[sdiy] They aren't sawtooths, they're ramps

[Harry Bissell]

Just to throw a monkey wrench into the works... some synths
(I don't remember which anymore) called a wave that rises
linearly to the right (on a graph) as a "ramp" and one that
falls linearly to the right as a "sawtooth"

The teeth "cut both ways" imho :^)

(they are sonically identical, but are different as LFO...)

H^) harry



----- Original Message -----
From: David G. Dixon <dixon at interchange.ubc.ca>
To: 'cheater cheater' <cheater00 at gmail.com>, 'Dave Manley' <dlmanley at sonic.net>
Cc: synth-diy at dropmix.xs4all.nl
Sent: Tue, 10 Nov 2009 20:46:33 -0500 (EST)
Subject: RE: [sdiy] They aren't sawtooths, they're ramps

In engineering, we tend to refer to y = e^-x as "exponential decay".


> Actually falling and rising exponential ramps are harmonically
> different and if you add both together you get, I think (by trying to
> picture in my imagination how the graph would look) some sort of
> parabolic thing - imagine x^2 from -1 to 1 repeated over and over - of
> course, mathematically, this is very wrong since it's e^x + e^-x, but
> it's similar enough... actually, now that I look at that formula, it's
> going to be exactly the hyperbolic cosine:
> 
> cosh(x) = 1/2(e^x + e^-x)
> 
> and you can get a hyperbolic sine this way:
> 
> sinh(x) = 1/2(e^x - e^-x)
> 
> which means that you need to flip the polarity of one of the waves.
> 
> And if you make a divider, then you can get tangens hyperbolicum also
> known as tanh(x).
> 
> So I say that there *IS* a point in making a distinction in names. I'm
> not so sure whether 'saw' and 'ramp' are the best names. I think that
> 'the expanding saw' is the best name that distincly describes y=e^x
> and y=-e^x, and 'the receding saw' best describes y=e^-x and y=-e^-x,
> by analogy with populations which expand exponentially and recede
> anti-exponentially. I suggest using 'ramp' for linear versions: y=x
> and y=-x. I don't think e^x looks like any ramp I've seen, the car
> would get stuck. I further suggest calling y=x^k (for x = 0..1) the
> accelerating saw and y = x^k (for x in -1..0) the decelerating saw
> (that all for k > 1 of course)

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-- 
Harry Bissell & Nora Abdullah 4eva