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2 iii) Simplify using Boolean algebra (A+B) (A+C). 2M (A+B)(A+C) = AA + AC+ AB + BC = A + AB + AC + BC ( Since A.A =A) = A (1+ B + C) + BC ( Since 1+B+C = 1) = A +BC (2 marks) iv) Draw symbol, truth table and logic equation of EX-NOR gate. 2M Symbol: Truth Table: Inputs A B Output Y = (Symbol: (1/2 mark, Truth Table: 1 mark, Logic Equation: ½ mark) Logic Equation: Y = v) Define minterm and maxterm. 2M Minterm: A minterm, denoted as mi, where 0 i < 2 n, is a product (AND) of the n variables in which each variable is complemented if the value assigned to it is 0, and uncomplemented if it is 1. Maxterm: A maxterm, denoted as Mi, where 0 i < 2 n, is a sum (OR) of the n variables (literals) in which each variable is complemented if the value assigned to it is 1, and uncomplemented if it is 0. (minterm and maxterm :1mark Page 2 of 25

3 vi) Draw symbol and truth table of T-flip-flop. 2M Symbol: (Symbol: 1 mark, Truth table: 1 mark) Truth Table: vii) What is the difference between edge Triggering and level Triggering (any 2)? 2M (1 mark Page 3 of 25

4 viii) State two specification of DAC. 2M Specifications of DAC: Resolution Settling time Linearity Accuracy (Any 2 specificati on: 1 mark (b) Attempt any two of the following: 8 Marks i) Perform the following operation a) using 1 s complement method. b) using 2 nd complement method. {**Note: Steps marking should be given marks**} a) using 1 s complement method: Step 1: Make the number of bits in the minuend and subtrahend equal (2 marks 1010 =01010 Step 2: Obtain 1 s complement of subtrahend (01010) 1 s complement of (01010) Step 3: Adding to Carry 1 1 = Step 4: Since final carry is generated, adding it to final answer and the result is positive = Carry 1 1 = b) using 2 nd complement method: The number of bits in minuend and subtrahend are equal Step I: Obtain 2 s complement of (11110) 1 s complement of (11110) Page 4 of 25

5 Adding 1to Carry 1 = Step II: Adding to Carry 1 = Step III: Since no carry is generated, the answer is negative and in 2 s compliment form Finding 2 s compliment of s compliment Carry = (-00100) ii) State and verify De-Morgan s first theorem using truth table. It states that the, complement of a sum is equal to product of their complements A B (Statement : 1 mark, Verificatio n: 3 marks) LHS RHS Page 5 of 25

6 iii) Compare R-2R and weighted resistor DAC any four points. Weighted resistor DAC R-2R ladder DAC (1 mark It requires more than two resistor values. It requires resistors of only two values. To get precise high value resistor is difficult Since same 2 value resistors are used, the precision of resistors is not a problem It requires one resistor per bit It requires two resistor per bit It is not possible to expand It can be easily expanded to handle more number of bits by adding resistors 2. Attempt any four of the following: 16 Marks a) Reduce the following logic expression using Boolean laws and De- Morgan s theorems. (Applying De Morgan s Theorems) (1 marks for each step) (By applying the Boolean law ) Page 6 of 25

7 b) Convert the following: i) (6AC) 16 = (?) 10 ii) (372) 8 = (?) 2 {**Note: Step marking should be given marks**} i) (6AC) 16 = 6x x x 16 0 = 6 x = (2 marks for each correct Conversio n) = 1708 (6AC) 16 = (1708) 10 ii)(372) 8 = (?) Binary equivalent of 3 is 011 Binary equivalent of 7 is 111 Binary equivalent of 2 is 010 (372) 8 = ( ) 2 c) Design Half adder using K-map and basic gates. Truth Table: (Truth table: 1mark, 1 mark each k map, 1mark circuit) K map K-map for Sum: Sum = Page 7 of 25

8 K-map for Carry: Carry = AB Half adder using Basic gates Page 8 of 25

9 d) Draw 8: 1 multiplexer using basic logic gates. {**Note: Equation optional**} (Diagram: 4 marks) e) Compare RAM and ROM any four point. RAM ROM Random Access Memory or RAM is a form of data storage that can be accessed randomly at any time, in any order and from any physical location, allowing quick access and manipulation Read-only memory or ROM is also a form of data storage that cannot be easily altered or reprogrammed. Stores instructions that are not necessary for re-booting to make the computer operate when it is switched off. They are hardwired (Four points:1 mark RAM allows the computer to read data quickly to run applications. It allows reading and writing. RAM is volatile i.e. its contents are lost when the device is powered off. The two main types of RAM are static RAM and dynamic RAM ROM stores the program required to initially boot the computer. It only allows reading It is non-volatile i.e. its contents are retained even when the device is powered off. The types of ROM include PROM, EPROM and EEPROM. Page 9 of 25

10 f) Draw logic diagram of JK-flop-flop and write its truth table. Logic Diagram: (Logic Diagram:2 marks, Truth Table :2 marks) Truth Table: Inputs Output J n K n Q n Q n Attempt any four of the following: 16 Marks a) Perform the following BCD arithmetic operation: i) (637) 10 +(463) 10 ii) (63) 10 + (19) 10 {**Note: Steps marking should be given marks**} i) (Correct answer: 2 marks Page 10 of 25

11 ii) b) What is an Universal gate? Prove NAND as an universal gate. NAND and NOR gates are called as Universal Gates because it is possible to implement any Boolean expression with the help of only NAND or NOR gates, Hence, a user can build any combinational; circuit with the help of only NAND gates or NOR gates. Proof: Using NAND gates, OR, AND and NOT gates may be constructed as shown below (Universa l gate: 2 marks, Proof: 2 marks) Page 11 of 25

12 c) Design 1:8 De-multiplexer using 1:4 demultiplexer. (Design ing: 4 marks) A,B,C are the select line inputs Fig: 1:8 De-multiplexer using 1:4 De-multiplexer d) Design 1 bit comparator using K-map and draw its logic diagram. 1. The one-bit comparator is combinational logic circuit with two inputs A and B and three Outputs namely A<B, A+B, A>B. 2. It compares two single bit numbers A and B and produces an output that indicates the result of comparison. Truth Table: INPUTS OUTPUTS A B Y1= A<B Y2= A=B Y3 = A>B (Truth table :1 mark, Imple mentati on using K maps :2 marks, Diagra m :1 mark) Page 12 of 25

13 e) With the help of block diagram explain the working of a ring counter. (Block diagra m: 2 marks, workin g: 2 marks) Fig: Ring Counter Block diagram Working: 1. Ring Counter is nothing but a SISO (Serial in Serial out) Shift register in which serial out i.e. Q 3 output of the fourth Flip flop is connected to D input of the first Page 13 of 25

14 Flip-Flop. Thus there is feedback from output to the input. 2. Clock pulses are given simultaneously to all flip-flops 3. Initially, count stored is D C B A = Clock pulses are applied. Count is rotated as shown in below table. f) Give four features of a dynamic RAM. 1. Data is stored in the form of charge on the capacitor. Hence each DRAM unit consists of one MOSFET and a capacitor. 2. Since Capacitors are used on input side, refreshing circuit is required. 3. Access time is more, hence it is slow in speed. 4. Memory cell per unit area are more than Static RAM. 5. Dynamic RAM are cheaper than SRAM. (Any 4 feature s: 1 mark 4. Attempt any four of the following: 16 Marks a) Compare between TTL and CMOS logic family (any four). Sr. No. Parameters TTL CMOS 1 Basic gates NAND NOR or NAND 2 Fan-in Fan-out 10 >50 4 Power dissipation per gate 10mW 0.01mW 5 Noise margin (immunity) 0.4V good 5V (excellent) 6 Propagation delay 10 ns 70 ns 7 Speed-power product Clock rate for flip-flop 35 MHz 10 MHz 9 Available function Very large Large 10 Packing Density Lower Larger 11 Cost Low Very low. (Any 4 points: 1 mark Page 14 of 25

15 b) Simplify following equation using Boolean Algebra and draw circuit diagram. Y = A (A + B) + B (A+B). (Simplifi cation: 2 marks, Circuit Diagra m: 2 marks) c) Design asynchronous mod-6 counter with its truth table. (1 mark for each Step) Page 15 of 25

16 {**Note: A reset logic using NAND gate may also be considered**} d) Minimize the following Boolean expression using K-map, Y = m (1, 3, 5, 7, 10, 11, 14, 15).Draw the logical circuit diagram of minimized expression using basic gates. (Simplifi cation: 2 marks, Logical Circuit diagram : 2 marks) e) What is race around condition? How can it be avoided? Race around Condition: The Race Around condition occurs when J=K=1 i.e. when the FF is in the toggle mode. In J-K Flip Flop, when J = 1 & K = 1 then the Output of J-K flip flop is Qn compliments of previous Output. Let Q = 0 & clock pulse is (Race around Conditi Page 16 of 25

17 applied. At a time interval t, the Output will change to Qn that means the Output now is Q = 1.Now we have J =1, K = 1 & Q = 1. After another time interval t, the Output will again change from 1 to 0 (Qn) due to feedback connection. Thus the output oscillates back and forth between 0 to 1 for the duration of the clock pulse. (C1K = 1) Hence at the end of the clock pulse, when Clock=0, the value of a Q Output is uncertain. This situation is called as race around condition. This race around condition can be avoided by using on: 2 marks, Avoida nce: 2 marks) 1) Master Slave I-K flip flop: Master Slave J-K flip flop is a cascade of two J-K flip flop. 2) Edge Triggered JK Flip Flop. f) Draw block diagram of ALU IC and explain function of each pin. (Diagra m: 2 marks, Function s of Pin: 2 marks) A 0 A 3 Operand Inputs (Active LOW) B 0 B 3 Operand Inputs (Active LOW) S 0 S 3 Function Select Inputs M Mode Control Input Cn Carry Input F 0 F 3 Function Outputs (Active LOW) A = B Comparator Output G Carry Generate Output (Active LOW) P Carry Propagate Output (Active LOW) C n+4 Carry Output Block diagram: ALU IC Page 17 of 25

18 5. Attempt any four of the following: 16 Marks a) Reduce the following expression and implement logic gates Y = AB+ ABC +AB (E+F) (Reducing Equation: 2 marks, Logic diagram: 2 marks) b) Simplify the following SOP expression with K-Map i) F(A, B, C, D) = m (0,1,3,4,5,7) ii) F ( A,B,C) = m (0,1,3,4,6). {**Note: Steps marking shall be considered**} i) (Simplifica tion: 2 marks ii) Page 18 of 25

19 c) Draw and explain the block diagram of IC decimal to BCD encoder. Write truth table. Explanation: IC is basically a 10:4 encoder or decimal to BCD encoder. A1 to A9 inputs are the active low inputs and A, B, C and D are the active low outputs.a1 has the lowest priority and A9 has the highest priority. In response to the inputs, chip produces inverted BCD code corresponding to the highest numbered Active input as shown in the truth table. Diagram: (Diagram: 1 mark, truth table :2 marks, explanatio n: 1 mark) Truth Table: d) Describe working of R-2R ladder type DAC. R-2R ladder DAC uses two resistors R & 2R. The input is applied through digitally controlled switches. (Explanati on: 2 marks, Diagram: 2 marks) Page 19 of 25

20 For example if the digital input is 001 Applying Thevenins theorem at XX Applying Thevenins theorem at YY Applying Thevenins theorem at ZZ Page 20 of 25

21 Similarly for digital input 010 and 100 the equivalent voltages are VR/2 2 and VR/2 1 respectively. The equivalent resistance is 3R in each case. So the simplified circuit of 3bit R-2R ladder DAC is : The analog output voltage for a given digital input is given by Vout = - ((RF/3R) VR x b0/2 3 + RF/3R VR x b1/2 2 + RF/3R VR x b2/2 1 ) = - (RF/3R) (VR/2 3 ) (2 2 b b b0) = - (RF/3R) (VR/2 3 ) (4b2 + 2b1 + b0) e) Differentiate between combination logic and sequential logic system. Combinational Logic Circuits Sequential Logic Circuits Output is a function of the Output is a function of clock, present inputs (Time Independent present inputs and the previous Logic) states of the system. Do not have the ability to store data (state). Have memory to store the present states that is sent as control input (enable) for the next operation. (Each Differenc e: 1mark (any 4)) Logic gates are the elementary building blocks. Independent of clock and hence does not require triggering to operate. Flip flops (binary storage device) are the elementary building unit. Clocked (Triggered for operation with electronic pulses). Page 21 of 25

22 Used mainly for Arithmetic and Boolean operations. Used for storing data (and hence used in RAM). It does not require any feedback. It simply outputs the input according to the logic designed. It involves feedback from output to input that is stored in the memory for the next operation. f) Draw circuit diagram of 3-bit SIPO shift register, right shift mode with the help of block diagram. Diagram: (Diagram: 2 marks, Truth table: 1 mark, Explanatio n: 1 mark) Truth Table: Explanation:- The serial input parallel output shift register is shown above (SIPO). It accepts the input data serially i.e. one bit at a time and outputs the stored data in parallel form. At the end of each clock pulse (-ve edge) a first data bit of higher significant bit (as LSB is entered first) enters into the Di i/p of FF1 and Q output of every FF gets shifted to the next FF on right side. Thus once the data bits are stored, each bit appears on its respective output line & all bits are available simultaneously at Q2 Q1 Q0 rather than a bit by bit basis with the serial output. D in=101 Page 22 of 25

23 6. Attempt any four of the following : 16 Marks a) Draw neat block diagram of Ramp ADC and explain its working. (Explanati on: 2 marks, Diagram: 2 marks) Fig: Ramp ADC Working the counter is reset to zero first by applying a reset pulse. Then after releasing the reset pulse, the clock pulse are applied to the counter through an AND gate Initially the DAC output is zero. Therefore the analog input voltage VA is greater than the DAC output V d i.e., VA> Vd The comparator output is high and the AND gate is enabled.thus the clock pulses are allowed to pass through the.and gate to the counter. The counter starts counting these clock pulse. Its output goes on increasing. As the counter output acts as input to DAC, the DAC output which is in staircase waveform also increase. As long as the DAC output Vd < VA this process will continue, as the comparator output remains high enabling the AND gate. However, when the DAC output is high than the input analog voltage i.e. Vd > VA, the comparator output becomes low so that AND gate is disabled and stop the clock pulse i.e. counting stops. Thus the digital output of the counter represents the analog input voltage va. When the analog input change to a new value, a second reset pulse is applied to the counter to clear it again the counting starts. b) State the application of shift register (any four). Application of Shift Registers 1. Delay line 2. Serial to parallel converter 3. Parallel to serial converter 4. Ring counter 5. Twisted Ring counter 6. Sequence generator (Any four applicatio n :1 mark Page 23 of 25

24 c) Implement following logical equation using multiplexer: Y (A, B, C) = m (0, 1, 2, 3, 6, 7). (Implemen tation: 4 marks) d) Perform the binary arithmetic i) ( ) 2 + ( ) 2 = (?) 2 (ii) ( ) 2 ( ) 2 = (?) 2 {**Note: Steps marking should be given marks**} ( Each conversion : 2 marks) Page 24 of 25

25 e) Draw symbol for 3 input OR gate with truth table and 3 input NAND gate with truth table. OR GATE: (Symbol: 1 mark each, Truth Table: 1 mark NAND GATE: f) Define the following specification of A-D Converter i) Conversion time (ii) Resolution. i) Conversion time: It is the total time required to convert the analog input signal into a corresponding digital output. This Conversion rate is also called as speed. This varies with analog voltage. ii) Resolution: Resolution is define as the maximum number of digital output codes. This is same as that of a DAC. Resolution is defined as the ratio of change in the value of the input analog voltage VA, required to change the digital output by 1 LSB. (Definition: 2 marks Page 25 of 25

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