The setup time of a clocked flip-flop is: A) The minimum amount of time that an
ID: 1302860 • Letter: T
Question
The setup time of a clocked flip-flop is:
A) The minimum amount of time that an input must remain stable after an active clock
transition.
B) The maximum amount of time that an output must remain stable after an active clock
transition.
C) The minimum amount of time that an output must remain stable before an active clock
transition.
D) The minimum amount of time that an input must remain stable before an active clock
transition.
The symbol for a flip flop has a small triangle - and no bubble - on its clock (CLK) input. The triangle indicates:
A) The FF is level active and can only change states when the CLOCK = 1.
B) The FF is edge-triggered and can only change states when the clock goes 0 to 1.
C) The FF is edge-triggered and can only change states when the clock goes from 1 to 0.
D) The FF is an active LOW device and can only change states when the CLOCK = 0.
A negative-edge-triggered J-K flip-flop is presently in the CLEAR state. Which of the following input conditions will cause it to change states?
A) CLK = NGT, J = 1, and K = 0
B) CLK = NGT, J = O, and K = 1
C) CLK = PGT, J = 1, and K = 0
D) CLK = PGT, J = O, and K = 1
In using the 2
Explanation / Answer
ANSWER:
The setup time of a clocked flip-flop is:
Option D:The minimum amount of time that an input must remain stable before an active clock
That is how the setup time is defined.
The symbol for a flip flop has a small triangle - and no bubble - on its clock (CLK) input. The triangle indicates:
B) The FF is edge-triggered and can only change states when the clock goes 0 to 1.
If a FF has a triangle, it means that it is sensitive to edges not levels. If it does not have a bubble, it means it is a positive edge triggered FF, i.e. it causes o/p to change when clk moves from low to high.
A negative-edge-triggered J-K flip-flop is presently in the CLEAR state. Which of the following input conditions will cause it to change states?
A) CLK = NGT, J = 1, and K = 0
Since the FF is negative edge triggered, it would 'react' only to negative edges. Hence CLK = NGT
When J = 1, and K = 0; D = Q+Q` = 1, which is different from the previous state.
When J = O, and K = ; D = 0, which is the same as the previous state.
Hence A is the answer.
In using the 2
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