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Thread: 2235H Project Help

  1. #16
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    Checked Your data

    I checked Your data. Your box should resonate at 35.6 Hz. I ran WinISD, selected the 2235H driver, 170 liter box volume and played a bit with the tuning: 32 Hz box resonate frequency (Your target) is a relative optimum.

    The ducts have one end flanged and one end free standing. The end corrections are:

    flanged: 0.85 R
    free: 0.613 R
    together: 1.463 R

    So the effective length of each tube should be (6 + 2.93) in = 8.93 in.

    Perhaps Your simple sine wave generator has a broken frequency pot?

    Your picture shows that You are using a tube amplifier. These things often do have a high output impedance. And Your passive xover will have a series inductor with maybe a high resistance. The following post shows how to calculate the resulting Qt for Your driver. This will have an influence on bass response.

    http://audioheritage.org/vbulletin/s...&postcount=105

    Ruediger

  2. #17
    Senior Member Loren42's Avatar
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    Quote Originally Posted by Ruediger View Post
    I checked Your data. Your box should resonate at 35.6 Hz. I ran WinISD, selected the 2235H driver, 170 liter box volume and played a bit with the tuning: 32 Hz box resonate frequency (Your target) is a relative optimum.

    The ducts have one end flanged and one end free standing. The end corrections are:

    flanged: 0.85 R
    free: 0.613 R
    together: 1.463 R

    So the effective length of each tube should be (6 + 2.93) in = 8.93 in.

    Perhaps Your simple sine wave generator has a broken frequency pot?

    Your picture shows that You are using a tube amplifier. These things often do have a high output impedance. And Your passive xover will have a series inductor with maybe a high resistance. The following post shows how to calculate the resulting Qt for Your driver. This will have an influence on bass response.

    http://audioheritage.org/vbulletin/s...&postcount=105

    Ruediger
    So, you are saying that the effective length of my ports are really 8.93" even though the physical length (with flange) is 6.0"?

    I published the link to my crossover design earlier, but here it is again so that you can see what I did:

    http://www.mdbq.net/pyramid/LEAP.pdf

    The main inductor is a Solen inductor I got here:

    http://www.madisound.com/catalog/pro...roducts_id=903

    Yes, that is a tube amplifier that I use for these cabinets.

  3. #18
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    End Correction and Generator Resistance

    Quote Originally Posted by Loren42 View Post
    So, you are saying that the effective length of my ports are really 8.93" even though the physical length (with flange) is 6.0"?
    1st of all: I may have made a mistake in my calculations, that's always possible

    Yes, the effective length of a tube is physical length plus an end correction. For the flanged end of a tube it's 0.85 * radius, and for a free standing end it's 0.613 * radius. For cases in between: interpolate.

    The 5.5 mH coil does have a DC resistance of 0.63 Ohms. Your amplifier will have an output impedance which cannot be neglected. Ask the manufacturer (easy) or measure it (not so easy). Often a "damping factor" is specified from which You can calculate the output impedance. Your cables will have a resistance (calculate). The sum of these 3 resistances is the "generator resistance" for the Qt-formula. Take this corrected Qt into account when calculating the box response.

    Ruediger

  4. #19
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    Quote Originally Posted by Ruediger View Post
    1st of all: I may have made a mistake in my calculations, that's always possible

    Yes, the effective length of a tube is physical length plus an end correction. For the flanged end of a tube it's 0.85 * radius, and for a free standing end it's 0.613 * radius. For cases in between: interpolate.

    The 5.5 mH coil does have a DC resistance of 0.63 Ohms. Your amplifier will have an output impedance which cannot be neglected. Ask the manufacturer (easy) or measure it (not so easy). Often a "damping factor" is specified from which You can calculate the output impedance. Your cables will have a resistance (calculate). The sum of these 3 resistances is the "generator resistance" for the Qt-formula. Take this corrected Qt into account when calculating the box response.

    Ruediger
    Useful knowledge. Thank you.

  5. #20
    Senior Member Loren42's Avatar
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    I have what I think are impedance plots for this cabinet:

    Here is the raw frequency plot data when plotted with a 829 Ohm resistor in series with the amp and the voltage measured at the speaker terminals:



    The program I am using (Fuzzmeasure) is very fuzzy about how to actually generate impedance plots from the frequency sweep and I did my best:



    The cabinet is tuned to 28 Hz, which is not too bad.

    I did a number of other more detailed sweeps with the cabinet in a number of orientations and I can say that the real rolloff is at 45 Hz. The room complicates things very much.

    I have a few other things to try, like reducing the internal fill in the cabinets and take more detailed measurements. I will keep you guys posted. THANKS!

  6. #21
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    Quote Originally Posted by Loren42 View Post
    I did a number of other more detailed sweeps with the cabinet in a number of orientations and I can say that the real rolloff is at 45 Hz.
    Maybe for the woofer but certainly not for the ports unless they are broken. If you are thinking that the woofer itself should be flat down to 28 Hz then you are mistaken. The bottom octave is handled by the port(s).

  7. #22
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    Quote Originally Posted by 4313B View Post
    Maybe for the woofer but certainly not for the ports unless they are broken. If you are thinking that the woofer itself should be flat down to 28 Hz then you are mistaken. The bottom octave is handled by the port(s).
    Good point. However, if the ports are supposed to be contributing to the total picture, "Where's the bass?" If I may coin a phrase.

    This is a carefully performed on-axis sweep with the microphone 1 meter from the front of the cabinet. The mic is centered on the cabinet.



    Should this not be the aggregate of the woofer and ports?

    When I put the mic at the listening position at the center of the sofa (magenta trace) and compare that with the 1 meter on-axis plot (green) I get this:



    I have a horrible 50 Hz antinode at that point in the room and both speakers have the sweep tone sent to them equally.

    An alternate arrangement for the speakers in the room (both speakers backed tight into the corners) smooths out that 50 Hz hump, but places a strange dip in the high end (blue trace).



    Hard to tell what is room artifact and what is really output from the speaker.

    Maybe there is more bass from the speaker than what the 1 meter plots reveal?

  8. #23
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    Underdamped response?

    Your frequency response can result from underdamping by a generator impedance which is too high. If I imagine that influence subtracted from the curve on the plot I see a -3dB point of 30+Hz.

    Test 1: short circuit the series inductor in Your xover. That will remove 0.63 Ohms, which is a lot. Then run the test again. Ignore what happens above 500 Hz.

    Test 2: get a proper transistor amp (that means: an amp without output transformers) with a high damping factor (200 or better). Keep the series inductor short circuited. Run the test again.

    Ruediger

  9. #24
    Senior Member Loren42's Avatar
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    Quote Originally Posted by Ruediger View Post
    Your frequency response can result from underdamping by a generator impedance which is too high. If I imagine that influence subtracted from the curve on the plot I see a -3dB point of 30+Hz.

    Test 1: short circuit the series inductor in Your xover. That will remove 0.63 Ohms, which is a lot. Then run the test again. Ignore what happens above 500 Hz.

    Test 2: get a proper transistor amp (that means: an amp without output transformers) with a high damping factor (200 or better). Keep the series inductor short circuited. Run the test again.

    Ruediger
    I have a solid state amp, but I doubt the damping factor is that high. Needless to say, I can give it a whirl. Thanks.

    Also, if I can man handle this thing outside I could run a sweep in the open yard or maybe face the cabinet up toward the sky, but I wonder about ground effects below the baffle step (210 Hz in this case)?

  10. #25
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    rear wall reflection compensates baffle step diffraction

    Quote Originally Posted by Loren42 View Post
    Also, if I can man handle this thing outside I could run a sweep in the open yard or maybe face the cabinet up toward the sky, but I wonder about ground effects below the baffle step (210 Hz in this case)?
    I think there is a common missunderstanding. Below a certain frequency the loudspeaker is radiating into full space (4 pi), above that frequency it is radiating into half space (2 pi).

    If the speaker is in a room and not in the free field then the sound which is radiated to the rear will be reflected by the wall behind the speaker.

    At 210 Hz the wavelength is 1.6 meter. Reflected sound will add to direct sound with quite a phase shift. At 100 Hz the wavelength is 3.4 meter, and the phase shift will be much smaller.

    At 50 Hz the wavelength is even 6.8 meter. Direct and reflected sound are almost in phase.

    This means destructive interference will happen around 210 Hz, and at low frequencies the interference will be constructive, "bass will not get lost".

    To my opinion it is wrong to boost the bass range to correct the baffle step diffraction. That would only make sense in free field. But then a reflecting panel behind the speakers is the more economic alternative.

    Ruediger

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