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- Newsgroups: rec.audio
- Subject: Re: Why filter D/A output?
- Message-ID: <1992Nov20.000415.1811@lugb.latrobe.edu.au>
- From: MATGBB@LURE.LATROBE.EDU.AU (BYRNES,Graham)
- Date: Fri, 20 Nov 1992 00:04:15 GMT
- Sender: news@lugb.latrobe.edu.au (USENET News System)
- References: <1992Nov8.222911.27702@doug.cae.wisc.edu>
- <1992Nov10.051755.21660@lugb.latrobe.edu.au> <PHILG.92Nov11173959@zug.ai.mit.edu> <1992Nov12.201144.28397@en.ecn.purdue.edu>
- Organization: La Trobe University
- In-Reply-To: syd@en.ecn.purdue.edu's message of 12 Nov 92 20:11:44 GMT
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- In <1992Nov12.201144.28397@en.ecn.purdue.edu> syd@en.ecn.purdue.edu writes:
-
- > philg@martigny.ai.mit.edu writes:
- >
- > >Driving a nonlinear, i.e. distorting, system, i.e. amplifier/speaker,
- > >with a single frequency produces an output made up only of that
- > >frequency and its harmonics. This can be shown easily by looking at a
- > >Taylor series expansion of a nonlinear function and seeing what
- > >happens to cosine squared and cosine cubed terms (use trig
- > >identities).
- >
- > This is [mathematically] not true in general.
- > Example: f(x) = arccos (x). The output of such a system
- > when x = cos(t) is simply t, which is not periodic, and
- > hence does not have a Fourier series expansion.
- >
- > One could be more obnoxious by letting f(x) = cos (1.5 arccos(x)).
- >
- > The Taylor series must converge at every point in order for this
- > argument to work. (Note that arccos (x) does not meet this requirement.)
- > A sufficient condition for this is that the nonlinearity be continuous.
- > (In the real world, this is probably a valid assumption.)
- >
- > --
- > Dennis Hilgenberg
- > syd@ecn.purdue.edu
- > LaRouche in '96
- I guess we can at least assume that the transfer function is bounded
- over any compact interval ? :-) One might even hope that it would be
- continuous....so we can use Taylor's theorem to get an arbitrarily
- accurate approximation over the domain (ie possible input values).
- Cheers, GB
- (Yeah, I know, what a wank)
-