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A_{3} A_{2} A_{1} A_{0}, B_{3} B_{2} B_{1} B_{0}Solution: from the circuit diagram (A=1010 ; B=1101)\begin{array}{l}A_{3}=A=1 \text { (bor } \Sigma_{3} \text { ) } \\B_{3}=1 \& \overline{a d d} / s_{u b}=1 \text { (always) Hence } \\\overline{B_{3} \oplus \text { add }} s_{u_{b}}=1\end{array}Now\begin{array}{l}A+B=1+1=0 \quad \frac{C \text { in }=1}{} \\\Sigma_{3}=0 \quad \text { Cout }=1\end{array}bor \varepsilon_{2} & cout2\begin{array}{l}A_{2}=0 ; \quad B_{2}=1 \quad \overline{\text { ada } / \text { sub }^{2}}=1 \\\overline{B_{2} \oplus \text { add } / \text { sub }}=1 \oplus 1=1 \\\text { Now } A_{2}+B_{2}=0+1=1 ; \text { cout }-2= \\1+\text { Coutz }_{2}=1+1=0 ; \text { cout } z=1\end{array}Hence \varepsilon_{2}=0 and Cout _{2}=1Now bos \Sigma_{1} and cout, Calculation :-A_{1}=1 ; B_{1}=0\overline{B_{1} \oplus \text { add } / s_{u_{b}}}=\overline{O \oplus 1}=0Now A_{1}+\overline{B_{1} \oplus \text { add } / \text { sub }}=1+0=11+\text { Cout }_{2}=1+1=0 ; \text { cout } 1=1Hence \Sigma_{1}=0 and C_{\text {out }}=1Now for \varepsilon_{0} and Cout -o\begin{array}{l}A_{0}=0 ; \quad B_{0}=1 \\\overline{B_{0} \oplus a d o / \mathrm{Sub}_{b}}=\overline{1 \oplus 1}=1 \\A_{0}+1=0+1=1 \\\Sigma_{0}=1 .\end{array}and Couto =1Hence output \varepsilon_{3} \Sigma_{2} \Sigma_{1} \varepsilon_{0}=000: 1\begin{array}{lllll}1 & 0 & 1 & 1 & 1 \\ \text { output }\end{array} of the sequence given circuit ...