daily quiz

The Widrow approximation is that quantization contributes additive noise of 1/sqrt(12) LSB. However, the validity of that approximation requires that the signal be at least a few LSBs in size. So your question is on the edge of being ill-posed.

However, if we assume that the approximation is valid, then for unipolar signals the dynamic range is 7*sqrt(12), and for bipolar sinusoidal signals it's reduced by a factor of about sqrt(8). The exact number is a bit arbitrary--you can argue about whether the maximum resolvable amplitude goes out to the edges of bins 0 and 7, or the centres, or wherever. As I said, it's on the edge of being ill-posed.

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 

hobbs at electrooptical dot net 
http://electrooptical.net

How do you define dynamic range? Ratio of max to min signal? In which case I'll guess the min is where the lsb is kicking on 1/2 the time and max is when 7 is kicking on 1/2 the time ... so 6.5/.5 = 13

George H.

No, it only supposes that the signal is uniformly distributed between quantization levels. And it isn't the Widrow approx., it's simply an elementary derivation of the noise power. What's your familiarity with Widrow?

Only in the sense that I didn't define dynamic range. I use it in the conventional manner. Which is ....

No reference or assumption of polarity is required. You're going astray.

You're confused. Though you're on the right track, considering noise and uncertainty, which influences the analysis.

Very remarkable that almost no one has thought about such a basic concept. It's a matter of defining the terms, then straightforward reasoning. But intuitional guessing won't do it -

--
Rich

bravo!

um, almost... keep trying -

I'll give a hint later, if necessary. Though it shouldn't be so - See? It IS a good interview question!

--
Rich

Not enough information supplied in the question.

Mitch

--
Terminal_Crazy 

Mitch - 1995 Z28 LT1 M6          terminal_crazy@sand-hill.freeserve.co.uk 
Lancashire England          http://www.sand-hill.freeserve.co.uk/terminal_crazy/

Looks to me the dynamic range would be from whatever the zero point is to the maximum signal that it will accept. The resolution would not be very much.

I took his DSP class at Stanford in about 1986. I'm repeating what he said about it. There's nothing elementary about it whatsoever.

Bipolar vs. unipolar is a 9 dB difference. You think that's not important?

I'm not confused, your question is poorly posed.

From what you say, I'm not convinced that you understand the issues.

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 

hobbs at electrooptical dot net 
http://electrooptical.net

Hmm well if Phil (and I) can't get it "right", then I'm not sure how good it is.

What's your answer?

George H.

To expand on this a bit: it's elementary that the RMS error of a many-bit quantization is LSB/sqrt(12). What's _not_ elementary is Widrow's theorem showing that under certain assumptions (mostly that the signal is at least several LSBs in size), quantization, which is nonlinear, can nevertheless be modelled accurately as a linear operation with additive noise of LSB/sqrt(12). That was widely doubted at the time, and so became Widrow's most famous contribution. Without it, linear systems theory would be much less useful.

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 

hobbs at electrooptical dot net 
http://electrooptical.net

I should add, apparently this definition is less well known, or accepted, than I assumed.

That clearly validates it.

Where did your 6.5 come from? Like the teacher said: show your work -

--
Rich

OK (I'll play) do you have a reference I can look up for that? My simple minded approach is LSB/2. Does the sqrt(12) need to assume some noise on the signal? (And then model the noise.)

And what's up with the distinction between bipolar and unipolar? Do I need to assume noise on both lines with bipolar?

George H. What's _not_ elementary is

Well forget about me, but if you are hiring someone to do some electronics, and your test eliminates Phil H. from your pool of candidates. Then (at least for me.) your test is a failure.

Didn't I say? That was when all the bits are getting turned on 1/2 of the time. Perhaps I'm over thinking your problem, But there looks to me that there is also a question of what's the maximum signal you can detect. If you read 7 all the time, then the input signal could be anything from 7 to

7,000... it's unknown. So something less than that.. I picked 1/2 a LSB. (If I didn't make it clear I'm only guessing.)

George H.

(Never mind I found this,

Geo

Assuming a locally uniform voltage PD, the RMS quantization error of an ideal ADC is

LSB*sqrt(integral (-1/2, 1/2) x^2 dx) = LSB * sqrt(2*0.5**3/3) = LSB/sqrt(12)

That's pretty simple. Widrow's theorem uses pretty deep probability theory--characteristic functions and so forth. (That's deep for me, anyway, and pretty nonobvious at the time he published it.)

A sine wave's p-p voltage has to fit within the ADC's range, so for RFish things the maximum RMS amplitude is 1/sqrt(8) of full scale, whereas for unipolar you can go right up to FS. That makes the dynamic range differ depending on what you're digitizing.

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 

hobbs at electrooptical dot net 
http://electrooptical.net