Path: csiph.com!usenet.pasdenom.info!news.etla.org!news.stack.nl!newsfeed.xs4all.nl!newsfeed4.news.xs4all.nl!xs4all!post.news.xs4all.nl!not-for-mail
Return-Path: <oscar.j.benjamin@gmail.com>
X-Original-To: python-list@python.org
Delivered-To: python-list@mail.python.org
X-Spam-Status: OK 0.023
X-Spam-Evidence: '*H*': 0.95; '*S*': 0.00; 'python.': 0.02; 'question?': 0.09; 'cc:addr:python-list': 0.11; 'posted': 0.15; 'cc:name:python list': 0.16; 'compute': 0.16; 'corresponds': 0.16; 'element:': 0.16; 'numpy': 0.16; 'peak': 0.16; 'peek': 0.16; '(you': 0.16; 'component': 0.16; 'index': 0.16; 'wrote:': 0.18; '>>>': 0.22; 'import': 0.22; 'cc:addr:python.org': 0.22; 'install': 0.23; 'question': 0.24; 'cc:2**0': 0.24; 'least': 0.26; 'header:In-Reply-To:1': 0.27; 'message-id:@mail.gmail.com': 0.30; "i'm": 0.30; 'this.': 0.32; 'probably': 0.32; 'could': 0.34; 'received:209.85': 0.35; 'received:209.85.220': 0.35; 'but': 0.35; 'received:google.com': 0.35; 'there': 0.35; 'received:209': 0.37; 'easily': 0.37; 'sure': 0.39; 'how': 0.40; 'duration': 0.60; 'new': 0.61; 'first': 0.61; 'maximum': 0.63; 'skip:n 10': 0.64; 'to:addr:gmail.com': 0.65; 'here': 0.66; 'lowest': 0.74; 'square': 0.74; 'wave': 0.74; 'oscar': 0.84; '2013': 0.98
DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20120113; h=x-received:mime-version:in-reply-to:references:from:date:message-id :subject:to:cc:content-type; bh=LK8gUr0Xxp29CVbVCjO2knSXQC9Iad/VFafvA8vkjiM=; b=I2R2r+dMfEOE5UL7HD1UXrnUYNW42nm2HFZzgSVqSDdNBbFeubkHCkrrVE1/zJ+G9p vx6PVDFofRvCkv0oUnXF/DUShsucghxUe8XUMm/XRWKjvJuxqPlC2UHegA6MIR9VaA4P EdstPCbn/49I/9KFT2pR9DlwxQ0a/4Grxlna7PY/lET5SQTr9c0+6xUvM2Q6h3xZvzos NDen6PDUKVIYyihPVBhK7c9lzG1oWkSuH8kSecL2tIQjtM6oYDo4gsHgTjVQ/w3IKpxU E03LWhbvap00FI3SwvE3GQy0VZudRU28BGXgvHCRW1bu0zJuhlhONqxFE9FacjG4dcET yVrQ==
X-Received: by 10.58.107.104 with SMTP id hb8mr5947233veb.50.1369003812970; Sun, 19 May 2013 15:50:12 -0700 (PDT)
MIME-Version: 1.0
In-Reply-To: <c841061c-2bae-4e41-bac2-f9cd9606c783@googlegroups.com>
References: <eb271b5d-ee83-4f0e-b8ea-5b129c7cb771@googlegroups.com> <mailman.1846.1368977695.3114.python-list@python.org> <c841061c-2bae-4e41-bac2-f9cd9606c783@googlegroups.com>
From: Oscar Benjamin <oscar.j.benjamin@gmail.com>
Date: Sun, 19 May 2013 23:49:52 +0100
Subject: Re: Harmonic distortion of a input signal
To: killybeard91@gmail.com
Content-Type: text/plain; charset=ISO-8859-1
Cc: Python List <python-list@python.org>
X-BeenThere: python-list@python.org
X-Mailman-Version: 2.1.15
Precedence: list
List-Id: General discussion list for the Python programming language <python-list.python.org>
List-Unsubscribe: <http://mail.python.org/mailman/options/python-list>, <mailto:python-list-request@python.org?subject=unsubscribe>
List-Archive: <http://mail.python.org/pipermail/python-list/>
List-Post: <mailto:python-list@python.org>
List-Help: <mailto:python-list-request@python.org?subject=help>
List-Subscribe: <http://mail.python.org/mailman/listinfo/python-list>, <mailto:python-list-request@python.org?subject=subscribe>
Newsgroups: comp.lang.python
Message-ID: <mailman.1850.1369003821.3114.python-list@python.org>
Lines: 46
NNTP-Posting-Host: 2001:888:2000:d::a6
X-Trace: 1369003821 news.xs4all.nl 15980 [2001:888:2000:d::a6]:34090
X-Complaints-To: abuse@xs4all.nl
Xref: csiph.com comp.lang.python:45569

On 19 May 2013 23:25,  <killybeard91@gmail.com> wrote:
> How can i at least find a peek in FFT spectrum of a square wave ?
> From there i could easily build formula. Sorry for bothering but i am new to Python.

Are you the same person who posted the original question?

You probably want to use numpy for this. I'm not sure if I understand
your question but here goes:

First import numpy (you may need to install this first):

>>> import numpy as np

Create a square wave signal:

>>> x = np.zeros(50)
>>> x[:25] = -1
>>> x[25:] = +1
>>> x
array([-1., -1., -1., -1., -1., -1., -1., -1., -1., -1., -1., -1., -1.,
       -1., -1., -1., -1., -1., -1., -1., -1., -1., -1., -1., -1.,  1.,
        1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,
        1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.])

Compute the magnitude spectrum:

>>> spect = abs(np.fft.fft(x)[:25])
>>> spect
array([  0.        ,  31.85194222,   0.        ,  10.67342282,
         0.        ,   6.47213595,   0.        ,   4.69726931,
         0.        ,   3.73254943,   0.        ,   3.13762901,
         0.        ,   2.7436023 ,   0.        ,   2.47213595,
         0.        ,   2.28230601,   0.        ,   2.15105461,
         0.        ,   2.06487174,   0.        ,   2.01589594,   0.        ])

Find the index of the maximum element:

>>> np.argmax(spect)
1

So the peak is the lowest non-zero frequency component of the DFT. In
Hz this corresponds to a frequency of 1/T where T is the duration of
the signal.


Oscar