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Edit:Value Exceeeds while dividing

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Pat
Pat 2012-11-29
I have 16 bit values and 14 bit values of same size
A=16 bit values ,B=14 bit values
i have sorted suxh a way that maximum value of 14 bit comes under max valu of 16 bit and so on,finally minimum of 14 bit comes min of 16 bit
used command as
[~,idx]=sort(A)
[~,idx1]=sort(idx)
C=B(idx1);
Dw=C./Cw
Dw=Dw*100;
the value of Dw exceeds 5 bit ,plese tell why i get this
suppose if max of 14 bit 8192/max of 16 bit 32767 we get 25 ,which is a 5 bit ,then why get more than 5 bit,please help
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Walter Roberson
Walter Roberson 2012-11-29
sort() and unique() work on negative values as well. Is the concern about the possibility that the max(16bit) might be negative and of small absolute value, but max(14bit) might be positive, leading to a large number divided by a number that is small absolute value?
Can you reproduce the large-result problem with a small demonstration vector that you could post?
Pat
Pat 2012-11-29
A1 =
-443 -221 -266 89 -45 -266 -177 -133 -266 -266 -177 140
B1 =
-66 -55 33 78 122 111 66 22 22 -33 -78 -78

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2012-11-29
[~,i1] = sort(A);
B1 = sort(B);
[~,i2] = sort(i1);
C = B1(i2);
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Pat
Pat 2012-11-29
Andre i have negative and unique values in it..my max(B) is 1540 and max(32768)
so 1540/32768 is 0.0470
but i get maximum 0.0995 after dividing C./A

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更多回答（1 个）

Walter Roberson
Walter Roberson 2012-11-29
Remember, negative divided by negative gives positive, so if -12 as a 14 bit number happened to get paired with -1 as a 16 bit number, the ratio would be -12/-1 which would be 12.
But you don't even need to take into account negatives for this. If 12 as a 14 bit number happened to get paired with 1 as a 16 bit number, the ratio would be 12 anyhow. Consider for example
14 bit: [12 30]
16 bit: [1 100]
then 12/1 = 12, 30/100 = 1/3, so the ratio of their maximums (30, 100) would be much smaller the internal ratio.

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