MACW wraparound

Tiger M · Apr 16, 2004, 08:22 PM

Greetings guys,

I wonder what is the PRACTICAL (I emphasize on that) use of the wraparound on overflow of the MACW and MACINTW instructions?

Thanks in advance.

Max M. · Apr 16, 2004, 11:18 PM

There're many...

here's for example demo vibrato/chorus code (i gave it once before, but that thread was lost):

Code:

name "delay line modulated by trangle lfo :)"
guid "d8dc1380-56ec-4098-960f-44e461f818d0"
           
control	freq = 0.2		; speed/frequency  
control phase = 1		; phase/panning/width
static  s  
temp b

input 	in1, in2
output 	out1, out2      
control depth = 0.1 
static 	osc1, osc2 

static  t1, t2
; delay lines
idelay write d1 at 	0
idelay read d11 at 	1000
idelay read d12 at 	1000 ; we need two delay taps per channel to get it 'smoother'     
idelay write d2 at 	1001 
idelay read d21 at 	2001
idelay read d22 at 	2001
itramsize 			2002 

const	maxtime = 0x1f3000	; maximum delay time -> (1000 - 4) * 0x800
; -4 used to prevent reading from 'alien' delay lines (when depth = 1)    
							
; freq control workaround  (as always)							
macs	b, 0, freq, 0.0008 

macwn	s, s, b, 1				; saw osc  
macw	osc2, s, phase, 1	; get shifted osc2
tstneg	osc1, s, s, 0			; osc1 saw->triangle
tstneg	osc2, osc2, osc2, 0;    ; osc2 saw->triangle         

; feed delay lines
macs	d1, 0, in1, 1         
macs	d2, 0, in2, 1 

; calculate an interpolation coeffs for mixing taps... 
; (or, in other words, 'fractional part of new address')
; (should be done one sample later, after writing t1/t2 to address regs)
macintw	t1, 0, t1, 0x100000 
macintw	t2, 0, t2, 0x100000   

; interpolate between taps and output (e.g. 'smoothing')
interp	out1, d11, t1, d12;     
interp	out2, d21, t2, d22; 

; applying depth 
macs  	t2, 0, maxtime, depth 
; modulation, calculates actual taps addresses (relative to base address of delays) and writes them   
macs	t1, &d1, t2, osc1		; delay base address + new 'modulated' time   
macints	&d11, t1, 0x800, 2		; first tap, time + 1 sample (should never be 0)
macints	&d12, t1, 0x800, 3		; second tap, time + 2 samples 
macs	t2, &d2, t2, osc2   
macints	&d21, t2, 0x800, 2
macints	&d22, t2, 0x800, 3        

end;

Well, yep, there're many uses of "wrap" besides above... In general, we can find use for any mad operation that can be ever invented ;)

Lex Nahumury · Apr 17, 2004, 04:02 AM

>>i gave it once before, but that thread was lost):
Yeah,..I still wonder who did that

..

/LeMury

Tiger M · Apr 17, 2004, 11:32 AM

Thanks very much. I've got another question if I may.

What is the purpose of the accumulator accesed via the MACMV instruction. I gues that it is a register that stores the result of the X*Y multiplication and according to the As10k manual it can be fetched with MACS, MACINTS or ACC3. Why is such a thing needed?

Would you tell me in your next reply Max if I'm bothering you with this type of questions.

Max M. · Apr 19, 2004, 02:01 PM

It's a register in the heart of execution unit and it "stores full-width result of the last instruction's execution".
Accumulator is typically used "when extra headroom (the accumulator guard bits) or precision (the LS 32 of the accumulator's 67 bits) need to be retained through the next multiply/accumulate instruction". Or, in other words, you use it when you need "non saturated", "non wrapped" and "non truncated" result of previous operation (instruction)...
Accumulator may be used only as A operand in any MAC** (except MACINT*) and ACC3.

--------

Example - two "mixers":

Code:

input 	in1, in2, in3, in4
control	lev1, lev2, lev3, lev4
output 	y
temp	t

macs	t, 0, in1, lev1
macs	t, t, in2, lev2
macs	t, t, in3, lev3
macs	y, t, in4, lev4
end

Code:

input 	in1, in2, in3, in4
control	lev1, lev2, lev3, lev4
output 	y

macs	0,     0, in1, lev1
macs	0, accum, in2, lev2
macs	0, accum, in3, lev3
macs	y, accum, in4, lev4
end

----

In general, accumulator availability is very significant feature...
- "Must-Have".

Tiger M · Apr 19, 2004, 03:36 PM

Thanks for the answer!

Russ · Mar 10, 2005, 12:56 AM

Quote:

Originally Posted by Max M.

...Or, in other words, you use it when you need "non saturated", "non wrapped" and "non truncated" result of previous operation (instruction)...

How would that work?

Registers only wrap around or saturate when the absolute value of that register > 1. Non wrapped or non saturated values would result in a floating point number, so how would the accum register handle that? I understand the extra precision, but not the "non wrapped" and "non saturated" part.

Tiger M · Mar 10, 2005, 07:34 AM

You brought up my first post in the forum again

From the 'As10k1' manual:

Quote:

Accumulator

The Emu's accumulator is 67 bits wide. Multiplication of two 32 bit 2's-compliment numbers has at most a 63 bit result. This leaves 4 guard bits for sucessive MAC operations where intermediate instructions may overflow.

bits 66-64 guard bits
62-31 HIGH ACCUM
31-0 LOW ACCUM

yes, there is a 1 bit overlap between HIGH ACCUM and LOW ACCUM, this is normal.

Russ · Mar 10, 2005, 02:09 PM

I have read that manual many times but did not make the connection before. So what is the maximum overflow (4 bits = 15)?

I had tried some examples and it wasn't working (the main reason for my question), but I must have been doing something wrong because I tried a couple more and it worked this time. Here are the examples I tried (for future reference), similar to Max's above, but with number's plugged in for easy verification of the results.

First Example:
-------------------------
static a = 0.9, b = 0.5, c = 0;
macs, 0, 0, a, b; (0.9 * 0.5 = 0.45)
macs, 0, accum, a, b; (0.45 + 0.45 = 0.9)
macs, 0, accum, a, b; (0.9 + 0.45 = 1.35)
macsn, c, accum, a, b; (1.35 - 0.45 = 0.9)
-------------------------
Result of above is 0x73333330 = 1932735280 / (2^31+1) = 0.89999999 = 0.9

Second Example:
-------------------------
static a = 0.9, b = 0
macs 0, 0, a, a; (0.9 * 0.9 = 0.81)
macs 0, accum, a, a; (0.81 + 0.81 = 1.62)
macs 0, accum, a, a; (1.62 + 0.81 = 2.43)
macsn 0, accum, a, a; (2.43 - 0.81 = 1.62)
macsn b, accum, a, a; (1.62 - 0.81 = 0.81)
-------------------------
Result of above is 0x67AE1478 = 1739461752 / (2^31+1) = 0.80999999 = 0.81

So, even though some of the intermediary results were floating point numbers, the final result was correct. Good stuff!

Thanks again,
-Russ

Apr 16, 2004, 08:22 PM	#1
Tiger M kX user Join Date: Apr 2004 Posts: 851 Rep Power: 0	MACW wraparound Greetings guys, I wonder what is the PRACTICAL (I emphasize on that) use of the wraparound on overflow of the MACW and MACINTW instructions? Thanks in advance.

Apr 19, 2004, 02:01 PM	#5
Max M. h/h member-shmember Join Date: Dec 2002 Location: Evil Empire Posts: 2,639 Rep Power: 69	It's a register in the heart of execution unit and it "stores full-width result of the last instruction's execution". Accumulator is typically used "when extra headroom (the accumulator guard bits) or precision (the LS 32 of the accumulator's 67 bits) need to be retained through the next multiply/accumulate instruction". Or, in other words, you use it when you need "non saturated", "non wrapped" and "non truncated" result of previous operation (instruction)... Accumulator may be used only as A operand in any MAC** (except MACINT) and ACC3. -------- Example - two "mixers": Code: input in1, in2, in3, in4 control lev1, lev2, lev3, lev4 output y temp t macs t, 0, in1, lev1 macs t, t, in2, lev2 macs t, t, in3, lev3 macs y, t, in4, lev4 end Code: input in1, in2, in3, in4 control lev1, lev2, lev3, lev4 output y macs 0, 0, in1, lev1 macs 0, accum, in2, lev2 macs 0, accum, in3, lev3 macs y, accum, in4, lev4 end ---- In general, accumulator availability is very significant feature... - "Must-Have". Last edited by Max M.; Apr 19, 2004 at 02:07 PM.*

Apr 17, 2004, 04:02 AM	#3
Lex Nahumury DH Senior Member Join Date: Jan 2003 Location: The Netherlands Posts: 1,932 Rep Power: 64	>>i gave it once before, but that thread was lost): Yeah,..I still wonder who did that .. /LeMury

Apr 17, 2004, 11:32 AM	Thread Starter #4
Tiger M kX user Join Date: Apr 2004 Posts: 851 Rep Power: 0	Thanks very much. I've got another question if I may. What is the purpose of the accumulator accesed via the MACMV instruction. I gues that it is a register that stores the result of the X*Y multiplication and according to the As10k manual it can be fetched with MACS, MACINTS or ACC3. Why is such a thing needed? Would you tell me in your next reply Max if I'm bothering you with this type of questions.

Apr 19, 2004, 03:36 PM	Thread Starter #6
Tiger M kX user Join Date: Apr 2004 Posts: 851 Rep Power: 0	Thanks for the answer!

Mar 10, 2005, 02:09 PM	#9
Russ HardwareHeaven Extreme Member Join Date: Jan 2005 Posts: 5,563 Rep Power: 62	I have read that manual many times but did not make the connection before. So what is the maximum overflow (4 bits = 15)? I had tried some examples and it wasn't working (the main reason for my question), but I must have been doing something wrong because I tried a couple more and it worked this time. Here are the examples I tried (for future reference), similar to Max's above, but with number's plugged in for easy verification of the results. First Example: ------------------------- static a = 0.9, b = 0.5, c = 0; macs, 0, 0, a, b; (0.9 * 0.5 = 0.45) macs, 0, accum, a, b; (0.45 + 0.45 = 0.9) macs, 0, accum, a, b; (0.9 + 0.45 = 1.35) macsn, c, accum, a, b; (1.35 - 0.45 = 0.9) ------------------------- Result of above is 0x73333330 = 1932735280 / (2^31+1) = 0.89999999 = 0.9 Second Example: ------------------------- static a = 0.9, b = 0 macs 0, 0, a, a; (0.9 * 0.9 = 0.81) macs 0, accum, a, a; (0.81 + 0.81 = 1.62) macs 0, accum, a, a; (1.62 + 0.81 = 2.43) macsn 0, accum, a, a; (2.43 - 0.81 = 1.62) macsn b, accum, a, a; (1.62 - 0.81 = 0.81) ------------------------- Result of above is 0x67AE1478 = 1739461752 / (2^31+1) = 0.80999999 = 0.81 So, even though some of the intermediary results were floating point numbers, the final result was correct. Good stuff! Thanks again, -Russ