4 minutes a 12 - bit ones complement ebcdic transmitter (tx) sends a 12 - bit twos complement ascii receiver…

4 minutes a 12 - bit ones complement ebcdic transmitter (tx) sends a 12 - bit twos complement ascii receiver (rx) the binary result below. if the result below was received with no errors, how will the rx represent the result internally for processing purposes? data ebcdic\nzone 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111\n0000 nul soh stx etx pf ht lc del rlf smm vt ff cr so si\n0001 dle dc1 dc2 tm res nl bs § can em cc oj1 £ ¥ ¤ ¦\n0010 ds sos fs byp lf etb esc sm cu2 enq ack bel\n0011 syn pn rs uc eot cus dc4 nak sub\n0100 sp\n0101 &\n0110 -\n0111\n1000 a b c d e f g h i\n1001 j k l m n o p q r\n1010 - s t u v w x y z\n1011\n1100 a b c d e f g h i\n1101 j k l m n o p q r\n1110 | s t u v w x y z\n1111 0 1 2 3 4 5 6 7 8 9\nascii\n0 nul 16 dle 32 48 0 64 @ 80 p 96 112 p\n1 soh 17 dc1 33! 49 1 65 a 81 q 97 a 113 q\n2 stx 18 dc2 34 \ 50 2 66 b 82 r 98 b 114 r\n3 etx 19 dc3 35 # 51 3 67 c 83 s 99 c 115 s\n4 eot 20 dc4 36 $ 52 4 68 d 84 t 100 d 116 t\n5 enq 21 nak 37 % 53 5 69 e 85 u 101 e 117 u\n6 ack 22 syn 38 & 54 6 70 f 86 v 102 f 118 v\n7 bel 23 etb 39 55 7 71 g 87 w 103 g 119 w\n8 bs 24 can 40 ( 56 8 72 h 88 x 104 h 120 x\n9 ht 25 em 41 ) 57 9 73 i 89 y 105 i 121 y\n10 lf 26 sub 42 * 58 : 74 j 90 z 106 j 122 z\n11 vt 27 esc 43 + 59 ; 75 k 91 107 k 123 {\n12 ff 28 fs 44, 60 < 76 l 92 \\ 108 l 124 |\n13 cr 29 gs 45 - 61 = 77 m 93 109 m 125 }\n14 so 30 rs 46. 62 > 78 n 94 ^ 110 n 126 ~\n15 si 31 us 47 / 63? 79 o 95 _ 111 o 127 del

4 minutes a 12 - bit ones complement ebcdic transmitter (tx) sends a 12 - bit twos complement ascii receiver (rx) the binary result below. if the result below was received with no errors, how will the rx represent the result internally for processing purposes? data ebcdic\nzone 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111\n0000 nul soh stx etx pf ht lc del rlf smm vt ff cr so si\n0001 dle dc1 dc2 tm res nl bs § can em cc oj1 £ ¥ ¤ ¦\n0010 ds sos fs byp lf etb esc sm cu2 enq ack bel\n0011 syn pn rs uc eot cus dc4 nak sub\n0100 sp\n0101 &\n0110 -\n0111\n1000 a b c d e f g h i\n1001 j k l m n o p q r\n1010 - s t u v w x y z\n1011\n1100 a b c d e f g h i\n1101 j k l m n o p q r\n1110 | s t u v w x y z\n1111 0 1 2 3 4 5 6 7 8 9\nascii\n0 nul 16 dle 32 48 0 64 @ 80 p 96 112 p\n1 soh 17 dc1 33! 49 1 65 a 81 q 97 a 113 q\n2 stx 18 dc2 34 \ 50 2 66 b 82 r 98 b 114 r\n3 etx 19 dc3 35 # 51 3 67 c 83 s 99 c 115 s\n4 eot 20 dc4 36 $ 52 4 68 d 84 t 100 d 116 t\n5 enq 21 nak 37 % 53 5 69 e 85 u 101 e 117 u\n6 ack 22 syn 38 & 54 6 70 f 86 v 102 f 118 v\n7 bel 23 etb 39 55 7 71 g 87 w 103 g 119 w\n8 bs 24 can 40 ( 56 8 72 h 88 x 104 h 120 x\n9 ht 25 em 41 ) 57 9 73 i 89 y 105 i 121 y\n10 lf 26 sub 42 * 58 : 74 j 90 z 106 j 122 z\n11 vt 27 esc 43 + 59 ; 75 k 91 107 k 123 {\n12 ff 28 fs 44, 60 < 76 l 92 \\ 108 l 124 |\n13 cr 29 gs 45 - 61 = 77 m 93 109 m 125 }\n14 so 30 rs 46. 62 > 78 n 94 ^ 110 n 126 ~\n15 si 31 us 47 / 63? 79 o 95 _ 111 o 127 del

Answer

Explanation:

Step1: Understand the receiver's format

The receiver is a 12 - bit two's complement ASCII device. It will process data in two's complement ASCII format.

Step2: No conversion needed for internal processing

Since the receiver is designed to work with two's complement ASCII, if the received binary result is already in a form that can be mapped to ASCII values (assuming it is a valid 12 - bit representation for an ASCII - related character or control code), it will represent the result internally in its native two's complement ASCII format for processing.

Answer:

The receiver will represent the result internally in 12 - bit two's complement ASCII format for processing purposes.