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21 Threads found on Unsigned Multiplication
The SIGNED/unsigned setting of lpm_add_sub actually matters, but only for the optional overflow output.
Hi, There are many ways to do it: An easy to understand method: * To find out sign of result: XOR both input sign bits * then make both inputs positive ( ==> only 3 bits unsigned) * multiply both values: 3 bits x 3 bits = 6 bits result * If the result is negative (output of XOR) then invert the multiplication result * form a 7 bit sign
multiplication will yield bit growth on more occasions then addition - but it isn't guarantied. Some examples: "0100" * "0010" = "1000" "1111" * "0001" = "1111" An example for what? It has nothing to do with the bit width of an unsigned VHDL multiply operation which doesn't depend on a particular number value. You'll fin
Hello, i have to design 32 bit alu unit that do normal unsigned multiplication, booths multiplication and unsigned division In binary format, i know that we multiply lsb by of the one number by the other number and then we save the result in some register , and then we get the bit next to the lsb we multiply and (...)
x must have an unsigned int16 type, otherwise you'll get an overflow in the multiplication. 164 means? x=input data(0 to 255)? correct or not? The input value x is assumed with a range of 0 to 398, as asked in your post. I suggest to check the method yourself by putting in different values.
I assume you are talking about unsigned multiplication. If you are, then you'll never get overflow under the stated conditions.
Nice explanation:: When you make an assignment or port connection form a signed data type to an unsigned data type, the signedness of original data is lost. Only the individual pattern of bits is transferred.
The following routine is a 10-bit X 10-bit assembly routine written by Martin Sturm for the PIC18F series: ; Multiply 10bit by 10bit unsigned ; by Martin Sturm 2010 ; Tested over full range of 10bit input combinations ; ; a (10bit) = aH:aL (not modified) ; b (10bit) = bH:bL (not modified) ; aH:aL * bH:bL --> rH:
To multiply two's complement, the sign bit must be treated differently. For any multiplication of two n-bit numbers, the least significant n-bits for two's complement and unsigned numbers are the same. If you use a 2n-bit result, the most significant n-bits will be different. You will get the correct results if you increase the number of bits
The error will be fixed like this: variable prod :t3:=(others => (others => 0 )); In your code you're trying to assign a numerical array type (like an unsigned or signed), but an integer is not an array of bits or std_logic. A few notes on the code: 1. variable i,j,k : integer:=0; - Not required, they are declared in the loops 2. I hope this is
Hi there, Recently I was trying to write a Verilog Code for multiplication by 3. Condition-My Input is variable-unsigned or Signed My Multiplier is fixed-3 So if i have -20 as input in binary my output should by -60. and 20 as input my output should be +60. I want to declare only one output that is product and depending o
VHDL has multiplication and divide functions for signed/unsigned numbers. Its as simple as writing the following code: op <= a * b; op <= a / b;
Depending on the referenced libraries, the "*" operator is only recognized for specific data types, e.g. signed, unsigned or integer. The support for "/" operator is generally limited and vendor dependant.
Hi friends, i have two doubts in microcontroller... 1) I am sending data from PC to 89S8252 microcontroller using VB6.0 The data format is in Hex.i am sending 5 words one after another. ex:mscomm1.output=word1(word1 contains value in hex formatg) result1=mscomm1.input mscomm1.output=word2(same as above)
hi, i am doing a project where i need to do 12 bit signed multiplication in VHDL, in ALtera platform. i have done it in two ways - 1) i have done first with unsigned library from IEEE, if the nos. are -ve do 2's complement, multiply tw unsigned no., then if the sign bit of multiplier and multiplicand are opposite, then do 2's complemen
hi friends am doing alu as my project . . so can any one send the verilog code for tat... its a multiplication of 8bit unsigned numbers...... pls
I have 3 8-bit std_logic_vectors A, B, C. If I include the unsigned.all library, would the code correctly perform the following?: C <= A + B; C <= A - B; C <= A and B; -- bitwise addition Also for multiplication, lets say I have a 16-bit vector called D. Would the following do what I want?: D <= A * B; C <= D(7 downto 0); -
Sounds like your compiler is not ANSI C compliant. It should automatically promote your two unsigned char values to int before it multiplies them. The CodeVisionAVR 1.24.7 user manual describes a compiler setting "promote char to int" for ANSI compatibility. Try it.
Hi Try this one. If we suppose that x0, x1, y0, & y1 are 32 bit then the 128-bit result is: z = x0.y0 + x1.y0<<32 + y1.x0<<32 + x1.y1<<64 only x1.y1 is signed all other are unsigned. Regards
Has some very nice code snippits. In your case I'd examine this library a bit closer: It has 8, 16 and 32 bit subtraction, multiplication, and division (signed and unsigned). Has good comments, should be worthwile... Happy coding!
Take a look here! multiplication is very simple to describe but it requires a large quantity of logic specially for a 32 bit magnitude a <= b * c; a b c signed or unsigned type or if you want std_logic_vector bye Stark.