Re: How can I subtract one frequency from another ???

Thanks!

That's a great help, I really appreciate it.
So, the next question is (now that I know what "kind" of device to use) what
is the best mixer for this application.

Is this a one IC solution, or do I have to design a circuit to my
specification?
Can I use the internal mixer of a max038, an XR2206, or even a 565 PLL?

What do you suggest for a simple low cost readily avalible mixer?
Can they be designed with op amps?

Thanks!
;)


"Andrew Holme" <andrew@xxxxxxxxxx> wrote in message
news:dpruk9$60v$1$830fa7a5@xxxxxxxxxxxxxxxxxxx
> Frank wrote:
> > Hi all,
> > A pretty basic question, but I seem to be having a brain block about
> > what approach to take for this application.
> >
> > I would like to take 2 different frequencies, between lets say DC to
> > 1000Hz, and subtract them from one another to get the output signal
> > frequency as the difference between the 2 signals.
> >
> > For example, 400Hz in one input, 410Hz in the 2nd input, the output
> > will be 10Hz.
> >
> > My first instinct was to use an op amp as a comparator, then I
> > thought of a differential amplifier, then, I stumbled across some
> > mixer schematics, and PLL schematics, and then some really complex
> > filter IC's, by that time I was well confused.
> >
> > I just want this to be as simple as possible, one IC if at all
> > possible and some periferal passives.
> >
> > Could one use an LM324 configured as a differential amp? What
> > confuses me is the CMMR part of it, and the fact that the
> > differential amp seems to only "differentiate" voltage differences,
> > and not "frequency" differences, however the common mode rejection
> > will reject like frequencies from both inputs.
> > I'm pretty rusty on my op amps, so I was hoping someone might be able
> > to at least point me in the right direction with this.
> >
> > Thanks!
> >
> > ;)
>
> You were on the right lines with mixers. Apply the two input signals to a
> balanced mixer, and you get output components at the sum and difference
> frequencies. A mixer is a multiplier, and can be understood by thinking
> about the following trig identity:
>
> 2 . cos At . cos Bt = cos (A+B)t + cos (A-B)t
>
> Given your example of 400 and 410 Hz inputs, the sum and difference
outputs
> would be 810 Hz and 10 Hz, and you would need a low pass filter to remove
> the unwanted sum product.
>
>
>


.

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