[sdiy] Frequency multipliers

[Joel B]

Wow, that is really interesting! Back to the books for me...

Sent from my iPhone

On Jun 4, 2010, at 4:21 PM, "David G. Dixon"  
<dixon at interchange.ubc.ca> wrote:

>> By what method does one convert an equation into a circuit?
>
> That's just about all we ever do, whether we realize it or not!  All  
> circuit
> elements have transfer functions which dictate the ratio of voltage  
> out to
> voltage in.  If you know these functions, and with Ohm's and  
> Kirchhoff's
> Laws, then you can design whole circuits just from equations.
>
> When you apply a gain, you are multiplying by a constant.  When you  
> use a
> summing opamp, you are adding and subtracting.  A multiplier  
> multiplies two
> voltages together.  A resistive divider divides a voltage by a  
> constant.  A
> transistor generates an exponential current from a linear base-emitter
> voltage.  A differential pair generates a current which is the  
> hyperbolic
> tangent of the differential voltages on the bases (approximately).   
> Filters
> and other reactive elements have transfer functions which are ratios  
> of
> polynomials in the frequency domain -- that's a whole lot of fun, if  
> you
> understand how to use it.  Since the frequency domain is more or  
> less the
> same thing as the Laplace domain, and since Laplace transforms are  
> used to
> solve differential equations, then filter circuits can be used to  
> solve
> differential equations.  Indeed, our humble state-variable filter  
> can be
> used to solve the classic mass-spring-dashpot problem of classical  
> physics.
> In fact, just about all the circuits we use in synthesizers were  
> originally
> developed for analog computing.  The operational amplifier was  
> invented to
> make analog computing easier.
>
> So, for this equation:
>
> cos(3z) = -3*cos(z) + 4*cos^3(z)
>
> presuming you already have a cos(z) from somewhere (a quadrature  
> oscillator
> or a VCO), then feed it to both inputs of a multiplier to get the  
> square,
> then feed the square and the original signal into another multiplier  
> to get
> cos^3.  Then, feed this to the + terminal of a summer with the correct
> resistors to give it a gain of 4 whilst applying the original signal  
> to the
> - terminal of the summer through the correct resistor to give it a  
> gain of
> -3.  The output of the summer will be cos(3z), the cosine at three  
> times the
> frequency of the original.*  Voila!
>
> * (I think, pending testing.)
>