CS302 Ass Sol

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CS302- Digital Logic Design

Assignment # 1

Spring 2010

Question_1:                                                                                                                     [marks: 4]

Convert each decimal fraction to binary using repeated-multiplication-by-2:

• 0.98

0.98         .98*2=1.96      .96*2=1.92    .92*2=1.84    .84*2=1.68    .68*2= 1.36                               .36*2=0.72   .72*2=1.44 .44*2=0.88      .88*2=1.76    .76*2=1.52    .52*2=1.04    .04*2=0.08          .08*2=.16

  keep going and you finally get
                          .11111010111000010100011

• 0.60

• 60        .6*2=1.2   .2*2=0.4     .4*2=0.8    .8*2=1.6   this is a repeating decimal  .10011001100110011000.60       

Question_2:                                                                                                                     [marks: 6]

Express each decimal number as an 8-bit number in the 2's complement form:

• +10
• +10   10/2=5 rem 0   5/2=2 rem 1   2/2= 1 rem 0    1/2 = 0 rem 1  =     1010   make 8 bits 00001010

• -125

        -125  125/2=62 rem 1   62/2=31 rem 0    31/2 = 15 rem 1   15/2=7 rem 1  7/2=3 rem 1   3/2=1rem 1  1/2 = 0 rem   1     = 1111101  make 8 bits    01111101      change to negative

•                                    flip bits                    10000010
                                    add 1                                   +1
                                                                    10000011

Question_3:                                                                                                                     [marks: 4]

Represent the following decimal numbers in BCD:

• 919

%         no                   rem

2          919                

2          459                 1

2          229                 1

2          114                 1

2          57                    0

2          28                    1

2          14                    0

2          7                      0

2          3                      1

2          1                      1

So we get  (1110010111)2=397 in BCD

Ans=397

• 321

%         no       rem

2          321    

2          160     1

2          80        0

2          40        0

2          20        0

2          10        0

2          5          0

2          2          1

2          1          0

So we get (101000001)2=  A1 in BCD

ANS;  A1

Question_4:                                                                                                                     [marks: 4]

What is the highest decimal number that can be represented by each of the following numbers of binary digits (bits) ?

      Digits        No.      Binary digits

two  =      2          0010

five  =       5          0101

nine=       9          1001

ten  =       10        1010

So ten  = 10        1010 is the highest no.

Question_5:                                                                                                                     [marks: 2]

Which input combination will produce logic 1 at the output of a 4-input NOR gate ?

  The NOR gate is a digital logic gate that implements logical NOR - it behaves A HIGH output (1) results if both the inputs to the gate are LOW (0).. NOR is a functionally complete operation—combinations of NOR gates can be combined to generate any other logical function. By contrast, the OR operator is monotonic as it can only change LOW to HIGH but not vice versa.

                        IN PUTS                                 OUT PUT

            A         B          C         D                     O/P

            0          0          0          0                      1

So we will get high 1 only when all inputs are zero.

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