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September 4 Information
RC filters, PWM etc
I don't understand the method described
here (second reply from top, by Olin Lathrop) where he increases the roll-off frequency in order to (paradoxically) increase roll-off and make it such that the combined attenuation of 3 filters is their attenuation cubed instead of attenuation times three (in order to have a more manageable PWM slice clock frequency.) How does that work? why do the three 8 kHz filters attenuate by 9dB, but three 24 kHz filters by 60 dB? I'd ask there but don't have enough karma
78.50.125.97 (
talk)
11:59, 4 September 2017 (UTC)reply
He says "We originally said our upper frequency of interest was 8 kHz, which implies we can tolerate that being 3 dB down unless we say otherwise. A single R,C filter will attenuate by 3 dB at the rolloff frequency, so we put it at right at 8 kHz. We can't have three filters at 8 kHz since they would attenuate by 9 dB there combined."
So he's saying that a filter at the frequency of interest (8 kHz) will attenuate by 3 dB (1⁄2 the power), and three such filters will attenuate by 9 dB (1⁄2 x 1⁄2 x 1⁄2 the power), which is undesirable.
He continues "So, we move the filters out by the number of poles (separate R,C filters in this case). The three R,C filters (three poles) are therefore at 24 kHz. It seems like we lost ground doing this, but the big advantage is that the frequencies above that are now attenuated by the ratio cubed instead of just the ratio as with a single pole."
Here, he's adding nothing new by talking about a cubed ratio, since with a three-pole filter the power ratio is always going to be cubed, as we just saw with the 8 kHz filter. He suggests moving the cutoff frequency out to 24 kHz, and if you look at the immediately preceding graph you'll see that the attenuation at 24 kHz is 20 dB. This is a power ratio of 100, so the total attenuation for three filters is 60 dB or 1⁄100 x 1⁄100 x 1⁄100, which is 1 millionth of the power.
To simplify, a moving out of the filter frequency increases the attenuation for one filter, and the effect of cubing (3 filters) greatly magnifies the effect.
Akld guy (
talk)
20:14, 4 September 2017 (UTC)reply
uh, ok, that makes sense. I overlooked the "as with a single pole" bit and didn't realize the comparison referred to three filters vs one filter rather than three 8 kHz filters vs three 24 kHz filters.
I still don't understand how feeding an 8 kHz signal into a 24 kHz LPF (any number of them) would attenuate it at all, never mind greatly (except smear out the edges a bit if it's rectangular)...
78.50.125.97 (
talk)
05:41, 5 September 2017 (UTC)reply
The gain-magnitude frequency response of a first-order (one-pole) low-pass filter.
Mr. Bode is rightly famed for his
useful lie about the frequency response of filters. The lie says that an LPF has no effect (i.e. has unity gain which in decibels is 0 dB) at frequencies below the cutoff frequency FC (angular frequency = 1 on the graph), and that above the cutoff frequency the gain falls in proportion to FC/f. The beauty of the lie is that it allows the entire frequency response of an RC LPF to be drawn as two straight blue lines on a log-log scaled graph shown. Without detracting from the Bode plot's value in clarifying how
filters work, for example in showing that the final roll-off rate of an n-pole LPF is -6n dB per octave i.e. -20n dB per decade (an RC circuit is a first-order filter with n = 1), it fails to show the true frequency response (red curve) near the cutoff frequency. The Bode plot will have us believe that a 24 kHz LPF does not affect an 8 kHz signal.
However look closely at the slight difference between the red and blue curves at angular frequency 0.33.. (corresponding to 8 kHz / 24 kHz ).
RC circuit#Gain and phase gives the accurate gain in dB as a function of .
frequency G
kHz wRC dB
8 .333.. -0.458
24 1 -3.010
48 2 -6.990
72 3 -10.00
240 10 -20.04
fair enough. that's still 1.374dB (3rd order 24kHz filter) for an 8 kHz PWM fundamental and 9dB for a 24kHz one, though...? I guess I don't understand what "moving out the frequency to 24kHz" means. is it resizing the filter or increasing the PWM fundamental? or adding stages? if a 3rd order 24kHz filter attenuates 240kHz by 60dB just like a 3rd order 8kHz filter attenuates 80kHz by 60dB, what exactly does "moving out..." accomplish?
80.171.95.62 (
talk)
15:42, 5 September 2017 (UTC)reply
The poster Olin Lathrop at
Stack Exchange adds to the explanation of a single RC filter an insight into how a 3-pole filter offers advantages. Since combining 3 of the original filter (-3 dB at 8 kHz) gives an unacceptable -9 dB at 8 kHz, he chooses a different filter that attenuates only 3x -0.458 dB at 8 kHz. On paper it is easy to design ("move to") any value of
Fcutoff = 1/(2 pi R C ) Hz
by selecting R (ohms) and C (farads). What the multi-pole filter achieves is twofold: less high-frequency loss in recovering the analog signal, and a 3-fold steeper high frequency rollof which permits a lower pulse frequency. This explanation for simplicity assumes that succeeding RC filters do not load preceding filters appreciably; this can be ensured by having buffer amplifiers between the filters.
Blooteuth (
talk)
18:27, 5 September 2017 (UTC)reply
I think I get it!11 The low attenuation is a good thing. The "frequency of interest" (8kHz) is _that which modulates the PWM_ (which in turn is much higher.) (I thought it was the actual PWM fundamental.) It is awkward to try and recover that with a 3rd order filter _for that exact frequency_ because it would attenuate _it_ by whole 9 dB. A 24kHz filter OTOH would leave it mostly intact. What he's basically saying is if one goes to higher order filters, don't put the filter at the frequency you try to recover because it would attenuate too much, instead, put it higher. Thank you everyone!
80.171.95.62 (
talk)
05:47, 6 September 2017 (UTC)reply
This thread follows the one at Stack Exchange that looks like this: The
original question contained the statement "PWM frequency is 10 kHz.". The answer by stevenvh focused on suppressing 10 kHz. Implicit but not mentioned is that pulses at 10 kHz cannot convey a signal bandwidth greater than 10/2 = 5 kHz, see
Nyquist–Shannon sampling theorem. Olin Lathrop revisits the design by taking signal bandwidth = 8kHz as a requirement and deducing what the lowest PWM frequency could be for 60 dB suppression: 8 MHz or 240 kHz with the two filters he considers. We are done here.
Blooteuth (
talk)
16:19, 6 September 2017 (UTC)reply
All light reflected from all shiny surfaces is linearly polarised to some degree, that degree increasing with the shallowness of the angle – see Section 4.2 of the
Specular reflection article already linked in the caption. It's possible to alter the amount of polarisation to some degree, and as the caption suggests, making cars look shinier is a thing manufacturers do, because it makes them look more attractive to a potential buyer. It may also contribute a little to road safety by making the cars more visible to other drivers. {The poster formerly known as 87.81.230.195}
90.204.180.96 (
talk)
01:19, 5 September 2017 (UTC)reply
Today I've stumbled upon
news (also reported in Russian:
[1],
[2],
[3]) that folks from the Massachusetts Institute of Technology found organic molecules in the Andromeda Galaxy. But my search didn't yield any major outlet reporting it.
Fake news or others just didn't catch up yet?
Brandmeistertalk20:26, 4 September 2017 (UTC)reply
The presence of organic compounds is unsurprising;
uniformitarianism is a scientific philosophy concept which comes originally through geology, the astronomical concept is called the
cosmological principle, but basically the concept that we do not hold a privileged position in the universe shows up all over scientific philosophy. Knowing that, we would expect that all galaxies should contain, at a first approximation, roughly the same chemical make-up as our own, especially at the resolution we can detect from this distance. Organic compounds are found all over our solar system and galaxy (see
Methane#Extraterrestrial methane), so we have no reason to be surprised to find them elsewhere. It's a
nothingburger, scientifically speaking. --
Jayron3201:46, 5 September 2017 (UTC)reply
KP is a tabloid and RenTV is huge into UFOs, ancient aliens and other such "fine energies." it all seems to be going back to one dodgy Bashkirian site. There's
nothing at MIT's press site or on Gogole News either (try other news aggregators you know.) A similar article on beforeitsnews.com from 2011 (no idea if it's related) mentions the name Cordiner. Googling it turns up a
paper also from 2011. They found polyaromatic cyclic hydrocarbons (PAH) but Ctrl-F'ing "life" or "living" turns up nothing.
I still wouldn't call it "fake news" though because it's, on the whole, harmless and also utterly pedestrian if we take "organic" to mean "anything with carbon in it." It is, at most, a
"vbros" (2.) :)
80.171.95.62 (
talk)
17:02, 5 September 2017 (UTC)reply
Multiple days per year
At the poles, there's one day per year, if I understand rightly — solar noon is at the summer solstice, and "midnight" at the winter solstice. How far from the poles do you have to go before there's more than one day each year?
Nyttend (
talk)
23:00, 4 September 2017 (UTC)reply
You have to define what "day" means for this to be answerable. To me there are 365 days per year even at the poles, because even though the Sun stays above the horizon for very long periods, it still rotates around the center of the sky every 24 hours.
Looie496 (
talk)
23:26, 4 September 2017 (UTC)reply
If "sets" means "midpoint of the Sun goes below the theoretical line of the horizon" then it is very close to the poles. You don't have to move very far from the poles before there is an extra sunset at the equinox.
Looie496 (
talk)
23:47, 4 September 2017 (UTC)reply
Sunset is the refracted disk disappearing below the geoid. Note that you can't just use 50 minutes of arc because that rule of thumb was made by Europeans and noticeably underestimates refraction in Arctic air.
Sagittarian Milky Way (
talk)
01:10, 5 September 2017 (UTC)reply
So it seems that according to their calculations the answer is around 11 or 12 minutes of latitude, or about the same number of nautical miles from the pole. However, I don't know what definition of "sunset" they use, and for that matter I don't know how
stable their computations are for positions where the sun sets at such a shallow angle. I also didn't try other longitudes, or positions near the South Pole, or years other than 2017. For what it's worth, I did find that at 0° E/W you have to go to 89°44' N before you get an additional sunrise and sunset near the spring equinox. --
69.159.60.147 (
talk)
01:22, 5 September 2017 (UTC)reply
I was just gobsmacked by a claim
[4][5] sourced to
this citing work back to 1992 that there are estimated to be 10 million to 1 billion black holes in the Milky Way. Somehow by the second source that became 100 million at a minimum. Which if I recall is about the number of stars in the
Milky Way, though according to that article now it might be up to 400 million.
Does all this mean that it is as like as not that the nearest star to the Sun is a black hole? And that we have absolutely no idea where it is?
Wnt (
talk)
23:52, 4 September 2017 (UTC)reply
<OOPS> ... shoulda noticed that. Sorry! It's the stuff ya think ya know that gets ya. There are 2000 stars within 50 light years, so that puts the black hole at around 50 light years away at an average. Which is rather less intimidating. ;)
Wnt (
talk)
00:59, 5 September 2017 (UTC)reply
Well, the articles say there's a huge disconnect between the 100 million black holes expected and the 60 that are known. I have no idea what distribution is expected for them, but the anthropic principle only says we wouldn't have been hit with one .... yet.
Wnt (
talk)
11:56, 5 September 2017 (UTC)reply
The original paper being referenced is an assumption based on an assumption based on an assumption. A separate paper used three long-time microlensing events to discuss the probability that one or more of them might be a black hole. The assumption is that one is most likely a black hole. The other two may or may not be. The second assumption is that there are far more microlensing events than the three mentioned and that most of those will also be black holes. The final assumption is that if we count up all the microlensing events and assume they are mostly black holes, there are a huge number of black holes floating around. Now, what if it is proven beyond doubt that none of the original three microlensing events were black holes? All the handwaving becomes silly talk.
209.149.113.5 (
talk)
14:26, 5 September 2017 (UTC)reply
I do not think that it is a plausible claim. Black holes are among the rarest objects. There should certainly be many more
neutron stars than black holes. Still the closest known neutron star is at about 400 light years. I do not even want to talk about
white dwarfs which population is
huge. Still the nearest star is not even a white dwarf.
Ruslik_
Zero18:45, 5 September 2017 (UTC)reply
But the very next star system still has a star more special than the Sun (hotter, brighter, faster, stronger, bigger, older, more metallic, massive) because the universe wants to make us feel insignificant.
Sagittarian Milky Way (
talk)
19:39, 5 September 2017 (UTC)reply
Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the
current reference desk pages.
September 4 Information
RC filters, PWM etc
I don't understand the method described
here (second reply from top, by Olin Lathrop) where he increases the roll-off frequency in order to (paradoxically) increase roll-off and make it such that the combined attenuation of 3 filters is their attenuation cubed instead of attenuation times three (in order to have a more manageable PWM slice clock frequency.) How does that work? why do the three 8 kHz filters attenuate by 9dB, but three 24 kHz filters by 60 dB? I'd ask there but don't have enough karma
78.50.125.97 (
talk)
11:59, 4 September 2017 (UTC)reply
He says "We originally said our upper frequency of interest was 8 kHz, which implies we can tolerate that being 3 dB down unless we say otherwise. A single R,C filter will attenuate by 3 dB at the rolloff frequency, so we put it at right at 8 kHz. We can't have three filters at 8 kHz since they would attenuate by 9 dB there combined."
So he's saying that a filter at the frequency of interest (8 kHz) will attenuate by 3 dB (1⁄2 the power), and three such filters will attenuate by 9 dB (1⁄2 x 1⁄2 x 1⁄2 the power), which is undesirable.
He continues "So, we move the filters out by the number of poles (separate R,C filters in this case). The three R,C filters (three poles) are therefore at 24 kHz. It seems like we lost ground doing this, but the big advantage is that the frequencies above that are now attenuated by the ratio cubed instead of just the ratio as with a single pole."
Here, he's adding nothing new by talking about a cubed ratio, since with a three-pole filter the power ratio is always going to be cubed, as we just saw with the 8 kHz filter. He suggests moving the cutoff frequency out to 24 kHz, and if you look at the immediately preceding graph you'll see that the attenuation at 24 kHz is 20 dB. This is a power ratio of 100, so the total attenuation for three filters is 60 dB or 1⁄100 x 1⁄100 x 1⁄100, which is 1 millionth of the power.
To simplify, a moving out of the filter frequency increases the attenuation for one filter, and the effect of cubing (3 filters) greatly magnifies the effect.
Akld guy (
talk)
20:14, 4 September 2017 (UTC)reply
uh, ok, that makes sense. I overlooked the "as with a single pole" bit and didn't realize the comparison referred to three filters vs one filter rather than three 8 kHz filters vs three 24 kHz filters.
I still don't understand how feeding an 8 kHz signal into a 24 kHz LPF (any number of them) would attenuate it at all, never mind greatly (except smear out the edges a bit if it's rectangular)...
78.50.125.97 (
talk)
05:41, 5 September 2017 (UTC)reply
The gain-magnitude frequency response of a first-order (one-pole) low-pass filter.
Mr. Bode is rightly famed for his
useful lie about the frequency response of filters. The lie says that an LPF has no effect (i.e. has unity gain which in decibels is 0 dB) at frequencies below the cutoff frequency FC (angular frequency = 1 on the graph), and that above the cutoff frequency the gain falls in proportion to FC/f. The beauty of the lie is that it allows the entire frequency response of an RC LPF to be drawn as two straight blue lines on a log-log scaled graph shown. Without detracting from the Bode plot's value in clarifying how
filters work, for example in showing that the final roll-off rate of an n-pole LPF is -6n dB per octave i.e. -20n dB per decade (an RC circuit is a first-order filter with n = 1), it fails to show the true frequency response (red curve) near the cutoff frequency. The Bode plot will have us believe that a 24 kHz LPF does not affect an 8 kHz signal.
However look closely at the slight difference between the red and blue curves at angular frequency 0.33.. (corresponding to 8 kHz / 24 kHz ).
RC circuit#Gain and phase gives the accurate gain in dB as a function of .
frequency G
kHz wRC dB
8 .333.. -0.458
24 1 -3.010
48 2 -6.990
72 3 -10.00
240 10 -20.04
fair enough. that's still 1.374dB (3rd order 24kHz filter) for an 8 kHz PWM fundamental and 9dB for a 24kHz one, though...? I guess I don't understand what "moving out the frequency to 24kHz" means. is it resizing the filter or increasing the PWM fundamental? or adding stages? if a 3rd order 24kHz filter attenuates 240kHz by 60dB just like a 3rd order 8kHz filter attenuates 80kHz by 60dB, what exactly does "moving out..." accomplish?
80.171.95.62 (
talk)
15:42, 5 September 2017 (UTC)reply
The poster Olin Lathrop at
Stack Exchange adds to the explanation of a single RC filter an insight into how a 3-pole filter offers advantages. Since combining 3 of the original filter (-3 dB at 8 kHz) gives an unacceptable -9 dB at 8 kHz, he chooses a different filter that attenuates only 3x -0.458 dB at 8 kHz. On paper it is easy to design ("move to") any value of
Fcutoff = 1/(2 pi R C ) Hz
by selecting R (ohms) and C (farads). What the multi-pole filter achieves is twofold: less high-frequency loss in recovering the analog signal, and a 3-fold steeper high frequency rollof which permits a lower pulse frequency. This explanation for simplicity assumes that succeeding RC filters do not load preceding filters appreciably; this can be ensured by having buffer amplifiers between the filters.
Blooteuth (
talk)
18:27, 5 September 2017 (UTC)reply
I think I get it!11 The low attenuation is a good thing. The "frequency of interest" (8kHz) is _that which modulates the PWM_ (which in turn is much higher.) (I thought it was the actual PWM fundamental.) It is awkward to try and recover that with a 3rd order filter _for that exact frequency_ because it would attenuate _it_ by whole 9 dB. A 24kHz filter OTOH would leave it mostly intact. What he's basically saying is if one goes to higher order filters, don't put the filter at the frequency you try to recover because it would attenuate too much, instead, put it higher. Thank you everyone!
80.171.95.62 (
talk)
05:47, 6 September 2017 (UTC)reply
This thread follows the one at Stack Exchange that looks like this: The
original question contained the statement "PWM frequency is 10 kHz.". The answer by stevenvh focused on suppressing 10 kHz. Implicit but not mentioned is that pulses at 10 kHz cannot convey a signal bandwidth greater than 10/2 = 5 kHz, see
Nyquist–Shannon sampling theorem. Olin Lathrop revisits the design by taking signal bandwidth = 8kHz as a requirement and deducing what the lowest PWM frequency could be for 60 dB suppression: 8 MHz or 240 kHz with the two filters he considers. We are done here.
Blooteuth (
talk)
16:19, 6 September 2017 (UTC)reply
All light reflected from all shiny surfaces is linearly polarised to some degree, that degree increasing with the shallowness of the angle – see Section 4.2 of the
Specular reflection article already linked in the caption. It's possible to alter the amount of polarisation to some degree, and as the caption suggests, making cars look shinier is a thing manufacturers do, because it makes them look more attractive to a potential buyer. It may also contribute a little to road safety by making the cars more visible to other drivers. {The poster formerly known as 87.81.230.195}
90.204.180.96 (
talk)
01:19, 5 September 2017 (UTC)reply
Today I've stumbled upon
news (also reported in Russian:
[1],
[2],
[3]) that folks from the Massachusetts Institute of Technology found organic molecules in the Andromeda Galaxy. But my search didn't yield any major outlet reporting it.
Fake news or others just didn't catch up yet?
Brandmeistertalk20:26, 4 September 2017 (UTC)reply
The presence of organic compounds is unsurprising;
uniformitarianism is a scientific philosophy concept which comes originally through geology, the astronomical concept is called the
cosmological principle, but basically the concept that we do not hold a privileged position in the universe shows up all over scientific philosophy. Knowing that, we would expect that all galaxies should contain, at a first approximation, roughly the same chemical make-up as our own, especially at the resolution we can detect from this distance. Organic compounds are found all over our solar system and galaxy (see
Methane#Extraterrestrial methane), so we have no reason to be surprised to find them elsewhere. It's a
nothingburger, scientifically speaking. --
Jayron3201:46, 5 September 2017 (UTC)reply
KP is a tabloid and RenTV is huge into UFOs, ancient aliens and other such "fine energies." it all seems to be going back to one dodgy Bashkirian site. There's
nothing at MIT's press site or on Gogole News either (try other news aggregators you know.) A similar article on beforeitsnews.com from 2011 (no idea if it's related) mentions the name Cordiner. Googling it turns up a
paper also from 2011. They found polyaromatic cyclic hydrocarbons (PAH) but Ctrl-F'ing "life" or "living" turns up nothing.
I still wouldn't call it "fake news" though because it's, on the whole, harmless and also utterly pedestrian if we take "organic" to mean "anything with carbon in it." It is, at most, a
"vbros" (2.) :)
80.171.95.62 (
talk)
17:02, 5 September 2017 (UTC)reply
Multiple days per year
At the poles, there's one day per year, if I understand rightly — solar noon is at the summer solstice, and "midnight" at the winter solstice. How far from the poles do you have to go before there's more than one day each year?
Nyttend (
talk)
23:00, 4 September 2017 (UTC)reply
You have to define what "day" means for this to be answerable. To me there are 365 days per year even at the poles, because even though the Sun stays above the horizon for very long periods, it still rotates around the center of the sky every 24 hours.
Looie496 (
talk)
23:26, 4 September 2017 (UTC)reply
If "sets" means "midpoint of the Sun goes below the theoretical line of the horizon" then it is very close to the poles. You don't have to move very far from the poles before there is an extra sunset at the equinox.
Looie496 (
talk)
23:47, 4 September 2017 (UTC)reply
Sunset is the refracted disk disappearing below the geoid. Note that you can't just use 50 minutes of arc because that rule of thumb was made by Europeans and noticeably underestimates refraction in Arctic air.
Sagittarian Milky Way (
talk)
01:10, 5 September 2017 (UTC)reply
So it seems that according to their calculations the answer is around 11 or 12 minutes of latitude, or about the same number of nautical miles from the pole. However, I don't know what definition of "sunset" they use, and for that matter I don't know how
stable their computations are for positions where the sun sets at such a shallow angle. I also didn't try other longitudes, or positions near the South Pole, or years other than 2017. For what it's worth, I did find that at 0° E/W you have to go to 89°44' N before you get an additional sunrise and sunset near the spring equinox. --
69.159.60.147 (
talk)
01:22, 5 September 2017 (UTC)reply
I was just gobsmacked by a claim
[4][5] sourced to
this citing work back to 1992 that there are estimated to be 10 million to 1 billion black holes in the Milky Way. Somehow by the second source that became 100 million at a minimum. Which if I recall is about the number of stars in the
Milky Way, though according to that article now it might be up to 400 million.
Does all this mean that it is as like as not that the nearest star to the Sun is a black hole? And that we have absolutely no idea where it is?
Wnt (
talk)
23:52, 4 September 2017 (UTC)reply
<OOPS> ... shoulda noticed that. Sorry! It's the stuff ya think ya know that gets ya. There are 2000 stars within 50 light years, so that puts the black hole at around 50 light years away at an average. Which is rather less intimidating. ;)
Wnt (
talk)
00:59, 5 September 2017 (UTC)reply
Well, the articles say there's a huge disconnect between the 100 million black holes expected and the 60 that are known. I have no idea what distribution is expected for them, but the anthropic principle only says we wouldn't have been hit with one .... yet.
Wnt (
talk)
11:56, 5 September 2017 (UTC)reply
The original paper being referenced is an assumption based on an assumption based on an assumption. A separate paper used three long-time microlensing events to discuss the probability that one or more of them might be a black hole. The assumption is that one is most likely a black hole. The other two may or may not be. The second assumption is that there are far more microlensing events than the three mentioned and that most of those will also be black holes. The final assumption is that if we count up all the microlensing events and assume they are mostly black holes, there are a huge number of black holes floating around. Now, what if it is proven beyond doubt that none of the original three microlensing events were black holes? All the handwaving becomes silly talk.
209.149.113.5 (
talk)
14:26, 5 September 2017 (UTC)reply
I do not think that it is a plausible claim. Black holes are among the rarest objects. There should certainly be many more
neutron stars than black holes. Still the closest known neutron star is at about 400 light years. I do not even want to talk about
white dwarfs which population is
huge. Still the nearest star is not even a white dwarf.
Ruslik_
Zero18:45, 5 September 2017 (UTC)reply
But the very next star system still has a star more special than the Sun (hotter, brighter, faster, stronger, bigger, older, more metallic, massive) because the universe wants to make us feel insignificant.
Sagittarian Milky Way (
talk)
19:39, 5 September 2017 (UTC)reply