Joel Kolstad wrote:
I'm looking at the form of the equation for the effective area of an antenna
and finding it a little non-intuitive...

We have D/Ae=4*pi/lambda^2, where D is the antenna's directivity and Ae is
the effective area. Hence, after rearranging, we find that Ae is
proportional to lambda^2 or 1/f^2 (f=frequency).

Is this really telling me that if I cut a receiving antenna in the form of a
half-wave dipole at 200MHz, it'll only intercept a quarter as much power as
a half-wave dipole at 100MHz?

Yes. It's half as long, so it won't intercept as much.

For instance... say the incident field has a power density of 1 W/m^2. A
half-wave dipole at resonance is ~73 ohms and its directivity is ~1.64
giving it an effective area of ~.131*lambda^2. At 100MHz, that's 1.18 m^2
and the antenna will produce 1.18 W. At 200MHz, I only get .295W... right?
If I attach a ~73 ohm transmission line to these antennas, equating
V^2/(2*R) to the powers gives me 13.13V (peak to peak) at 100MHz and 6.56V
at 200MHz. This is going to make it twice as hard for me to detect some
signal at 200MHz, won't it??? How does the antenna noise scale with
frequency -- do I still have the same SNR in both cases?

Thermal noise has a uniform frequency distribution, so over a given
bandwidth, the noise power is the same at any frequency. You'll have
better S/N ratio with the 100 MHz dipole than the 200 MHz dipole.
Atomospheric noise is usually more important at HF, and it does vary a
great deal with frequency, but at VHF and above, thermal noise generally
dominates.

Roy Lewallen, W7EL

Roy Lewallen wrote:
Joel Kolstad wrote:
Is this really telling me that if I cut a receiving antenna in the form
of a half-wave dipole at 200MHz, it'll only intercept a quarter as much
power as a half-wave dipole at 100MHz?

Yes. It's half as long, so it won't intercept as much.

Well that certainly seems like something of a raw deal!

I like that way of thinking about it intuitively (it's shorter), although I
believe that -- if you match it properly -- if I take that 200MHz half-wave
dipole and operate it at 100MHz (effectively becoming a full-wave dipole)
it'll actually still collect about the same power as the 100MHz half-wave
dipole. (Since its directivity is still about the same... Looking at it the
other way, for dipoles ~0.2*lambda, the effective area is ~.119*lambda^2,
and it doesn't really matter if it's 0.1*lambda or 0.05*lambda -- although
the later antenna here will be harder to efficienctly match to since it'll
have a noticeably smaller radiation resistance).

I suppose those remote controlled aircraft they build that have the arrays
of dipoles and diodes on their bellies to intercept microwave radiation to
power the motor only use microwaves, then, because they're so much easier to
focus than, e.g., a VHF or lower frequency?

Thermal noise has a uniform frequency distribution, so over a given
bandwidth, the noise power is the same at any frequency. You'll have
better S/N ratio with the 100 MHz dipole than the 200 MHz dipole.

OK, gotcha.

So...in summary... if the cell phone guys had just stuck with AMPS (FM
signaling) when they went from 900MHz to 1.8GHz, the SNR of the incoming
signal would have dropped by 6dB, correct (assuming the antenna was still
the same electrical length)? Bummer...

---Joel

Joel Kolstad wrote:
I like that way of thinking about it intuitively (it's shorter), although
I believe that -- if you match it properly -- if I take that 200MHz
half-wave dipole and operate it at 100MHz (effectively becoming a
full-wave dipole)

This should be 'effectively becoming a QUARTER WAVE dipole,' of course.
Sorry. :-) I stand by the rest of the posting.

---Joel

Of course, you can always increase the size of the antenna to keep the
power density constant!!

That's what all those dishes do...

DD, W1MCE

Joel Kolstad wrote:

I'm looking at the form of the equation for the effective area of an antenna
and finding it a little non-intuitive...

We have D/Ae=4*pi/lambda^2, where D is the antenna's directivity and Ae is
the effective area. Hence, after rearranging, we find that Ae is
proportional to lambda^2 or 1/f^2 (f=frequency).

Is this really telling me that if I cut a receiving antenna in the form of a
half-wave dipole at 200MHz, it'll only intercept a quarter as much power as
a half-wave dipole at 100MHz?

For instance... say the incident field has a power density of 1 W/m^2. A
half-wave dipole at resonance is ~73 ohms and its directivity is ~1.64
giving it an effective area of ~.131*lambda^2. At 100MHz, that's 1.18 m^2
and the antenna will produce 1.18 W. At 200MHz, I only get .295W... right?
If I attach a ~73 ohm transmission line to these antennas, equating
V^2/(2*R) to the powers gives me 13.13V (peak to peak) at 100MHz and 6.56V
at 200MHz. This is going to make it twice as hard for me to detect some
signal at 200MHz, won't it??? How does the antenna noise scale with
frequency -- do I still have the same SNR in both cases?

Thanks for the help,
---Joel Kolstad

Joel Kolstad wrote:
Roy Lewallen wrote:

Joel Kolstad wrote:

Is this really telling me that if I cut a receiving antenna in the form
of a half-wave dipole at 200MHz, it'll only intercept a quarter as much
power as a half-wave dipole at 100MHz?

Yes. It's half as long, so it won't intercept as much.


Well that certainly seems like something of a raw deal!

Yeah, the laws of physics can sure be unfair sometimes. Petition the
deity of your choice and maybe you can talk her into fixing it.

I like that way of thinking about it intuitively (it's shorter), although I
believe that -- if you match it properly -- if I take that 200MHz half-wave
dipole and operate it at 100MHz (effectively becoming a full-wave dipole)
it'll actually still collect about the same power as the 100MHz half-wave
dipole. (Since its directivity is still about the same... Looking at it the
other way, for dipoles ~0.2*lambda, the effective area is ~.119*lambda^2,
and it doesn't really matter if it's 0.1*lambda or 0.05*lambda -- although
the later antenna here will be harder to efficiently match to since it'll
have a noticeably smaller radiation resistance).

That's another real bummer about physics, and one of the major ways it
differs from, say, philosophy. Thinking about it a different way just
doesn't change it a bit. That's a lesson a lot of the magic-antenna
inventors don't seem to pick up on.

I suppose those remote controlled aircraft they build that have the arrays
of dipoles and diodes on their bellies to intercept microwave radiation to
power the motor only use microwaves, then, because they're so much easier to
focus than, e.g., a VHF or lower frequency?


That's one good reason. Another is the one you pointed out above -- even
though a one foot antenna intercepts as much power at 1 MHz as at 1 GHz,
it's impossible to practically extract the power at 1 MHz.

Thermal noise has a uniform frequency distribution, so over a given
bandwidth, the noise power is the same at any frequency. You'll have
better S/N ratio with the 100 MHz dipole than the 200 MHz dipole.


OK, gotcha.

So...in summary... if the cell phone guys had just stuck with AMPS (FM
signaling) when they went from 900MHz to 1.8GHz, the SNR of the incoming
signal would have dropped by 6dB, correct (assuming the antenna was still
the same electrical length)? Bummer...

Yip. Bummer indeed. But there's no point cursing the darkness (unless
you have an "in" with a deity who's able and willing to make an
exception for you). Like they say, if I had some ham, I'd make a ham
sandwich. If I had some bread.

Roy Lewallen, W7EL