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From: lou@caber.valid.com (Louis K. Scheffer)
Newsgroups: comp.dsp
Subject: Re: 48k to 44.1k sample rate conversion
Message-ID: <481@valid.valid.com>
Date: 22 May 91 02:58:31 GMT
References: <5826@media-lab.media.mit.edu.MEDIA.MIT.EDU> <1991May13.173129.18295@bernina.ethz.ch> <1991May15.105543.12165@bernina.ethz.ch> <1991May16.052920.21060@netcom.COM>
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mfvargo@netcom.COM (Michael Vargo) writes:

>2. An interpolation can be done using a function similar to the following.
>I wrote this to downsample Mac sounds before converting them to mu-law PCM
>at 8kHz.

[...code omitted ...]

>Now my question:

>Does the act of interpolation actually do some sort of low pass filtering?
>I think not but maybe it could be looked at as some sort of averaging?

Yes - interpolation is a low-pass filter.  The easiest way to see this is
to note that interpolating is exactly equal to convolving with a time function
that looks like:

	 *                  <- 1
	* *
       *   *
      *     *
******       *************  <- 0

 |   |   |   |   |   |   |  <input sample rate

Since a convolution is the same as a filter, this is a filter, in this case
a low pass.  If you want to know the exact response, you look at the
transform of the triangle function.  This is most easily derived by noting
that a triangle is the convolution of a rectangle with itself.  Since a
rectangle has response sin(f)/f, where the first 0 occurs at the half
sampling rate, then the triangle above has response sin^2(f)/f^2, with the
first 0 at the input half sampling rate.  You can also look at this as
averaging with the above function as a weighting factor, and get the same
answer.

  -Lou Scheffer