{"id":188782,"date":"2025-02-07T09:38:04","date_gmt":"2025-02-07T09:38:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188782"},"modified":"2025-02-07T09:38:06","modified_gmt":"2025-02-07T09:38:06","slug":"write-the-excess-3-equivalent-codes-of-610-7810-and-35710","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/07\/write-the-excess-3-equivalent-codes-of-610-7810-and-35710\/","title":{"rendered":"Write the excess-3 equivalent codes of (6)10, (78)10 and (357)10"},"content":{"rendered":"\n

Write the excess-3 equivalent codes of (6)10, (78)10 and (357)10, all in 16-bit format.0011001100111001, 0011001110101011, 0011011010001010Determine the Gray code equivalent of (10011)2 and the binary equivalent of the Gray code number110011.A 16-bit data word given by 1001100001110110 is to be transmitted by using a fourfold repetitioncode. If the data word is broken into four blocks of four bits each, then write the transmitted bitstream.1001100110011001100010001000100001110111011101110110011001100110<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Let\u2019s go step by step to clarify each of your queries.<\/p>\n\n\n\n

Excess-3 Code (XS-3)<\/strong>:<\/h3>\n\n\n\n

Excess-3 (XS-3) is a binary-coded decimal (BCD) code where each decimal digit is encoded by adding 3 (0011) to the binary representation of the digit. You provided decimal values (6), (78), and (357) and asked for the 16-bit Excess-3 equivalent.<\/p>\n\n\n\n

    \n
  1. (6)\u2081\u2080 in Excess-3 (16-bit format)<\/strong>:<\/li>\n<\/ol>\n\n\n\n
      \n
    • First, we encode 6 using Excess-3 by adding 3 to it: (6 + 3 = 9), which in binary is 1001<\/code>.<\/li>\n\n\n\n
    • So, the Excess-3 code for 6 (in a 16-bit format) is 0000000000001001<\/code>.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. (78)\u2081\u2080 in Excess-3 (16-bit format)<\/strong>:<\/li>\n<\/ol>\n\n\n\n
          \n
        • The first digit, 7, in Excess-3: (7 + 3 = 10) (binary 1010<\/code>).<\/li>\n\n\n\n
        • The second digit, 8, in Excess-3: (8 + 3 = 11) (binary 1011<\/code>).<\/li>\n\n\n\n
        • So, the Excess-3 code for 78 (in a 16-bit format) is 0000101000101011<\/code>.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. (357)\u2081\u2080 in Excess-3 (16-bit format)<\/strong>:<\/li>\n<\/ol>\n\n\n\n
              \n
            • The first digit, 3, in Excess-3: (3 + 3 = 6) (binary 0110<\/code>).<\/li>\n\n\n\n
            • The second digit, 5, in Excess-3: (5 + 3 = 8) (binary 1000<\/code>).<\/li>\n\n\n\n
            • The third digit, 7, in Excess-3: (7 + 3 = 10) (binary 1010<\/code>).<\/li>\n\n\n\n
            • So, the Excess-3 code for 357 (in a 16-bit format) is 0000011000101000101010<\/code>.<\/li>\n<\/ul>\n\n\n\n

              Gray Code<\/strong>:<\/h3>\n\n\n\n

              Gray code is a binary numeral system where two successive values differ in only one bit. This encoding is used in various error correction applications because only one bit changes at a time.<\/p>\n\n\n\n

                \n
              1. Gray code of (10011)\u2082<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                  \n
                • The first bit (most significant) is the same in Gray code as in the original binary number. So the first bit of the Gray code is 1.<\/li>\n\n\n\n
                • The subsequent bits of the Gray code are generated by XORing each pair of consecutive bits of the binary number:\n
                    \n
                  • Gray code bit 2: (1 \\oplus 0 = 1)<\/li>\n\n\n\n
                  • Gray code bit 3: (0 \\oplus 0 = 0)<\/li>\n\n\n\n
                  • Gray code bit 4: (0 \\oplus 1 = 1)<\/li>\n\n\n\n
                  • Gray code bit 5: (1 \\oplus 1 = 0)<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                  • Therefore, the Gray code equivalent of (10011)\u2082 is 11010<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                      \n
                    1. Binary equivalent of the Gray code 110011<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                        \n
                      • The first bit of the Gray code (most significant bit) is the same as the binary number: (1).<\/li>\n\n\n\n
                      • To decode, XOR each subsequent Gray code bit with the previous binary bit:\n
                          \n
                        • The first binary bit is 1.<\/li>\n\n\n\n
                        • The second binary bit: (1 \\oplus 1 = 0)<\/li>\n\n\n\n
                        • The third binary bit: (0 \\oplus 0 = 0)<\/li>\n\n\n\n
                        • The fourth binary bit: (0 \\oplus 0 = 0)<\/li>\n\n\n\n
                        • The fifth binary bit: (0 \\oplus 1 = 1)<\/li>\n\n\n\n
                        • The sixth binary bit: (1 \\oplus 1 = 0)<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                        • So, the binary equivalent of Gray code 110011 is 100100<\/strong>.<\/li>\n<\/ul>\n\n\n\n

                          Fourfold Repetition Code<\/strong>:<\/h3>\n\n\n\n

                          In a fourfold repetition code, each bit in the data is repeated four times to increase reliability. The given 16-bit data word is:<\/p>\n\n\n\n

                          (1001100001110110)<\/p>\n\n\n\n

                          To apply fourfold repetition, each 4-bit block is repeated four times. We break the 16-bit word into four 4-bit blocks:<\/p>\n\n\n\n

                            \n
                          1. (1001 \\rightarrow 1001 1001 1001 1001)<\/li>\n\n\n\n
                          2. (1000 \\rightarrow 1000 1000 1000 1000)<\/li>\n\n\n\n
                          3. (1110 \\rightarrow 1110 1110 1110 1110)<\/li>\n\n\n\n
                          4. (0110 \\rightarrow 0110 0110 0110 0110)<\/li>\n<\/ol>\n\n\n\n

                            Therefore, the transmitted bitstream is:<\/p>\n\n\n\n

                            1001100110011001100010001000100001110111011101110110011001100110<\/strong><\/p>\n\n\n\n

                            Summary:<\/h3>\n\n\n\n
                              \n
                            • Excess-3 Code<\/strong>: We converted each decimal number into its corresponding Excess-3 code and formatted it in 16 bits.<\/li>\n\n\n\n
                            • Gray Code<\/strong>: We found the Gray code of a binary number and converted a Gray code back to binary.<\/li>\n\n\n\n
                            • Fourfold Repetition<\/strong>: We applied a fourfold repetition code to each 4-bit block of the data word, resulting in a longer but more redundant transmission for error detection.<\/li>\n<\/ul>\n\n\n\n

                              This explains the conversions and the steps in detail! Let me know if you have further questions.<\/p>\n","protected":false},"excerpt":{"rendered":"

                              Write the excess-3 equivalent codes of (6)10, (78)10 and (357)10, all in 16-bit format.0011001100111001, 0011001110101011, 0011011010001010Determine the Gray code equivalent of (10011)2 and the binary equivalent of the Gray code number110011.A 16-bit data word given by 1001100001110110 is to be transmitted by using a fourfold repetitioncode. If the data word is broken into four blocks […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-188782","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188782","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=188782"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188782\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188782"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188782"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188782"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}