# 2nd freq to tuned ant

N

#### Nathan Minos

Jan 1, 1970
0
Can anyone tell me what happens if you power a dipole antenna with a
continuous sinewave at its resonant frequency and then apply a second
sinewave that is slightly offset?

To quantify, let's say the resonance is 1Mhz and the second frequency
is 100Hz less. Would a 100Hz beat be transmitted. Does the tuned
antenna act as a modulator or is the product more complex?

Thank you for any insights.

Nathan

B

#### Bruce Varley

Jan 1, 1970
0
Nathan Minos said:
Can anyone tell me what happens if you power a dipole antenna with a
continuous sinewave at its resonant frequency and then apply a second
sinewave that is slightly offset?

To quantify, let's say the resonance is 1Mhz and the second frequency
is 100Hz less. Would a 100Hz beat be transmitted. Does the tuned
antenna act as a modulator or is the product more complex?

Thank you for any insights.

Nathan

An antenna is a linear device ignoring extreme conditions, so it deals with
each frequency component independently. If a component is not at the
resonant frequency of the antenna, then the antenna will appear as a
reactive load at that frequency, and radiation efficiency will be lower. You
wouldn't notice 100 Hz difference in 1 MHz, it's tiny, so if you looked at
the waveform of what's transmitted using a scope and a receive antenna, what
you'd see would be the same as the waveform fed to the transmitting antenna.
HTH

B

#### [email protected]

Jan 1, 1970
0
Can anyone tell me what happens if you power a dipole antenna with a
continuous sinewave at its resonant frequency and then apply a second
sinewave that is slightly offset?

To quantify, let's say the resonance is 1MHz and the second frequency
is 100Hz less. Would a 100Hz beat be transmitted. Does the tuned
antenna act as a modulator or is the product more complex?

The antenna itself is a linear device, so the two frequencies
shouldn't mix and and you should not see any 100Hz difference tone if

The circuits driving the antenna probably aren't going to be all that
linear, so you may get some mixing in their output stages, and your
receiver may also have a slightly non-linear response (unless you over-
drive it, in which case you will see loads of mixing).

J

#### Jeroen Belleman

Jan 1, 1970
0
The antenna itself is a linear device, so the two frequencies
shouldn't mix and and you should not see any 100Hz difference tone if

However, the signal is also exactly identical to a 0.999950 MHz
carrier, double side-band modulated with 50 Hz. An AM receiver with
the classical diode detector would produce a distorted 100 Hz, while
one with a product detector would let you hear a pure 50 Hz tone.

The antenna doesn't care. That much is true.

Jeroen Belleman

D

#### Don Bowey

Jan 1, 1970
0
However, the signal is also exactly identical to a 0.999950 MHz
carrier, double side-band modulated with 50 Hz. An AM receiver with
the classical diode detector would produce a distorted 100 Hz, while
one with a product detector would let you hear a pure 50 Hz tone.

No, regardless of the type of detector, both receivers will produce the 100
Hz. Difference signal.

Don

T

#### Tom Bruhns

Jan 1, 1970
0
No, regardless of the type of detector, both receivers will produce the 100
Hz. Difference signal.

Don

Looks like you didn't carefully read between the lines in Jeroen's
posting. Clearly, he had in mind a product detector driven by a
1MHz-50Hz LO. On the other hand, he better be careful about the
phases of the signals. If the product detector is driven by a LO of
sin(2*pi*999950*t), and if the 1MHz is sin(2*pi*1e6*t), the difference
is a cosine: A*cos(2*pi*50*t). If the 999.99kHz signal is
sin(2*pi*999900*t) the difference is the same cosine term, so the
total output is 2*A*cos(2*pi*50*t). But if the 999.99kHz signal is
the opposite polarity, -sin(2*pi*999900*t), then the outputs from the
two cancel and you get zero. I suppose most receivers that use
product detectors either have some carrier to lock their LO to, or
just receive the signal as a single sideband and suppress the other--
or just detect the two sidebands independently.

Although mixing in the output amplifiers is tough to avoid if you just
blindly combine the signals, you can use a circuit to keep the
signals out of the alternate amplifiers. One such circuit is a
Wilkinson combiner. Problem: it wastes half the power. It could be
done with filters, but Qu would have to be very high (incredibly
high?) to avoid significant power loss and get good isolation, given
such close frequency spacing.

Cheers,
Tom

D

#### Don Bowey

Jan 1, 1970
0
Looks like you didn't carefully read between the lines in Jeroen's
posting. Clearly, he had in mind a product detector driven by a
1MHz-50Hz LO.

He stated "...the signal is also exactly identical to a 0.999950 MHz
carrier, double side-band modulated with 50 Hz." The OP's signal is NOT
identical to what he described.

On the other hand, he better be careful about the
phases of the signals. If the product detector is driven by a LO of
sin(2*pi*999950*t), and if the 1MHz is sin(2*pi*1e6*t), the difference
is a cosine: A*cos(2*pi*50*t). If the 999.99kHz signal is
sin(2*pi*999900*t) the difference is the same cosine term, so the
total output is 2*A*cos(2*pi*50*t). But if the 999.99kHz signal is
the opposite polarity, -sin(2*pi*999900*t), then the outputs from the
two cancel and you get zero. I suppose most receivers that use
product detectors either have some carrier to lock their LO to, or
just receive the signal as a single sideband and suppress the other--
or just detect the two sidebands independently.

Although mixing in the output amplifiers is tough to avoid if you just
blindly combine the signals, you can use a circuit to keep the
signals out of the alternate amplifiers. One such circuit is a
Wilkinson combiner. Problem: it wastes half the power. It could be
done with filters, but Qu would have to be very high (incredibly
high?) to avoid significant power loss and get good isolation, given
such close frequency spacing.

beyond the topic. Good, but not relevant.

Don

N

#### Nathan Minos

Jan 1, 1970
0
An antenna is a linear device ignoring extreme conditions, so it deals with
each frequency component independently. If a component is not at the
resonant frequency of the antenna, then the antenna will appear as a
reactive load at that frequency, and radiation efficiency will be lower. You
wouldn't notice 100 Hz difference in 1 MHz, it's tiny, so if you looked at
the waveform of what's transmitted using a scope and a receive antenna, what
you'd see would be the same as the waveform fed to the transmitting antenna.
HTH

OK, what if the antenna was non-linear? I am thinking here of a plasma
tube, fed with two non-earth referenced sinewaves (single wire each)
from either end. If one frequncy is resonant and one is not will they
modulate, and to what extent?

Or, each frequency could be equidistant from the resonance. For
example, if the resonance is 1MHz, one would be 100Hz more and the
other 100Hz less.

Nathan

D

#### Don Bowey

Jan 1, 1970
0
OK, what if the antenna was non-linear? I am thinking here of a plasma
tube, fed with two non-earth referenced sinewaves (single wire each)
from either end. If one frequncy is resonant and one is not will they
modulate, and to what extent?

OK, playing "what if," if the antenna were non-linear, multiplication would
generate, among lesser amplitude signals. 200 Hz. This would not propagate
well at all from a 1 MHz antenna.

Have you been lurking on s.e.b? You're Radium, aren't you?
Or, each frequency could be equidistant from the resonance. For
example, if the resonance is 1MHz, one would be 100Hz more and the
other 100Hz less.

The detected or demodulated signal of the two frequencies would simply be
200 Hz.

You should go away and study Amplitude Modulation.

Don

Pl

L

#### LVMarc

Jan 1, 1970
0
Don said:
OK, playing "what if," if the antenna were non-linear, multiplication would
generate, among lesser amplitude signals. 200 Hz. This would not propagate
well at all from a 1 MHz antenna.

Have you been lurking on s.e.b? You're Radium, aren't you?

The detected or demodulated signal of the two frequencies would simply be
200 Hz.

You should go away and study Amplitude Modulation.

Don

Pl
non linear system, if you put in two sine wave you get a plethura of sum
s and differences.

Marc

J

#### Jeroen Belleman

Jan 1, 1970
0
Don said:
On 5/16/07 6:10 AM, in article [email protected], "Jeroen

[...]
To quantify, let's say the resonance is 1MHz and the second frequency
is 100Hz less. [...]
However, the signal is also exactly identical to a 0.999950 MHz
carrier, double side-band modulated with 50 Hz. An AM receiver with
the classical diode detector would produce a distorted 100 Hz, while
one with a product detector would let you hear a pure 50 Hz tone.
No, regardless of the type of detector, both receivers will produce the 100
Hz. Difference signal.

Don
Looks like you didn't carefully read between the lines in Jeroen's
posting. Clearly, he had in mind a product detector driven by a
1MHz-50Hz LO.

He stated "...the signal is also exactly identical to a 0.999950 MHz
carrier, double side-band modulated with 50 Hz." The OP's signal is NOT
identical to what he described.

But it is!

sin(1MHz) + sin(1MHz-100Hz) = 2 * sin(1MHz-50Hz) * cos(50Hz)

Elementary mathematics. I stand by my original statement.

Jeroen Belleman

O

#### oopere

Jan 1, 1970
0
Nathan said:
Can anyone tell me what happens if you power a dipole antenna with a
continuous sinewave at its resonant frequency and then apply a second
sinewave that is slightly offset?

To quantify, let's say the resonance is 1Mhz and the second frequency
is 100Hz less. Would a 100Hz beat be transmitted. Does the tuned
antenna act as a modulator or is the product more complex?

Thank you for any insights.

Nathan

The antenna is fully linear. As a consequence, this antenna would
transmit the two frequencies -nothing else: you end up with two tones in
the transmitted spectrum. There are no mixing effects (at least not for
any reasonable transmitted power)

Pere

D

#### Don Bowey

Jan 1, 1970
0
Don said:
On 5/16/07 6:10 AM, in article [email protected], "Jeroen

[...]
To quantify, let's say the resonance is 1MHz and the second frequency
is 100Hz less. [...]
However, the signal is also exactly identical to a 0.999950 MHz
carrier, double side-band modulated with 50 Hz. An AM receiver with
the classical diode detector would produce a distorted 100 Hz, while
one with a product detector would let you hear a pure 50 Hz tone.
No, regardless of the type of detector, both receivers will produce the 100
Hz. Difference signal.

Don

Looks like you didn't carefully read between the lines in Jeroen's
posting. Clearly, he had in mind a product detector driven by a
1MHz-50Hz LO.

He stated "...the signal is also exactly identical to a 0.999950 MHz
carrier, double side-band modulated with 50 Hz." The OP's signal is NOT
identical to what he described.

But it is!

sin(1MHz) + sin(1MHz-100Hz) = 2 * sin(1MHz-50Hz) * cos(50Hz)

Elementary mathematics. I stand by my original statement.

Jeroen Belleman

Yours is only usable elementary math after you have clipped the posts to
eliminate thee correct relevant information.

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