ࡱ> wyv}7 (bjbjUU 7|7|Y$lZZZZD666(7$L7-f688"9929C;a; m;eeeeeee$cg ieu;!;"C;u;u;euFZZ929euFuFuFu;Z8929euFu;euF uFOr8XT8Y298 9999 then -- returns x"9999". -- Input: binary - 16 bit unsigned value. -- Returns: BCD unsigned. -- Ussage: BCD = to_bcd(binary); ----------------------------------------------------------------------- function to_bcd(binary: std_logic_vector) return STD_LOGIC_VECTOR is variable rsr : std_logic_vector(15 downto 0); -- Shift reg for BCD result begin if CONV_INTEGER(binary) > 9999 then rsr := x"9999"; else -- Enter your code to complete this function end if; return rsr; end to_bcd; end func_to_bcd; -- tb_to_bcd.vhd Version 1.1 by Dr. Jerry H. Tucker -- Test bench for to_bcd.vhd -- Created 10/12/07 library IEEE; use IEEE.STD_LOGIC_1164.ALL; use IEEE.STD_LOGIC_ARITH.ALL; use IEEE.STD_LOGIC_UNSIGNED.ALL; use work.all; use func_to_bcd.all; -- uses to_bcd in package func_to_bcd.vhd entity tb is end tb; architecture test of tb is signal bi, bcd: std_logic_vector(15 downto 0) := x"ffff"; signal int: integer; begin -- Use linear feedback shift register to -- generate random binary values. bi <= (bi(5) xor bi(3) xor bi(2) xor bi(0)) &bi(15 downto 1) after 5 ns; int <= conv_integer(bi); -- Get integer value of binary bcd <= to_bcd(bi); end test;  Divide by 3  EMBED Equation.DSMT4  Algorithm: Sum = 0 P = x If P = 0 then exit P = P/4 Sum = Sum + P GOTO 3 Example 256/3 = 85 or 0101 0101 X = 100000000 PSum100000000000000000010000000100000000010000010100000000010001010100000000000101010 Homework #10 Due Wednesday 10/8/08 For this homework you and your lab partner may work together and turn in a joint solution. Do not discuss the homework with anyone else other than the instructor. The file pkg_func.vhd is provided. You are to complete the functions in this package and develop a separate test bench for each function. Turn in all source code and simulation waveforms. For each problem write a short paragraph describing how you concluded from the test bench that the function was working correctly. Do not turn in solutions that do not compile. Using the algorithm we developed in class, complete the function to_bcd and develop a test ench. Assume the input and returned value are both 16 bits. Using an algorithm you developed, complete the function log2 and develop a test ench. Write a short paragraph describing your algorithm. Assume the input is 16 bits, and the returned value is 8 bits. Return the integer of log base 2 of the input, unless log base 2 is negative then return 0. Using the algorithm we developed in class, complete the function div_by_3 and develop a test ench. The function must work for all input lengths greater than 2. assume the return length is the same as the input length. Hint: You may need the use some of the arithmetic conversions shown below. slv_1 <= CONV_STD_LOGIC_VECTOR(n, 3); u_1 <= CONV_UNSIGNED(n, 3); s_1 <= CONV_SIGNED(n, 3); i_1 <= CONV_INTEGER(slv); ----------------------------------------------------------------------- --Function: to_bcd - Converts 16 bit input to BCD. -- If input > 9999 then returns x"9999". -- Input: binary - 16 bit unsigned value. -- Returns: BCD unsigned. -- Ussage: BCD <= to_bcd(binary); ----------------------------------------------------------------------- function to_bcd(binary: std_logic_vector) return STD_LOGIC_VECTOR is variable rsr : std_logic_vector(15 downto 0); -- Shift reg for BCD result variable bsr : std_logic_vector(15 downto 0); -- Shift reg for binary input variable correction : std_logic_vector(15 downto 0); -- Add to do BCD correction begin if CONV_INTEGER(binary) > 9999 then rsr := x"9999"; else -- Enter your code here end if; return rsr; end; -------------------------------------------------------------- -- Function: LOG2 - Returns integer of 16 bit input unless LOG2 -- is negative then 0 is returned. -- Inputs: x - a 16 bit std_logic_vector -- Returns: 8 bit std_logic_vector containg integer of log base 2 -- Ussage: log <= Log2(X); ---------------------------------------------------------------- function log2(x: std_logic_vector) return std_logic_vector is variable log: std_logic_vector(7 downto 0); begin -- Enter your code here return(log); end; --------------------------------------------------------------- -- Function: Div_by_3 - Divides input by 3 -- Inputs: X - a std_logic_vector -- Returns: a std_logic_vector -- Note that both standart logic vectors must be of the form -- std_logic_vector(n downto 0), and n is the same for -- both the input and the output. -- Do not make assumptions about the value of n. -- The function must work for any n > 2. ---------------------------------------------------------------- function Div_by_3(x: std_logic_vector) return std_logic_vector is variable Sum, P: std_logic_vector(x'length - 1 downto 0); begin -- Enter your code here return(Sum); end; end pkg_func; PAGE  PAGE 8  EMBED Word.Picture.8  *+;<MN^_op )*EFab} *+ab 0?Zivw ! 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TuckertC:\Documents and Settings\Jerry Tucker\Application Data\Microsoft\Word\AutoRecovery save of LHO 11 binary to BCD.asdDr.Jerry H. TuckertC:\Documents and Settings\Jerry Tucker\Application Data\Microsoft\Word\AutoRecovery save of LHO 11 binary to BCD.asdDr.Jerry H. TuckertC:\Documents and Settings\Jerry Tucker\Application Data\Microsoft\Word\AutoRecovery save of LHO 11 binary to BCD.asdDr.Jerry H. Tucker3P:\public_html\f08-egre365\LHO 11 binary to BCD.docDr.Jerry H. 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