This assignment consists of converting decimal numbers to binary, and also converting BCD (base 16) number to binary and octal. Additionally, it involves adding binary numbers using 2's complement for negative numbers, and designing an odd parity generator and checker for error detection in a 3-bit data.
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Student ID:Name: Computer Architecture and Organization I CMPE 263 Spring 2018 March 2018 Dept. of Computer Science and Engineering College of Engineering Qatar University Homework #1 – Spring 2018 Digital Logic Circuit and Data Representation Question #1(4 marks): Given the Boolean function: F(A,B,C,D)=∑m(0,2,3,5,7,8,10,11,13,15) 1)[2 marks]Drive the simplified Boolean function in sum-of-products form by means of a four-variable Karnaugh map. 2)Draw the logic diagram with a.[1 mark]AND-OR gates; b.[1 mark]NAND gates. Solution: Page1of12 50
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Student ID:Name: Question #2(2 marks): Given a JK flip flop, show how can you convert it into a D flip-flop. Page2of12
Student ID:Name: Question #3(6 mark): Design a priority encoder with three inputs X, Y, Z and three outputs A, B, C. The truth table for the circuit is shown below. a)[3 marks]Drive the minimized logic equations for A, B, and C (use Karnaugh maps). b)[3 marks]Draw the logic circuit diagram for A, B, and C. XYZABC 000 001 010 011 100 101 110 111 000 001 011 011 101 101 101 101 Page3of12
Student ID:Name: Question #4(6 mark): Given a sequential circuit (shown in the figure below) that has two D flip-flopsAand B, two inputsXandY, and one outputZ. a)[2 marks]Write the logic function for the flip-flops inputsDAandDB, as a function ofX,Y,AandB. b)[1 mark]Write the logic function for the outputZ. c)[3 marks]If thepresent stateofAandBareA = 1andB = 0, and the circuit inputs areX = 0andY = 0. Then, what will be thenext statevalues for the flip-flopsAandB? What will be the value of outputZ, in this case? Page4of12
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Student ID:Name: Question #5(3 marks): The content of a 4-bit register is initially 1011. The register is shifted six times to the right, with the serial input (SI) being 110011. What is the content of the register after each shift? (Fill the table below.) Q3 Q2 Q1 Q0 Initial content of register1011 Content after first shift1101 Content after second shift1110 Content after third shift0111 Content after fourth shift1011 Content after fifth shift1101 Content after sixth shift1110 Question #6(3 marks): Construct a 16-to-1 line multiplexer with two 8-to-1 line multiplexers and one 2-to-1 line multiplexer. Use block diagrams for the three multiplexers. Solution: Multiplexer is one of the basic building units of a computer system which in principle allows sharing of a common line by more than one input lines. It connects multiple input lines to a single output line. At a specific time one of the input lines is selected and the selected input is passed on to the output line. Page6of12
Student ID:Name: Question #7(4 marks): The following memory units are specified by the number of words times the number of bits per word. How many address lines and input-output data lines are needed in each case? a)4K x 16 b)32K x 8 c)16M x 32 d)4G x 64 Solution: a)4K x 16 : Address line = 4k=4*1024 =2^12 : 12 Bits Data line = 16 Bits b)32K x 8 Address line = 32k=32*1024 =2^15 : 15 Bits Data line = 8 Bits c)16M x 32 Address line = 16M=16*1024*1024 =2^24 : 24 Bits Data line = 32 Bits d)4G x 64 Address line = 4G=4*1024*1024*1024 =2^32 : 32 Bits Data line = 64 Bits Page7of12
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Student ID:Name: Question #8(3 marks): Convert the following (unsigned) binary numbers to decimal. (Note that the numbers have a fractional part): a)101110.1 b)101010.01 c)100111.101 Solution a)101110.1 ( 101110 )2= ( 46 )10 Explanation Step 1: Start at the rightmost digit. Take that digit and multiply with 20(20= 1). Multiple second digit with 21, third with 22... In this example we have: 0*20=0*1=0 1*21=1*2=2 1*22=1*4=4 1*23=1*8=8 0*24=0*16=0 1*25=1*32=32 Step 2: Add together all products 0 + 2 + 4 + 8 + 0 + 32 = 46 Smilarly the Fractional part of Number will be converted to decimal by division by 2 ( 101110.1 )2= ( 46.5 )10 b)101010.01 ( 101010 )2= ( 42 )10 Explanation Step 1: Start at the rightmost digit. Take that digit and multiply with 20(20= 1). Multiple second digit with 21, third with 22... In this example we have: 0*20=0*1=0 1*21=1*2=2 0*22=0*4=0 1*23=1*8=8 0*24=0*16=0 1*25=1*32=32 Step 2: Add together all products 0 + 2 + 0 + 8 + 0 + 32 = 42 Similarly .01 = 0/2 + ¼=0.25 ( 101010.01 )2= ( 42.25 )10 c)100111.101 ( 100111 )2= ( 39 )10 Page8of12
Student ID:Name: Explanation Step 1: Start at the rightmost digit. Take that digit and multiply with 20(20= 1). Multiple second digit with 21, third with 22... In this example we have: 1*20=1*1=1 1*21=1*2=2 1*22=1*4=4 0*23=0*8=0 0*24=0*16=0 1*25=1*32=32 Step 2: Add together all products 1 + 2 + 4 + 0 + 0 + 32 = 39 Similarly .101 = ½+0/4+1/8=0.625 ( 100111.101 )2= ( 39.625 )10 Question #9(4 marks): Convert the following decimal numbers to binary: 133.375. Show the steps of your work. Solution: The integral part of the decimal number is converted to binary by successive division by 2 Therefore 133 will be successively divided by 2 ( 133 )10= ( 10000101 )2 Explanation Step 1: Write down the decimal number and continually divide by 2 to give a result and a remainder. The remainder is either a1or a0. In this example we have: 133/2result66remainder1 66/2result33remainder0 33/2result16remainder1 16/2result8remainder0 8/2result4remainder0 4/2result2remainder0 2/2result1remainder0 1/2result0remainder1 Step 2: Read the remainders from bottom to top. Similarly Fractional part will be converted to binary by successive multiplication by 2 0.375*2=0.75Take 0 0.75*2=1.5Take 1 0.5*2=1Take 1 Therefore ( 133.375 )10= ( 10000101.011 )2 Question #10(3 marks): Page9of12
Student ID:Name: Convert the BCD (base 16) number FA35D1 to binary (base 2) and octal (base 8). Show the steps of your work. Solution: Given Number = FA35D1 We can convert this to Binary by writing binary equivalent of each digit in a bunch of 4 bits FA35D1 = 111110100011010111010001 Binary to octal conversion is easy and we can do that by grouping 3 bits for each digit 111110100011010111010001 = 111 110 100 011 010 111 010 001=(76432721)o Question #11(8 marks): Add the following numbers in binary using 2’s complement to represent negative numbers. Use word length of 6 bits (including sign bit), and show if overflow occurs. a)12+ (-5) b)(-12) + 11 c)22+ 11 d)(-11) + (-21) Solution: Page10of12
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