Hi.  A useable circuit for beginners exploring the 40 khz range would be:

http://www.electronic-circuits-diagrams.com/audioimages/audiockt10.shtml

It's the only (simple) one I've found operating in that range that is variable frequency.
Another possible implimentation might be found at
http://coecsl.ece.uiuc.edu/ge423/spring05/group9/ultrasonic.html

Sincerely,
Mike Nolley

Hi Hans,

Quote

I ask, because its actually possible to make a modulator that allows a _stereo_ PC sound card to produce _any_ arbitrarily complex sound in any 30KHz band (i.e., 20-50KHz, 15-45KHz, etc.).  So if AM isn't really what you want, you have other options. 

I am really out of my depth on this one. Can you or someone please explain to me how this works, perhaps a circuit diagram etc.

Hans von Lieven

I'm happy to...  I can explain this using trig identities or complex number arithmetic.  The latter is more modern and less complicated, but it's a little more esoteric, so I guess I'll use the trig identities.

Start with 2 carrier signals at frequency w, in radians/sec, offset by 90 degrees, so you have:

c(t) = cos(wt) and d(t) = sin(wt)

Your stereo PC sound card has 2 channels, each good to at least 15KHz, i.e., they can each generate an arbitrary waveform made by summing frequencies <= 15KHz.  let's call the signals you output from these channels x(t) and y(t), and lets program the sound card to generate signals in quadrature with arbitrary amplitude A, frequency f and phase offset g, so you have:

x(t) = Acos(ft+g) and y(t) = Asin(ft+g)


Now, the modulator you want uses 2 multipliers and a difference to produce output:

o(t) = c(t)*x(t) - d(t)*y(t)

Lets see what comes out for the input we have programmed:

o(t) = cos(wt)*Acos(ft+g) - sin(wt)*Asin(ft+g)

= A (cos(wt)*cos(ft+g) - sin(wt)*sin(ft+g))

= A cos(wt+ft+g) , by trig identity. see, for example: http://www.clarku.edu/~djoyce/trig/identities.html

= A cos( (w+f)*t + g)

And that is a pure tone, with our arbitrary amplitude A and phase offset g, but at frequency w+f!  If our quadrature carriers are at, say, w = 30KHz, and our quadrature sound card outputs are at f = 15KHz, then the output is a pure tone at f+30KHz = 45KHz, with whatever amplitude and phase you like.

You can make pure tones up to 15KHz lower than the carrier, too.  Lets try:

x(t) = Acos(ft+g) and y(t) = -Asin(ft+g)

so

o(t) = A (cos(wt)*cos(ft+g) + sin(wt)*sin(ft+g))

= A (cos(wt)*cos(-ft-g) - sin(wt)*sin(-ft-g))

= A cos( (w-f)*t - g)

Finally, note that when you add any two pairs of signals from the sound card together, the output at the end of the modulator is simply the sum of the outputs you would get from each pair individually.  So, if the output you want is the sum of arbitrary tones between 15 and 45KHz, use 30KHz carriers and have the sound card produce the simple sum of the quadrature signals that produce those tones.


So, for an actual modulator circuit, you need 2 good multipliers just like the one you have in that AM modulator you posted, you need a difference, which you can find in any op-amp cookbook, and you need carriers in quadrature.

The easiest way to make quadrature carriers is to use a digital crystal oscillator at 4X the frequency into a divide by 4 counter.  The counter bits cycle

00
01
10
11

instead of the last bit, use the XOR of the two bits, so you have

00
01
11
10

And that is 2 signals at the carrier frequency in quadrature.  Pass them through identical filters to make them nice and sinusoidal, and your carriers are done.

Good luck,

Mr. Entropy

P.S.  Your soundcard may not be able to reproduce frequencies < 20Hz or so, in which case there will be a 40Hz band around the carrier that you can't use.  You should have no problem putting the carrier at a frequency you can afford to drop -- 40Hz is quite narrow.

Hi.  A useable circuit for beginners exploring the 40 khz range would be:

http://www.electronic-circuits-diagrams.com/audioimages/audiockt10.shtml

It's the only (simple) one I've found operating in that range that is variable frequency.
Another possible implimentation might be found at
http://coecsl.ece.uiuc.edu/ge423/spring05/group9/ultrasonic.html

Sincerely,
Mike Nolley

ATTENTION

In 1 linked circuit is somthig WRONG in receiver schematics  !!

R13 R14 /D2 D4
cant produce the needed voltage level to the IC.
It need  some parts mor in schematic to do that.

G.Pese

So you will find out if you will use this schema !

ADD:

Circuits like this was constructONLY for use as remote-control
for Ultro sound cleaner or vaporizer  it give not enough power.

G.P

Thanks for the information, now I have to find someone who can use this information to design a circuit for me :-)

Yeah, I'm afraid I can't help you there -- I couldn't confidently design it without actually building it.  I could help you muddle through if you have a scope and a solderless breadboard, but since electronics isn't your thing, I'm guessing you don't.

Cheers,

Mr. Entropy

Hi KT

As I'm sure you are aware, that cct is effectively one half of Lawtons circuit, but as such, and with preset frequencies, not very versatile.

No offence intended, but I had to chuckle when I looked at Mr Entropy's post somewhere above. Hans had just stated that he was a little out of his depth with electronics and so, as if to rub it in, Mr Entrophy follows with a post containing quite detailed formulas and equations.  Why go for the less complicated, 'complex number arithmetic' when you can have 'trig identities', eh!  Perhaps it's just me, but I thought it was funny!

However, this is why having no technical or scientific background can make progress difficult, in that if you do stumble across something that works, you won't necessarily know why it works or how to explain it.

Farrah Day

No offence intended, but I had to chuckle when I looked at Mr Entropy's post somewhere above. Hans had just stated that he was a little out of his depth with electronics and so, as if to rub it in, Mr Entrophy follows with a post containing quite detailed formulas and equations.  Why go for the less complicated, 'complex number arithmetic' when you can have 'trig identities', eh!  Perhaps it's just me, but I thought it was funny!

Ah, but there was no electronics in that post, miss Day --  just math.  Plain ol' algebra and trigonometry that you _should_ remember from high school.  If Hans can get someone to design and/or build a circuit along those lines, he will need to know that math in order to program his sound card correctly.

I expect that Hans can do the math.  If not, he can research it a bit and do it tommorrow.  He has, after all, done a lot of research in his time.

Cheers,

Mr. Entropy

Hello Hans,

spricht Du deutsch ?

Anyway here is something which I found a few years ago. This might be of great interest for you :

Macrosonixs corp. Beginning in 1998
http://www.americanscientist.org/template/AssetDetail/assetid/21006/page/1?&print=yes

and today :

http://www.macrosonix.com/motors/motorapps.htm

I am pretty short in time and will not participate in this thread but will drop by once and awhile to see
how this topic develops.

Your concept of a asymetic-rotation-driven-pendulum is very interesting.

Regards

Kator

G'day Fritz,

This is an interesting and ambitious project you have in mind. I don't want to give an answer to this off the cuff as it were, so please give me a day or two to put something together that might be helpful.

Hans von Lieven