help with bass reflex port response

Hello to everyone, it is been a while since i last postet on this forum. I recently build new speaker cabinet for my seas excel drivers. I just need your opinion on the port frequency responce. I use the W26fx001 woofer in 60l cabinet with 70mm diameter and 260mm port lenght (front port). The port measurement shows that the port is working just fine in the low end but there is something at 200-500 hz. It doesn't affecting the overall speaker responce but what is that? port resonances? The woofer near responce dont show significant cabinet resonances. Or just the port noise due to the air blowing at the mic? You can see the port responce with the crossover filter aplied also. the cross frequency is at 400Hz LR24db and at 1600Hz LR24db. The other speakers are the W15ch001 and the T29cf002 tweeter. The speakers is active with electronic crossover using IIR for the filters and EQ and FIR for the LR filter linearization. And of course delays are also applied on the mid and tweeter. The speaker sounds great but just wanna know where the port energy at 200-500Hz is coming from.(can't sleep at nights 🙂 )
 

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That looks like pipe resonance all right, with the half-wave resonance at 500 Hz and the full-wave resonance at 1 kHz, although 260 mm is short for those frequencies -- I'd expect 360 mm (speed of sound = 343 m/s so 1 kHz wavelength = 343 mm).

It will somewhat mess with your frequency response below crossover since the port output is only about 7 or 8 dB down at the crossover point, so about -13 or -14 dB from passband sensitivity.
 
So one solution seem to be lower the cross frequency? But i really do not want to do that. Stuffing the speaker cabinet lowers the overall port output, i have tried it. I have one EQ filter at about 500 hz with gain -2dB to lower the peak, that seems to work just fine. Are there any other solutions?
 
I don't know, I tend to cross over well below port resonance, about as useful to you as pork-fat dancing shoes.

Once the cabinet is built the options for fixing port problems are limited. EQ is one, stuffing the port to make a sealed box and moving it closer to a room corner (for bass reinforcement) is another.
 
Note the word 'Begin...'. It's the starting point, not the end point! One starts with this optimum condition and works backwards towards a practical design, but without losing sight of it. Poorly designed ports are are often the scourge of an already intrinsically flawed concept, but which is often the most practicable solution.
My take on this matter is: 'Who would wilfully design poorly damped resonance into an acoustic reproduction system?'.
 
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The speaker sounds great but just wanna know where the port energy at 200-500Hz is coming from.
Greets!

What's tuning (Fb)? Regardless, the vent's (open cylinder) 1/2 WL resonance is < ~661 Hz once all end corrections are accounted for, ergo its output is between these two F's based on varying box acoustic pressure. A 7 cm dia vent = 7^2*pi/4 = ~38.485 cm^2 and the woofer's (Sd) = 330 cm^2, so with a good ROT of a min. 4:1 ratio unless there's only 'flea' power available your ~8.575:1 vent is apparently acting as a variably tuned whistle at typical box pressures. :cuss: :headbash:
 
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It will somewhat mess with your frequency response below crossover since the port output is only about 7 or 8 dB down at the crossover point, so about -13 or -14 dB from passband sensitivity.
Thanks. I have designed a woofer but wont be building it until I know more about the design process etc. I came across a video about beginners faults and port resonance was mentioned. Needs to be above the cross over region. Turns out that the slot port I have derived for a low bass frequency response resonates ~500hz. You've given me another way of looking at the problem. So far I've used winisd but have mixed feelings about that.

Another thread on gated frequency response measurements helps a lot too.
 
Why would that rule eliminate pipe resonances?

Looking at pipe resonance here, ignoring wall losses the relevant equation is: Q = c^2/(2 π A f^2) where c is the speed of sound, A is the area, and f of course the frequency. We can restate this for the halfwave resonance, where the length L is c / (2 f), as Q = L^2/(8 π A). Introducing the aspect ratio of length / diameter as z and noting area A = (π/2) diameter^2, we go through Q = L^2/(8 π (π/2) d^2) to obtain Q = z^2/(4 π^2), or Q = (z/(2 π))^2 for the halfwave resonance. Following the same procedure for the fullwave resonance yields Q = (z/π)^2, the 3/2 wave resonance as Q = (3 z/(2 π))^2, and so on.

However, resonance can be partly alleviated by rounding over the edges, preferably at both ends. Two 3/4 inch (19 mm) roundovers can be seen as a single 1.5 inch (38 mm) roundover, about 1/8 wavelength at 1100 Hz, which will spread resonant energy over a significant bandwidth.

This derivation is consistent with observing that duct resonance is an issue mostly below about 1 kHz except for difficult cases.
 
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Thanks. I have designed a woofer but wont be building it until I know more about the design process etc. I came across a video about beginners faults and port resonance was mentioned. Needs to be above the cross over region. Turns out that the slot port I have derived for a low bass frequency response resonates ~500hz. You've given me another way of looking at the problem. So far I've used winisd but have mixed feelings about that.

Another thread on gated frequency response measurements helps a lot too.

As I mention above, putting big ol' roundovers on both ends of a pipe goes a fair way to mitigating resonances. Hefty bevels on a slot port would fill the same function.

If the design allows then putting the port on the back to one side is also a useful trick. Why to the side? If the resonance does diffract around the edges of the cabinet to the front then placing the port asymmetrically means the wavefront will take two paths to reach the front, and if the path difference is half the wavelength then the first resonance will cancel on the listening axis. Conveniently the path length difference is the same as the pipe length, so offsetting the duct from the center line by half the pipe length should produce the desired effect if the cabinet is wide enough. If not then any asymmetry would be helpful -- you don't want the resonance to reinforce itself on-axis. Slapping some acoustic foam behind the box will trap resonances radiating to the back while allowing wanted duct output to pass unharmed since a moderate thickness of foam absorbs almost nothing in the bottom end.

I've had good experiences with WinISD. Nulls on cone movement at resonance have been pretty well on the money for a number of different designs and frequency responses seem close to predicted.
 
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Hefty bevels on a slot port would fill the same function.
I could do that even slow rads to a thin edge. Another tip I picked up on building with a slot vent was assume you will need to adjust it. Height suggested.

I've had good experiences with WinISD.
😉 You may have heard of a fitted kitchen. I need a fitted woofer for the lounge. A rough out finished ~800mm tall, not much wider than an Dayton RS225 and 500 odd mms deep. Logical place to put the speaker is near the top. -3db ~25hz. SPL ~83db. Fluke - oddly similar to some speakers I want to use with it. The bass shelf is 1 and a bit db down at 30+hz, That roll of starts at ~250hz,
I would ideally like to run it through something simple to use that accounts for the box sizes. I'm not keen on spending money. They unfortunately didn't finish that off in WinISD.

🙂 Yes I know there are other complication especially matching it up to the others. I could rad the outer edges of the front which I vaguely recollect reading helps with edge diffraction. Not sure where.
 
Looking at pipe resonance here, ignoring wall losses the relevant equation is: Q = c^2/(2 π A f^2) where c is the speed of sound, A is the area, and f of course the frequency. We can restate this for the halfwave resonance, where the length L is c / (2 f), as Q = L^2/(8 π A). Introducing the aspect ratio of length / diameter as z and noting area A = (π/2) diameter^2, we go through Q = L^2/(8 π (π/2) d^2) to obtain Q = z^2/(4 π^2), or Q = (z/(2 π))^2 for the halfwave resonance. Following the same procedure for the fullwave resonance yields Q = (z/π)^2, the 3/2 wave resonance as Q = (3 z/(2 π))^2, and so on.

However, resonance can be partly alleviated by rounding over the edges, preferably at both ends. Two 3/4 inch (19 mm) roundovers can be seen as a single 1.5 inch (38 mm) roundover, about 1/8 wavelength at 1100 Hz, which will spread resonant energy over a significant bandwidth.

This derivation is consistent with observing that duct resonance is an issue mostly below about 1 kHz except for difficult cases.

So basically Q = c^2/(2 π A f^2) shows a relation between d and f, and decreasing d or decreasing f (increasing L) causes a rise in Q. I did not know that! I still don't understand why there is anything special about 2:1 limit of length to diameter. The equation looks like it would result in a smooth gradient of Q when changing that aspect ratio. I know all the port testing I did in my box construction thread, it just so happens most ports were 2" diameter and 4.25" long. They had the typical port resonance. I doubt that extra .25" created this resonance; or if I shortened .it 25" it would disappear, or even meaningfully change.

About the roundover, would the implication be that if you had a port with no straight wall, but was a curve of constant radius (referring to Roozen's paper), this port should have much reduced port resonance compared to straight, or straight+flared end port? I didn't find that to be the case.

Not doubting the theory, I'm just trying to connect the theoretical principles cited above with the actual measurements I made when I was testing port designs. I just never never saw the behavior that's been predicted in this thread.
 
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