Test 3

Question 1

assuming enough resources
(2-input AND gates) are available, ALL partial products (PPs) are calculated

Answer 1

concurrently

Question 2

in a binary multiplication, which bits of the final product are calculated first?

Answer 2

the LSbits

Question 3

Binary multiplication can be implemented in custom hardware (HW) or (SW)

Answer 3

true

Question 4

when is the value of the PP zero?

Answer 4

when the associated multiplier bit is 0

Question 5

the most time consuming operation in a binary multiplication is

Answer 5

the multi-operand addition of the PPs

Question 6

in the SSRM discussed in class, the PH-reg is initialized to

Answer 6

0

Question 7

in the SSRM discussed in class the M-reg stores the value

Answer 7

of the original Multiplicand

Question 8

in the SSRM described in class a new PP is generated using

Answer 8

2-to-1 muxes

Question 9

in the SSRM discussed in class, which sequence of operations are repeated?

Answer 9

Add- shift-right

Question 10

in the SSRM mPL-Reg is used to store

Answer 10

the remaining bits of the multiplier and the incoming LSbits of the FP

Question 11

How many different data sets can the PSRM described in class process concurrently?

Answer 11

4

Question 12

the PSRM improves

Answer 12

throughput

Question 13

in each stage of the PSRM which operations are executing during every machine cycle?

Answer 13

add and shift right

Question 14

in each PSRM stage the PP is generated using

Answer 14

2-input AND gates

Question 15

on average, division takes longer than multiplication

Answer 15

true

Question 16

the generate function indicates that a carry-out of ‘1’ will be generated by the stage

Answer 16

no matter what the carry-in value is

Question 17

compared to RCA, a CLA uses:

Answer 17

more hardware resources

Question 18

what is the RCAs biggest performance problem

Answer 18

the propagation of the carry values from one state to the next

Question 19

how many gate-level propagation delays are in a generic Sum-Of-Products (SOP)?

Answer 19

2

Question 20

if the propagate function p = ‘1’ then the carry-out value will be ‘1’ if:

Answer 20

the carry-in value is ‘1’

Question 21

at the cost of additional and acceptable propagation delay, the second-level carry-look-ahead reuses computation results in the form of:

Answer 21

Gj and Pj

Question 22

in an attempt to tradeoff cost/performance we can combine carry look-ahead and ripple-carry in the same adder unit

Answer 22

true

Question 23

what is a practical limit for the number of inputs to a (static) CMOS gate?

Answer 23

4

Question 24

how many gate-level propagation delays does it take to compute the sum bit of a stage in a RCA or CLA from the moment the cin to that stage is available

Answer 24

1

Question 25

comparing a n-bit CLA to a n-bit RCA

Answer 25

the CLA is faster, but uses sign9ificant more hardware resources

Question 26

in a positional number system, the position of a digit

Answer 26

represents the weight it carries

Question 27

to convert a number to an IEEE floating point format, the number is first normalized. the integer part of the normalized value:

Answer 27

is not stored, but implied when the value of the number is being reconstructed

Question 28

in the IEEE floating formats, the stored exponent is stored in an excess form. this means that the actual exponent value is equal to:

Answer 28

the stored exponent value minus excess values

Question 29

to be able to use floating points numbers in manipulation, these have to be:

Answer 29

unpacked -> manipulated -> packed

Question 30

in a weighted number system, the value of a number is equal to:

Answer 30

the Sum-Of-Products between each digit value and the radix (or base) raised to the power of its position

Question 31

the behavioral description of the SSRM CU uses if-else and/or case constructs. which of the following is NOT used in the condition evaluations of these constructs?

Answer 31

DONE

Question 32

the SSRM CST captures ONLY se3quential events

Answer 32

false

Question 33

why is the external user control input START necessary?

Answer 33

so that the SSRM knows when a new set of input operands is valid and ready to be manipulated

Question 34

if the multiplicand is n-bits wide and the multiplier is m-bits wide, how many times is the add-shift sequence repeated in the SSRM?

Answer 34

m

Question 35

in the lecture SSRM ASM chart we initialized n = 4 and counted down to 0. if we would initialize n = 0, we would need to

Answer 35

count up to 4