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Old 13th September 2007, 07:11 AM   #11
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Quote:
Originally posted by Svante
OT: am I also right in that the port is not positioned very close to the driver? (Testing an hypothesis of my own here)
Looks like it's right around 40 cm long, too.
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Old 13th September 2007, 09:36 AM   #12
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Tube is 340/66mm and in opposite side of box.
Charles: What ripple you mean?

How can you see location and length of tube from these curves?

BR, Tipo
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Old 13th September 2007, 10:26 AM   #13
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Charles: What ripple you mean?
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Old 13th September 2007, 10:48 AM   #14
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Charles: What ripple you mean?
The "distortion" that you were complaining about.

Regards

Charles

Edit: sorry, had a nervous finger. Maybe a mod can remove above post.
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Old 13th September 2007, 07:56 PM   #15
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Quote:
Originally posted by tipo1000
Tube is 340/66mm and in opposite side of box.
Charles: What ripple you mean?

How can you see location and length of tube from these curves?

BR, Tipo
Length of tube is easy: you have a peak at 410 and another at 880. Ducts are open at both ends, so they tend to resonate at multiples of a half-wavelength. Imagine a sine half-wavelength with a node at the middle of the tube and the maximum at either end: that's the first resonance. Now visualize a sine with a maximum at either end and a full wavelength within the tube. That's the second resonance.

Let's look at the second resonance, the full wavelength: 340 meters/second / 880 Hz = 38 cm. That's a bit longer than your duct, so there might be some end effects going on.
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Old 13th September 2007, 11:01 PM   #16
Svante is offline Svante  Sweden
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Quote:
Originally posted by tipo1000

How can you see location and length of tube from these curves?
If the vent had been placed near the driver, the driver null would have been shifted towards lower frequencies due to crosstalk from the vent. Now it coincides with the resonance of the vent at ~21 Hz, which means that the crosstalk is negible.
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Old 13th September 2007, 11:10 PM   #17
Svante is offline Svante  Sweden
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Quote:
Originally posted by DSP_Geek


And you can even tell it's around 500 Hz by figuring out the period of the step response ripples. They're about 1.8 milliseconds (ms), indicating a problem around 550 Hz (period of 2 ms). Lo and behold, the response graph shows a narrow and deep null there, indicating a high-Q resonance. I wouldn't worry about it, since the subwoofer crossover should be many dB down by then.
Hmm, in the early part if the step response, there is a ringing with a period of ~0.5 ms. This is consistent with the peak at slightly less than 2 kHz.

At the null there is not a resonance, but an anti-resonance, and these typically does not result in oscillations (I might be wrong here), at least not as powerful as a resonance with the same Q.

I would guess that the 2 ms ripple comes from the sharp corner at 550 Hz, not the null at 650 Hz.
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Old 14th September 2007, 01:43 AM   #18
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Quote:
Originally posted by Svante


Hmm, in the early part if the step response, there is a ringing with a period of ~0.5 ms. This is consistent with the peak at slightly less than 2 kHz.

At the null there is not a resonance, but an anti-resonance, and these typically does not result in oscillations (I might be wrong here), at least not as powerful as a resonance with the same Q.

I would guess that the 2 ms ripple comes from the sharp corner at 550 Hz, not the null at 650 Hz.
Actually, anti-resonances will cause ringing in a step response. Here's a hand-waving explanation:

Look at a second-order transfer function equation set for unity gain, where w = 1:

Hunity(s) = (s^2 + s/Q + 1)/(s^2 + s/Q + 1)

The bandpass function is:

Hbandpass(s) = (s/Q)/(s^2 + s/Q + 1)

The bandstop function is:

Hunity(s) = (s^2 + 1)/(s^2 + s/Q + 1)

Note the bandstop is unity - bandpass. Now, set Q quite high. The bandpass function will ring on transients such as a step function. But wait! Since the bandstop is unity - bandpass, that will ring too, but in opposite phase from the bandpass. The ringing is identical for equivalent Q, only the sign changes.
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Old 14th September 2007, 07:11 AM   #19
Svante is offline Svante  Sweden
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Quote:
Originally posted by DSP_Geek


Actually, anti-resonances will cause ringing in a step response. Here's a hand-waving explanation:

Look at a second-order transfer function equation set for unity gain, where w = 1:

Hunity(s) = (s^2 + s/Q + 1)/(s^2 + s/Q + 1)

The bandpass function is:

Hbandpass(s) = (s/Q)/(s^2 + s/Q + 1)

The bandstop function is:

Hunity(s) = (s^2 + 1)/(s^2 + s/Q + 1)

Note the bandstop is unity - bandpass. Now, set Q quite high. The bandpass function will ring on transients such as a step function. But wait! Since the bandstop is unity - bandpass, that will ring too, but in opposite phase from the bandpass. The ringing is identical for equivalent Q, only the sign changes.

Mm, you are forgetting one thing. The bandpass only lets a narrow band through. This means that the signal has little energy.

The bandstop removes very little.

The amplitude if the ringings in these two signals will be the same, as you say.

But here is the twist; if these two signals are normalised, so they have the same amount of energy, the amplitude of the ringing will become larger in the bandpassed signal.

In real life the systems are not as simple as these, but my impression is that a peak of 10 dB causes much more ringing than a dip of 10 dB.
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Old 14th September 2007, 05:08 PM   #20
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Originally posted by Svante

Mm, you are forgetting one thing. The bandpass only lets a narrow band through. This means that the signal has little energy.

The bandstop removes very little.

The amplitude if the ringings in these two signals will be the same, as you say.

But here is the twist; if these two signals are normalised, so they have the same amount of energy, the amplitude of the ringing will become larger in the bandpassed signal.

In real life the systems are not as simple as these, but my impression is that a peak of 10 dB causes much more ringing than a dip of 10 dB.
That's the virtue of the step and impulse functions: they contain all frequencies so any resonant modes will get excited, no matter the bandwidth.

Ears are relatively insensitive to time domain phenomena such as ringing; they mostly work in the frequency domain. A 10 dB peak is more obvious because it increases narrow-band energy, which the ear notices more readily than a narrow-band lack of energy.

I still suspect the ringing will be the same, and I'm willing to throw a 'scope on a resonant system to find out, but things are _really_ busy for me right about now. Let's talk about this some time in November.


Cheers,
Francois.
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