradian, which increases as r^2. Hence,
along any radial line, power density always follows an inverse square law,
and the field intensity, an inverse linear law.

The lobe pattern merely defines the radiated power in a particular
direction, ie watts per steradian along a *radial* line.

> 
> So measure it.

Agreed.

-- 
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

On Sat, 05 May 2012 15:14:59 -0700, Winston wrote:

> Winston wrote:
> 
> (...)
> 
>> So, the on-axis 'near field' of an antenna with a gain of 10 is actually
>> more like 50 wavelengths - rather than say 5 wavelengths - away?
>>
>> I conjecture that I *should* be calculating watts / cm to fall inversely
>> with distance (rather than as the inverse cube...)
> 
> Er. make that '(rather than as the inverse square...)'
> 

No. A lot of people get this wrong, in one way or the other.

Power density (watts per unit area) falls inversely as the square of the
distance.

Field intensity (volt per unit length) falls inversely as the distance.

-- 
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

On Sat, 05 May 2012 14:21:58 -0700, Winston wrote:

> The example implied that my receiver is *always* on-axis with the main
> lobe on both planes, so I understand that the field strength would measure
> the same as if the transmitter were 1000 W with an isotropic radiator, all
> else being equal.

Huh?

-- 
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

On Sat, 05 May 2012 15:02:20 -0700, Winston wrote:

> Fred Abse wrote:
>> On Sat, 05 May 2012 11:42:57 -0700, Winston wrote:
>>
>>> This'll be a 1/4 wave ground plane so I assume it'll receive equally
>>> well or poorly in all directions.
>>
>> Unfortunately no. There's no such thing in reality as the theoretical
>> isotropic radiator. The E-plane polar pattern will be similar to half a
>> half-wave dipole.
>>
>> End-on, the gain/sensitivity will be zero, with maximum at right angles
>> to the antenna axis. Imagine, if you will, a half bagel, lying
>> cream-cheese-side-down on the ground plane:-)
> 
> Ah. I stand corrected. (Orthopedic sneakers).
> 
> For my purpose, that 'half bagel' pattern will work just fine. (I assume
> that the 'reception pattern' for a receiver
>   exhibits the same geometry as does the 'radiation pattern' for a
>   transmitter for a given antenna, yes?)

Yes.


> 
> 
>>> Five wavelengths at 1000 MHz is about 1.5 m. I'm interested in field
>>> intensity at that radius and somewhat beyond, so the far-field answer
>>> will work for me.
>>
>> Five wavelengths is certainly getting there. The trick is to do the math
>> for a uniform isotropic, then adjust for the antenna gain in the
>> direction of interest.
> 
> So, the on-axis 'near field' of an antenna with a gain of 10 is actually
> more like 50 wavelengths - rather than say 5 wavelengths - away?

No, near field is dominated by induction, rather than radiation, which is
not necessarily a function of antenna gain, From (now quite distant) memory, the
inductive field is 3dB down at about 2/3 wavelength. I'll have to check
this.

>
> I conjecture that I *should* be calculating watts / cm to fall inversely
> with distance (rather than as the inverse cube...) if my 1 GHz receiver
> is within say 45 meters of the (gain of 10) antenna?  Holy Moley!

There's no such thing as watts/cm. Power density is watts/area, ie.
watts/square meter, or watts/square cm (if you must).

Power density (watts/square meter) falls off as the inverse square of the
distance.

Field intensity (volts/meter) falls off as the inverse distance.

> 
>> It's easier with receiving antennas to use volts/meter, rather than
>> watts/square meter.
> 
> My application is EMC / agency compliance related, where W/cm^2 is the
> lingua franca. So I want to be able to express field intensity using that
> radix, even if it means doing the involved math to convert from W/m^2   :)

What involved math?. For example 10mW/cm^2 is 100W/m^2. You can do that in
your head, just multiply by 100^2 ;-)

Any regulatory standards in the last 30 years should be in SI units,
anyway. Last thing I had to do in this area (which admittedly was for a
European directive in about 1990), levels were quoted in W/m^2.

> 
>> For example, the EMF induced in a half-wave dipole is usually taken to
>> be:
>>
>> e = E * lambda / pi, where e is the induced emf and E is the field
>> strength in volts per meter.
>>
>> Using your antenna's stated isotropic gain of 1dB, and the theoretical
>> gain of a half wave dipole (2.15dB), that means that your antenna will
>> be 1.15dB down on E lambda / pi.
>>
>> This is (open circuit) EMF. For a perfectly matched antenna, the
>> voltage at the receiver terminals will be half that.
> 
> 'Sounds as if I need to build a prototype and calibrate
>   it against a leveled 1 GHz source of known wattage (and known
>   radiation pattern).


That's how it's done. That's why we have expensive measuring antennas with
NIST traceable calibration.


I'd recommend that you get a copy of Kraus's "Antennas", if you really
want to get into this stuff. I guess most university bookstores still have
it, or can get it.

-- 
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

On Sun, 06 May 2012 08:17:54 -0700, Fred Abse <excretatauris@invalid.invalid>
wrote:

>On Sat, 05 May 2012 15:34:56 -0700, Jim Thompson wrote:
>
>> As soon as you have an anisotropic radiator, all "nice" equations are
>> meaningless.
>
>Some of them are.
>
>> 
>> Your roll-off with distance will be somewhere between 1/r and 1/r^2
>> depending on the lobe pattern, and how much you are off axis.
>
>I disagree. At any given angle, the radiated watts per steradian are
>constant, irrespective of distance. What changes with radial distance is
>the surface subtended by one steradian, which increases as r^2. Hence,
>along any radial line, power density always follows an inverse square law,
>and the field intensity, an inverse linear law.
>
>The lobe pattern merely defines the radiated power in a particular
>direction, ie watts per steradian along a *radial* line.
>
>> 
>> So measure it.
>
>Agreed.

Picky!  Picky!  Picky!  ;-)
 
                                       ...Jim Thompson
-- 

| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.

Fred Abse wrote:
> On Sat, 05 May 2012 15:14:59 -0700, Winston wrote:
>
>> Winston wrote:
>>
>> (...)
>>
>>> So, the on-axis 'near field' of an antenna with a gain of 10 is actually
>>> more like 50 wavelengths - rather than say 5 wavelengths - away?
>>>
>>> I conjecture that I *should* be calculating watts / cm to fall inversely
>>> with distance (rather than as the inverse cube...)
>>
>> Er. make that '(rather than as the inverse square...)'

Yes, I meant W/cm^2 not W/cm.  My bad.

> No. A lot of people get this wrong, in one way or the other.
>
> Power density (watts per unit area) falls inversely as the square of the
> distance.

Yes, in the far field.  I've measured that to be the case.
That ain't my question.

1) The old tale about power density falling inversely (not as the
    inverse square) within the *near field* is true, yes?
    (I know. 'Just Measure It'.)  :)

2) The center lobe of a directional antenna with a gain of say 10
    would be expected to have a near field border about 10 x further
    from it than would an isotropic, all else being corrected, yes?
    (That sounds perfectly reasonable, but I don't want to assume.)

Thanks for your help!  :)

--Winston