Page 271 - FUNDAMENTALS OF COMPUTER

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NPP Number System, Boolean Algebra and Logic Circuits 271

Example of shift and add method of {eâQ> H$amo d Omo‹S>mo {d{Y H$m CXmhaU (AZgmBÝS>
multiplication (Unsigned numbers) g§»`mE§)

Consider following example of multi- {ZåZ CXmhaU H$mo XoI| {Og_| Xmo g§»`mAm| H$m JwUm
plication of two unsigned numbers: H$aZm h¡:
1011 × 1010
multiplicand = 1011 Bg_| 1011 _ëQ>rpßbH|$S> h¡ VWm 1010 (n = 4)
multiplier = 1010 (n = 4) _ëQ>rßcm`a h¡, CnamoŠV JwUZ\$c àmßV H$aZo Ho$ {cE {eâQ>
Now applying shift and add method: H$amo d Omo‹S>mo {d{Y cJmZo na:

Step 1: n = 4, since multiplier is a 4-bit ñQ>on 1: n = 4 h¡ Š`m|{H$ _pëQ>ßcm`a 1010 _| 4-
number. {~Q>| h¢Ÿ&

Step 2: LSB of multiplier is 0. thus, par- ñQ>on 2: _pëQ>ßcm`a H$s LSB eyÝ` h¡Ÿ& AV: Am§{eH$
tial product = p.p = 0000 JwUZ\$c = p.p. = 0000,

n = n – 1 = 3.
Step 3: Next LSB of multiplier is 1. ñQ>on 3: AJcr LSB "1" h¡Ÿ& AV: P.P. H$mo amB©Q>
Therefore shifting partial product to the {eâQ> H$aHo$ _pëQ>pßbH|$S> H$mo Omo‹S>Zo na Z`m P.P. Bg
right and adding multiplicand to get new àH$ma àmßV H$a|Jo:
partial product:

00000
+ 10 1 1
P.P. = 1 0 1 1 0

Step 4: ñQ>on 4:
n = n – 1 = 2
Step 5: since next LSB is “0” shif t par- ñQ>on 5: AJcr LSB 0 h¡Ÿ& AV: P.P. H$mo EH$ ~ma
tial product to the right by one bit position: amB©Q> {eâQ> H$aHo$ Z`m P.P. àmßV H$a|JoŸ:

P.P. = 010110
Step 6: ñQ>on 6:
n = n – 1 = 1
Step 7: Since next LSB is 1. ñQ>on 7: M±y{H$ AJcr LSB 1 h¡,
Shift right and add, multiplicand. AV: amB©Q> {eâQ> H$aHo$ _pëQ>pßbH|$S> Omo‹S>Zo na:
0 0 101 1 0
+ 10 1 1
P.P. = 1 1 0 1 1 1 0

Step 8: ñQ>on 8:
n = n – 1 = 0,