Multiply the signed binary numbers 01010011 and 11001101.

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Accepted Answer

Step 1: The MSBs are different therefore the product will be negative.
Step 2: Change the negative number to its true (uncomplemented) form.

11001101 becomes 00110011

Step 3: Multiplication of the magnitude bits:

[latex]\begin{array}{r} \begin{array}{r} \ ext{Step 4:} \ \ \ \& \ \ \ \ \ \ \ \ \ \end{array} &\begin{array}{c} &&&&&&1&0&1&0&0&1&1 \ &&&&&× &0 &1 &1 &0 &0 &1 &1 \ \hline &&&&&&1&0&1&0&0&1&1 \ &&&&&1& 0& 1& 0& 0& 1& 1 \ \hline &&&&&1&1&1&1&1&0&0&1 \ &&&& 0& 0& 0& 0& 0& 0& 0 \ \hline &&&&0&1&1&1&1&1&0&0&1 \ &&&0& 0& 0& 0& 0& 0& 0 \ \hline &&&0&0&1&1&1&1&1&0&0&1 \ &&1& 0& 1& 0& 0& 1& 1 \ \hline &&1&1&0&0&0&1&0&1&0&0&1 \ &1& 0& 1& 0& 0& 1& 1 \\hline 1&0&0&0&0&1&0&0&0&1&0&0&1 \ 0 &0 &0 &0 &0 &0 &0 \ \hline 1&0&0&0&0&1&0&0&0&1&0&0&1\end{array}& \begin{array}{r} \ \ 1^{\mathrm{st}} ext{ partial product} \ 2^{\mathrm{nd}} ext{ partial product} \ \ 3^{\mathrm{rd}} ext{ partial product} \ \ 4^{\mathrm{th}} ext{ partial product} \ \ 5^{\mathrm{th}} ext{ partial product} \ \ 6^{\mathrm{th}} ext{ partial product} \ \ 7^{\mathrm{th}} ext{ partial product} \ ext{ Final product} \quad \quad \end{array} \end{array}[/latex]

Step 5: Since the product MSB is 1, take the 2’s complement of the final product and attach the sign bit.

1000010001001 becomes 0111101110111

and with the sign bit, the product of the given numbers is

10111101110111.

If you care to check, you will find that the equivalent decimal numbers were 83 and −51 giving a product of −4233.

Question 25.22: Multiply the signed binary numbers 01010011 and 11001101.

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