Quadrature Filter+Multiplier="Clean" Octave Up?

[lovekraft0]

I've been doing a lot of reading, here and elsewhere, and I need to know how much of the following is true:
If one implements a quadrature (Dome? Bode?) filter such that it outputs two identical signals with a relative phase relationship of 90º, and apply these outputs respectively to the X and Y inputs of a 4 quadrant multiplier, when a harmonically complex signal (ie, not a sine wave) is input, the multiplier's output will be a fairly faithful reproduction of the original signal at twice the frequency, with little or no non-harmonic content (a "clean" octave). Will it work - more specifically, if the input signal is steeply bandpassed in the optimal shift range of the quadrature filter, will the output be fairly musical? The math ( and I'm not very good at math, so I could be way wrong) appears to indicate that it will work. Any suggestions before I sail off into the darkness without a compass?  

[puretube]

with a sine, it`ll work! (IIRC...)

for all else: the trick is, to keep either X or Y at a constant level, to keep e2 (dunno how to write this - I`m not good in math, either...), i.e. quadratic distortions limited, or even don`t let`em get into the scheme...

(in the past, Paul Perry, others, and me have talked the subject before, so maybe "searching" will help you...

[Paul Perry (Frostwave)]

If you have a sine, and run it into both the X and the Y inputs of a multiplier, then you get a sine at twice the frequency. No need for the futzing with dome filters etc.

cos 2x = 1 - 2sin2 x
I think that is how it works, you actually get a constant term (blocked by the capacitor) and a cos instead of a sin, but what is a cos but a sine with a phase shift :wink:

Where you DO need a dome filter, is if you want to shift all the harmonics in an audio signal by a fixed amount (the classic case, is single sideband radio). But, adding a fixed freq to each harmonic isn't going to be 'musical', in fact it just gives you half what a ring modulator does.
If you search for "Bode Frequency Shifter" all may be revealed.

[Johnny G]

just nit picking but that is meant to be cos^2(x) = 1 - sin^2(x) isnt it?

[lovekraft0]

Thanks, guys, that was quick! I'd like to think that I kinda know how the Bode Frequency Shifter works, but I'm talking about using the same signal (with a 90º phase relationship) for both the input and the carrier fequency - the equation that I've come across a couple of places is this one.


Am I totally clueless, or does that not appear to imply a special case? Maybe I just don't get it... :?

[puretube]

lots of people been tryin`to wear that t-shirt over the past 40 years - mine isn`t that dirty...

[SeanCostello]

If you can get your guitar into a computer, you can download Pd and try this for yourself. Pd comes with a nice hilbert~ module (8th order phase differencing network), that you can run both outputs into a multiplier and try it for yourself. Pd runs on Linux, OS X and Windows, and is free, so odds are that the computer you are reading this on will be able to run the software.

I think I have tried this before in the past, and it is not particularly memorable. It might be a bit "purer" of an octave up signal, in that it is a bit more sinusoidal. However, it still sounds like an Octavia, as opposed to a pitch-shifted signal, or a guitar signal that has been sped up, or playing the note an octave higher.

Digital systems are great for testing these Hilbert-based circuits. I need to get a good Hilbert network into VisualAudio (a visual DSP tool, vaguely similar to PD or MAX/MSP, that I have been working on for the past few years). The phase differencing network poles can easy be translated via the bilinear z-transform into a network of 1st order allpasses, that can also be rearranged into 2nd order biquad sections for maximum computational efficiency.

Sean Costello

[Paul Perry (Frostwave)]

http://www.freepatentsonline.com/4144581.html

A guitar fx patent, usign a multiplier chip, with the maths.
I think my trig identity posted before was an irrelevancy. :oops:

[rocket]

A possible analog way to get clean octave up is splitting the signal into bands by means of BP filters.
Then rectify or square each band and filter it again with a BP of twice the frequency of the first BP, then add up all the bands again.
(somehow like a vocoder works)
Using enough band should give a reasionable quality.
I've seen a circuit with 4 bands for speech (donald duck/mickey mouse voice).

[Transmogrifox]

Here's why it isn't just as simple as multiplying one cosine with a sine:

A guitar signal can be described as a sum of many cosines of different phases, amplitudes, and frequencies are all related by an integer multiple, say, it's something like:

a*cos (w) + b*cos(2w) + c*cos(3w)....(stopped at 3rd harmonic for simplicity)

Let's take the simple case and assume that all the frequencies started at exactly the same time and have 0 phase.  We shift 90 degrees and multiply:

[a*cos (w) + b*cos(2w) + c*cos(3w)]  * [a*sin (w) + b*sin (2w) + c*sin (3w)]

for simplicity sake, let's take each of the internal terms and assign a variable to it to do the multiplication:

(A + B + C) * (D + E + F)
= AD +AE + AF + BD + BE + BF + CD + CE + CF

Notice all the cross-terms.  Now you have many terms to the effect of sin(2w)*cos(w)

and that's only up to the 3rd harmonic!

This is why it is hard to get an octave up without adding extra harmonic (and other non-harmonic and even non-musical) content to the signal.

This is why everybody has been talking about this filtering thing. If you filter well enough and do this with multiple bands, then you're attenuating the cross-terms from the multiplication, and just adding a bunch of doubled frequencies at the end of it.

A 16- band network with 6th order filters would probably do the trick for you.

[Paul Perry (Frostwave)]

A 16- band network with 6th order filters would probably do the trick for you.

Ah, if only DSP were as easy as ABC!

[Transmogrifox]

Yeah.  With DSP you have to know the alphabet to Z

I think I remember a thing or 2 about z transforms and the e^(jw) circle....

[brett]

My advice is to take a multiplier like the MC1496 and feed the signal into both the X and Y inputs.  One of the ring modulator schematics is set up for something very close to this (Maestro??).

Other multipliers, like diode-based ring modulators, generally have too much gating and/or DC bias to be really good for guitars, which work over such a wide dynamic range.

In the end, the Octavia/Bobtavia/Neoctavia circuits are workable solutions to octave, and RG's mu-doubler should work well, too (but I had problems getting enough gain out of mine - probably needed J201s).

cheers