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Digital Electronics [ Question No: 1 To 20 ] :
Number of flip-flops needed to construct a shift register capable of storing decimal numbers upto 32 are
For \(I_{OLmax}\) = 16 mA and \(I_{ILmax}\) = 0.0016 A, the fan-out (low) is
The amount of ROM needed to implement a 4 bit multiplier is
Logic gates required to built up a half adder circuit are :
On what does the output of a Mealy machine depend ?
The minimal expression of A + A'BA is
Which logic gate is called a universal gate ?
From the circuit of the following logic gates the basic logic gate obtained is ................
The following digital circuit acts as a :
Output values of Moore type FSM are determined by its ..............
The write pulse width (twp) is the time for which :
The minimum memory size to implement a BCD to seven segment decoder will be :
Boolean algebra does not apply to the relation :
How many entries will be there in the truth table of a 3 input NAND gate ?
For a NAND gate having two inputs the output is low (0) only :
In the fission of U-235 nuclei, it is observed that the fission fragments decay by emission of negatively charged beta particles and attain state of stable nuclei.
The reason for emission of negatively charged beta particles is that the fission fragments :
Consider a digital display system (DDS) shown in the figure that displays the contents of register X. A 16-bit code word is used to load a word in X, either from S or from R. S is a 1024-word memory segment and R is a 32-word register file. Based on the value of mode bit M, T selects an input word to load in X. P and Q interface with the corresponding bits in the code word to choose the addressed word. Which one of the following represents the functionality of P, Q, and T?
(Figure: the code word has three fields, M, S-address and R-address. The S-address field feeds P, whose output goes to S; the R-address field feeds Q, whose output goes to R. The outputs of S and R are the two inputs of T, and M is connected to T. The output of T goes to X, which feeds the DDS.)
Which of the following options represents the simplified form of the Boolean equation \(Y = \bar{A}\bar{B}\bar{C} + \bar{A}B\bar{C} + A\bar{B}\bar{C} + AB\bar{C}\) ?
Consider the Boolean operator # with the following properties:
\(x\#0 = x\), \(x\#1 = \bar{x}\), \(x\#x = 0\) and \(x\#\bar{x} = 1\). Then \(x\#y\) is equivalent to
What is the degree of the following function?
F(A,B,C,D) = A`B`C`D`
Total number of questions : 20
Number of answered questions : 0
Number of unanswered questions : 20
Digital Electronics [ Question No: 1 To 20 ] :
Number of flip-flops needed to construct a shift register capable of storing decimal numbers upto 32 are
Your Answer : (Not Answered)
Correct Answer : A
Six flip-flops are needed, option (A). To store a decimal number up to 32 in binary, we have to be able to hold the value 32, which is 100000 and needs six bits. A shift register stores one bit per flip-flop, so six flip-flops are required. Five flip-flops, option (C), can hold the values from 0 to 31 only, because \(2^{5} = 32\) gives 32 combinations. Four, option (B), can hold up to 15, and two, option (D), only up to 3. The reading that the number 32 itself must be stored is what makes the answer six.
For \(I_{OLmax}\) = 16 mA and \(I_{ILmax}\) = 0.0016 A, the fan-out (low) is
Your Answer : (Not Answered)
Correct Answer : B
The low-state fan-out of a logic gate is the largest number of gate inputs it can drive in the low state. It is the ratio of the maximum current the output can sink to the current each input needs to supply in the low state:
\[ N = \frac{I_{OL\,max}}{I_{IL\,max}} \]
\[ N = \frac{16\ \text{mA}}{1.6\ \text{mA}} = 10 \]
The figure of 16 comes from forgetting that 0.0016 A is 1.6 mA, so that the ratio of 16 to 1.6 is taken as 16 to 1. The value 1.6 is the input current itself, and 1 would apply only if the sink and source currents were equal. The answer is 10.
The amount of ROM needed to implement a 4 bit multiplier is
Your Answer : (Not Answered)
Correct Answer : D
The ROM needed is 2 kbits.
A 4-bit by 4-bit multiplier has 8 input bits, which are the ROM address, and its product needs up to 8 bits, which is the ROM word. So
\[ \begin{aligned} 2^{8} \times 8 &= 256 \times 8 \\ &= 2048\ \text{bits} = 2\ \text{kbits} \end{aligned} \]
1 kbits would come from a 4-bit word, too small for products up to \(15 \times 15 = 225\). 64 and 128 bits use far too few addresses.
Logic gates required to built up a half adder circuit are :
Your Answer : (Not Answered)
Correct Answer : B
A half adder needs an EX-OR gate and AND gate.
A half adder adds two bits A and B. The sum bit is 1 when exactly one input is 1, which is A ⊕ B (EX-OR), and the carry bit is 1 only when both are 1, which is A · B (AND).
An OR gate would give a carry when either input is 1, which is wrong, and NOR or the EX-NOR and NAND pair produce the complements of the needed outputs.
On what does the output of a Mealy machine depend ?
Your Answer : (Not Answered)
Correct Answer : C
A Mealy machine's output depends on the current input and the present state.
In a Mealy machine, outputs are attached to transitions, so the output is a function of both the state the machine is in and the input it is receiving. A change in input can change the output at once, without waiting for a state change.
Output that depends only on the present state defines a Moore machine, the usual contrast. Output from the input alone would make it a combinational circuit, not a state machine, and the clock only times the state changes; it does not determine the output value.
The minimal expression of A + A'BA is
Your Answer : (Not Answered)
Correct Answer : A
The minimal expression is A.
The second term contains both A and its complement, which makes it 0:
\[ \begin{aligned} A + A'BA &= A + (A'A)B \\ &= A + 0 \cdot B \\ &= A + 0 = A \end{aligned} \]
A + B would be the answer to the different expression A + A'B, which is the likely trap. AB and B do not follow, since the term A'BA contributes nothing for any input. A quick check: if A = 1 the expression is 1, and if A = 0 every term contains A or is A, so it is 0; the output always equals A.
Which logic gate is called a universal gate ?
Your Answer : (Not Answered)
Correct Answer : C
The universal gate is NAND.
A gate is called universal when every other logic function can be built from it alone. Tying the inputs of a NAND gate together gives NOT; a NAND followed by a NOT gives AND; and NOT-ing the inputs of a NAND gives OR (by De Morgan's theorem). From NOT, AND and OR any Boolean function follows. NOR is the other universal gate.
AND, OR and XOR cannot do this on their own: none of them can produce inversion (NOT) from its inputs, so they are not universal.
From the circuit of the following logic gates the basic logic gate obtained is ................
Your Answer : (Not Answered)
Correct Answer : D
The circuit is a NAND gate.
Follow the signals through. The NOT gates give \(\bar{A}\) and \(\bar{B}\). The NOR gate then gives, by De Morgan's theorem,
\[ \overline{\bar{A} + \bar{B}} = A\cdot B \]
The final NAND gate takes \(AB\) and \(B\):
\[ Y = \overline{(AB)\cdot B} = \overline{AB} \]
since \(AB\cdot B = AB\). So \(Y\) is A NAND B.
The output depends on both inputs, so it is not a NOT gate, and it is the complement of AND, so it is neither AND nor OR: for \(A = B = 0\) it gives 1, which an AND or OR gate would not.
The following digital circuit acts as a :
Your Answer : (Not Answered)
Correct Answer : A
The circuit is a half-subtractor.
Reading the gates:
\[ \begin{aligned} X_{out} &= A\bar{B} + \bar{A}B = A \oplus B \\ Y_{out} &= \bar{A}B \end{aligned} \]
For the subtraction A - B of two bits, the difference is \(A \oplus B\) and a borrow is needed only when A = 0 and B = 1, which is \(\bar{A}B\). So \(X_{out}\) is the difference and \(Y_{out}\) the borrow.
A half-adder has the same XOR output but its second output, the carry, is AB, not \(\bar{A}B\). Nothing in the circuit switches between adding and subtracting, so it is neither a half-adder/subtractor nor a 2's complement adder/subtractor.
Output values of Moore type FSM are determined by its ..............
Your Answer : (Not Answered)
Correct Answer : B
In a Moore machine the outputs are determined by the current state.
A Moore FSM attaches its outputs to the states themselves, so the output depends only on which state the machine is in. The inputs affect the output only indirectly, by deciding the next state at the clock edge. That makes Moore outputs change only after a clock edge and keeps them free of glitches from the inputs.
In a Mealy machine, by contrast, outputs depend on both the current state and the present inputs, which is why 'input values' describes Mealy rather than Moore. The clock only times the state changes, and 'output values' would be a circular answer.
The write pulse width (twp) is the time for which :
Your Answer : (Not Answered)
Correct Answer : B
The write pulse width is the time for which \(\overline{\mathrm{WR}}\) goes low.
The write control line of a memory chip or peripheral is normally active low, which is why it is written with a bar. A write is performed while \(\overline{\mathrm{WR}}\) is held low, and the timing parameter \(t_{wp}\) is the minimum duration of that low pulse needed for the data to be written reliably.
\(\overline{\mathrm{WR}}\) high is its inactive state, and an active-high WR is not the signal the timing diagram describes, so the other options describe the idle condition or the wrong signal.
The minimum memory size to implement a BCD to seven segment decoder will be :
Your Answer : (Not Answered)
Correct Answer : B
The minimum memory size is 10 bytes, as the key states.
A BCD to seven-segment decoder can be built as a look-up table in ROM. The input is a 4-bit BCD digit and the output is the seven segment lines a to g, which fit in one byte. Only the digits 0 to 9 are valid BCD, so ten locations of one byte each are enough: 10 bytes.
A physical ROM with four address lines has 16 locations, so a chip would be 16 bytes (option A); the six codes 1010 to 1111 are never used in BCD, which is why the minimum is 10. 8 bytes could hold only the digits 0 to 7, and 1 k byte is far larger than needed.
Boolean algebra does not apply to the relation :
Your Answer : (Not Answered)
Correct Answer : D
Boolean algebra does not apply to 0 - 1 = 1.
Boolean algebra defines only OR (written +), AND (written as multiplication) and NOT. There is no subtraction operation, so a relation written with a minus sign is not a Boolean relation at all.
The other three are valid Boolean OR relations: 0 + 0 = 0, 0 + 1 = 1, and 1 + 1 = 1. The last looks wrong by ordinary arithmetic, but in Boolean algebra it means 1 OR 1 = 1, which is correct; it is the classic trap the question sets.
How many entries will be there in the truth table of a 3 input NAND gate ?
Your Answer : (Not Answered)
Correct Answer : D
The truth table has 8 entries.
Each input can be 0 or 1, so n inputs give 2n combinations. For three inputs, 23 = 8 rows, from 000 to 111. For a NAND gate the output is 0 only in the row 111 and 1 in the other seven.
3 counts the inputs rather than the combinations, 6 is 2 × 3, and 9 is 32, none of which is the number of binary input combinations.
For a NAND gate having two inputs the output is low (0) only :
Your Answer : (Not Answered)
Correct Answer : A
The output of a two-input NAND gate is low only when both inputs are high.
NAND is AND followed by NOT: \(Y = \overline{A \cdot B}\). The AND of the inputs is 1 only when A = 1 and B = 1, so the NAND output is 0 only in that one case and 1 for the other three input combinations.
| A | B | Y |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Both inputs low, or one high and one low, all give a high output, so the other options describe cases where the output is 1.
In the fission of U-235 nuclei, it is observed that the fission fragments decay by emission of negatively charged beta particles and attain state of stable nuclei.
The reason for emission of negatively charged beta particles is that the fission fragments :
Your Answer : (Not Answered)
Correct Answer : D
The fragments emit β− particles because they are rich in neutrons.
U-235 has a neutron-to-proton ratio of about 1.55. The fission fragments, of mass number roughly 90 to 145, inherit nearly that ratio, but stable nuclei of that size have a ratio of only about 1.2 to 1.4. Each fragment therefore has too many neutrons, and it moves toward stability by a chain of β− decays, in each of which a neutron becomes a proton.
Proton-rich nuclei decay by positron emission or electron capture, not β−. Prompt neutron emission does happen, but it only partly relieves the excess and is not the cause of beta decay, and spin has no bearing on the sign of the beta particle.
Consider a digital display system (DDS) shown in the figure that displays the contents of register X. A 16-bit code word is used to load a word in X, either from S or from R. S is a 1024-word memory segment and R is a 32-word register file. Based on the value of mode bit M, T selects an input word to load in X. P and Q interface with the corresponding bits in the code word to choose the addressed word. Which one of the following represents the functionality of P, Q, and T?
(Figure: the code word has three fields, M, S-address and R-address. The S-address field feeds P, whose output goes to S; the R-address field feeds Q, whose output goes to R. The outputs of S and R are the two inputs of T, and M is connected to T. The output of T goes to X, which feeds the DDS.)
Your Answer : (Not Answered)
Correct Answer : C
P is a 10:\(2^{10}\) decoder, Q is a 5:\(2^{5}\) decoder and T is a 2:1 multiplexer.
S holds 1024 = \(2^{10}\) words, so its address field has 10 bits, and P must turn those 10 bits into one select line out of 1024: that is a 10-to-\(2^{10}\) decoder. R holds 32 = \(2^{5}\) words, so Q turns 5 address bits into one of 32 select lines, a 5-to-\(2^{5}\) decoder. T has two data inputs, the word from S and the word from R, and the mode bit M chooses which one is passed to X; choosing one of two inputs under a select signal is a 2:1 multiplexer.
A 10:1 multiplexer picks one of ten inputs, which does not address 1024 words. An encoder converts one active line into a code, which is not what T does. A 1:10 de-multiplexer routes one input to ten outputs, again not an address decoder for 1024 locations.
Which of the following options represents the simplified form of the Boolean equation \(Y = \bar{A}\bar{B}\bar{C} + \bar{A}B\bar{C} + A\bar{B}\bar{C} + AB\bar{C}\) ?
Your Answer : (Not Answered)
Correct Answer : D
The expression simplifies to \(\bar{C}\).
Every term contains \(\bar{C}\), so it can be taken out as a common factor. What is left is the sum of all four combinations of \(A\) and \(B\), which is always 1:
\[ \begin{aligned} Y &= \bar{C}(\bar{A}\bar{B} + \bar{A}B + A\bar{B} + AB) \\ &= \bar{C}\,[\bar{A}(\bar{B} + B) + A(\bar{B} + B)] \\ &= \bar{C}(\bar{A} + A) = \bar{C} \end{aligned} \]
So \(Y\) is 1 whenever \(C = 0\), whatever \(A\) and \(B\) are. \(A\bar{B}\) and \(\bar{A}B\bar{C}\) cover only one or two of the four minterms, and \(\bar{B}C\) requires \(C = 1\), where every term of \(Y\) is 0.
Consider the Boolean operator # with the following properties:
\(x\#0 = x\), \(x\#1 = \bar{x}\), \(x\#x = 0\) and \(x\#\bar{x} = 1\). Then \(x\#y\) is equivalent to
Your Answer : (Not Answered)
Correct Answer : A
\(x\#y\) is equivalent to \(x\bar{y} + \bar{x}y\), the exclusive-OR.
The given properties are exactly those of XOR: \(x \oplus 0 = x\), \(x \oplus 1 = \bar{x}\), \(x \oplus x = 0\) and \(x \oplus \bar{x} = 1\). Check the candidate at \(y = 0\): \(x \cdot 1 + \bar{x} \cdot 0 = x\); at \(y = 1\): \(0 + \bar{x} = \bar{x}\). Both match.
\(xy + \bar{x}\bar{y}\) is XNOR, which gives \(x\#0 = \bar{x}\). \(x\bar{y} + \bar{x}\bar{y} = \bar{y}\) and \(\bar{x}y + xy = y\) ignore \(x\) entirely and fail \(x\#0 = x\).
What is the degree of the following function?
F(A,B,C,D) = A`B`C`D`
Your Answer : (Not Answered)
Correct Answer : A
The degree of the function is 4. The function F(A,B,C,D) = A`B`C`D` is a single product term in which each of the four variables appears once, in complemented form, so the term has four literals and its degree is 4. The values 8, 0 and 5 do not match the number of variables in the term, so only 4 is correct. Counting each variable in the product A`B`C`D` once gives the number 4, which is why the other numbers fail.
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Total number of questions : 20
Number of answered questions : 0
Number of unanswered questions : 20