2nd freq to tuned ant

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

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 your drivers and receiver are all close to ideal

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).

-- Bill Sloman, Nijmegen

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

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

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.

Yes, but if you re-read the original post, you will see your comments are beyond the topic. Good, but not relevant.

Don

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

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

But it is!

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

Elementary mathematics. I stand by my original statement.

Jeroen Belleman

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

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