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** 32 X 32 Bit
Multiplication Using Urdhva Triyakbhyam (Vertically and crosswise) for two
Binary numbers**

Consider two ‘32’ bit numbers A and B respectively as follows, where each number has ‘23’ bit floating point part; ‘8’ bit integer part and ‘1’ bit sign bit.

A=
**a _{31 }a_{30}a_{29}a_{28}a_{27}a_{26}a_{25}a_{24}a_{23
}a_{22}a_{21}a_{20}a_{19}a_{18}a_{17}a_{16}a_{15}a_{14}a_{13}a_{12}a_{11}a_{10}a_{9}a_{8
}a_{7}a_{6}a_{5}a_{4}a_{3}a_{2}a_{1}a_{0}**

(X_{2}) (X_{1})
(X_{0})

B=
b_{31 }**b _{30}b_{29}b_{28}b_{27}b_{26}b_{25}b_{24}b_{23
}b_{22}b_{21}b_{20}b_{19}b_{18}b_{17}b_{16}b_{15}b_{14}b_{13}b_{12}b_{11}b_{10}b_{9}b_{8}b_{7}b_{6}b_{5}b_{4}b_{3}b_{2}b_{1}b_{0}**

(Y_{2})
(Y_{1}) (Y_{0})

Thus the above multiplication process can be viewed as

X_{2} X_{1} X_{0}

* Y_{2} Y_{1} Y_{0}

--------------------------

SABC

Where,
CP= C= X_{0}Y_{0}

CP= B= X_{1}Y_{0}+X_{0}Y_{1}

CP= A= X_{1}Y_{1}

S=X_{2 }(xor) Y_{2}

** BLOCK DIAGRAM**

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Figure 4.5

A block diagram representing *8-*bit
multiplication

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