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Convert the 32-bit 2’s complement number 111111111111111111111111111110012 to decimal.  (HINT: if you think about what sign extension is you should be able to do this calculation in your head in under 30sec.)

Given the fields shown what operation is being performed DA = 01, WE = 1, SA1 = 00, SA2 = 11, MODE = 11 [1]

In a fixed-point binary number the digit immediately to the left of the binary point has a weight of [1]. The digit immediately to the right of the binary point has a weight of [2].

Complete a LC-3 instruction to zero R3 [1][2][3]  

Remember you can refer to your LC-3 Programmer’s Reference Sheet When bit 5 of a(n) [type] instruction is 1 the 2nd operand is referred to as [addrmd] Bit 5 of the instruction directly controls what signal in the LC-3 datapath [sig]  

Create an assembler directive that will allocate one word of memory with the label TEN that is initialized to decimal value ten. [1][2][3]

The data word size of a memory dictates:

Match the register name to its definition

To negate a 2’s comp number we [step1] then we [step2].  This procedure works for [works]. We use sign extension to transform a 2’s comp number into another that [SE_purpose].  To sign extend a 2’s comp number we [SE_meth].

 What statement can be made about these adders