preemp.f,v

lpc10-15为美军2400bps语音压缩标准的C语音源代码。

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1.3
date	96.03.14.23.16.29;	author jaf;	state Exp;
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next	1.2;

1.2
date	96.03.11.23.23.34;	author jaf;	state Exp;
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1.1
date	96.02.07.14.48.48;	author jaf;	state Exp;
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desc
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1.3
log
@Just added a few comments about which array indices of the arguments
are used, and mentioning that this subroutine has no local state.
@
text
@********************************************************************
*
*	PREEMP Version 55
*
* $Log: preemp.f,v $
* Revision 1.2  1996/03/11  23:23:34  jaf
* Added a bunch of comments to an otherwise simple subroutine.
*
* Revision 1.1  1996/02/07 14:48:48  jaf
* Initial revision
*
*
********************************************************************
*
*   Preemphasize speech with a single-zero filter.
*  (When coef = .9375, preemphasis is as in LPC43.)
*
* Inputs:
*  NSAMP  - Number of samples to filter
*  INBUF  - Input speech buffer
*           Indices 1 through NSAMP are read.
*  COEF   - Preemphasis coefficient
* Input/Output:
*  Z      - Filter state
* Output:
*  PEBUF  - Preemphasized speech buffer (can be equal to INBUF)
*           Indices 1 through NSAMP are modified.
*
* This subroutine has no local state.
*
	subroutine preemp(inbuf, pebuf, nsamp, coef, z)

*       Arguments

	integer nsamp, i
	real inbuf(nsamp), pebuf(nsamp), coef, z

*       Local variables
*       
*       None of these need to have their values saved from one
*       invocation to the next.

	real temp

*
*       Logically, this subroutine computes the output sequence
*       pebuf(1:nsamp) defined by:
*       
*       pebuf(i) = inbuf(i) - coef * inbuf(i-1)
*       
*       where inbuf(0) is defined by the value of z given as input to
*       this subroutine.
*       
*       What is this filter's frequency response and phase response?
*       
*       Why is this filter applied to the speech?
*       
*       Could it be more efficient to apply multiple filters
*       simultaneously, by combining them into one equivalent filter?
*       
*       Are there ever cases when "factoring" one high-order filter into
*       multiple smaller-order filter actually reduces the number of
*       arithmetic operations needed to perform them?

*       When I first read this subroutine, I didn't understand why the
*       variable temp was used.  It seemed that the statements in the do
*       loop could be replaced with the following:
*       
*           pebuf(i) = inbuf(i) - coef * z
*           z = inbuf(i)
*       
*       The reason for temp is so that even if pebuf and inbuf are the
*       same arrays in memory (i.e., they are aliased), then this
*       subroutine will still work correctly.  I didn't realize this
*       until seeing the comment after PEBUF above that says "(can be
*       equal to INBUF)".

	do 10 i = 1, nsamp
	    temp = inbuf(i) - coef*z
	    z = inbuf(i)
	    pebuf(i) = temp
10	continue

	return
	end
@


1.2
log
@Added a bunch of comments to an otherwise simple subroutine.
@
text
@d6 3
d21 1
d23 1
a23 1
* In/Out:
d27 3
d32 3
d36 6
a41 1
	real temp, inbuf(nsamp), pebuf(nsamp), coef, z
d43 1
a43 2
*       The only local variable is temp, and its value need not be saved
*       from one call to the next.
d45 1
@


1.1
log
@Initial revision
@
text
@d5 4
a8 1
* $Log$
d27 35
@