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Poles & filters
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| xmotleyx |
If someone has a moment, could you explain what poles are in filters, and how they work. I am the type of person that needs to understand to use and I cannot get my head around this completely.
Thanks .. .
Cheers ... |
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| Centra Spike |
| It's basically how steep the filter curve is. Clicky. |
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| 404 Science |
Ok, I'll put my electrical engineering degree to use
So basically, a filter in the time domain is represented by a differential equation. If you take the Laplace transform of this differential equation, you get a new equation of variable 'omega' which is the angular frequency expressed in radians. This equation will give you what is called the 'transfer function' which means that for any given frequency, you know the input/out relationship. In other words, if you input a certain frequency into the system (filter), you get a certain output. So the transfer function tells you the amount of amplification or attenuation for each frequency when it is passed through the system.
Knowing basic complex variable calculus helps by the way...
A filter is such a system, and based on the type, it attenuates a certain band of frequencies and amplifies others. Well, if you consider the transfer function again, the poles are the roots of the transfer function. In other words, if you factor the quadratic equation you get, (or equation of whatever degree), the poles will be the values for omega which make the transfer function go to zero. In the time domain, what happens is the behaviour of the differential equation goes to infinity around the poles, hence the name "pole" (if you takje a 3d graph of all the inputs to the differential equation, you see a spike around the poles)
I'm severly roughing this so any math majors, please excuse any mistakes...
So basically when you have a pole at a certain frequency, what happens in a filter, is you get anattenuation of roughly 3dB. This has to do with 20*log of the absolute value of the complex number at the pole frequency. You get 20*log of 0.707 at the poles, which gives -3dB. Try it in your calculator. If you have more poles at a certain frequency, for example you can have double poles, you get a faster attentuation slope for the frequencies above or below the pole frequency (low pass or high pass)...
So in the end, you get a steeper rolloff...
Poles also contribute to stability, so essentially your filter is more robust for larger pole filters...
Pick up a book on Signals & Systems theory if you want to know more.
I know this probably made no sense, I tried... |
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| farris |
| quote: | Originally posted by 404 Science
Some technical story |
What the...:wtf:.
Good read though.
Most simple is what Centra Spike said.
Have a read on that sound on sound link.
To give an example of what both said:
It's basically how steep the filter curve is.
And
So in the end, you get a steeper rolloff...
Basically 1 pole is the same as 6dB.
Now look at the dB (12dB = 2 poles, 24dB = 4 poles and 48dB = 8 poles ) in the picture
and then look at the corresponding curves/rolloff.
The more poles, the steeper.
Hope you understand it a bit more now.
- farris |
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| xmotleyx |
Cheers for the replies.
Of course one went beyond my knowledge in math. I didn't get further than first year calculus with my molec degree.
Oh yea, I tried a search for poles etc at SOS and I didn't get anything. I guess I need to tweak my search skills.
I ended up printing off the whole "synth secret" series. Dam good read ... |
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| emc^2 |
| quote: | Originally posted by 404 Science
Ok, I'll put my electrical engineering degree to use
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Going forward, please DON'T. :wtf: Even my 2 years of E.E. didn't help. |
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| Centra Spike |
| quote: | Originally posted by xmotleyx
Oh yea, I tried a search for poles etc at SOS and I didn't get anything. I guess I need to tweak my search skills. |
You could start by searching my first reply for the link :p |
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