# Re: Number truncation, round off and saturation in Fixed Point Arithmetic

*From*: steve <bungalow_steve@xxxxxxxxx>*Date*: Wed, 20 Jun 2007 10:06:47 -0700

On Jun 19, 3:10 am, Abhishek <abhisheksgum...@xxxxxxxxx> wrote:

Hi,

I have the following specific doubts regarding fixed point arithmetic.

1) If I have Q(1.15) meaning one sign bit and 15 fractional bits, and

If I also have another unsigned 8 bit number represented in a 16 bit

number with leading 8 zeros in it, I then multiply these two treating

the second number as a signed 16 bit number or Q(16.0) format (.i.e. 1

sign bit and 15 intger bits. here sign bit is zero).

Whats the result size and what would be its format

after multiplication? Can anyone explain whats happening here clearly?

Now once it is multiplied, the author says it is in Q17.15

format(which you need to explain how he got it plzzz). Then he

converts Q17.15 result of multiplication to 16 bit signed integer by

right shifting the number by 15 bits and extracting the lower 16 bits

of the right shifted result.

Can anyone explain in detail whats happening here and exactly why did

he follow these steps?

Suitable examples and a clear explanation would be highly

appreciated :-)

Thanks

With Regards,

Abhishek S

see section 2 of

http://www.superkits.net/whitepapers/Fixed%20Point%20Representation%20&%20Fractional%20Math.pdf

But I don't think any explanation is going to make you happy until you

get a bit piece of paper and do the multiply yourself. Its somewhat

akin to explaining to a kid why 5 + 5 = 10, you just have to get the

kid 10 apples and work it out himself....

.

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