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

On 1/19/06 11:27 AM, in article RKRzf.177269$tl.24738@pd7tw3no, "Technician"
<technician@xxxxxxx> wrote:

> Hi all,
> Sorry to still be going on about this seemingly simple problem, but there
> have been so many suggestions with so many approaches, it has left me more
> confused than ever, especially since a couple of them did not work.
>
> What I'm going to settle on it using an AD633 mixer IC, as it seems very
> simple to use as far as peripheral components are concerned anyway.
> Being inexperienced with this type of electronics (audio) I've never
> configured a mixer before, so forgive me if my questions sound borderline
> rediculouse for one who supposedly knows something about electronics.
>
> That said, I would like to know how to input a 600Hz sine wave to one input,
> and a 610Hz sine wave to another input, to get an output of 10Hz on one of
> the pins, this being the difference of the 2 input frequencies.
> I would like the output frequency to always be the difference of whatever 2
> input frequencies are inputed, inputs will never exceed 1KHz.
>
> It was previously mentioned that this mixer IC would give me the sum and the
> difference of the 2 input frequencies. What I am not clear on is if the
> difference and sum frequencies will appear on 2 different output pins, or in
> the same output wave on one pin which needs to be filtered with a low pass
> filter such as a butterworth filter.
>
> I have a couple AD633, and the data ***, but I just cannot seem to grasp
> it's mathematics in order to get a firm idea of how to configure this.
> Any help would be appreciated.
>
> Thanks!
> ;)
>
>
>
> "jgreimer" <jgreimer@xxxxxxxxxxxxx> wrote in message
> news:WZEwf.57592$q%.26972@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>> If you're going to be using pure tones and can amplify them enough so
>> they're clipped to resemble square waves, by far your best and simplest
>> solution is to use a D flip flop as the mixer. Don't use TTL logic for
> this
>> and shunt excess signal voltage to ground and the positive supply with
>> diodes. The advantage of a D flip flop is that it produces only the
>> difference frequency so there's no need to filter the output. Input one
>> signal on the D input and the other on the C input. You can take the
> output
>> at either Q or Q bar. If you send music to it however, you're going to
> get
>> a strange output waveform.
>>
>> - jgreimer
>>
>> "Jamie" <jamie_ka1lpa_not_valid_after_ka1lpa_@xxxxxxxxxxx> wrote in
> message
>> news:z8hwf.559$gN1.141@xxxxxxxxxxx
>>> 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!
>>>>
>>>> ;)
>>>>
>>>>
>>> LM1496 chip maybe?
>>>
>>>
>>> --
>>> Real Programmers Do things like this.
>>> http://webpages.charter.net/jamie_5
>>>
>>
>>
>
>

You need to apply Figure 3. As you can see, the entire output mix is at a
single pin (7), where you need to place the low-pass filter.

Don


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