Using the knowledge in computational language in LC-3 Assembly it is possible to write a code that replaces the value in R0 with its absolute value
<h3>Writting the code </h3>
.ORIG x0200
START
AND R0, R0, #0 ; copy R0 to itself to set the condition codes based on R0;
; i.e performing addition operation with Zero option to set the flags
BRzp DONE ; if R0 is NON-NEGATIVE, skip the negation (already correct);
; Branch to DONE if number is poistive
NOT R0, R0 ; R0 is negative, so negate it i.e taking 2's complement
ADD R0, R0, #1 ; R0 = -R0 is performed successfully
DONE BR START
.END
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Answer: Time division multiplexing
Explanation: because a slot is equivalent to 125ms
And each position inside a slot is for each signal and a single bit frim each voice conversation is sent during each 125ms slot
Answer:
Explanation:
a) taking 00000005, it is a 13th bit which is 0, and therefore, we'd call it an ordered chunk
b) taking 00000005 again, it is 14th (B) and 15th (E) bits are 0 and so we'd call it a middle fragment
c) 00000015 is equal to 21, it is not a multiple of 4, and as such, it needs 3 padding bytes
d) 00000005 is equal to 5 making it a TSN. So the TSN is 5
e) taking 0003, we can then say the SI is 3
f) taking 000A, the SSN is then 10
g) the message is 48656C6C
Answer:
1. volatile data
2. temporary data
3. persistent data
seems like you already got it correct
Explanation:
Answer:
Since there are only two digits in binary, there are only two possible outcomes of each partial multiplication: If the digit in B is 0, the partial product is also 0. If the digit in B is 1, the partial product is equal to A.
Explanation:
