Large scale Matrix-vector multiplication fixed-point

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Giulio
Giulio on 6 Mar 2012
Hi,
I have a strange behavior on a large matrix-vector product in fixed-point precision. The matrix is block banded, so it has few non-zero elements in each row. That is what happen
K>> size(A)
ans =
250 550
K>> A.numerictype
ans =
DataTypeMode: Fixed-point: binary point scaling
Signedness: Signed
WordLength: 32
FractionLength: 22
K>> A.fimath
ans =
RoundMode: floor
OverflowMode: saturate
ProductMode: KeepMSB
ProductWordLength: 32
SumMode: KeepMSB
SumWordLength: 32
CastBeforeSum: true
K>> size(x)
ans =
550 1
K>> x.numerictype
ans =
DataTypeMode: Fixed-point: binary point scaling
Signedness: Signed
WordLength: 32
FractionLength: 22
K>> x.fimath
ans =
RoundMode: floor
OverflowMode: saturate
ProductMode: KeepMSB
ProductWordLength: 32
SumMode: KeepMSB
SumWordLength: 32
CastBeforeSum: true
Lets consider just the first row of A in which only the first 11 entries are non-zero while vector x has entries all equal 1(x=ones(250,1)). If I perform the product:
K>> A(1,1:11)*x(1:11)
ans =
0.5742
DataTypeMode: Fixed-point: binary point scaling
Signedness: Signed
WordLength: 32
FractionLength: 8
K>> A(1,:)*x
ans =
-1
DataTypeMode: Fixed-point: binary point scaling
Signedness: Signed
WordLength: 32
FractionLength: 2
K>> A(1,12:550)*x(12:550)
ans =
0
DataTypeMode: Fixed-point: binary point scaling
Signedness: Signed
WordLength: 32
FractionLength: 2
Can someone explain this strange behavior? Thanks.

Answers (2)

DSP Masters
DSP Masters on 30 Apr 2012
Hi Giulio, you can replace the matlab 'fi' by ur own script, then I feel your problem would be solved
  1 Comment
Giulio
Giulio on 30 Apr 2012
Thanks for replaying!
However I had lot of trouble figuring out the more appropriate "fimath" to use for a given DSP processor.
I will really appreciate whether you can help me on this.
Thanks

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Hong Ye
Hong Ye on 1 Aug 2012
Hi Giulio, The behavior you observed is due to the fimath setting you picked for A and x. Here is an explanation for the multiply and accumulate operation: 1) the full precision multiply of two s32,22 number result in s64,44, ProductMode keepMSB and productWordLength 32 indicate that the most significant (or from left side) 32 bits of the the 64 bits result will be kept, so you will get a product of s32,12 (with 20 bits Integer WL) for example,
>> A(1, 1)*x(1)
ans =
0.8147
DataTypeMode: Fixed-point: binary point scaling
Signedness: Signed
WordLength: 32
FractionLength: 12
2) SumMode is also keepMSB with SumWordLength as 32, so adding two s32,12 numbers will lead to result with s32,11 data type. the integer WordLength grows from 20 to 21 (so the fractionLength decrease by 1 bit to 11 given the fixed WordLength) to avoid overflow. for example
>> A(1, 1:2)*x(1:2)
ans =
1.7773
DataTypeMode: Fixed-point: binary point scaling
Signedness: Signed
WordLength: 32
FractionLength: 11
>> A(1, 1:11)*x(1:11) % Integer WL grow from 20 to 20+ceil(log2(11))(=24) bits
ans =
6.4102
DataTypeMode: Fixed-point: binary point scaling
Signedness: Signed
WordLength: 32
FractionLength: 8
For experimenting purpose, you can first remove fimath of variable x by x.fimathislocal = false, then vary A's fimath settings to see different numerical result and data type being returned. e.g., set A.ProductMode = 'FullPrecision' and A.SumMode = 'FullPrecision', and see what you will get for A(1,1)*x(1) and A(1,1:2)*x(1:2)

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