Clarify Frequency Multiplication

May 18, 2012 63 Replies

As usual, you hypocritical oaf, you take a conceptual sketch and try to denigrate its usefulness while stating that yours should be free from criticism because they're only conceptual sketches. Did it never occur to you that the input _periods_ could be measured to whatever precision required, the multiplication done, and the output synthesized while new periods were being measured? Probably not, but then you're hardly one to own up to having egg on your face.

Of course I did. I do picosecond-resolution period measurements in several of my products. But that's not what you said. And the whole thing is silly.

John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators

No. The output of a multiplier is the sum and difference of the input sine waves. The two inputs don't make it through. It should be obvious why.

A normal radio-type unbalanced mixer, or a simple diode mixer, will propagate the inputs, because it's not a real multiplier.

John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators

I had this nice HP3314A which could output a frequency which was N* or N/ by the input frequency. Basically it would lock its VCO using an external reference frequency.

Failure does not prove something is impossible, failure simply indicates you are not using the right tools... nico@nctdevpuntnl (punt=.) --------------------------------------------------------------

OK, I get the picture now. No real world process to "multiply" two frequencies. Another nomenclature trap for the aspiring genius.

So then, 3,240,000Hz (?) Seems strange, but no doubt true. Hard for me to picture the implied dynamic though.

Thanks again to all for helping to sort out my thoughts.

Klaus Jensen

Tim said nothing "common, useful". It's pretty straightforward to make a circuit that would output 1800Hz when you input 30Hz and 60Hz- just glomp together a few AD537s (two as F-V and one as V-F) and an AD633 (as a voltage multiplier) inbetween. It would be much cheaper in that frequency range to do it in the digital domain, of course. Now, _why_ one would want to do it is the question, and because there's no good answer for that, it follows that it's not a common requirement.

He's saying if you want 1800Hz from 60Hz and 30Hz, mathematically you effectively have to have a constant of 1 second multiplied by the product of the frequencies, since Hz * Hz * seconds = Hz.

Best regards, Spehro Pefhany

"it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com

I can't imagine why, since the multiplier will exhibit time-variable gain resulting in amplitude modulation with the carrier, the local oscillator, and both sidebands appearing at its output.

Any neophyte without an axe to grind would have grasped the equivalence without having to be led explicitly down that road.

Oh John, you just set yourself up as the straight man in a comedy team. Mikek :-)

If you mean analog multipliers, they multiply *voltages*, not frequencies.

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

You've described an analog voltage multiplier, not a frequency multiplier, which is something different, like a harmonic multiplier.

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

Bill posted the relevant trig identity. There are no fundamentals in the output.

But you can just think about it. Input F2 periodically flips the phase of F1, symmetrically. So F1 is present in the output in equal amounts of straight-up and inverted. Those components cancel.

You're not an EE, and don't understand the theory, so why do you keep asserting stuff that's wrong?

John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators

No. 60Hz is 60 cycles per second. 30Hz is 30 cycles per second. (60Hz)* (30Hz) is 1800 cycles squared per second squared -- it's not a rate, it's an acceleration. And it's meaningless, because what the heck is a cycle^2?

Depending on the problem, (some number of radians per second) * (some other number of radians per second) could possibly make dimensional sense

-- but only because a radian is a convenient name that we give to a dimensionless ratio; in that case whatever the answer is would be in seconds^(-2), it would be an angular acceleration (or a rate of frequency ramping), and there had better be some physical process to resolve what the "radians" were that got tossed out the window.

My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com

But where does the additional seconds come from?

a

Sorry, an analog multiplier (Gilbert cell) is double balanced, neither input appears at the output (except in terms of unbalance or nonlinearity).

?-(

Actually any non-linear device will do. Exponential devices like diodes and transistors do pretty well at this, even square law devices like = tubes do a good job; and none are double balanced.

?-/

It is defined into existence, becasue he wants to have Hz as the output.

If you wanted to represent power with the output of a V-F converter (maybe to transmit it across galvanic isolation), you'd need a constant that has units of Hz/watt. If it was measuring displacement, then you'd need a contant of Hz/femtometer or whatever.

In this case, the constant needs units of Hz/Hz^2 or seconds.

Best regards, Spehro Pefhany

"it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com

Good question, and ultimately what I was trying to get the OP to consider.

If it doesn't come from some physically meaningful aspect of the problem, you're just pulling numbers out of your assumptions.

My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com

diodes

tubes

You are welcome.

?-))

Square law is good. It minimizes high order intermodulation products.

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

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