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4017 + resistors = sorta sine

Just a note:

There was a longish set of threads here recently about building a wide- band audio-frequency VCO. The last time I looked the OP had taken a suggestion (from Phil Hobbs?) to use a 4017 and ten cunningly-sized resistors. He was struggling with the harmonics on his output, because he needed something like a 3:1 range.

Well, _I_ needed to whip together an audio-range sine wave generator recently, because my Beckman Industrial signal generator just wasn't stable enough to really wring out a PLL I'm building.

But I have an RF signal generator (Heathkit, with toobs). So I used the

4017 + resistors trick, and a crappy old two-stage RC filter -- and it works pretty good! The signal still wanders a bit, so I either need to treat my lab to a modern signal generator or improve the RF generator to flog a bit more stability out of it, but I have a much better idea of how my PLL works.

So, thanks Phil (if it was you) for the suggestion, even if you didn't make it to me.

--
www.wescottdesign.com
Reply to
Tim Wescott
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James suggested the demuxed counter -> resistors first, I just suggested the 4017, which is one of my favourite chips--I've never run a display off it, or that sort of thing, but it works great for simple sequencing things. Sine waves are only one of their many talents. ;)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510
845-480-2058

hobbs at electrooptical dot net
http://electrooptical.net
Reply to
Phil Hobbs

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Dat be me,

Here's a shot of the output after a 6 pole Butterworth filter at ~200kHz.

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And the signal before the filter,

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There's a bit of 'crud in the middle of each step... I assume a ground issue. The biggest error is the over-shoot at the smallest resistor value... the largest negative signal. This is 'least' bothersome when I run the MC14017 from a +15Volt supply. I guess caused by the capacitance to ground of the inverting opamp stage that I run the resistors into. (I can make it worse by adding more C.) 2nd harmonic is down about 65dB. The 'goal' was to look at sidebands when the VCO is modulated. Here's modualtion at 1 KHz,
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And then higher drive levels,
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You can do the ratio of 160 mV, 256 and 341 and compare them to the first three Bessel function zeros, 2.405, 3.832 and 5.136 And it all looks just fine!

I got a couple of analog switches in the other day and I may try the trick of a switched +1/-1 gain stage (using all the resistors twice with more reduction of the 2nd harmonic, in theory).

George H.

Reply to
George Herold

Very pretty!

Another Fourier-ish trick that FM is good for is modulation-generated carrier interferometry. If you modulate a diode laser, and adjust the frequency deviation and the time delay so that the modulation index (= peak phase deviation) of the detected carrier is about 2.6, you can make the first and second sidebands equal in amplitude, and they contain about 85% of the RF power.

The reason this is useful is that the two are 90 degrees out of phase, so you can get I/Q detection of the optical signal with just two amplitude measurements. I invented this independently about 20 years ago, unfortunately about 2 years after somebody else published it. I still think it's a pretty neat idea.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510
845-480-2058

hobbs at electrooptical dot net
http://electrooptical.net
Reply to
Phil Hobbs

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Reply to
George Herold

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Gee.. Thanks. Pretty pictures are important sometimes.

Hmm, How were you modulating the laser, an AOM? I'm confused by the time delay part. I've only modulated a diode laser by wiggling the current. The sidebands were used to 'calibrate' a confocal Fabry-Perot cavity. (A nice trick, but I assume as old as the hills)

Interesting I didn't know about the phase relation between sidebands.

OK googling 'modulation-generated carrier interferometry', got me a picture of a dead bug prototype from your book. Do you talk about it somewhere else in the book? (Building Electro-Optical Systems...)

George H.

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Reply to
George Herold

You modulate the bias current, which of course causes some grossly huge modulation index on the actual light, since the deviation sensitivity is generally very high. The interferometer performs a finite difference on the phase, though--for an interferometer with a round-trip path difference T, the RF phase is

phi_RF = phi_optical(0)-phi_optical(-T)

which reduces the modulation index a lot. (If the interferometer had zero path difference, you'd be frequency insensitive--i.e. you'd have white light fringes.)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510
845-480-2058

hobbs at electrooptical dot net
http://electrooptical.net
Reply to
Phil Hobbs

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Oh, You were changing the interferometer path length, to see different sidebands 'appear'? I'm still a bit confused, but I have to do something to really understand it.

Abrupt change of subject. So my boss was looking at NMR signals from a uniform roughly linear sample with a magnetic field gradient, along the linear dimension of the sample*. We pulse the sample and look at the signal. It's a linear array all going at a slightly different frequencies. There are beats in the 'scope signal. The FFT shows a 'horned' shaped spectrum. Largest at the two extreme frequencies and smaller in the middle.

My 'silly george' theory is that the middle is being 'interfered with' from both ends. While the ends are only 'interfered with' from one middle.

If true is there a name for this?

Or is it something else?

George H.

*The sample filled the middle 2/3rds of the pickup coil. We could move it around without chaning the horned shape 'too much'. At the ends one or the other side would disappear.

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Reply to
George Herold

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