From 40ed06977206f3708be9b2f694e165cdc5217bc3 Mon Sep 17 00:00:00 2001
From: lshprung <lshprung@yahoo.com>
Date: Mon, 12 Oct 2020 11:00:43 -0700
Subject: Post-class 10/12

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+[\<- Verilog basics](8.md)
+
+---
+
+# Number systems and adders
+
+## Hexadecimal numbers
+
+### Binary and hexadecimal
+
+- Binary is base-2, digits range from 0->1
+- Decimal is base-10, digit range: 0->9
+- Hexadecimal (hex) is base-16, 0->15
+	- Need digits for 11-15
+	- Use, A, B, C, D, E, F
+- Hex range in binary is exactly 4 bits
+	- Used as a convenience for expressing 4 bit groupings for values that are a large number of bits
+
+### Comparison
+
+- Convention is to preced hex with 0x
+	- e.g., 0x10 means 16, not 10
+
+|Decimal|Binary|Octal|Hexadecimal|
+|-------|------|-----|-----------|
+|00     |00000 |00   |00         |
+|01     |00001 |01   |01         |
+|02     |00010 |02   |02         |
+|03     |00011 |03   |03         |
+|04     |00100 |04   |04         |
+|05     |00101 |05   |05         |
+|06     |00110 |06   |06         |
+|07     |00111 |07   |07         |
+|08     |01000 |10   |08         |
+|09     |01001 |11   |09         |
+|10     |01010 |12   |0A         |
+|11     |01011 |13   |0B         |
+|12     |01100 |14   |0C         |
+|13     |01101 |15   |0D         |
+|14     |01110 |16   |0E         |
+|15     |01111 |17   |0F         |
+|16     |10000 |20   |10         |
+|17     |10001 |21   |11         |
+|18     |10010 |22   |12         |
+
+---
+
+## Converting binary to hex and vice versa
+
+- No need to understand base-16 arithmetic
+	- Easy conversion from one form to another
+- Example binary to hex
+	- 0011011010001110 would compress to
+		- 0011 -> 3
+		- 0110 -> 6
+		- 1000 -> 8
+		- 1110 -> E
+		- 368E or 0x368E
+- Going the other way, hex to binary
+	- 0x407B would expand out to 0100000001111011
+
+---
+
+## Adding binary numbers, and the Full Adder (FA) circuit
+
+### The Process of Addition
+
+- A 2N-input truth table? Ouch
+- Let's decompose instead
+	- Add numbers in the same position, including the "carry-in" -> 3 inputs
+	- Result is a sum and carry-out -> 2 outputs
+
+![diagram](9.1.png)
+
+### The Full Adder (FA) Circuit
+
+![diagram](9.2.png)
+![diagram](9.3.png)
+![diagram](9.4.png)
+
+---
+
+## The ripple carry adder
+
+- Use full adders and connect the carry-out to the carry-in of the next position
+	- The carry ripples thru the addrs
+	- Set c0 to 0 (unless we want to add 1...)
+- This is one approach; many ways to add
+
+![diagram](9.5.png)
+
+---
+
+## 4-bit adder, abstraction and hierarchical Verilog description
+
+### A 4-bit adder
+
+- 4-bit inputs => 4-bit output
+	- A hardware contruct, with wires and gates
+
+![diagram](9.6.png)
+
+### A 4-bit adder in verilog
+
+```
+module adder4(carryin, x3, x2, x1, x0, y3, y2, y1, y0, s3, s2, s1, s0, carryout);
+	input carryin, x3, x2, x1, x0, y3, y2, y1, y0;
+	output s3, s2, s1, s0, carryout;
+
+	fulladd stage0(carryin, x0, y0, s0, c1);
+	fulladd stage1(c1, x1, y1, s1, c2);
+	fulladd stage2(c2, x2, y2. s2, c3);
+	fulladd stage3(c3, x3, y3, s3, carryout);
+
+endmodule
+
+module fulladd(Cin, x, y, s, Cout);
+	input Cin, x, y;
+	output s, Cout;
+
+	assign s = x ^ y ^ Cin;
+	assign Cout = (x & y) | (x & Cin) | (y & Cin);
+
+endmodule
+```
-- 
cgit