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multiplier-accumulator(vhdl)
- 用VHDL语言描述和实现乘法累加器设计,4位的被乘数X和4位的乘数Y输入后，暂存在寄存器4位的寄存器A和B中，寄存器A和B的输出首先相乘，得到8位乘积，该乘积再与8位寄存器C的输出相加，相加结果保存在寄存器C中。寄存器C的输出也是系统输出Z。(原创，里面有乘法部分和累加部分可以单独提出来，很好用) -With the VHDL language to describe the design and realization of multiplier-accumulator, four of
disanci
- 5位的操作数X和Y输入后暂存在寄存器A和B中，两位的操作控制码control暂存在寄存器C中，按照control码的不同，分布实现下列操作： 00控制X+Y 01控制X-Y 10控制X and Y 11控制 X xor Y 运算结果暂存在寄存器D中，然后输出。 -5 of the operand X and Y after the temporary importation of A and B in the register, the two operational c
logic
- 5位的操作数X和Y输入后暂存在寄存器A和B中，两位的操作控制码control暂存在寄存器C中，按照control码的不同，分布实现下列操作： 00控制X+Y 01控制X-Y 10控制X and Y 11控制 X xor Y 运算结果暂存在寄存器D中，然后输出。 -5 of the operand X and Y after the temporary importation of A and B in the register, the two operational c
MIPS1CYCLE
- MIPS single-cycle processor design in verilog.Instruction memory to the design and initialise it with your assembly code-a. Load the data stored in the X and Y locations of the data memory into the X and Y registers. b. Add the X and Y registers an
AssignmentP4
- Assignment 4: 1. Analyze and simulate the following code lists (code1 and code 2) with the same input signals shown below by presenting POW and OL. If the data type of “a, b, c, d, u, v, w, x, y, z” is declared as std_logic, what will the simulatio
vhdl-all-english
- A major obstacle that stands in the way of efficient test response compaction are the unknown values (x-values) captured by scan cells during testing. If test responses with s and the correctness of the compactor inputs cannot be verified at
基于FPGA的反余弦算法代码
- 通过crodic算法迭代通过对常数C的设定旋转迭代x的分量使它等于常数C从而求出输入值c的反余弦arccosc的角度值