[sdiy] Frequency multipliers

[Joel B]

By what method does one convert an equation into a circuit?

Sent from my iPhone

On Jun 4, 2010, at 3:11 PM, "David G. Dixon"  
<dixon at interchange.ubc.ca> wrote:

>>> Are there some relatively simple analog circuits that can take an
>>> input sine frequency of x and output a multiple of x?
>>
>> If you've got a single sine wave or a quadrature signal, you can  
>> use a
>> multiplier to get double the frequency (or a square root circuit to
>> half it).  Higher multiples are possible, but more cumbersome and  
>> less
>> effective due to additional products you would have to cancel.  Or
>> maybe you don't need to cancel them, since these by-products are also
>> multiples of the input frequency.
>
> The most straightforward way to get a sine wave of twice the  
> frequency is to
> take the square of the sine wave with a multiplier, and then apply a  
> gain of
> 2 and a negative dc bias of half the original amplitude to the  
> output.  You
> will end up with a sine wave at twice the frequency.  This is an  
> electronic
> manifestation of the following trigonometric identity, known as a
> "half-angle formula" (one of three):
>
> cos(z/2) = +/- sqrt[(1 + cos(z))/2]
>
> or, put another way,
>
> cos(z) = 2*cos^2(z/2) - 1
>
> This is equation 4.3.21 on page 72 of Abramowitz and Stegun,  
> Handbook of
> Mathematical Functions, Dover, New York, 1965 (just about the best  
> $15 I
> ever spent!).
>
> Obviously, this operation may also be done in reverse using a square- 
> root
> circuit.
>
> (Hey, this has the makings of an excellent final exam question for
> second-year EE students!)
>
> Now, if you're really clever, you can design a circuit to get a sine  
> wave at
> three times the frequency of the original.  Here's the trigonometric
> identity (ibid., equation 4.3.28):
>
> cos(3z) = -3*cos(z) + 4*cos^3(z)
>
> Recognizing that cos^3 = cos*cos^2, this can be done with two  
> multipliers
> and a summer, or one additional multiplier and summer.  No dc level  
> shifting
> is involved.
>
> This sounds like so much fun, I think I'll have to do it myself!
>
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