1. Our aim is to find 9 bit binary till 301.
All approximation we considered as 2n
.
If you consider 512 , binary form is 1000000000 – it is 10 digit.
We want only 9 digit. In 2n
combination 20
=1, 21
=2,…., 28
=256. 29
=512.
So till 28
we need to consider.
256 – 100000000. – 9 digit. Is n.
Decrement by one bit ie 2n-1
and compare the value with 301.
If (256+2n-1
)>301 consider that bit as 1. Move to next bit.
Use the First (256+2n-1
)<301 value.
(256+2n-1
)<301 consider that bit as 0.
Use the same last (256+2n-1
)<301 value .
Find the corresponding binary. Find the value till 301. More exp. below
2. Explanation for above one. n=9,8….0
1. 256 – binary no 301(Vin)>256(Vax) .
So this 256 I will use until I get next minimum value than 301.
2. 256(I used here because < 301.)+2n-1
(128)=384. 384>301. So I
used 1. Moved to next bit.
So this 256 I will use until I get next minimum value than 301.
3. 256+2n-1
(64)=320. 384>301. So I used 1. Moved to next bit
256 I will use until I get next minimum value than 301.
4. 256+2n-1
(32)=288. 288<301. So I used 0. Moved to next bit.
288 I will use from next bit until I get lesser one because
288<301.
5. 288+2n-1
(16)=304, so 1 I used moved to next
288 I will use here. It is the last low value than 301.
6. 288+2n-1
(8)=296, so I used 0 moved to next
From next bit 296 I will use bcs last lower one than 301.
7. 296+2n-1
(4)=300, so I used 0 moved to next
From next bit 300 I will use bcs last lower one than 301.
8. 300+2n-1
(2)=300, so 1 I used moved to next
300I will use here. It is the last low value than 301.
9. 300+2n-1
(1)=301, got the value which I required . These binary
technique used in successive approximation.
