Problem 1: Convert the following numbers. (5mins)


    1.1) Decimal    11      to 8-bit Binary:                    00001011

    1.2) Decimal   -11      to 8-bit Binary:                    11110101

    1.3) Decimal    11.875  to Binary (as many bits as needed): 1011.111

    1.4) Decimal   -11.875  to IEEE 754 Single Precision (Show in 
hexadecimal): 0XC13E0000.


Problem 2:   Add these two numbers:
       A = 1.5
       B = 1.5
    2.1)  Convert A to IEEE 754 Single precision //
          0X 3FC0 0000

    2.2)  Use module fp_add(sum, exp_overflow, a_original,b_original)
             from the Notes 10  to add A and B;
          Show the output by using these lines:
            $display(" sum = %h\n",sum);
            $display(" sum = %d\n",sum);
          Sumbit the output and the testbench program(around 10 lines).
    2.3)  Decimal output for the second line is not 3,although 1.5 + 1.5 = 
3.0.
          Why?        =>  IEEE 754 Format so  40400000.

Problem 3: To fix the 2.3 problem, Write a conversion module which will take 
input in
       IEEE 754 format  and produce output in decimal format.
          dec_conversion(dec_sum,sum);
   3.1) Show the output by using these lines:
          $display(" sum = %h\n",sum);
          $display("decimal sum = %d\n",dec_sum);

=> sum =40400000
=> decimal sum = 4

################################
//assume positive integer number...
//sum is greater than 1;
module dec_conversion(dec_sum,sum);
output dec_sum;
input [31:0] sum;
reg [23:0]frac;
reg [31:0] sum1;
reg [7:0] exp,diff;
reg sign;
real dec_sum;
   always @(sum)
      begin
        exp = sum[30:23];
        frac = {1'b1,sum[22:0]};
        sign = sum[31];

          diff = exp;
        if(diff >=127)   // greater than 1;
              begin
                   diff = diff -127;
                   frac = frac >>(23-diff);
              end

       $display( "decimal sum = %d\n",frac);

      dec_sum = frac;
      end
endmodule
########################################




   3.2) Sumbit the output and the code for the conversion module.
   3.3) Name your program hw33.v and leave it on your account.





Problem 4: Add these two numbers:
          A = 1.5;
          B = 1.5;
     Use module module fp_add_seq(sum, exp_overflow, 
ready,a_original,b_original,start,clk)
           from Notes 10 to add A and B;
     Use clk which has #1 delay.
              // #1;
              // clk =~clk;

    4.1)  Show the output by using this line:
            $display(" sum = %h\n",sum);       =>sum = 40400000
    4.2)  Sumbit the testbench program( around 25 lines).
########TESTBENCH###########
module test();
   integer i,j;
    reg ck;
    wire ready;
    reg start;

    wire [31:0] sum;
    wire exp_overflow;
    reg [31:0] a_original,b_original;

//fp_add     fpadd(sum, exp_overflow, a_original,b_original);

fp_add_seq  fpadd(sum, exp_overflow, ready,a_original,b_original,start,ck);




    initial  ck = 0;
    initial begin
             i = 0;
             for(i = 1; i<= 33 ;i = i+ 1) begin
              #1; i = i + 1;
             ck = ~ck;
             if(ready)   $display("i = %d, sum = %h\n",i,sum);


              end
             end
    initial begin
             a_original = 31'h3fc00000;
             b_original = 31'h3fc00000;    //3.0 = 40400000

             start = 1;
             $monitor(" sum = %h\n",sum);
            end

endmodule








    4.3)  True or False. =>False.
                         =>It has 4 steps inside the program and each step 
takes only                              1cycle.
            since A and B are 32 bits long, the sequential floating point 
adder
             ,fp_add_seq(sum, exp_overflow, 
ready,a_original,b_original,start,clk),
           will take at least  32 clks to produces right output.


    4.4)  Bonus Points Prblem( 7 pts):
             the sequential floating point adder,
             fp_add_seq(sum, exp_overflow, 
ready,a_original,b_original,start,clk),
             will produce right output after (  ) cycles.
                      => 4  steps and each step takes 1cylce