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Help fixing massive 20dB peak!!

Discussion in 'Home Cinema Speakers' started by Dom H, Oct 2, 2003.

  1. Dom H

    Dom H
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    Moved into a new house setup the speakers and run some tones to test the response and setup the sub.

    the sub gave a very smooth response but the mains produced a massive nearly 20dB peak @ 80Hz follwed by quite a large null. (the 20dB was without any correction on the RS spl meter)

    The mains are X-overed at 80Hz, even setting the x-over to 120Hz still produces the peak.

    Turning the sub off does nothing so it's not that combining with the mains. Surprisingly moving the speakers also has very little effect, maybe knocking off 3-4 dB of the peak.

    Any ideas or is my only option equalization?
     
  2. Apocalypse

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    Sounds more like an acoustic problem rather than an equalisation problem, bass traps would be ideal to tame peaks created by the mains, if it were caused by the sub then the BFD would easily cope with it. Moving the speakers around won't alter the room characteristics obviously, your room like every other room has mathematical dimensions which excite certain frequencies. In my room I have a huge peak at 50-60Hz no matter where I place the sub.
     
  3. Dom H

    Dom H
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    thanks, does a BFD not work on mains aswell?

    Any idea how I find which dimensions are causing the problem?
     
  4. Apocalypse

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    The BFD works on whatever is connected to it at line level and in most cases people hook only the sub to it, it would for instance work on active speakers.

    To find out what your room behaves like use the calculator in this link : http://www.harman.com/wp/index.jsp?articleId=131
     
  5. alexs2

    alexs2
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    A pity that moving the speakers didnt seem to help your problem,and maybe the link above will help you to find out why.
    Bass traps and other room treatment systems(such as ASC's Tube traps)can be very effective,but also very expensive,and unfortunately the other option of active equalisation/room correction is also currently expensive.
    Moving soft furnishings around in the room can be surprisingly effective,depending on where the room reflections etc are.
     
  6. lowrider

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    Other than EQ:

    If your speakers have ports, try stuffing them with foam...

    Also, walk around the room, you will find out that in some places, probably the thirds of the room, you wont hear that boom, then place your coach there... :rolleyes:
     
  7. Dom H

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    I've found the problem

    yikes

    3 modes all around 80Hz on all 3 dimensions.
    Never realised the room was similar in length/width.

    I dont think the height is a problem as my head is closer to the centre where gain is minimum (if im reading the graph right)


    This leaves the 2 secondary modes on the length/width. If I sit to one side of the sofa and lean my head back :rotfl: thus putting me in one of the thirds of the length/width with my head the boom totally dissapears (as the graph predicts).


    While this isn't very practical (nor comfortable!) I think I can get rid of the width mode because there is a staircase behind the sofa, if I made the railings solid this would effectively reduce the width of the room a couple of feet, removing one of the 80Hz modes leaving only the lenght mode which seems to be cured by sitting off centre.


    One question I have is, what could I put on the railings to reduce the width of the room, ive tried some wood (my wife thinks ive finally lost it ;)) about 10mm thick which didn't seem to do anything, is that not thick enough?
     
  8. lowrider

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    Remove the railings and seat on the stairs to listen to music... :laugh:

    I don´t think you can solve your problem puting whatever on the railings... :rolleyes:

    One solution for square rooms is placing the speakers diagonally, probably not pratical, I know... :eek:

    I think that 80 hz has a wave length of 53 cms, are your speakers that distance from any wall, or the floor, if they are, try to change that... :rolleyes:
     
  9. avanzato

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    Lowrider: I worked out the wave length to be 14.75ft (about 4.25m)

    Speed of sound 1180 ft sec divided by frequency 80hz.

    Dom: The wood on the staircase will not be a solid enough 'wall' to effect the standing waves. If you can, move the speakers into a null of 80hz in your room then it will stop driving that mode and the mode will disappear. The null is the dip in the black line on your graphs for length or width.
     
  10. lowrider

    lowrider
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    OK, thanks... :blush:

    So it is not the wave length, (my knowledge is not that technical), what I think is meaningfull, (I read it somewhere), is that 53 cms is the maximum distance your sub, (and speaker woofers, of course), should be from walls and floor in order to reinforce all frequencies up to 80 hz...

    So, if Dom H keeps his speakers away from that distance, it might help a bit, IMHO... :rolleyes:
     
  11. NicolasB

    NicolasB
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    Rather the reverse, actually: the speakers should be at a point where the sound is loudest when the speakers are against the wall.

    One good way of breaking up standing waves is to use diffusors. Most standing waves depend on large, hard, flat, parallel surfaces. If you can make the surfaces very much not-flat then they can't form in the same way. For example: if you have a bookshelf with a wide mixture of books of lots of different sizes, then that makes that part of the wall no longer flat. Convex surfaces are also good. I don't know if anyone has tried stacking shelves with lots of different types of large bottles (preferably full, or they'll rattle) but I expect that would work too. (There was a craze at one point during the 70s for covering the walls in eggboxes - same principle).
     
  12. avanzato

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    NicholasB can you point me in the direction from where you got that positioning information, I always willing to learn.

    My advice comes from the Harman White paper Getting the Bass right and backed by a quote from a Terry Montlick on AVS forums.

    The logic of your positioning argument is that you would put your speakers in the corners of the room where the bass is loudest and expect them to sound good. Axial modes end at the corners and sides of a room that's why the sound is loudest. Also the loudness of different frequencies will be at different points along the wall which ones are you suggesting putting the speakers at?

    I think I've got the maths for low frequency diffusors somewhere, I'll have a look.
     
  13. NicolasB

    NicolasB
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    Well, I'm specifically talking about reducing the influence of standing waves - there may be other considerations when it comes to positioning speakers.

    Generally, the closer you put the speaker to the node of the standing wave, the larger the vibration will end up being at the antinode. Ask a violinist: how do you adjust the volume of a note (played with the bow)? The answer is that you move the bow slightly towards or away from you. The string is fixed at both ends, and the note the string produces is the first-harmonic standing wave - nodes at the ends of the string, and an antinode in the centre. If the bow is farther along the string towards the centre, the vibration is less - quieter note. If it is brought towards the bridge it's greater - louder node.

    So to minimise standing waves in a room you need to place the speaker at or near the standing wave antinode. The closer the speaker is to a node (to the wall, for example) the more significant the standing wave will be. A good way of finding an antinode is to put the speaker in a position where it is generating as big a standing wave as possible (e.g. in the corner) then listen for where in the room the resonance is at its loudest - that will be the antinode, and potentially a good place to put the speaker.

    (Obviously you have to be a little careful with this, as there isn't only the first harmonic to worry about. The best way to minimise the first harmonic might be to put a speaker exactly in the centre of the room - first harmonic antinode - but that's also a node for the second harmonic.)

    I think possibly you thought I meant you should arrange the speakers so that the sound is at its loudest. What I actually meant was that you should find the listening position where the sound is loudest, and then put the speaker there.
     
  14. avanzato

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    I think we are talking at cross purposes here.

    That is what I said. Position the speakers at the distances that the spreadsheat calculates for the anti node/null of the second mode and the standing wave will not be energised.

    That doesn't sound right to me. A speaker in the corner will energise all room modes so when you listen for the loudest resonance you are hearing the point of highest sound pressure. A standing wave not the null.

    Two speakers in phase either side of the first mode putting out effectivly mono bass will cancel the first mode out. The polarity of the sound being -ve on one side of the null and +ve on the other. If in addition you position the speakers at the nulls of the second mode as I suggested they will then not energise that mode either. The third mode will also cancel as one speaker is in an area of +ve phase and the other is in an area of -ve phase.

    Sorry I don't understand what you are saying there, what listening position?
     
  15. Dom H

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    Thanks guys, can someone explain why moving the speakers affects the standing waves, im not sure I understand.

    Secondly the graph I linked to, am I correct saying the amplitude of the standing waves are least at the dips and highest at the peaks? i.e I could reduce the height mode if I were directly half way between the roof and the floor? same width/length ways but a 3rd of the way?

    Many thanks

    Dom
     
  16. avanzato

    avanzato
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    The speaker is the 'engine' for the standing wave. The mode is where the sound reflections reinforce themselves and the null is where they cancel out. The closer to the null you move the speaker the more energy at that frequency is cancelled and the smaller the standing wave becomes. Move towards the mode and the reverse happens.

    Yes the graph has the peak SPL at the top and the minimum SPL at the bottom. You have to try and imagine the graphs in 3D. This is where I get a headache.

    You reduce the peak by moving closer to the null. Half way between the floor and ceiling you would be in the null of the first and third modes but in the peak of the second and forth. When you say '3rd of the way' do you mean the null of the 2nd mode or a little bit further along? A bit further than the null along width and length would probably halve the 2nd mode but only knock off 10-20% of the 4th mode.

    Ultimately the speadsheet won't match your room exactly as the rooms construction will alter where the standing waves appear. Plasterboard ceilings and wooden floors flex and absorb bass for example.

    The fourth mode is into where simple acoustic treatments would work. You just need to sort out the lower modes. :D which if it were that easy we'd all do it.
     
  17. NicolasB

    NicolasB
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    The term "null" refers to a node not an antinode. It is therefore the worst possible place to put a (single) speaker.

    The situation with a pair of speakers is more complex. If you're talking about reducing a resonance generated across the width of the room, then yes, a pair of speakers in phase either side of a node will not excite a standing wave. But that doesn't help with standing waves generated along the length of the room, nor with higher order harmonics acorss the width, which have an even number of nodes between the two speakers.


    A bit of standing wave theory for newbies:

    Standing waves happen in situations where a wave passes repeatedly along the same path, for example when it bounces back and forward between two points. What happens is that at certain points (the "nodes" or "nulls") there is no oscillation at all, while at others (the "antinodes") there is lots of oscillation.

    A good experimental demonstration: get a piece of light rope and tie it to a door handle. Hold on to the other end and pull it reasonably tight. Now swing the other end gently to and fro, and vary the rate at which you swing it. You'll find that there's one particular rate that you can swing it at which will result in the rope swinging to and fro through a large distance - the centre of it moving through a much larger distance than your hand. You've now generated a standing wave in the rope. The nodes are at the point where the rope is tied (which obviously can't move at all) and a point just behind your hand. (More about that in a sec). The antinode is the point in the middle of the rope which moves through a long distance.

    Points to note:

    1) If you try moving the rope a bit faster or a bit slower then there will be much less movement. At one particular frequency there is lots of movement. This is an example of what's called "resonance": the system has a natural vibration frequency, and if it is driven at that frequency you get a much larger oscillation than you do at other frequencies.

    2) If you try waggling the rope at double the rate that produces the standing wave we were previously talking about then you fill find that, again, you get lots of oscillation. But this time the pattern is different. Now you have point where there is no oscillation in the centre of the rope as well as the far end and (effectively) behind your hand. And you'll note that, on either side of that point in the centre, the rope is moving in opposite directions. There are now two points which have maximum oscillation.

    This is what's called the "second harmonic" frequency, and it's a characteristic of standing waves that if the system exhibits first harmonic resonance at a particular frequency it will also exhibit resonance at multiples of that frequency.

    If you waggle the rope really quickly (but not through a large distance) you may also be able to see a third harmonic (three antinodes and four nodes) at three times the original frequency, and so on.

    Standing waves in a room work very like this, except that in a room you've effectively got the rope tied at both ends, and you are waggling it at a point in between. (A standing wave is formed under these conditions because the fact that rope can't move to and fro at the ends causes a wave moving along the rope to be reflected at the ends and travel back along the rope). The walls of the room correspond to where the rope is tied, and the speaker corresponds to your hand moving the rope to and fro.

    Strictly speaking sound is a "longitudinal" wave while a wave moving along a rope is a "transverse" wave. This means that in the rope the direction the wave travels - along the rope - is at right angles to the oscillation - the rope swinging to and fro. In a sound wave the movement of the air is to and fro along the direction of the wave. Sound reflects from the walls because the air molecules cannot move to and fro through the wall.

    But this isn't too important. What matters is that you have the same pattern of nodes - no movement - and antinodes - lots of movement - in the same relative positions. There are nodes at the walls of the room. The first harmonic has an antinode in the middle of the room. The second harmonic has a node exactly in the middle (and at the walls), and antinodes 1/4 and 3/4 of the way across the room. The third harmonic has nodes at the walls, 1/3, and 2/3 of the way across the room, and antinodes 1/6, 1/2, and 5/6 of the way across, and so on.

    So, if you have a sound in the room whose frequency matches the resonant frequency, that sound will be amplified - a small vibration of the speaker will make a very big vibration in the air, in the same way as a small vibration in your hand makes a big vibration in the rope. You will get the biggest amount of amplification at the antinode of the standing wave. The effect spoils the sound, because it means that certain specific frequencies sound much louder than others, even if they are supposed to sound the same volume in the original recording.

    Now, think about the rope tied at both ends, stretched reasonably tight, with you waggling it through a small distance. Whereabouts would you put your hand in order for the resonant vibration to generate the biggest possible resonant vibration for the smallest possible movement of your hand? The answer is that your hand needs to be right next to the node (for the first harmonic, right next to the end of the rope). If you put your hand almost in the middle of the rope and waggle it at the first harmonic frequency the rope will only move about the same amount as your hand does. But if you waggle it right at the end, the middle of the rope moves through a much bigger distance.

    So if you want to reduce the extent to which a subwoofer excites standing waves (i.e. the influence on the sound that the standing wave has) what you need to do is to put the subwoofer at or near the antinode of the standing wave that is causing a problem. That way the standing wave will be much less excited than it will be if you put the subwoofer next to a node. Putting the sub against the wall is the worst possible place, because all the harmonics have a node at the wall.

    Putting it exactly in the middle of the room will result in hardly any first harmonic - but of course the middle of the room is a node for the second harmonic, so the exact room centre may not be the best place. The best place will be a compromise - you'll have to accept a certain amount of some harmonics because reducing them will result in other harmonics being made stronger.

    The business of putting a pair of speakers either side of a node has to do with the way that the rope is moving in opposite directions on either side of a node - if you make it move in the same direction then a node cannot form at that point, and that harmonic drops out.

    Finally: in a room, of course, things are a bit more complicated, because you're working in three dimensions where a wave moving along a rope is working in just one :) . In a room you can get standing waves forming between the walls across the room, along the room, between floor and ceiling, and also potentially along much more complicated paths - the only thing that's required is that a sound wave can end up travelling through the same point, headed in the same direction. It doesn't matter how many times it has to be reflected on the way. So you can actually get standing waves bouncing around between multiple surfaces at weird angles. But the basic node/antinode behaviour is still the same.
     
  18. lowrider

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    Why don´t you go back to basics, place the subwoofer where you are going to seat, if you have alternative seating positions, try them too...

    Then play just the sub and walk around the room, until you find the place you hear less boom, place the sub there...

    Then try to place the front speakers on the nulls for the 80hz width node to cancel at least those...

    Good luck... :suicide:
     
  19. dunkyboy

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    Thanks NicB, that's a fantastic explanation of standing waves - I've always had a vague idea of how they work, but (as you are no doubt aware) I was often mistaken about the details. Now I have a much clearer idea of how they work. :smashin:

    Dunc
     
  20. avanzato

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    My bad. By that point I had decided you must be using Node the way Auralex does on it's LENRD page 'The resonant bump'.

    Now we agree on what the null/node is and with your description I know where you have gone wrong.

    A standing wave cannot have a node at the wall.

    The node is where the soundwave changes polarity. If the node is at the wall the reflected wave will be in opposite polarity to the incident wave and they will cancel. This is easy to test put on a CD and walk to the wall. The sound at the wall will be louder, if there were nodes at the wall the sound would get quieter.

    Speakers are pressure devices. The movement of the cone pressurises the air as it moves. A speaker placed at an antinode will be in a high pressure area of the mode and can easily transfer it's energy into the mode. Placed at a node or near a node these being low pressure areas the speaker will inefficiently transfer the energy. Try listening to a speaker in space where there is no air pressure.

    As Dom wants to reduce the energy in the modes an inefficient transfer is what we want.


    The sound is 'amplified' at certain points in the room because the in phase reflections enhance each other and the out of phase reflections cancel. This is why the sound level varies across the room. Different parts of the waveform at different amplitudes and different phase interacting with each other.

    As I've demonstrated there aren't nodes at the wall and putting the sub at the antinode will increase the energy transfer to the mode. The walls have a large effect on the mode because the sub will be in an high pressure area, high pressure because that area has 'in phase' reflections for the longest time as the wave reflects.

    I was not talking about a single speaker in my first suggestion. The best for a single speaker is to be as close to the node/null of all the modes you don't want to energise. There by minimising as much as possible the excitation of each one. This is usually gets in the way of the TV in my experience. :D This also applies to multiple speakers but Dom had a specific problem at 80hz. Positioning also has to take into account where the furnture is and spacing for best imaging so the best modal position is probable not the best practical position.

    All this is in the Harman PDF I linked to earlier in the thread.
     
  21. dunkyboy

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    Argh, just when I thought I had it all worked out, avanzato, you go and confuse me again! Where's that bashing-head-against-the-wall smiley! :p

    Dunc
     
  22. avanzato

    avanzato
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    Dunc it took me a while of staring dumbly at the diagrams and reading books before I got it.

    I finally figured it out when I realised that the standing waves which aren't moving are made up of sound waves that are moving. When the reflected sound waves move over each other 'in phase' they reinforce and created an area of higher pressure. The standing wave happens when the 'in phase' reflections always overlap at the same place in a room. The string example doesn't take in to account that the wave is moving forward and reflecting.

    Don't know if any of this is helping Dom :rolleyes: :D
     
  23. NicolasB

    NicolasB
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    I'm not surprised. I'm afraid quite a lot of what Avanzato is saying here is simply wrong. I'll try and explain why, but I'm spending way too much time posting to this thread already, so I may have to skip over a lot of this, and this is certainly the last time I'm going to repeat a lot of it. From now on, if you don't believe me you'll just have to look it up yourself.

    First, a general point: Avanzato needs to be much more careful to distinguish between acoustic phenomena that are the result of standing waves, and ones that are not.

    OK, let's go.

    No. It can, and (if it's a standing wave formed between two parallel walls) it invariably does.

    No. The node is the point in a standing wave where there is no oscillation.

    If you have a node in the middle of the room somewhere then it is true that, at any given moment in time, the oscillation on one side of that node will not be in the same direction as the oscillation the other side of it. The problem is that Avanzato is using the word "polarity" to mean several different things, and assuming that, because he is using the same word for all of them, they are actually all the same.

    To illustrate what he means by "polarity" here, think about a standing wave in a rope or a string, a transverse wave with the string moving from right to left; you could say that, at any given moment and at any given point, the rope must either be moving left or right, and that if it is moving left one one side of a node then it is moving right on the other side.

    However, all of this is of no relevance whatsoever when it comes to the issue of whether you can have a node at a wall.

    Okay, this is completely wrong.

    We've dealt with the first thing Avanzato is using "polarity" to mean. Let's try and explain the second.

    When a wave is reflected it can happen in two possible ways. It can be reflected without changing phase, or it can be reflected with a 180 degree phase change. (The "phase" of a wave means what the wave's medium is doing at any one point. If you consider a wave from peak to trough to peak, one full cycle, that's described as "360 degrees" of phase. "Phase difference of 180 degrees" therefore means a change from a peak to a trough, or vice versa. If you imagine a transverse wave, like the ripples on water, any given point on the water's surface is actually moving up and down, not in the direction of the wave - how far up or down it is at any one point is the phase of the wave.)

    Now, imagine one of those "slinky" springs, stretched out, with one end fixed. You waggle the other end and send a "pulse" along the spring. When it reaches the other end the pulse is reflected off the fixed end and travels back along the spring again. If your pulse consists of the spring moving to the right and then back to the centre again, then the reflected pulse will also consist of the spring moving to the right and then back to the centre. If you now try the same thing with the other end of the sping free to move, then your pulse will still be reflected back, but this time the pulse will be inverted - the spring moving to the left and recovering.

    (The reflection happens at the loose end because the end of the spring moves to the right, swings back, overshoots to the left, and thus initiates a moving-left pulse travelling in the opposite direction).

    In general, if you send a wave form along the spring, and it is reflected off a fixed end, then what any part of the spring was doing immediately before the wave was reflected is the same as what it is doing after reflection. The peak of a wave remains the peak, the trough of a wave remains a trough. It's just that the wave is now travelling in the opposite direction. But when the wave reflects off a loose end then everything is reversed - a peak becomes a trough and vice versa.

    What determines whether there is a phase change or not is whether the wave is bouncing off an "open" or "closed" end - i.e. whether the wave is bouncing because it has encountered a region where it is much more difficult for the wave to propagate (fixed end, hard to move) or much more easy (loose end with no coils the other side of it to hold it still, easy to move). In acoustic terms, a wall is a "closed" end - the sound can't make the wall compress and uncompress nearly as easily as the air. Something like the open end of an organ pipe acts as an "open" end, and causes sound coming up the pipe to be reflected back down with a 180 degree phase shift.

    Okay, so we've established that, at a wall, there is no phase shift when the sound is reflected. So, the vibration that results from a wave immediately before it hits the wall will be the same as the vibration from the same wave immediately after it hits the wall. These two vibrations add together in a way that's called "constructive interference" to make a larger vibration (a bit like two ripples passing through one another - at a point where the peaks from both ripples coincide, there is a higher peak). Hence, as Avanzato says:

    HOWEVER, this analysis is only valid when you are considering a travelling wave, i.e. NOT a standing wave.

    So when Avanzato goes on to say:

    this does not follow at all.

    What happens in a standing wave is that you start off with an ordinary travelling wave. It gets reflected back along its original path in the opposite direction, then gets reflected again so that it comes back along the original path in the same direction again, and so on, to and fro. Thus, the same wave is not just passing through the point next to a wall twice (once before reflection and once after) it is passing through it many times. If you go through the maths for this, and add up all the waves that are interfering with each other, then you will find out that the result will either be pretty unrecognisable, or it will the familiar standing wave pattern of nodes and antinodes. (Which it is depends on how the wavelength of the wave - the distance from peak to peak - compares with the distance between the walls - that's why only certain wavelengths form standing waves).

    (Continued in next post)
     
  24. NicolasB

    NicolasB
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    (continued)


    Now, in a sense, Avanzato is using the word "polarity" to mean "phase" - but to do so is misleading. Phase is not something that has two discrete states, it is a smoothly-varying quantity, The phase of a travelling wave varies across the cycle from peak to trough to peak. The phase of a standing wave is a rather more slippery concept, but it could be said to be varying smoothly between a node, an antinode, and the next node. But either way the fact that a travelling wave does not change phase by 180 degrees when it reflects off a wall simply doesn't have anything to do with whether or not you can have a standing wave node there. (Well, it has, but not in the way that Avanzato suggests).

    And the amplification of a travelling wave at a wall has nothing to do with standing waves. If you measure the intensity (loudness) of a standing wave then you will indeed find that it is quiet next to the wall, and loud in the middle of the room (or wherever the antinode is).

    Think about a vibrating string, fixed at both ends - this is a standing wave. The ends of the string must be nodes because they cannot move! The wave in the string is reflected off the ends without a phase change, because thye are "closed" ends. Hence, clearly you can have a standing wave node at a point where a wave is reflected without a phase change. The ends of the string are exactly analogous to the walls of the room.

    True; but he goes on to talk about "high pressure", and the effects of a vacuum, which suggests that he doesn't really understand what a sound wave is. A sound wave is created by a variation in pressure from higher to lower.

    Think about the stretched out slinky spring again, but this time instead of waggling the end of the spring from side to side, imagine pushing it forwards and then pulling it back. This will send a different sort of "pulse" along the spring: instead of the coils moving from side to side (at right angles to the direction of the pulse) what is now happening is that they move to and fro along the length of the spring (parallel to the direction the pulse is travelling). Before the "peak" of a wave consisted of a point where the spring coils had moved sideways by the maximum amount. Now the "peak" is the place where the coils are compressed together as closely as possible, and the "trough" is where they are farther apart. In both cases the coils themselves are vibrating to and fro around the same point, but the place where the coils' position its at is biggest deviation from the normal position moves along the spring (this is what a wave is).

    This is called a "longitudinal" or "compression" wave. It behaves much the same way as a "transverse" (side to side) wave, and can be described in the same way. The distance between regions of maxmimum compression is the wavelength. How much closer together or farther apart the coils are is the "amplitude" of the wave - analogous to the "height" of a transverse wave.

    A sound wave is a compression wave of this sort. The speaker cone moves the air at the cone backwards and forwards. As it moves forwards it compresses the air in front of it (gives it a higher pressure); as it moves backwards it gives it a lower pressure. The air molecules act like the coils of the spring, so the point in the room at which the air molecules are relatively squeezed together (high pressure), or relatively far apart (low pressure) moves through the room; that's what the wave is.

    As explained, this is not correct. At any one point the wave produces a cycle from higher-than-normal pressure to lower-than-normal pressure. The antinode is the point at which the difference between the maximum pressure and the minimum pressure is at its greatest, not where the absolute pressure is highest - sound waves do not generate static areas of high pressure.

    Again, simply wrong. Think about a rope fixed at both ends and waggled through a small distance to and fro at the first harmonic frequency. Does the middle of the rope move more when you hold the rope in the middle, or when you hold it near one end?

    True, but (again) irrelevant. As we've discussed, sound waves are not about the level of pressure but about the level of pressure variation.

    Correct. But putting the speaker at the antinode is the way to achieve this.

    Yes, but, as discussed, that argument really only applies to travelling waves.

    Yes, in the sense that a standing wave node/anti-node pattern is, ultimately, the result of interference between a wave and its reflection, but it is the product of a large number of reflections, not just one.

    You haven't.

    There are.

    No, it will decrease it.

    Well, I could go on, but you get the general idea....


    Already acknowledged.

    {quote]The best for a single speaker is to be as close to the node/null of all the modes you don't want to energise. There by minimising as much as possible the excitation of each one. [/quote]
    No, it's best at the antinode.

    As the antinode for the first harmonic is in the middle of the room, yes, this is a problem. :(
     
  25. Nobber22

    Nobber22
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    Hell's Bell's Nic! :eek: I have a few truths about gravity, the shape of the Earth and the Dodo I might need you to check up for me. :D

    Seriously, these highly technical explanations are fantastic, just what some of us are looking for. Doesn't mean I understand it all, but learning new stuff is what this hobby is all about. Thank-you. :smashin:
     
  26. avanzato

    avanzato
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    I can see that NicolasB can't be pursuded to change his mind and like him I will just be repeating what I've already said. He is wrong and when he relaxes his grip on peices of string he might get somewhere.

    I will just point people again to the Harman White Papers on speaker positioning and standing waves and in particular NicolasB to the bottom 2 pictures on page 11 of the Loudspeakers&roomPt3.pdf. The papers are produced by one of the largest audio companies in the world and by Dr. Floyd E. Toole a world leader in acoustic science. Ignoring any technical arguments the positioning advice in the paper is the same as I recommend not as NicolasB recommends.

    For anyone interested in understanding their room acoustics I also recommend buying or borrowing the 'Master Handbook of Acoustics' about £20 from online bookstores.
     
  27. Dom H

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    wow thanks alot guys.

    I haven't the time to read it all now (i will later promise)

    Just like to add, with the use of the graph i've managed to record a new curve from about 20Hz - 160Hz which is +/-9dB, a big change from >20dB peaks i'm sure you'll all agree!

    1 more question, my front door is inline with my seating position, does this mean the standing waves across the width will be weak at the seating position as energy is lost through the wooden door? in effect the room ceases to be 13ft as there is no brick wall on one side.

    Much apreciated

    Dom.
     
  28. NicolasB

    NicolasB
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    I'm actually slightly miffed that you should regard them as "highly technical" - they're designed for someone who doesn't have much grasp of physics, hence all the analogies about strings and springs.

    This isn't exactly degree-level. In my day it was in the Physics O-Level syllabus. I would imagine it's still in the GCSE syllabus now.
     
  29. avanzato

    avanzato
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    Dom I'm glad all this has improved your sound.

    Good one with the door. The door will have an effect due to it's differing acoustic properties. Calculating it would be complex I think and it's surface area in relation to the brick wall will be small. So it won't throw out the calculations too much. Changes in air pressure/temperature between days will have more effect on the calculations as sound travels at different speeds depending on the air pressure.

    Incidentaly opening the front door would act as a perfect bass trap as any sound wave hitting the opening can't reflect and will dissappear outside. It would be cold in winter :D
     
  30. NicolasB

    NicolasB
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    Ahhhh.... I see what the problem is: you're talking about variations in pressure, where I'm talking about the movement of air. (Not surprising I couldn't figure out what you were talking about with all that gibberish about "polarity" and vacuums).

    Yes: areas of high pressure variation correspond to areas of low oscillation of air molecules, and vice versa. So, in air movement terms the wall is a node - the air is stationary at that point. (It has to be because the molecules can't flow backwards and forwards through the wall). But in pressure variation terms it's an anti-node.

    What a subwoofer does is to cause air to move to and fro. To cause it to excite as little standing wave as possible, you need to put it where the air is moving the most - at an air-movement anti-node. (As I've already explained at some length). But in pressure variation terms that's a node.

    Either way, you're actually talking about the same point! As far as the first harmonic is concerned, the centre of the room is a pressure node and a movement antinode, so, whichever way you choose to describe it, putting the subwoofer in the middle of the room is the best possible place to minimise the first harmonic.

    And the rest of what has been said previously also applies: the centre of the room is the optimum position for reducing odd-numbered harmonics, but it's the worst possible position for the even-numbered harmonica, so there are going to be trade-offs....
     

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