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Abstract:
A nanometer is one billionth of a meter. If you blew up a baseball to the size of the earth, the atoms would become visible, about the size of grapes. Some 3- 4 atoms fit lined up inside a nanometer. Nanotechnology is about building things atom by atom, molecule by molecule. The trick is to be able to manipulate atoms individually, and place them where you wish on a structure.

Nanotechnology uses well known physical properties of atoms and molecules to make novel devices with ex ....etc

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Space-Time Adaptive Processing: Fundamentals


ABSTRACT


Space-time Adaptive Processing (STAP) is a signal processing technique most commonly used in radar systems. It involves adaptive array processing algorithms to aid in target detection. Radar signal processing benefits from STAP in areas where interference is a problem (i.e. ground clutter, jamming, etc.). Through careful application of STAP, it is possible to achieve order-of-magnitude sensitivity improvements in target detection.STAP involves a twodimensional filtering technique ....etc

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BANDWIDTH ESTIMATION FOR IEEE 802.11 BASED ADHOC NETWORK
Abstract: Since 2005, IEEE 802.11-based networks have been able to provide a certain level of quality of service (QoS) by the means of service differentiation, due to the IEEE 802.11e amendment. However, no mechanism or method has been standardized to accurately evaluate the amount of resources remaining on a given channel. Such an evaluation would, however, be a good asset for bandwidth-constrained applications. In multihop ad hoc networks, such evaluation becomes even more difficult. C ....etc

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A proof of the Heine-Borel Theorem
Theorem (Heine-Borel Theorem). A subset S of Ris compact if and only if S is closed and bounded.
Proof. First we suppose that S is compact. To see that S is bounded is fairly simple: Let In = (−n, n).
Then
1
[
n=1
In = R.
Therefore S is covered by the collection of {In}. Hence, since S is compact, finitely many will suffice.
S  (In1 [ · · · [ Ink ) = Im,
where m = max{n1, . . . , nk}. Therefore |x|  m for all x 2 S, and S is bounded.
Now we will show that S is closed. Suppose not. The ....etc

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