75th or 150th harmonics?
A bandlimited squarewave of 8 kHz @ 44.1 kHz samplerate is a sinewave. 3 *
8 kHz is already outside of the bandwidth.
This means that the basic frequency must be pretty low to get a square wave
shape.
- Uli
2018-06-13 20:40 GMT+02:00 robert bristow-johnson
:
>
>
>
I guess I have the solution now for my question by Steffan' answer.s
ap := IFFT(FFT(lp)/FFT(p))
It's simply a complex division in frequency domain.
Thanks
Uli
2016-12-07 22:14 GMT+01:00 Stefan Stenzel <stefan.sten...@waldorfmusic.de>:
>
> > On 7 Dec 2016, at 13:10 ,
2016-08-09 10:14 GMT+02:00 James McCartney :
>
>
> The original formula created a smoothed sawtooth wave only when 'a' was an
> odd integer.
> The new formula creates a smoothed sawtooth wave for any real 'a' greater
> than 1.0.
>
>
Please apologize my misunderstanding. It is
2016-08-09 8:49 GMT+02:00 James McCartney <asy...@gmail.com>:
>
>
> On Aug 8, 2016, at 23:43, Uli Brueggemann <uli.brueggem...@gmail.com>
> wrote:
>
> 2016-08-09 4:05 GMT+02:00 James McCartney <asy...@gmail.com>:
>
>>
>>
>> On Tue, Jul 5,
2016-08-09 4:05 GMT+02:00 James McCartney :
>
>
> On Tue, Jul 5, 2016 at 2:42 PM, James McCartney wrote:
>
>> In the same vein: a family of smoothed sawtooth waves
>>
>> f(x) = x - x^a
>>
>
> changing this to :
>
> f(x) = x - sgn(x)*abs(x)^a
>
> allows 'a' to
Maybe I miss the real question of the topic but I have played around with
creating a FIR filter:
1. generate white noise of a desired length
2. window it with an exponentially decaying envelope
3. apply some gain, e.g. 0.5
4. add a Dirac pulse at the first sample
The result is sprectrally not flat
I listened to the example at http://www.srmathias.com/huggins-pitch/ and I
hear the tones.
But a deeper inspection shows that taking the differences of the magnitude
responses after FFT results in quite big deviations, even > 10 dB.
So it seems that the allpass delays are not really allpasses
Of course I have tried to match the resistors. But then you will also
recognize that there are gain differences between the channels. So I also
ended up with further trials of matching including trimpots. At the end I
could fine-tune by a trimpot and minimize the sum signal.
Then I tried to
Theo,
this reminds me on a simple test where I have never got a desired result.
Take a digital signal (a sine wave or your saw wave), send it thru a DAC.
For the second channel take the inverse wave. Add the DAC outputs e.g. by a
resistor network and try to get zero.
The digital signals add
http://web.stanford.edu/class/ee104/shannonpaper.pdf is a reprint from 1949
2015-06-19 14:00 GMT+02:00 STEFFAN DIEDRICHSEN sdiedrich...@me.com:
According to the german Wikipedia, Shannon published it here:
Proc. IRE. Vol. 37, No. 1, 1949
And Nyqvist published his theorem here:
Harry Nyquist
STransform, see e.g. http://djj.ee.ntu.edu.tw/S_Transform.pdf
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Thanks so far for the different proposals.
I must admit that I do not have much skills with Matlab.
And it seems the proposals require all steep learning curve.
Maybe the problem I like to solve can be described with an example:
Let's assume a series of two biquad peaking filters, first with
theo verelst wrote:
I didn't get the part where the frequency response is a given, I mean, how
is that given (and where does that come from). And if indeed a certain
response is to be implemented as a digital filter, why does phase not
matter, too ?
Assume a frequency response of a speaker
Hello Greg,
I've unsuccessfully tried to find more about FDLS.
Can you please give me a tip or even send me some info by PM?
- Uli
2014-03-04 17:37 GMT+01:00 gjberc...@charter.net:
On Tue, 04 Mar 2014 11:28:37 -0500, robert bristow-johnson wrote:
as far as i know there is the prony method.
Hello music-dsp,
I like to decompose an arbitrary frequency response by biquads. So I'm
searching for an algorithm or paper on how to run an iterative
decomposition. In my imagination it should be possible to
a) find a first set of biquad parameters with a best fit frequency response
in
Today I have played a bit with a peaking filter of data
A: fc = 1000 Hz, Q = 1 and gain = 6 dB
Then I have created another filter
B: fc = 1000 Hz, gain = 3 dB and played with different Q values.
As far as I understand the Rane paper the proprtional Q shall result in a
constant skirt.
So a Q ~
Hi Denis,
long time no see :-)
In the meantime I have developed the formula by myself.
Anyway thanks for your tip.
Uli
2013/4/26 Denis Sbragion d.sbrag...@infotecna.it:
Hello Uli,
On Fri, April 26, 2013 14:51, Uli Brueggemann wrote:
...
Is there a formula?
you'll find the exact
We may ask: what is the difference between a plot in dBFS, dBu, dBV or
dBxx scale?
Indeed we will find some kind of factor between the units, something
like 1 full scale = f1 * voltage = f2 * sound pressure level = f3 *
another reference (f1..= factors).
Now in the logarithmic domain a value x
My simple point of view about dBFS (full scale):
The full scale FS of a 16 bit soundcard is 2^15=32768, of a 24 bit
soundcard it is 2^23 = 8388608.
The dB number Y of a value X represents a relation to the full scale:
Y [dBFS] = 20*Log10(X/FS)
So with X=32786 and a 16 bit soundcard you get Y =
/fowler%20personal%20page/EE521_files/IV-05%20Polyphase%20FIlters%20Revised.pdf
[2] https://ccrma.stanford.edu/~jos/sasp/Multirate_Filter_Banks.html
--
João Felipe Santos
On Tue, Jan 18, 2011 at 5:46 AM, Uli Brueggemann
uli.brueggem...@gmail.com wrote:
Hi,
a convolution of two vectors
[mailto:music-dsp-boun...@music.columbia.edu] On Behalf Of Uli Brueggemann
Sent: 19 January 2011 14:56
To: A discussion list for music-related DSP
Subject: Re: [music-dsp] Factorization of filter kernels
Hi,
thanks for the answer so far.
A polyphase filter is a nice idea but it does not answer
Hi,
a convolution of two vectors with length size n and m gives a result
of length n+m-1.
So e.g. two vectors of length 512 with result in a vector of length 1023.
Now let's assume we have a vector (or signal or filter kernel) of size
1024, the last taps is 0.
How to decompose it to two vectors
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