# What is the width?



## Joppnl (Nov 2, 2010)

Hello,

I 'sweeped' an analogue eq and for 1 freg (1k5) this is the result:










My question is, what is the bandwidth for this eq:

Center freq = 1500Hz,

+/- 3dB points are at 1962 / 1142 Hz

So, what is the calculation now?

What I see is that the difference is 820Hz (1962-1142)= 54% from the center freq.


But what is the Q or the bw/oct? how do you calculate both??

Thanks!


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## Joppnl (Nov 2, 2010)

Well,

I found some explanation on the internet and in fact calculating the Q is very simple, calculating the bw/oct is not but can be made easy.

The Q:

simply devide the center freq by the difference at +/- 3dB

So, in this case: Q = 1500 / (1962-1142) = 1500/ 862 =1.8

As you can see, the higher the Q, the smaller the bandwidth of the filter.

Calculating the bw/oct is much more difficult, but, attached, there is a handy Excell sheet where you can fill in the freq's and the result shows the Q and bw/oct.

And here you can find all the info: 

http://www.rane.com/note170.html


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## Moonfly (Aug 1, 2008)

Good post, thanks for sharing :T


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## Joppnl (Nov 2, 2010)

thank you for your comment.


A small addition:

To visualize:

Q=0 means that there is no filtering at all, a straight line


Q=1 means that you have your +-3dB point exactly half-ways on the Fcenter (so id Fc = 1500, Q=1 then Flow(-3dB) = 750Hz

Q=max as possible (infinite) means that you have a peak at Fcenter as high as possible


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## JohnM (Apr 11, 2006)

Joppnl said:


> The Q: simply devide the center freq by the difference at +/- 3dB


This is only true for some EQ filter implementations. Whilst all use Q = bandwidth in Hz/centre frequency, the point at which bandwidth is measured depends on the implementation. Another common implementation is to use the bandwidth at the half-gain points, which has some advantages - for example, with the Rane definition how would you express the bandwidth of a filter that has 2dB of boost? 

The REW help lists the bandwidth definitions used by the various equalisers it emulates (see Equaliser Selection), for more info Robert Bristow-Johnson wrote a paper that you can find at musicdsp.org.


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## Joppnl (Nov 2, 2010)

Thank you for explaining this to me John.

Last days I did some extensive testing of the DEQ2496 I own and with the '-3dB rule' the width which the DEQ says on the front panel does not correspond to the measurements....

With your explanation and taking the half dB points it does almost perfectly, I'll post them later in this post.

But one more question: 

I thought I could calculate the center-freq by drawing a line half-ways the dB amplification (in fact, is doesn't matter where you draw it) taking the 2 freq's on the +dB line, add them together and devide them by 2.

i.e.: I have a filter 2kHz, +10dB, at +5dB I find 1500Hz and 2500Hz, together 4000, divide by 2 = 2000 which is indeed the centerfreq.

This theory works but with wider filters in practice it does not work.









Fcenter= 2000Hz, BW/OCT = 1/1 Gain = 10dB I measure (see the green line in the pic):

Fmin = 1415,9
Fmax= 2824

Fcenter = (Fmin+Fmax)/2 = 2119,95


This looks strange to me. I thought it had to do with the logarithmic scale so I changed the view to lineair but that was not correct as well.

So why is the center freq not the 2 freq's together divided by 2?

It even get worse with wider filters.



Thanks,


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## JohnM (Apr 11, 2006)

Joppnl said:


> So why is the center freq not the 2 freq's together divided by 2?


It is because of the logarithmic nature of the frequency axis for EQ filters, the centre is at sqrt(flow*fhigh) - it is a geometric rather than arithmetic average.


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## Moonfly (Aug 1, 2008)

This thread got really interesting. I never stop being impressed by Johns knowledge :T


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## Joppnl (Nov 2, 2010)

wow, that's a quick answer, again thanks, I will update the Excell sheet I've made with your formula to see if I did measure right)

But... when I change the scale from log to lin in REW calculating the center freq is still not possible. Am i correct in this? If so, I do not understand but may-be it's to much math for me, I'll try your formula....

I was still editing the post for the results of the DEQ2496 measurement and adding a picture but I'll give the info in this post :

OK, here are the results I got for measuring the BW/oct:

Fcenter=2kHz, gain=+10dB

BW(oct)....Bw/oct......=1/
1/10.........0,11.........8,97
1/9...........0,12.........8,21
1/8...........0,14.........7,39
1/7...........0,16.........6,42
1/6...........0,17.........5,72
1/5...........0,21.........4,84
1/4...........0,26.........3,91
1/3...........0,34.........2,97
1/2...........0,50.........1,99
3/4...........0,75.........1,33
1/1...........1,00.........1,00
3/2...........1,49	
2.............1,99	
3.............2,98	
4.............3,96	
5.............4,91	
6.............5,88	
7.............6,78	

(pfffft, making a decent table is not easy (spaces are ignored so had to do it with dots....  )

As you can see, the 1/10 BW/oct is actually almost 1/9, 1/9 = 1/8 and so on up to 1/6, here the filters are more or less what they should be.

Thanks for reading,


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## Moonfly (Aug 1, 2008)

If you want to add charts, it might be easier to do one in Excel or similar, do a screen capture and paste it into paint, then attach the saved image to your post :T


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