Digital Design Interview Questions - v1.2

Q.1) Design a combinational logic circuit which doubles the frequency of input clock. (Output clk freq. = 2*  input clk freq.) (Frequency multiplier(*2)).

Answer) To solve these kind of questions, first draw waveforms of given input and output clocks and then try to add one or more waveforms to input which when applied to a combinational gate  will give the output 

Consider the time period of input clock is T. If the input clock is delayed by T/4 time period and if this delayed clock along with original input clock is given to a combinational gate then one would see that only XOR gate fits in, to produce output with double the frequency of input clock.

Q.2) What kind of circuit is this :                                                            A and B are inputs to an AND gate. AND gate's output goes to one input of OR gate. The other input of OR gate comes from an EX-OR gate. Inputs to the Ex-OR gate are C and the output of the OR gate. 

         A) Combinatorial /Sequential?

         B)  Synchronous / Asynchronous?

Answer) Combinatorial and Asynchronous

Q.3) What is the function of a D FF whose complemented output ( Qbar ) is connected to its input,D. What is the maximum clock frequency that can be used for it?

Answer) Suppose that there was some value at the output Q('0') of the flip flop, so when rising edge of clock comes, it samples the inverted value of output and retains it until next clock. When next rising edge comes, flop samples inverted value of output and so on. This behaviour of circuit shows that output will be half of the clock.

The circuit behaves as a frequency divider(/2).

Q.4) To convert a 2-input NAND gate into inverter.

Answer) Two ways to convert a 2-input NAND gate into inverter :-

               a) Short both inputs of NAND gate

               b) Tie one of the inputs to ‘1’.

Q.5) If A ? B = C and A?C = B, then what is the operator "?"

Answer) Starting with OR gate,

               A or B = C ; A or C = B

               Replacing value of B in first equation,

               A or (A or C) =C ----> This is false, hence OR is not the answer.

                AND gate,

                A and B = C; A and C = B

                Replacing value of B in first equation,

                A and   (A and C) = C ---> This is again false, hence AND is not the answer.

                XOR gate,

                A xor B = C; A xor C = B

                Replacing value of B in first equation,

                A xor (A xor C) = C,

                A xor (A.not(C) + C.not(A)) = C,

                A.not(A.not(C) + C.not(A)) + not(A).(A.not(C) + C.not(A)) = C,

                A(not(A).not(C) + A.C) + not(A).C = C,

                0 + A.C + not(A).C = C,

                C = C, -----> This is true, hence XOR is the answer

                  

                These type of questions are usually solved using hit and trial method.