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Direct relationship

Our first step was to generate a Y1 with random ones and zeros and see if there is any direct relationship to v1. The following demonstrates the code used to produce random binary values (impedances for Y1), while Fig 1 shows the output of the comparison.

for c = 1:m, if rand>0.5, y1(c) = one;else y1(c) = 0;end endY1 = diag(y1); v1 = nfc(Y1, S0);subplot(2,1,1) stem(1:20, y1, 'r')title('Magnitude of Y1') subplot(2,1,2)stem(1:20, abs(v1), 'b') title('Magnitude of v1')

Y1 and the corresponding v1
The diagonal of Y1 and the corresponding v1 values.

From these graphs, there was no obvious correlation between the “0’s” and “1’s” in Y1 and the values of v1. From here, a linear relationship was tested.

Testing for linearity

To determine if there was a linearity in computing the values of v1, we decided to compute the values of v1 for a Y1 matrix that had a diagonal of 11000000000000000000. As such, we first computed a set of v1 where there was a single 1 in the first port of Y1, and then a second set of v1 where there was a single 1 in the second port of Y1, and then added those two sets of values together. We were hoping to see if the result would produce a set of values of v1 that is identical to or similar to that which would be produced by a matrix with a diagonal of two 1's followed by 0's. The results of this test are shown in Fig 2 and Fig 3.

The separate v1 values.
The separate v1 values.

Comparison for linearity
Comparison for linearity.

Looking at Fig 3. shows that the methods produce similar looking graphs, but each port has a slightly different magnitude. From these results we were able to conclude that the relationship between Y1 and v1 was non-linear.

Fourier transform

Our next approach was to look at the Fourier transform of several different patterns, to see if a Fourier transform of the values of v1 could provide any insight into a relationship. The following set of graphs shows the values of the diagonal of the Y1 matrix, and the corresponding v1 values computed from it.

Pattern 1
Pattern 1 and the Fourier transform of the corresponding v1 values.
Pattern 2
Pattern 2 and the Fourier transform of the corresponding v1 values.
Pattern 3
Pattern 3 and the Fourier transform of the corresponding v1 values.
Pattern 4
Pattern 4 and the Fourier transform of the corresponding v1 values.
Pattern 5
Pattern 5 and the Fourier transform of the corresponding v1 values.
Pattern 6
Pattern 6 and the Fourier transform of the corresponding v1 values.

From these graphs, once again, there is no evidence that there is any relationship between any of the patterns and the values of v1, nor any relationship between Y1 and the Fourier transform of the values in v1.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
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sure. what is your question?
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I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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Commplementary angles
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im all ears I need to learn
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
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Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
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what is system testing?
preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
I'm interested in Nanotube
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Near field communication simulation & Identification. OpenStax CNX. Dec 19, 2011 Download for free at http://cnx.org/content/col11398/1.1
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