David Craig :
> Hi, thanks for that. I've made the following changes:
>
> NFFT = 100*60*10# Linked to window size
> Fs = stream[0].stats.sampling_rate
> win = np.hanning(NFFT)
> overlap = NFFT/2
> power, freq = plt.psd(data, NFFT, Fs, win, overlap)
>
> but it returns the following error:
>
Le lundi 30 janvier 2012 à 13:34 +, David Craig a écrit :
> Hi, thanks for that. I've made the following changes:
>
> NFFT = 100*60*10# Linked to window size
> Fs = stream[0].stats.sampling_rate
> win = np.hanning(NFFT)
> overlap = NFFT/2
> power, freq = plt.psd(data, NFFT, Fs, win, ov
Hi, thanks for that. I've made the following changes:
NFFT = 100*60*10# Linked to window size
Fs = stream[0].stats.sampling_rate
win = np.hanning(NFFT)
overlap = NFFT/2
power, freq = plt.psd(data, NFFT, Fs, win, overlap)
but it returns the following error:
Traceback (most recent call las
Le lundi 30 janvier 2012 à 11:45 +, David Craig a écrit :
> Hi I have some data for a 24hr period with a sample rate of 100
> samples/second. I want to create a power spectrum using matplotlibs
> function psd. I want it to have 10 minute windows with a 50% overlap,
> but cant seem to get the sy
Hi I have some data for a 24hr period with a sample rate of 100
samples/second. I want to create a power spectrum using matplotlibs
function psd. I want it to have 10 minute windows with a 50% overlap, but
cant seem to get the syntax right. My code is as follows:
NFFT = len(data)
Fs = 100
window=n
Hi,
I spent a fair amount of time today debugging what I thought was a bug
in the scaling of psd() when I was using NFFT to specify zero-padding.
This was a mis-use of the code on my part, where I should have been
using pad_to to get zero-padding. Most embarassing about this is that
I'm basically
That was it
Thanks :)
John Hunter-4 wrote:
>
> On Tue, Mar 2, 2010 at 5:13 AM, Dunx wrote:
>>
>> Hi,
>>
>> I know about the difference between pylab.psd and mlab.psd, but in theory
>> the only difference between the returned values is:
>>
>> pylab.psd = 10*log10(mlab.psd)
>>
>> Except this is
On Tue, Mar 2, 2010 at 5:13 AM, Dunx wrote:
>
> Hi,
>
> I know about the difference between pylab.psd and mlab.psd, but in theory
> the only difference between the returned values is:
>
> pylab.psd = 10*log10(mlab.psd)
>
> Except this is not true, there is noticeable difference:
>
> pylab.psd = 23
Hi,
I know about the difference between pylab.psd and mlab.psd, but in theory
the only difference between the returned values is:
pylab.psd = 10*log10(mlab.psd)
Except this is not true, there is noticeable difference:
pylab.psd = 23.4962 (all these pylab.psd values are read from graph)
10*log1
On Wed, Jul 29, 2009 at 1:39 AM, 3togo wrote:
>
> mlab.psd and pyplot.psd are different
The plotting command pyplot.psd scales the numbers in decibels (the dB
in the ylabel in pyplot.psd). mlab.psd just returns the unscaled
power. So in your example in subplot(3,1,3), call before plotting
mlab.psd and pyplot.psd are different
http://www.nabble.com/file/p24713274/psd_testpng.png
http://www.nabble.com/file/p24713274/psd_testpng.png psd_testpng.png
#This program below do reveal that mlab.psd and pyplot.psd are different
import numpy as np
import matplotlib.pyplot as plt
import
matplotlib-users-boun...@lists.sourceforge.net
[mailto:matplotlib-users-boun...@lists.sourceforge.net] On Behalf Of
Fago, Matt - AES
Sent: Wednesday, January 14, 2009 11:36 AM
To: matplotlib-users@lists.sourceforge.net
Subject: Re: [Matplotlib-users] psd question
> I'm trying to use
> I'm trying to use the matplotlib psd function to plot some data. The
> result from MATLAB's pWelch function looks vastly different.
What is the default window type for Matlab? For Matplotlib it is the
Hanning window. If Matlab defaults to a rectangular window the
results will look quite differ
Lewis, Ambrose J. wrote:
> Hi All:
>
> I’m trying to use the matplotlib psd function to plot some data. The
> result from MATLAB’s pWelch function looks vastly different.
>
> Any suggestions/recommendations would be greatly appreciated!
>
What version of matplotlib are you running? I made som
Hi All:
I'm trying to use the matplotlib psd function to plot some data. The
result from MATLAB's pWelch function looks vastly different.
Any suggestions/recommendations would be greatly appreciated!
THANXS
amb
Here is the MATLAB code:
Fs = 13e6;
fid = fopen('C:\asp\roseRT\ics_output.bin
S(f) of a sine wave is a delta function!
Joseph Park <[EMAIL PROTECTED]>
Sent by: [EMAIL PROTECTED]
26/10/2007 11:50 AM
To
cc
matplotlib-users@lists.sourceforge.net
Subject
Re: [Matplotlib-users] PSD amplitudes
spectral density is by convention a 1Hz binwidth, not an arbitrary on
Subject
Re:
[Matplotlib-users] PSD amplitudes
.sourceforge.net
Subject
Re: [Matplotlib-users] PSD amplitudes
is the suggestion that the matplotlib algorithm is correct in computing
PSD amplitudes?
btw, increasing nFFT increases the number of points used in the FFT, which
increases the spectral frequency resolution (smaller binwidth) bu
Subject
Re:
[Matplotlib-users]
ect
Re: [Matplotlib-users] PSD amplitudes
Shouldn't the PSD for a simple sine wave tend to infinity
the spectral resolution will impact the amplitude, if you
are not dealing with a density. by definition a spectral density
has applied the bandwidth resolution correction. the PSD ampli
To
matplotlib-users@lists.sourceforge.net
cc
Subject
[Matpl
Please try the attached script.
The answer should be ~0 dB for each of the frequencies.
Most likely a simple scaling issue/parameter of which i'm ignorant.
--
##
## Name: psd_scale.py
##
## Purpose
If one creates a 1V RMS sine wave, e.g. with a peak value of 1.414 or
peak-peak of 2.83, then computes psd, the resulting amplitude is around
24 dB, yet classic theory dictates the answer ought to be 20 log(1) = 0
dB.
This offset seems consistent across various frequencies/sample
rates/amplitud
23 matches
Mail list logo
