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5. Error Analysis How did your volumetric ratios compare with the mole ratios in the
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Router(config-router)# Router(config-router)# Router(config-router)# Router(config-router)# network network network network 172.16.1.0 172.16.2.0 192.168.1.64 192.168.1.128
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Fig. 1.48
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Data Table 1
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Figure 2.26 A phasor diagram of OOK modulation with
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The Zoom Tool is the fourth tool located on the Toolbox, is clearly marked by its magnifying glass icon, and can be used to zoom in and zoom out of a page. When you ve chosen the Zoom Tool, the Property Bar (see Figure 7-3) displays buttons plus a drop-down selector that provides just about every common degree of magnification you could ask for (if you re currently not comfortable manipulating the Zoom Tool s cursor manually). The following list describes the purpose of these options on the Property Bar:
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and buffer status can all have a serious impact on the overall IP-over-satellite performance curves. Another problem is the asymmetry encountered in back channel bandwidth, which in many IP-over-satellite systems is only a fraction of the forward channel bandwidth. Because TCP emerged in advance of IP-over-satellite, TCP often gets mistakenly identified as the source of the problem. It is the operating environment and not TCP alone that is the problem. The design of TCP is not optimized for conditions encountered during satellite transmissions. The standard GEO round trip hop of 44,600 miles creates problems with the inherent transmission-related delay, breaking down the flow of packets between the sender and receiver, especially at very high transmission speeds. TCP works well as a generalpurpose protocol for a congested environment, such as the Internet. However, TCP perceives a satellite delay and bit error as congestion and inherently slows down.
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1.9 DESIGN CONSIDERATIONS
Range This control can also be thought of as the motion detector s
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As with StreamWriter, in some cases, you can open a file directly using StreamReader. To do so, use this constructor: StreamReader(string filename) Here, filename specifies the name of the file to open, which can include a full path specifier. The file must exist. If it doesn t, a FileNotFoundException is thrown. If filename is null, then an ArgumentNullException is thrown. If filename is an empty string, ArgumentException is thrown. IOException and DirectoryNotFoundException are also possible.
sin(1 + e) + cos 2 sin(1 e) + sin 2 cos(1 + e) + cos 2 cos(1 + e) cos 2 cos(1 + e) + cos 2
0 1 2 3 4 5 6 7 8 9
Now let s deal with the assumption that the energy of the molecule is the same regardless of how we arrange the unwound bases. One of the ways we can deal with this assumption is by melting fewer base pairs when we need extra enthalpy to break the extra stacking interactions. If we do this, we are not so much removing the assumption but actually strengthening the basis for it. The technique is this: We hold the energy of the molecule constant by melting fewer base pairs in the case of those arrangements that would otherwise require more energy to break the additional stacking interactions. By melting fewer base pairs we can keep the energy constant. In this way we can assume that the energy of the molecule is constant for all of the different arrangements, with no doubts about that assumption. If so, then all of the various arrangements (for a given energy level) truly represent different ways of achieving the same energy state of the molecule, and therefore truly represent the entropy state of the molecule. So, for example, if the molecule has enough enthalpy to overcome 300 stacking interactions, we can unwind 300 contiguous bases at one end of the molecule. If, as another arrangement, we unwind ten separate segments (instead of one), then we have ten additional stacking interactions to overcome. Given the same amount of enthalpy, we can only unwind 290 base pairs. So we would unwind 10 segments of 29 base pairs each (instead of 10 segments of 30 base pairs each). The same argument applies as we increase the number of segments. In the extreme case we can melt a total 150 base pairs in 150 separate segments.
150 150
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Initial values of i and j: 10 20 Swapped values of i and j: 20 10
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