Question 1 |

The figure below shows a multiplexer where S_1 \; and \; S_0 are the select lines, I_0 \; to \; I_3 are
the input data lines, EN is the enable line, and F(P, Q, R) is the output, F is

PQ+\bar{Q}R | |

P+Q\bar{R} | |

P\bar{Q}R+\bar{P}Q | |

\bar{Q}+PR |

Question 1 Explanation:

Output,F=\bar{P}\bar{Q}R+P\bar{Q}R+PQ\, \, \, \,

F=\bar{Q}R+PQ

F=\bar{Q}R+PQ

Question 2 |

A four-variable Boolean function is realized using 4x1 multiplexers as shown in the figure.

The minimized expression for F(U,V,W, X) is

The minimized expression for F(U,V,W, X) is

(UV+\bar{U}\bar{V})\bar{W} | |

(UV+\bar{U}\bar{V})(\bar{W}\bar{X}+\bar{W}X) | |

(U\bar{V}+\bar{U}V)\bar{W} | |

(U\bar{V}+\bar{U}V)(\bar{W}\bar{X}+\bar{W}X) |

Question 2 Explanation:

Output of the first multiplexer can be expressed as,

F_{1}=\bar{U} V+U \bar{V}

Output of the second multiplexer can be expressed as,

\begin{aligned} F &=\bar{W} \bar{X} F_{1}+\bar{W} X F_{1}=\bar{W} F_{1} \\ &=(\bar{U} V+U \bar{V}) \bar{W} \end{aligned}

Question 3 |

A programmable logic array (PLA) is shown in the figure.

The Boolean function F implemented is

The Boolean function F implemented is

\bar{P}\bar{Q}R + \bar{P}QR + P\bar{Q}\bar{R} | |

(\bar{P}+\bar{Q}+R) (\bar{P}+Q+R) (P+\bar{Q}+\bar{R})
| |

\bar{P}\bar{Q}R + \bar{P}QR + P\bar{Q}R | |

(\bar{P}+\bar{Q}+R) (\bar{P}+Q+R) (P+\bar{Q}+R) |

Question 3 Explanation:

F=\bar{P} \bar{Q} R+\bar{P} Q R+P \bar{Q} R

Question 4 |

Figure I shows a 4-bits ripple carry adder realized using full adders and Figure II shows the circuit of a full-adder (FA). The propagation delay of the XOR, AND and OR gates in Figure II are 20 ns, 15 ns and 10 ns respectively. Assume all the inputs to the 4-bit adder are initially reset to

At t=0, the inputs to the 4-bit adder are changed to X_{3}X_{2}X_{1}X_{0}=1100,\; Y_{3}Y_{2}Y_{1}Y_{0}=0100 \; and \; Z_{0}=1. The output of the ripple carry adder will be stable at t (in ns) = ___________

At t=0, the inputs to the 4-bit adder are changed to X_{3}X_{2}X_{1}X_{0}=1100,\; Y_{3}Y_{2}Y_{1}Y_{0}=0100 \; and \; Z_{0}=1. The output of the ripple carry adder will be stable at t (in ns) = ___________

60 | |

65 | |

50 | |

75 |

Question 4 Explanation:

In this question inputs to be added are :

\begin{aligned} X_{3} X_{2} X_{1} X_{0}&=1100 \\ Y_{3} Y_{2} Y_{1} Y_{0}&=0100 \text { and } Z_{0}=1 \end{aligned}

For this combination of addition, total minimum delay depends on the addition of most-significant two bits (since least significant two bits are zeros they does not cause any change in Z_{1} and Z_{2}).

So, in the process of addition of given two digits, waveforms at Z_{1} and Z_{2} become stable at t=0 itself.

In the above diagram the waveform at A and B become stable at t = 0 itself, as the applied input combinations does not cause any change.

So, for he given combination of inputs, output will settle at t = 50 ns.

\begin{aligned} X_{3} X_{2} X_{1} X_{0}&=1100 \\ Y_{3} Y_{2} Y_{1} Y_{0}&=0100 \text { and } Z_{0}=1 \end{aligned}

For this combination of addition, total minimum delay depends on the addition of most-significant two bits (since least significant two bits are zeros they does not cause any change in Z_{1} and Z_{2}).

So, in the process of addition of given two digits, waveforms at Z_{1} and Z_{2} become stable at t=0 itself.

In the above diagram the waveform at A and B become stable at t = 0 itself, as the applied input combinations does not cause any change.

So, for he given combination of inputs, output will settle at t = 50 ns.

Question 5 |

Consider the circuit shown in the figure.

The Boolean expression F implemented by the circuit is

The Boolean expression F implemented by the circuit is

\bar{X}\bar{Y}\bar{Z}+XY+\bar{Y}Z | |

\bar{X}Y \bar{Z}+XY+\bar{Y}Z | |

\bar{X}Y \bar{Z}+XY+\bar{Y}Z | |

\bar{X}\bar{Y}\bar{Z} +XY+\bar{Y}Z |

Question 5 Explanation:

\begin{aligned} F_{1} &=\bar{X} Y \\ F &=\bar{Z} F_{1}+Z \bar{F}_{1} \\ &=(\bar{X} Y) \bar{Z}+(\bar{X} Y) Z \\ &=\bar{X} Y \bar{Z}+(X+\bar{Y}) Z \\ F &=\bar{X} Y \bar{Z}+X Z+\bar{Y} Z \end{aligned}

Question 6 |

For the circuit shown in the figure, the delays of NOR gates, multiplexers and inverters are 2 ns, 1.5 ns and 1 ns, respectively. If all the inputs P, Q, R, S and T are applied at the same time instant, the maximum propagation delay (in ns) of the circuit is __________

2 | |

4 | |

6 | |

8 |

Question 6 Explanation:

When, T=\text{logic} 0, the path followed by the circuit would be,

NOR gate \rightarrow \text{MUX} 1 \rightarrow \text{MUX} 2

\Rightarrow 2 \mathrm{ns} \rightarrow 1.5 \mathrm{ns} \rightarrow 1.5 \mathrm{ns}

\Rightarrow 5 \mathrm{ns}

When, T=\text{logic} 1, the path followed by the circuit would be,

NOR gate \rightarrow MUX 1 \rightarrow NOR gate \rightarrow MUX 2

\Rightarrow 1 \mathrm{ns} \rightarrow 1.5 \mathrm{ns} \rightarrow 2 \mathrm{ns} \rightarrow 1.5 \mathrm{ns}

\Rightarrow 6 \mathrm{ns}

\therefore \quad Maximum propagation delay is 6 ns

NOR gate \rightarrow \text{MUX} 1 \rightarrow \text{MUX} 2

\Rightarrow 2 \mathrm{ns} \rightarrow 1.5 \mathrm{ns} \rightarrow 1.5 \mathrm{ns}

\Rightarrow 5 \mathrm{ns}

When, T=\text{logic} 1, the path followed by the circuit would be,

NOR gate \rightarrow MUX 1 \rightarrow NOR gate \rightarrow MUX 2

\Rightarrow 1 \mathrm{ns} \rightarrow 1.5 \mathrm{ns} \rightarrow 2 \mathrm{ns} \rightarrow 1.5 \mathrm{ns}

\Rightarrow 6 \mathrm{ns}

\therefore \quad Maximum propagation delay is 6 ns

Question 7 |

A 4:1 multiplexer is to be used for generating the output carry of a full adder. A and B are the bits to be added while C_{in} is the input carry and C_{out} is the output carry. A and B are to be used as the select bits with A being the more significant select bit.

Which one of the following statements correctly describes the choice of signals to be connected to the inputs I_{0},I_{1},I_{2} \; and \; I_{3} so that the output is C_{out}?

Which one of the following statements correctly describes the choice of signals to be connected to the inputs I_{0},I_{1},I_{2} \; and \; I_{3} so that the output is C_{out}?

I_{0}=0,I_{1}=C_{in},I_{2}=C_{in} and I_{3}=1 | |

I_{0}=1,I_{1}=C_{in},I_{2}=C_{in} and I_{3}=1 | |

I_{0}=C_{in},I_{1}=0,I_{2}=1 and I_{3}=C_{in} | |

I_{0}=0,I_{1}=C_{in},I_{2}=1 and I_{3}=C_{in} |

Question 7 Explanation:

In case of a full adder,

\begin{aligned} \therefore \quad I_{0}&=0 \\ I_{1}&=C_{\text {in }} \\ I_{2}&=C_{\text {in }} \\ I_{3}&=1 \end{aligned}

\begin{aligned} \therefore \quad I_{0}&=0 \\ I_{1}&=C_{\text {in }} \\ I_{2}&=C_{\text {in }} \\ I_{3}&=1 \end{aligned}

Question 8 |

Identify the circuit below.

Binary to Gray code converter | |

Binary to XS3 converter | |

Gray to Binary converter | |

XS3 to Binary converter |

Question 8 Explanation:

The truth table of the circuit is shown below,

\begin{array}{ccc|ccc} X_{2} & X_{1} & X_{0} & Y_{2} & Y_{1} & Y_{0} \\ \hline 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 1 \end{array}

As Per the truth table none of the options given in the question are correct. However, by making some (minor) changes in the circuit, the answer could be obtained as option (A).

As per GATE official answer Marks to ALL.

\begin{array}{ccc|ccc} X_{2} & X_{1} & X_{0} & Y_{2} & Y_{1} & Y_{0} \\ \hline 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 1 \end{array}

As Per the truth table none of the options given in the question are correct. However, by making some (minor) changes in the circuit, the answer could be obtained as option (A).

As per GATE official answer Marks to ALL.

Question 9 |

A 1-to-8 demultiplexer with data input D_{in}, address inputs S_{0}, S_{1}, S_{2} (with S_{0} as the LSB) and \bar{Y}_{0} \; to \; \bar{Y}_7 as the eight demultiplexed outputs, is to be designed using two 2-to-4 decoders (with enable E input and address inputs A_{0} \; and \; A_{1}) as shown in the figure. D_{in} , S_{0}, S_{1} \; and \; S_{2} are to be connected to P, Q, R and S, but not necessarily in this order. The respective input connections to P, Q, R, and S terminals should be

S_{2},D_{in},S_{0},S_{1} | |

S_{1},D_{in},S_{0},S_{2} | |

D_{in},S_{0},S_{1},S_{2} | |

D_{in},S_{2},S_{0},S_{1} |

Question 9 Explanation:

Consider a 1 \times 8 demultiplexer

\begin{aligned} \text{So,}\quad \bar{Y}_{0} &=\left(\overline{D_{\text {in }} \cdot \bar{S}_{0} \bar{S}_{1} \bar{S}_{2}}\right) \\ \bar{Y}_{0} &=\left(\bar{D}_{\text {in }}+S_{0}+S_{1}+S_{2}\right) \\ \bar{Y}_{1} &=\left(\bar{D}_{\text {in }}+\bar{S}_{0}+S_{1}+S_{2}\right) \\ \bar{Y}_{2} &=\left(\bar{D}_{\text {in }}+S_{0}+\bar{S}_{1}+S_{2}\right) \\ \bar{Y}_{3} &=\left(\bar{D}_{\text {in }}+\bar{S}_{0}+\bar{S}_{1}+S_{2}\right) \\ \bar{Y}_{4} &=\left(\bar{D}_{\text {in }}+S_{0}+S_{1}+\bar{S}_{2}\right) \\ \bar{Y}_{5} &=\left(\bar{D}_{\text {in }}+\bar{S}_{0}+S_{1}+\bar{S}_{2}\right) \\ \bar{Y}_{6} &=\left(\bar{D}_{\text {in }}+S_{0}+\bar{S}_{1}+\bar{S}_{2}\right) \\ \bar{Y}_{7} &=\left(\bar{D}_{\text {in }}+S_{0}+S_{1}+S_{2}\right) \end{aligned}

From the circuit given in question we can see that

\begin{array}{l} \bar{Y}_{0}=\left(1 A_{0}+1 A_{1}+\bar{E}\right) \\ \bar{Y}_{0}=(R+S+P+Q) \end{array}

Similarly,

\bar{Y}_{1}=\left(1 \bar{A}_{0}+1 A_{1}+1 \bar{E}\right)=(P+Q+\bar{R}+S)

\bar{Y}_{4}=\left(2 \bar{A}_{0}+2 A_{1}+2 \bar{E}\right)=(R+S+P+\bar{Q})

So comparing, we get

\begin{array}{l} P={D}_{\text {in }} \\ Q=S_{2} \\ R=S_{1} \\ S=S_{0} \end{array}

\begin{aligned} \text{So,}\quad \bar{Y}_{0} &=\left(\overline{D_{\text {in }} \cdot \bar{S}_{0} \bar{S}_{1} \bar{S}_{2}}\right) \\ \bar{Y}_{0} &=\left(\bar{D}_{\text {in }}+S_{0}+S_{1}+S_{2}\right) \\ \bar{Y}_{1} &=\left(\bar{D}_{\text {in }}+\bar{S}_{0}+S_{1}+S_{2}\right) \\ \bar{Y}_{2} &=\left(\bar{D}_{\text {in }}+S_{0}+\bar{S}_{1}+S_{2}\right) \\ \bar{Y}_{3} &=\left(\bar{D}_{\text {in }}+\bar{S}_{0}+\bar{S}_{1}+S_{2}\right) \\ \bar{Y}_{4} &=\left(\bar{D}_{\text {in }}+S_{0}+S_{1}+\bar{S}_{2}\right) \\ \bar{Y}_{5} &=\left(\bar{D}_{\text {in }}+\bar{S}_{0}+S_{1}+\bar{S}_{2}\right) \\ \bar{Y}_{6} &=\left(\bar{D}_{\text {in }}+S_{0}+\bar{S}_{1}+\bar{S}_{2}\right) \\ \bar{Y}_{7} &=\left(\bar{D}_{\text {in }}+S_{0}+S_{1}+S_{2}\right) \end{aligned}

From the circuit given in question we can see that

\begin{array}{l} \bar{Y}_{0}=\left(1 A_{0}+1 A_{1}+\bar{E}\right) \\ \bar{Y}_{0}=(R+S+P+Q) \end{array}

Similarly,

\bar{Y}_{1}=\left(1 \bar{A}_{0}+1 A_{1}+1 \bar{E}\right)=(P+Q+\bar{R}+S)

\bar{Y}_{4}=\left(2 \bar{A}_{0}+2 A_{1}+2 \bar{E}\right)=(R+S+P+\bar{Q})

So comparing, we get

\begin{array}{l} P={D}_{\text {in }} \\ Q=S_{2} \\ R=S_{1} \\ S=S_{0} \end{array}

Question 10 |

A 16-bit ripple carry adder is realized using 16 identical full adders (FA) as shown in the figure. The carry-propagation delay of each FA is 12 ns and the sumpropagation delay of each FA is 15 ns. The worst case delay (in ns) of this 16-bit adder will be ______.

210 | |

195 | |

185 | |

165 |

Question 10 Explanation:

Full adder is a combinational circuit that performs the sum of three bits i.e. add two bits and a carry from previous addition and produces output as sum and carry.

Given carry propagation delay of each FA= 12\mathrm{nsec}.

Sum-propagation delay of each FA= 15\mathrm{nsec}.

As can be seen from above diagram, full adder is not received from previous full adder.

So, in 16-bit ripple carry adder, worst case delay of this 16-bit adder will be

\begin{aligned} &= (15 \times 12) \mathrm{nsec}. + 15 \mathrm{nsec}.\\ &= 195 \mathrm{nsec}. \end{aligned}

Given carry propagation delay of each FA= 12\mathrm{nsec}.

Sum-propagation delay of each FA= 15\mathrm{nsec}.

As can be seen from above diagram, full adder is not received from previous full adder.

So, in 16-bit ripple carry adder, worst case delay of this 16-bit adder will be

\begin{aligned} &= (15 \times 12) \mathrm{nsec}. + 15 \mathrm{nsec}.\\ &= 195 \mathrm{nsec}. \end{aligned}

There are 10 questions to complete.