DISSOCIATING WATER, A KEELY Project

[miken277]

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

[hansvonlieven]

G'day Entropy and all,

The carrier is fine.  It's just amplitude modulation that's weird.  I'd expect a system of resonators to create a simple superposition of the resonator frequencies.  When you do amplitude modulation, you actually change the frequencies produced.  Putting frequencies A and B through a multiplier outputs frequencies A+B and A-B.

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

[Mr.Entropy]

Hi Hans,

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.

[hansvonlieven]

HAPPY BIRTHDAY MIKE !

Hans

[pese]

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