Digital Logic Test 4

Result

Digital Logic Test 4

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A sequential circuit counts from 0 to 255 using JK flip-flop. If the propagation delay of each flip flop is 50 ns, the maximum clock frequency that can be used is _____ MHz

A
2.5-2.5

B
2.54-2.51

C
2.57-2.56

D
2.59-2.52

Solution

MOD 256 counter counts 0 to 255
Number of Flip Flops = log₂ 256 = 8
Propagation delay Td = 50 n sec
Total propagation delay = 8 × 50 = 400 n sec.
The maximum clock frequency that should be used is:
\( F = \frac{1}{{400 \ ns}} = 2.5\;MHz\)
Question 2
1 / -0

An Up-counter with 10 flip-flops is initially set zeros in all the flip flops; after 1050 clock pulses the binary number in the counter is

A
0000011010

B
0000011011

C
0000011100

D
0000010010
Solution
- Up to 1023 clock pulses the binary number in counter goes from 000000000 to 1111111111
- Then on 1024 resets to all Flip flop at 0
- At 1050 clock pulse the number in the counter is binary of 26 i.e 0000011010
Question 3
1 / -0

What is the minimum number of flip flops needed to construct a binary modulo 1026 counters?

A
11-11

B
114-111

C
117-116

D
119-112

Solution

Concept:
Binary modulo N counter can count up to 0 to N – 1
Therefore, binary modulo 256 counters will count from 0 to 1025 (order may vary)
Formula:
Using J-K flip in asynchronous counter.
Minimum number of flip flop needed = ⌈log₂ N⌉
Calculation:
Minimum number of flip flop needed = ⌈log₂ 1026⌉ = 11
Question 4
1 / -0

A new flipflop with inputs X and Y, has the following property
Inputs
Current state
Next state
X
Y
0
0
Q
1
0
1
Q
Q̅
1
1
Q
0
1
0
Q
Q

Which of the following expresses the next state in terms of X, Y, current state?

A
(X̅ ∧ Q̅) ∨ (Y̅ ∧ Q)

B
(X̅ ∧ Q) ∨ (Y̅ ∧ Q̅)

C
(X ∧ Q̅) ∨ (Y ∧ Q)

D
(X ∧ Q̅) ∨ (Y̅ ∧ Q)

Solution

Concept –
For finding the expression of Next state, a truth table needs to be constructed.

Current State
Next State
X
Y
Q
Q_n+1
0
0
0
0
1
1
0
0
1
1
2
0
1
0
1
3
0
1
1
0
4
1
0
0
0
5
1
0
1
1
6
1
1
0
0
7
1
1
1
0

Equation for Q_n+1is made by writing all values of X, Y, Q for which Q_n+1= 1
\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar X\;\bar Y\;\bar Q}} + {\rm{\;\;\bar X\;\bar YQ}} + {\rm{\;\bar X\;Y\;\bar Q}} + {\rm{\;X\;\bar YQ}}\)
\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar X\;\bar Y\;}}\left( {{\rm{\bar Q}} + {\rm{Q}}} \right) + {\rm{\;\bar X\;Y\;\bar Q}} + {\rm{\;X\;\bar YQ}}\)
\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = \;{\bf{\bar X}}{\rm{\;}}{\bf{\bar Y}}{\rm{\;}} + {\rm{\bar XY\bar Q}} + {\rm{\;}}{\bf{X}}{\rm{\;}}{\bf{\bar YQ}}\)
\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar Y}}\left( {{\rm{\bar X}} + {\rm{XQ\;}}} \right) + {\rm{\bar XY\bar Q}}\)
\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\bf{\bar Y}}\;{\bf{\bar X}} + {\rm{\bar YQ}} + {\bf{\bar XY\bar Q}}\)
\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\bar X}}.{\rm{\bar Y}} + {\rm{\bar X}}\overline {.{\rm{Q}}} + {\rm{\bar YQ}}\)
Consensus theorem: ab + a̅c + ac = ab + a̅c
In a = Q̅, b = Y̅, c = Z̅
\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\bar X\bar Q}} + {\rm{\bar YQ\;\;\;\;}}\)
Q_n+1 = (X̅ ∧ Q̅) ∨ (Y̅ ∧ Q)
Important Point:
. → AND → ∧
+ → OR → ∨ )
Minimization could have been done using K - Map