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If you go to the group delay plot and then select "Generate Minimum Phase" from the Controls menu (right side and just above the plot area), you should be able to see the "excess group delay plot" if you check the box below the plot.  From that plot, you can see how much more delay you need in your case. For instance, for a 1979 Cornwall, the following excess group delay plot shows a 0.732 ms (732 microsecond).  See cursor position at intersection with the vertical axis at the bottom left) lag of the midrange horn/driver and woofer relative to the tweeter.  That's how much delay to add to the tweeter channel to get it into time alignment:

 

1175651186_CornwallExcessGroupDelay.jpg.d898f965171942988ac4cfe8faac2dd9.jpg

 

Dude, you need to add something like 1 ms to the delay that you've already got. (I don't have the excess group delay plot to make that exact call, however, but you do.) You can read it directly from your excess group delay plot after you place the cursor over the approximate average value of the lower frequency driver (a midrange or woofer in your case).  Voila! Time alignment. 

 

Now what this might do is to create a polarity issue at crossover, leading to a drop out in SPL response around the crossover frequency due to phase cancellations between the higher frequency driver and the lower frequency driver.  If this happens, simply flop the polarity of the lower frequency driver in the output channel menu (midrange or woofer, whichever you're using).  Then take another sweep to see if that helped to eliminate the cancellation.  You may have to fine tune the delay in the 0.1 ms or 0.01 ms range to get rid of the cancellations at crossover.  This is normal, and it is a result of finally getting close to time alignment of the drivers--the cancellations become more pronounced just before time alignment and the correct polarity are found.

 

Chris

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By the way, in the plot above, the 0.732 ms delay corresponds to a 9.94 inch movement backwards of the tweeter in the stock 1979 Cornwall in order to achieve physical time alignment, i.e.,

 

speed of sound at room temperature = 13584 inches/second

 

.732 ms = 0.000732 seconds

 

13584 * 0.000732 = 9.94 inches

 

Chris

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Looking at your intersection line, I see it sets at the 5.39k and can see the 732u you referenced.  How or why are these points determined? Sorry if I am not following, but to me it seems these are random locations.  The 2 reasons I am guessing you chose these, 1 the lowest point at 100 hz where I assume there is a crossover or cutoff of some sort, 2 the 5.39 is roughly where the bass to mid is crossover ed?

image.thumb.png.15ab2750a9031531fadc08d2c9f5a6e6.png

 

 

 

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Cornwall Excess Group Delay.jpg

 

In the above example, I was referring to the tweeter (a CT-125 in this instance) and K-600/K-55-V midrange horn/driver. They crossover at ~5.4 kHz in my measurements, shown above at the point where I placed the cursor.   It is pretty well known that the tweeter leads the midrange horn in terms of time alignment.  That's the focus of this example. 

 

Looking at the excess group delay curve--starting on the right hand side above 5.4 kHz, the total group delay and the excess group delay are zero (0).  Then at the crossover frequency (moving from right to left), the excess group delay rises to 0.732, down to approximately 1100-600 Hz, where then woofer kicks in.  So the correct place to set the cursor is at ~0.73 ms.  So you first read the positive excess group delay of about 0.73 ms curve at ~5.4 kHz, and down toward 600-1000 Hz where the overlap between the midrange and woofer occur (i.e., the midrange-bass bin "crossover interference band").  This is the contribution from the midrange relative to the tweeter.  You can set the cursor at the approximate average (eyeballing it) position on the horizontal excess group delay curve where the midrange excess group delay curve lies, and this the time delay you need to add to the tweeter channel to bring it back to midrange time alignment with the midrange.

 

Also, in this particular example, the Cornwall woofer actually leads the midrange horn/driver because the woofer is direct radiating and the midrange horn/driver is approximately 13.5 inches long from the mouth of the metal horn to the back end of the K-55 driver.  So the woofer leads the midrange by a small amount.  Adding ~0.4 ms to the woofer channel delay overall will time align the woofer to the midrange. 

 

In this example, the midrange is the "long pole in the tent", i.e., it has the most delayed output in terms of time alignment.  So it will be the reference for adding time delays to the other channels.  Since the bass bin and the midrange are almost time aligned (not quite within 1/4 wavelength), it will usually be okay to run the woofer channel without delay.  If you get within 1/4 wavelength of time alignment at the crossover points, then the drivers will acoustically couple and combine in-phase.  If, however, you're more than 1/4 wavelength out--say at 1/2 wavelength misalignment, you will get phase cancellation at the crossover frequency, and therefore a polarity reversal of one driver will be required in order to keep the two channels from destructively interfering with each other. 

______________________________________________________________

 

In almost all other instances, the woofer/bass bin channel usually has the longest time delay, so it is usually the "datum" by which the other driver channels are delayed to match the channel with the longest delay. (We can't add negative delay, but we can delay the other channels to match the one channel with the longest delay.) 

 

This is especially true for horn-loaded bass, where the K-horn bass bin lags the midrange by about 4.8 ms, and the tweeter by about 6 ms (give or take, and working only from memory).  So you'd add 6 ms delay to the tweeter channel, and 4.8 ms delay to the midrange channel in the Khorn case, and you'd be pretty close.  The dialing-in process will clean up any polarity issues at its two crossover interference bands (~400 Hz and ~4.9 kHz). 

 

Chris

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In your case, you show the following excess group delay curve, and I assume you're running a two-way loudspeaker:

 

image.png

 

Below 400 Hz, the excess group delay actually begins to decrease--an odd situation that you'll have to explain by either describing the loudspeaker you're measuring more completely or by posting a picture. 

 

Above 400 Hz, the excess group delay is also decreasing toward the 1500-2000 Hz point, where it flattens out to zero group delay.  There are two peaks, one at ~3.1 kHz, and the other at ~1.2 kHz.  These are areas where there are local discontinuities in the phase response of the loudspeaker, which causes the group delay curve to spike.  These could be in-cabinet or interior driver resonances or reflections, and or effects of lobing of the midrange-to-woofer or midrange-to-tweeter patterns, and these cause local disturbances in the phase response.  These will likely move around if you slightly reposition the microphone, since they are showing the effects of 3-D radiation pattern irregularities (i.e., non-spherical radiation patterns ). 

 

In your case, I'd crank in an additional 1-1.5 ms and take a measurement to see where the excess group delay curve moves to.  If the curve moves above the midrange excess group delay value, i.e., lagging group delay, above your crossover point...(where was that?  600 Hz?) where the HF driver/horn is radiating by itself, then you can back off the tweeter delay value until it is at zero excess group delay again.

 

Chris

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Below 400 Hz, the excess group delay actually begins to decrease--an odd situation that you'll have to explain by either describing the loudspeaker you're measuring more completely or by posting a picture. 
 


Belle bass bin (diy) with crossover to k402 with lp at 420

K402 with BMS 4590 coaxial crossed hp 470 to Belle bass bin

BMS is crossed at 6300 lp/hp from mid to high.


Sent from my SM-G986U using Tapatalk

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19 hours ago, The Dude said:

According to Google 1 MM = 1MS. If so, than 11 U.S. = 3.8 MM or 3.8 MS of delay.

 

My post-it note says that sound travels approx 345mm in 1 millisecond, or 1.13 ft. in 1/1000 second, or 13.58 inches per millisecond. A K-402 horn / tweeter on a Belle bass cabinet should only need about (one) 1 millisecond of delay.

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1 hour ago, Khornukopia said:

A K-402 horn / tweeter on a Belle bass cabinet should only need about (one) 1 millisecond of delay.

Using what order of crossover filters? Zeroth? First? Fourth?

 

Crossover filter sets (one low pass, one high pass) introduce 90 degrees of phase lag on the lower frequency drivers relative to the higher frequency drivers for every order:

 

First order set = 90 degrees

Second order set = 180 degrees

Third order set = 270 degrees

Fourth order set = 360 degrees

.

.

.

Eighth order = 720 degrees

______________________________________________

 

If you're crossing at ~500 Hz, the respective crossover-filter-induced time delays are as follows:

zeroth order = 0 ms

1st order = 0.5 ms

2nd order = 1 ms

3rd order = 1.5 ms

4th order = 2 ms

.

.

.

8th order = 4 ms

 

Chris

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In the US customary measurement system, the speed of sound in air is as follows:

Temperature Speed of sound
T (°F)  c (feet/s) 
95 1154
86 1145
77 1135
68 1126
59 1116
50 1106
41 1097
32 1087
23 1077
14 1067
5 1056
-4 1046
-13 1036

 

I use 1132 ft/s as the speed of sound at room temperature (at about 72-73 °F), which is 13584 inches/second. 

 

The time alignment of drivers is a function of temperature.  In really cold or hot atmospheric conditions, temperature compensation of time alignment is usually required for a loudspeaker array.

 

Chris

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6 hours ago, Chris A said:

Using what order of crossover filters? Zeroth? First? Fourth?

 

Crossover filter sets (one low pass, one high pass) introduce 90 degrees of phase lag on the lower frequency drivers relative to the higher frequency drivers for every order:

 

First order set = 90 degrees

Second order set = 180 degrees

Third order set = 270 degrees

Fourth order set = 360 degrees

.

.

.

Eighth order = 720 degrees

______________________________________________

 

If you're crossing at ~500 Hz, the respective crossover-filter-induced time delays are as follows:

zeroth order = 0 ms

1st order = 0.5 ms

2nd order = 1 ms

3rd order = 1.5 ms

4th order = 2 ms

.

.

.

8th order = 4 ms

 

Chris

 

 

Yes, you are correct to point out this important information.

 

The numbers I posted for @The Dude were a generalization to counter some numbers he found on google about the speed of sound that appeared to be way off standard reference, which could lead to some confusion for people like me who are still learning and sometimes using a tape measure while trying to visualize the correlation between time and distance.

 

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10 hours ago, Khornukopia said:

y post-it note says that sound travels approx 345mm in 1 millisecond

Im not sure I truly understand what ms and ms are Millisecond or Microsecond.  If millisecond and 345 mm is equal to 1 millisecond, then I was really low on my delay, so I have it bumped to 1003 U.S. or 344 mm which should be right at 1 millisecond.  My peaks at the 3.1k and 1.2k are gone, but now a massive peak at the actual crossover point. I have tried inverting which made it worse than what the graph shows below. 

 

image.png.619b655c23c2f9a9933960bc6bed83fd.png

image.thumb.png.e3bb068ceba220709c0355552e225bd3.png

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In the measurement, the delay was added to the woofer/bass bin.  This time around I added it to the Mid/high section.  

image.png.e78bb782cf8e99faf466040e3672bfb7.png

 

 

image.thumb.png.3441250a9faa1946c1dc064eb42f6761.png

 

 

Chris, if I recall, did you mention that the excess group delay would tell me what I need for delay? If so, why is when I add that amount, It almost seems to go the wrong direction?  As far as your qeustion in regards to the type of crossover, if my understanding of the Hypex software is correct.  I have second order LR at 420 lp and 470 HP.  From the Hypex manual, it states LR only has 2, 4, and 8th orders.  

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Ok, now I am confused as I thought that if the GD looked better (straighter line under 2 ms), then the Spectrogram would look better.  I cascaded another set of crossovers which should give me 4th order LR at the crossover points. The GD looks better (to my understanding), but the Spectrogram looks worse.

image.thumb.png.99c882c407903cc9848daef6143de546.png

 

image.png.d51f7eb421ec1a7ad85f91c92e5f192f.png

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16 minutes ago, The Dude said:

1003 U.S. or 344 mm which should be right at 1 millisecond.

 

I see now that you are using capitol letters, when microsecond is usually represented by the symbols "us" with the "u" having a little tail (when available on the keyboard). The main thing is that you are selecting the proper category or denomination that corresponds with the desired adjustment, since there is a huge difference between "us" and "ms". 

 

As for specifics on your system, @Chris A is the expert and his guidance and recommendations are the best.

 

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This is getting somewhere:

 

image.png

 

It still says that you could add about 1 more millisecond delay to the high frequency channel (not low frequency) to match the low frequency delay, but this is really close because the spikes in excess group delay are gone.   You are probably really close to being on target here with the time delay. 

 

What does the SPL response and phase response plot look like?  If they are flat and smooth, then adding one more millisecond delay will bring time alignment, but you will be out of phase, i.e., you are now close enough on time delay to where the bass channel polarity reversal might be necessary to get the bass bin response to be in-phase with the high frequency--if you add one more millisecond of delay to the high frequency channel. 

 

Chris

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I have to be upfront, I did something between my last GD and this gd that introduced a bump again at the 6k area. So I went ahead and added another millisecond of delay, but things went nuts around the 400 region, so I inverted the HF side of the coaxial driver.  Now I have this.   

 

image.thumb.png.41e520c038fe6f722d19b737d362a080.png

 

8 minutes ago, Chris A said:

What does the SPL response and phase response plot look like?

image.png.a98f3854404b42c3442a0cf64d27173d.png

 

Current delay settings, oddly everything is inverted. 

image.png.189815300df58682fcb15fa867965b40.pngimage.png.952265b99f6a75010dc239c72281da1c.pngimage.png.470d0c0c49f62bcd8a572db7f255caad.png      

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