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Debora [2.8K]
3 years ago
6

Develop an sec (single error correction) code for a 16-bit data word. generate the code for the data word 0101000000111001. show

that the code will correctly identify an error in data bit 4.
Computers and Technology
2 answers:
natita [175]3 years ago
7 0

Answer:

010100000001101000101

Explanation:

When an error occurs in data bits, the SEC code is used to determine where the error took place. 5 check-bits are needed to generate SEC code for 16-bits data word. The check bits are:

C16=0, C8=0, C4=0,C2=0,C1=1

Therefore the SEC code is 010100000001101000101

jarptica [38.1K]3 years ago
4 0

Answer:

<em>The code is given as = 010100000001101000101</em>

<em>Explanation:</em>

<em>The steps take is shown below, </em>

<em>The SEC code is used to ascertain where errors had occurred. Such as the occurrence of errors in data  bits. </em>

<em>The inequality given is: 2^k - 1 >= M + K </em>

<em>Where M is =16 </em>

<em>For the arrangement of the data bits and checking of the bits, the following steps is taken below </em>

<em>   Bit position              Number             Check bits             Data Bits </em>

<em> 21                                   10101 </em>

<em> 20                                  10100 </em>

<em> </em>

<em>The bits is therefore checked in a way up to the bit position 1 </em>

<em>Therefore, the code is then written as follows: 010100000001101000101 </em>

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elena55 [62]

Answer:

It does all of these

Explanation:

The stub is able to do all of what we have in this question. It can locate computer connections, locate ports on a computer, transit message to server etc.

in distributed computing, a stub is a piece of code that does the work of converting parameters that are passed between the client and the server at a time of remote procedure call.

The main objective of an RPC is to allow the client to call procedures remotely on another server.

3 0
2 years ago
Create a function that will perform linear interpolation from a set of measured data stored in a list or array. The function sho
wel

Answer:

This question is incomplete, here is the complete question:

Python

a) You should create a function that will perform linear interpolation from a set of measured data. The function should take as input a list of values at which samples were taken, and then another list giving the measurements (you can assume each measurement is a single value) at those values. It should also take in a query value, and should give the best estimate it can of the value at that query. Be sure to handle values that are outside of the range, by extrapolating. You should write a program that allows you to test your function by reading the lists from a file where each line of the file is a pair of numbers separated by spaces: the value where the sample was taken, and the measurement at that value. Your program should ask the user for the name of the file and for a query value. Important: The two lists will correspond to each other: i.e. for the i-th value in the first list, the measurement will be the i-th element of the second list (these are called parallel lists or arrays). But, you should not assume that the input values are in increasing/decreasing order. That is, the values in the first list can be in any random ordering, not necessarily from smallest to largest or largest to smallest. You will have to account for this in your program, and there is more than one way to do so. You should discuss what options you can think of to handle the data arriving in any order like that, and decide what you think the best option for handling it is.

Explanation:

from __future__ import division

from cStringIO import StringIO

import numpy as np

from scipy.interpolate import RectBivariateSpline

np.set_printoptions( 1, threshold=100, edgeitems=10, suppress=True )

   # a file inline, for testing --

myfile = StringIO( """

# T P1 P2 P3 P4

0,   80,100,150,200

75, 400,405,415,430

100, 450,456,467,483

150, 500,507,519,536

200, 550,558,571,589

""" )

   # file -> numpy array --

   # (all rows must have the same number of columns)

TPU = np.loadtxt( myfile, delimiter="," )

P = TPU[0,1:] # top row

T = TPU[ 1:,0] # left col

U = TPU[1:,1:] # 4 x 4, 400 .. 589

print "T:", T

print "P:", P

print "U:", U

interpolator = RectBivariateSpline( T, P, U, kx=1, ky=1 ) # 1 bilinear, 3 spline

   # try some t, p --

for t, p in (

   (75, 80),

   (75, 200),

   (87.5, 90),

   (200, 80),

   (200, 90),

   ):

   u = interpolator( t, p )

   print "t %5.1f p %5.1f -> u %5.1f" % (t, p, u)

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Answer:

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Explanation:

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levacccp [35]

Answer:

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Explanation:

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Kindly check the attached image below to get the step by step explanation to the above question.

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