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  1. Re:Correct Units? on Intel Announces Laser Breakthrough · · Score: 3, Informative
    Wrong. Here is a good explanation in lay terms. You can find much more detailed explanations with a bit of digging.
    That site is totally wrong about the origin of coherence in laser light. I sent the following to the site maintainer:
    RE: LASERS EMIT COHERENT LIGHT, BUT NOT BECAUSE THE ATOMS EMIT IN-PHASE LIGHT WAVES

    I ran across your site, and I would like to point out that your explanation for the coherence of laser light is incorrect. Coherence and phase are intrinsically-related. Measurements of the temporal or spatial coherence of light are in fact measurements of the relative phase of two different samples of a light wave. The fact that stimulated-emission results in an emitted photon that is exactly in-phase with the incident photon does explain spatial coherence. This is because the laser beam originates as a single (or relatively few) spontaneously-emitted photon(s). That photon is amplified by the stimulated-emission process to form the laser beam. Because the path of the initial photon is generally not along the optical axis of the laser cavity, and because it can be coherently scattered as it propagates through the gain medium, it traverses the gain medium along many paths. Thus the entire laser beam inherits its phase from the initial spontaneously-emitted photon, and is therefore fully spatially-coherent. Even if we consider the laser beam to originate from several spontaneously-emitted photons, the result is that beam is the superposition of several fully spatially-coherent beams, which can be shown to be a fully spatially-coherent beam.

    You are correct that starlight becomes more spatially-coherent by propagating long distances, but that cannot the mechanism for the spatial-coherence of a laser, as I will explain with a fairly simple counter-example. With Q-switching, it is trivial to switch a laser on and off within 10 nanoseconds, in which time light travels about 3 meters in vacuum. Yet the laser pulse can be measured to be as or more coherent than starlight, even though its propagation distance is on the order of meters rather than light years.

    You are correct that the pure color (monochromaticity) of laser light is due to the mirrors (which form a Fabry-Perot or some other resonant cavity), but I would argue that the explanation for the pure color for laser light is at a less advanced level (third-year physics undergraduate) than the explation for its coherence. Coherence is an advanced-undergraduate to graduate-level topic, as a proper analysis of coherence requires Fourier transforms, and the coherence of stimulated emission is a topic in quantum electrodynamics. The most readable but rigorous treatment of optical coherence that I am aware of is _Statistical Optics_ by Joseph Goodman, but even that is written at the advanced-undergraduate to graduate level.
  2. Re:Correct Units? on Intel Announces Laser Breakthrough · · Score: 2, Informative
    The reason laser light is coherent is because it travels an enormous distance before being emitted. This gives the individuals waves time to become coherent.
    This explanation for why laser light is coherent is WRONG. The coherence properties of laser light are due to the properties of the stimulated emission process, and therefore localized in time and space to the emission event.

    The best explanation I've seen for the coherence of stimulated emission is "Rereading Einstein on Radiation" by Daniel Kleppner in this month's issue of Physics Today. The explanation is that light-matter interaction can be modeled as a driving force applied to an oscillator (like a pendulum or spring). In the presence of a driving force, an oscillator absorbs or emits energy depending on its current phase with respect to the phase of the driving force. A simple example is pushing someone on a swing (which is a pendulum, and therefore an oscillator). If you push them at the right times, they swing higher and higher--the oscillator absorbs energy. If you pull on the swing at those times, they swing less and less--the oscillator loses energy.

    In light-matter interaction, the electromagnetic attraction between an electron and an atomic nucleus can be modeled as a spring, and the driving force is an incident electromagnetic wave, i.e. incident light. In a stimulated emission process, an atom (oscillator) loses energy in the presence of an incident photon (driving force). If the energy is emitted as a photon exactly out-of-phase with the incident photon (fully anti-coherent), the two photons would destructively interefere, reducing the net energy of the system and therefore violating conservation of energy. In fact, if the emitted photon is anything other than exactly in-phase with the incident photon, conservation of energy is violated. Thus the emitted photon must be exactly in-phase with the incident photon, and is therefore fully coherent with the incident photon.
  3. Re:But it's not a laser on Intel Announces Laser Breakthrough · · Score: 2, Informative
    However, my point is that this device can't convert non-optical energy into optical energy. Furthermore, since it's a non-linear optical process, you can only get the necessary intinsity to drive this process from a coherent source. Therefore you must have an actual laser to start this process.
    So I guess all those argon-pumped Ti:sapphire oscillators and diode-pumped Nd:YAG oscillators aren't real lasers either. The radiative mechanism for this device is different than for a direct-bandgap semiconductor material, but the terminal behavior isn't much different than most modern optically-pumped solid-state lasers. You illuminate the gain medium with pump light at some wavelength, and you get coherent light at some longer wavelength out. I suppose you could define the term "laser" to exclude such devices if you wanted (there is some precedent, as optical parametric oscillators are traditionally distinguished from lasers), but be careful not to make the definition so narrow as to exclude devices based on stimulated emission!
  4. Re:The University of Virginia - BME MAJOR! on MIT Certifies Biological Engineering Major · · Score: 1

    I would argue that Biological Engineering, as outlined at MIT, is a superset of Biomedical Engineering (BME). BME, as implemented in most BME programs, primarily concerns the application of engineering technology to medical problems. As described, Biological Engineering is both interested in the application of engineering technology to biological (including medical) problems, AND the application of biological knowledge to engineering problems (e.g. biomemetics). The latter is usually considered a specialization with a traditional engineering program (e.g. biomemetic materials as a specialization in materials science and engineering, or biomemetic robots as a specialization in electrical engineering)

    Personally, I think MIT has a more rational approach to BME than most other universities. Many problems in BME can be solved with specialized knowledge of a single engineering discipline supplemented with some basic biological information (usually just physical properties of biological tissue). For example, medical imaging is primarily physics or electrical engineering supplemented by knowledge of the radiological properties of tissue. Your tissue repair example is primarily chemistry or material science supplemented by some knowledge of immune compatibility. This is consistent with making BME a minor to supplement a traditional engineering degreee. The recent craze of establishing degree programs in "biomedical engineering" is a gold rush capitalizing on a grant-giving organization called the Whitaker Foundation, which recently began handing out money left and right to anything containing the words "biomedical engineering".

  5. Re:Not the first school to have this program on MIT Certifies Biological Engineering Major · · Score: 1

    Define "quite some time." I can't find any information about the history of the Biological Engineering program at Guelph, but the University of California, San Diego has had a Bioengineering program since 1966. Furthermore, research and teaching in fields that would currently be classified as bio/biological/biomedical engineering have been going on at MIT in various departments for "quite some time," they just never bothered to formalize it as its own department or degree program. Personally, I think it is better to specialize in some biological or engineering field first, then work out the applications to bio/biological/biomedical engineering, but recently a grant-giving organization called the Whitaker Foundation has been handing out money left and right to anyone placing the strings "bio" and "engineering" in close proximity to each other, so I can hardly blame MIT for formalizing their biological engineering activity now.

  6. Re:Um, hello! on The Death of the Music CD · · Score: 1
    Have you actually every read Nyquist's theorem? It does in fact say that you capture all of the information about anything at a frequency of half the sample rate or below. You can reconstruct the signal exactly by using sinc functions as suggested.
    I would modify the second sentence to read "It does in fact say that you CAN capture all of the information about anything at a frequency of half the sample rate or below" because there is a pathological case, which I describe below.
    So, it's true that if you have a really crazy waveform repeating at 20KHz, you won't be able to reconstruct it with samples at 40KHz. However, that's because the really crazy waveform actually has higher-frequency components.
    Actually, that's not true, and Anonymous Coward almost pointed out the correct counter-example. If I have a 20 kHz sinusoid, its only frequency content is at 20 kHz. Suppose I sample this sinusoid at 40 kHz. If I am very lucky, I collect samples at the peaks of the sinusoid, in which case I recover the frequency, amplitude, and phase of the sinusoid exactly, satisfying the "CAN reconstruct exactly" criterion of the sampling theorem. However, if the sampling is not matched to the phase of the sinusoid, I can only recover the frequency of the original signal--the recovered amplitude will be less than the original amplitude, and the phase will be mismatched. In the worst case, I collect samples at the nulls of the sinusoid, in which case the recovered amplitude is zero, and therefore it is meaningless to consider the frequency or phase of the recovered signal. If the phase mismatch is a random variable uniformly distributed between 0 and pi/2, the best and worst cases occur with infinitessimal probability, so I can expect to recover the frequency correctly almost always and can estimate the amplitude and phase.
  7. Re:Um, hello! on The Death of the Music CD · · Score: 1

    I think you mis-typed your comment. when you said " (i.e. connect-the-dots, like you describe), the reconstructed signal is a sine wave." I think you meant to say triangle wave.

    Yes, that was a typo--linear interpolation reconstructs a triangle wave, not a sine wave.

    Anyway, I beleive I was correct in what I said.

    And I disagree with you, as I'll explain below.

    You are describing a method to 'recover' information that isn't there. Its like on CSI when they convert a crappy low res image into a super hi-res image -- purely fiction. You can not recover what isn't there. You can only guess (interpolate). Applied to the picture analgy you would end up with an interpolated image with either blocky or blurry data between the original pixels; depending on the interpolation method used -- but at no time is any new source information introduced.

    You've got linear interpolation and sinc interpolation backwards: linear interpolation creates information that is not present in the sampled data, whereas sinc interpolation estimates only using the information content of the sampled data. Since we're already considering sinusoidal decompositions of signals, we might as well introduce Fourier analysis and apply it to the problem at hand. The one-line summary of Fourier analysis is that mathematical functions (and physical signals) can be decomposed into sinusoidal components. The motivation for this is that it may be easier to analyze a known class of functions (e.g. sinusoids) than the space of all possible mathematical functions. Sinusoids are convenient because they are differentiable everywhere to any arbitrary degree, whereas finite-order polynomials, triangle waves, or square waves are not. They are also convenient because their information content is minimal--a sinusoid is fully characterized by its frequency, amplitude, and phase. There are other advantages to sinusoids, but they involve concepts like "completeness", "orthogonality", and "eigenfunctions of linear, shift-invariant systems" that I'd rather not get into. :-p

    In the domain of Fourier analysis, it takes infinitely many sinusoids to represent a triangle wave--you need a sinusoid at the fundamental frequency of the triangle, and you need sinusoids at some subset of the harmonics (integer multiples of the fundamental frequency) to reconstruct the straight segments and corners of the triangles. Thus by reconstructing a triangle wave, you have introduced information in the Fourier domain that was not present in the original signal.

    What you are really describing is considered a filter.

    This is actually an advantage for sinc interpolation. In theory, digital-to-analog conversion by sinc interpolation is easily implemented by a ideal low-pass filter. (In practice, we have to settle for the best approximation to the ideal low-pass filter that we can build and/or afford) In any case, an analog-to-digital converter requires a lowpass filter before the sampling stage in order to avoid aliasing effects in the reconstruction (more on aliasing below).

    In addition to the above you would have to first resample the signal to a higher sample rate to provide the extra time slices for you to interpolate into -- this also changes the Nyquist cut-off frequency.

    Actually, once you've collected the samples of the signal, the information content in the Fourier domain is fixed. Even if I upsample the data to a higher sampling frequency by using sinc interpolation, the information content of those new samples in the Fourier domain is unchanged, and therefore, the Nyquist criterion is unaffected. However, if I use linear interpolation, I create new information at harmonics of the fundamental frequency.

    In fact, I know the additional information at these frequencies could not have been collected from the original signal because of

  8. Re:Um, hello! on The Death of the Music CD · · Score: 1
    Sorry but, you couldn't be any more wrong.

    Take a piece of paper and follow along..

    If you sample a 1hz sine wave
    (ok now draw one cycle of a sine wave)

    at a sample rate of 2hz - nyquist therom
    (draw two 'sample' points. one 90 deg and the other at 270 -- the peaks of the sine wave).

    now connect the dots with a line. using your imagination, knowing that the sine is a continuous function, you will realize that by connecting these samples taken at nyquist generates a triangle wave -- not a sine wave.

    NYQUIST tells us the ABSOLUTE minimum sample rate needed to 'detect' a frequency component -- as you can see it poorly represents it. It makes sense that the absolute minimum sample rate would generate a triangle wave -- shortest distance between two points being a line and all.

    Also note in the experiment above -- the samples were taken at the optimum phase angle. If you were to shift the sample points 30deg you'll notice the amplitde of the triangle wave will be greatly affected. NYQUIST is no good for representing amplitude at it's limit.

    So in review we learned that NYQUIST is an ABSOLUTE MINIMUM sample rate that is neither good at representing the original signal or even detecting the correct amplitude at its limit.
    Actually, you're not entirely correct, either. If you use linear interpolation (i.e. connect-the-dots, like you describe), the reconstructed signal is a sine wave. However, if you use a more sophisticated interpolation scheme, you can recover a better estimate of the original signal. In particular, if you use a correctly-scaled sinc (sin x / x) function to weight the contribution of each sample to each point of the reconstructed signal, you can recover the original sine wave EXACTLY, provided you capture the correct samples, as you point out at the end of your comment. Fortunately, in most real-world signals, most of the signal energy and information is contained in the lower frequencies--many schemes of perceptual audio and image coding use this property to achieve high compression ratios.
  9. Re:Interesting... on Cellphone Drivers Drive Like Drunks · · Score: 1

    Yes, but latency != throughput. ;-)

  10. Re:Interesting... on Cellphone Drivers Drive Like Drunks · · Score: 1
    There is also the classic paper
    Redelmeier DA, Tibshirani RJ. Association between cellular-telephone calls and motor vehicle collisions. N Engl J Med 1997;336:453-458. ( Abstract/Full Text)

    They found a statistically significant-correlation between cell-phone usage and automobile accidents, and also found that hand-free units do not reduce the risk of an accident.
  11. Re:It's not true, u dont need a test, its just not on Inside the iPod, Past and Present · · Score: 1

    I think the point is that frequency response is not the only metric for audio fidelity. Someone mentioned linearity, distortion, and jitter as being other important metrics, which then makes jitter measurements relevant.

  12. Re:ARRRG! He gets Nyquist WRONG! on Audio Compression Primer · · Score: 1
    Still, would it be true to say that it would be impossible to reconstruct the *amplitude* of the 10k sine, as you have no way of telling where on it's curve you had taken the samples?
    That is correct. I think pure sinusoidal signals give Nyquist sampling more trouble than general signals, so in fact most sampling schemes use a sampling rate just slightly higher than the Nyquist rate. For example, I think audio signals are bandlimited to 0-20kHz before being sampled at 44.1 kHz (for recording to a CD, for example), which is just marginally higher than the Nyquist rate of 40 kHz. (Also, there is generally less signal energy at higher frequencies, so accurate reconstruction of those amplitudes is not of critical importance.)
  13. Re:ARRRG! He gets Nyquist WRONG! on Audio Compression Primer · · Score: 1
    It does not matter if the samples are take at the peaks or the zero crossing (with 10k sine, 20k sample rate), the result is a line in either case, and will always be a straight line, no matter where on the sine the sample is taken. (If it's at zero then the result is a 'line' at zero).
    What you describe is sampling a 10kHz sine wave at 10kHz. In that case, only one point per period of the sine wave is sampled, and the reconstruction is a straight line. At a sampling rate of 20kHz, I can capture two samples per period from a 10kHz sine wave. In the best case, I capture a sample from the positive peak and a sample from the negative peak; in the worst case, I capture samples from the nulls. I can reconstruct a 10kHz sine wave from samples taken at the peaks, but I can only reconstruct a straight line from samples taken at the nulls.
  14. Re:more algorithms on Audio Compression Primer · · Score: 2, Informative
    Any other resources that Slashdot readers can recommend for those who are interested in the subject of audio compression and representation?
    • An older but good technical survey of digital audio compression, including MP3, is Davis Yen Pan, "Digital Audio Compression," Digital Technical Journal (Spring 1993). (PDF)
    • Some other technical reference material on MP3 is also available on the Digital Audio Systems website.
    • A more recent survey of perceptual coding of audio, which covers more recent formats like AAC, is Painter and Spanias, "Perceptual Coding of Digital Audio," Proc. IEEE (April 2000). (PDF)
    • Ogg Vorbis is documented on the Xiph.org website, but I found the documentation to be lacking when read from a signal processing perspective. Christopher Montgomery provides a better description from that perspective in a Slashdot interview from 2000. I found another good description in this thread in the hydrogenaudio forums--it hyperlinks a good block diagram of the encoding process.
  15. Re:ARRRG! He gets Nyquist WRONG! on Audio Compression Primer · · Score: 1
    You are wrong as well. You need MORE than at least twice as many samples as the largest bandwidth of the signal to correctly reconstruct it. That's why you can't sample a 10k sine with a 20k sample rate, as you need at least three samples to reconstruct a sine.
    You can sample and reconstruct a 10kHz sine wave with a 20kHz sampling rate, but the samples have to be taken at the peaks of the sine wave for the reconstruction to be accurate. In the worst case, the samples could be taken at the zeros of the sine wave, in which case the reconstruction is a straight line. That is why it is necessary, in general, to oversample, so that you collect enough data points to ensure that the reconstructed signal is a reasonable approximation of the original signal.
  16. Re:How do mobile phones affect people on UK Report Suggests Dangers In Cell Phone Use · · Score: 1
    I always thought the main danger to radiation (low-intensity, long-term at any rate) was that it was known to damage DNA.
    That's true for ionizing electromagnetic radiation (UV, x-rays, etc.). Microwaves do not have sufficient energy per photon to damage DNA.
  17. Re:I can't believe the article didn't mention on The Future of Holograms · · Score: 1
    The cheapest way to make a hologram: http://www.amasci.com/amateur/holo1.html
    How exactly are you going to hand-scratch 30 holograms per second for holographic video?
  18. Re:Already been discussed over at HF on Making Holograms In The Kitchen · · Score: 1

    I will have to check out Saxby and determine whether or not to add it to my library. My current references on holography are Goodman Introduction to Fourier Optics, Hariharan Optical Holography, and Collier, Burckhardt, and Lin Optical Holography, and might reflect the fact that I work in a research laboratory in holography and optical information processing... ;-)

  19. Re:Already been discussed over at HF on Making Holograms In The Kitchen · · Score: 1

    Ah, I just looked up "embossed holograms" in Hariharan's Optical Holography and read all about nickel-plating holograms recorded on photoresists. I guess the mention of gelatin was just a red herring. :-p

  20. Re:Hand-drawn holograms on Making Holograms In The Kitchen · · Score: 1

    An hologram isn't necessarily an image of something. Loosely speaking, a hologram is a recording of the shape as well as the brightness of a light wave, and a side effect of this is that if the wave that is recorded is the light that's reflected by an object, reconstruction the hologram forms an image (actually two images, in general) of the object. But in response to your question, a diffraction grating can be considered to be a hologram of a particular type of object. A sinusoidal grating is a hologram of a plane wave, or a very distant line (or point) source. A square-wave grating is a hologram of a periodic array of line (or point) sources of specific brightnesses.

  21. Re:++informative on Making Holograms In The Kitchen · · Score: 1

    Actually, you don't need to use the same wavelength in reconstruction that was used in recording, but the reconstructed image will be magnified or demagnified by the ratio of the two wavelengths.

  22. Re:Hand-drawn holograms on Making Holograms In The Kitchen · · Score: 1

    He wasn't the first to make holograms by scribing them. Ruling engines for making diffraction gratings (which ARE holograms) have been around for a century. Nowadays, many diffraction gratings are made holographically, because holographic gratings are more precise and cheaper and easier to make.

  23. Re:Already been discussed over at HF on Making Holograms In The Kitchen · · Score: 1

    I believe the holographic medium you used was dichromated gelatin, and the holograms that you recorded were actually volume holograms, which have several optical advantages over film holograms (including color fidelity and white-light viewability). The advantage of film holograms (and in particular this kit) is that they are easier to develop.

  24. Re:Better kits on Making Holograms In The Kitchen · · Score: 1

    Given a little knowledge of optics, I don't see any reason these kits can't be used to record white-light viewable rainbow holograms. It would take two pieces of film to create the white-light hologram, but I don't see any physical reason why it couldn't be done.

  25. Re:It's not going to work. on Making Holograms In The Kitchen · · Score: 1

    I was wondering when someone was going to address the issue of the mechanical stability required for interferometry (of which holography is a special case). Those requirements can be relaxed if you shorten the exposure times (which generally means increasing the power of the light source, for example by using a pulsed laser), but that does not apply to this kit (which essentially uses a laser pointer as the light source).