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The counter itself must count in Gray code, or if the counter runs in binary then the output value from the counter must be reclocked after it has been converted to Gray code, because when a value is converted from binary to Gray code, [nb 1] it is possible that differences in the arrival times of the binary data bits into the binary-to-Gray ...
This table illustrates an example of an 8 bit signed decimal value using the two's complement method. The MSb most significant bit has a negative weight in signed integers, in this case -2 7 = -128. The other bits have positive weights. The lsb (least significant bit) has weight 1. The signed value is in this case -128+2 = -126.
For example, a 32-bit integer can encode the truth table for a LUT with up to 5 inputs. When using an integer representation of a truth table, the output value of the LUT can be obtained by calculating a bit index k based on the input values of the LUT, in which case the LUT's output value is the k th bit of the integer.
See Hamming code for an example of an error-correcting code. Parity bit checking is used occasionally for transmitting ASCII characters, which have 7 bits, leaving the 8th bit as a parity bit. For example, the parity bit can be computed as follows. Assume Alice and Bob are communicating and Alice wants to send Bob the simple 4-bit message 1001.
In a synchronous counter, the clock inputs of the flip-flops are connected, and the common clock simultaneously triggers all flip-flops. Consequently, all of the flip-flops change state at the same time (in parallel). For example, the circuit shown to the right is an ascending (up-counting) four-bit synchronous counter implemented with JK flip ...
An n-bit LUT can encode any n-input Boolean function by storing the truth table of the function in the LUT. This is an efficient way of encoding Boolean logic functions, and LUTs with 4-6 bits of input are in fact the key component of modern field-programmable gate arrays (FPGAs) which provide reconfigurable hardware logic capabilities.
The left figure below shows a binary decision tree (the reduction rules are not applied), and a truth table, each representing the function (,,).In the tree on the left, the value of the function can be determined for a given variable assignment by following a path down the graph to a terminal.
For example, a run of 4 bits such as 0000 2 using NRZI encoding contains no transitions and that may cause clocking problems for the receiver. 4B5B solves this problem by assigning the 4-bit block a 5-bit code, in this case, 11110 2. There are eight 5-bit codes that have 3 consecutive 0s: 00000, 00001, 00010, 01000, 10000, 00011, 10001, 11000.