Digital Electronics-Digital Arithmetic Operations and Circuits
Digital Electronics-Digital Arithmetic Operations and Circuits
41. An input to the mode pin of an arithmetic/logic unit (ALU) determines if the function will be:
- one's-complemented
- arithmetic or logic
- positive or negative
- with or without carry
42. Could the sum output of a full-adder be used as a two-bit parity generator?
- TRUE
- FALSE
43. In VHDL, what is a GENERATE statement?
- The start statement of a program
- Not used in VHDL or ADHL
- A way to get the computer to generate a program from a circuit diagram
- A way to tell the compiler to replicate several components
44. Binary subtraction of a decimal 15 from 43 will utilize which two's complement?
- 101011
- 110000
- 011100
- 110001
45. Which of the following is the primary advantage of using binary-coded decimal (BCD) instead of straight binary coding?
- Fewer bits are required to represent a decimal number with the BCD code.
- BCD codes are easily converted from decimal.
- the relative ease of converting to and from decimal
- BCD codes are easily converted to straight binary codes.
47. The carry propagation delay in full-adder circuits:
- is normally not a consideration because the delays are usually in the nanosecond range.
- decreases in a direct ratio to the total number of FA stages.
- is cumulative for each stage and limits the speed at which arithmetic operations are performed.
- increases in a direct ratio to the total number of FA stages but is not a factor in limiting the speed of arithmetic operations.
48. What is the difference between a full-adder and a half-adder?
- Half-adder has a carry-in.
- Full-adder has a carry-in.
- Half-adder does not have a carry-out.
- Full-adder does not have a carry-out.
49. The summing outputs of a half- or full-adder are designated by which Greek symbol?
- omega
- theta
- lambda
- sigma
50. The BCD addition of 910 and 710 will give initial code groups of 1001 + 0111. Addition of these groups generates a carry to the next higher position. The correct solution to this problem would be to:
- ignore the lowest order code group because 0000 is a valid code group and prefix the carry with three zeros
- add 0110 to both code groups to validate the carry from the lowest order code group
- disregard the carry and add 0110 to the lowest order code group
- add 0110 to the lowest order code group because a carry was generated and then prefix the carry with three zeros