ssrs 2016 barcode The average power received in a binary polar transmission is 10 in Software

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Example 10.1 The average power received in a binary polar transmission is 10
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mW, and the bit period is 100 s. If the noise power spectral density is 0.1 J, and optimum filtering is used, determine the bit error rate.
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Solution
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From Eq. (10.17): Eb 5 10 3 1023 3 100 3 1026 5 1026J
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Digital Signals
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BER versus (Eb/N0) for baseband signaling using a binary polar NRZ waveform. The curve also applies for BPSK and QPSK modulated waveforms.
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and Eb N0 10 10 10
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erf( 210) > 0.9999923 Combining Eqs. (10.18) and (10.19): BER 0.5(1 3.9 0.9999923) 10
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Equation (10.18) is sometimes expressed in the alternative form Pe Qa 2Eb b N 0 (10.20)
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Here, the Q( ) function is simply an alternative way of expressing the complementary error function, and in general erfc(x) 2Q( 22x) (10.21)
These relationships are given for reference only and will not be used further in this book. An important parameter for carrier systems is the ratio of the average carrier power to the noise power density, usually denoted by [C/N0]. The [Eb/N0] and [C/N0] ratios can be related as follows. The average carrier power at the receiver is PR W. The energy per symbol is therefore PR/Rsym J, with Rsym in symbols per second. Since each symbol contains m bits, the energy per bit is PR/mRsym J. But mRsym Rb, and therefore, the energy per bit, Eb, is Eb PR Rb (10.22)
As before, let N0 represent the noise power density. Then Eb/N0 PR/RbN0. But PR/N0 is the carrier-to-noise density ratio, usually denoted by C/N0, and therefore, Eb N0 C/N0 Rb (10.23)
Rearranging this and putting it in decibel notation gives c C d N0 c Eb N0 d [Rb] (10.24)
It should be noted that whereas [Eb/N0] has units of decibels, [C/N0] has units of dBHz, as explained in App. G.
Example 10.2 The downlink transmission rate in a satellite circuit is 61 Mb/s, and
the required [Eb /N0] at the ground station receiver is 9.5 dB. Calculate the required [C/N0].
Solution
The transmission rate in decibels is [Rb]
10 log(61
106)
77.85 dBb/s
Hence s C N0 t 77.85 9.5 87.35 dBHz
Digital Signals
The equations giving the probability of bit error are derived on the basis that the filtering provides maximum signal-to-noise ratio. In practice, there are a number of reasons why the optimal filtering may not be achieved. The raised-cosine response is a theoretical model that can only be approximated in practice. Also, for economic reasons, it is desirable to use production filters manufactured to the same specifications for the transmit and receive filter functions, and this may result in some deviation from the desired theoretical response. The usual approach in practice is that one knows the BER that is acceptable for a given application. The corresponding ratio of bit energy to noise density can then be found from Eq. (10.18) or from a graph such as that shown in Fig. 10.17. Once the theoretical value of Eb/N0 is found, an implementation margin, amounting to a few decibels at most, is added to allow for imperfections in the filtering. This is illustrated in the following example.
Example 10.3 A BPSK satellite digital link is required to operate with a bit error
5 rate of no more than 10 , the implementation margin being 2 dB. Calculate the required Eb /N0 ratio in decibels.
The graph of Fig. 10.17 shows that Eb /N0 is around 9 dB for a BER of 10 5. By plotting this region to an expanded scale, a more accurate value of Eb/N0 can be obtained. This is shown in Fig. 10.18. from which [Eb/N0] is seen to be about 9.65 dB. This is without an implementation margin. The required value, including an implementation margin, is 9.65 2 11.65 dB.
Solution
To summarize, BER is a specified requirement, which enables Eb/N0 to be determined by using Eq. (10.18) or Fig. 10.17. The rate Rb also will be specified, and hence the [C/N0] ratio can be found by using Eq. (10.24).
1 10 4
Pe(x)
1 10 5
1 10 6 9
9.4 9.6 xdB(x)
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