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Figure 14.23 (a) SPEC transmitter; (b) SPEC receiver. (Courtesy of Sciulli and Campanella, 1973. Copyright 1973, IEEE.)
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for that channel, and a logic 0 means that the sample was unchanged. At the receiver, the sample assignment word either directs the new (unpredictable) sample into the correct channel slot, or it results in the previous sample being regenerated in the reconstruction decoder. The output from this is a 4.096-Mb/s PCM multiplexed signal which is demultiplexed in the PCM decoder. By removing the redundant speech samples and silent periods from the transmission link, a doubling in channel capacity is achieved. As shown in Fig. 14.23, the transmission is at 2.048 Mb/s for an inputoutput rate of 4.096 Mb/s. An advantage of the SPEC method over the DSI method is that freezeout does not occur during overload conditions. During overload, sample values which should change may not. This effectively leads to a coarser quantization and therefore an increase in quantization noise. This is subjectively more tolerable than freeze-out.
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14.7.11 Downlink analysis for digital transmission
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As mentioned in Sec. 14.6, the transponder power output and bandwidth both impose limits on the system transmission capacity. With TDMA, TWT backoff is not generally required, which allows the transponder to operate at saturation. One drawback arising from this is that the uplink
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station must be capable of saturating the transponder, which means that even a low-traffic-capacity station requires comparatively large power output compared with what would be required for FDMA. This point is considered further in Sec. 14.7.12. As with the FDM/FDMA system analysis, it will be assumed that the overall carrier-to-noise ratio is essentially equal to the downlink carrier-to-noise ratio. With a power-limited system this C/N ratio is one of the factors that determines the maximum digital rate, as shown by Eq. (10.24). Equation (10.24) can be rewritten as [Rb] c C d N0 c Eb N0 d (14.29)
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The [Eb/N0] ratio is determined by the required BER, as shown in Fig. 10.17, and described in Sec. 10.6.4. For example, for a BER of 10 5 an [Eb/N0] of 9.6 dB is required. If the rate Rb is specified, then the [C/N0] ratio is determined, as shown by Eq. (14.29), and this value is used in the link budget calculations as required by Eq. (12.53). Alternatively, if the [C/N0] ratio is fixed by the link budget parameters as given by Eq. (12.53), the bit rate is then determined by Eq. (14.29). The bit rate is also constrained by the IF bandwidth. As shown in Sec. 10.6.3, the ratio of bit rate to IF bandwidth is given by Rb BIF 1 m
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where m 1 for binary phase-shift keying (BPSK) and m 2 for QPSK and is the rolloff factor. The value of 0.2 is commonly used for the rolloff factor, and therefore, the bit rate for a given bandwidth becomes Rb mBIF 1.2 (14.30)
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Example 14.6 Using Eq. (12.53), a downlink [C/N0] of 87.3 dBHz is calculated for
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5 a TDMA circuit that uses QPSK modulation. A BER of 10 is required. Calculate the maximum transmission rate. Calculate also the IF bandwidth required assuming a rolloff factor of 0.2.
Solution
From Fig. 10.18 which is applicable for QPSK and BPSK. [Eb/N0] 5 for a BER of 10 . Hence [Rb] 87.3 9.65 77.65 dBb/s
9.65 dB
This is equal to 58.21 Mb/s.
Satellite Access
For QPSK m
2 and using Eq. (14.30), we have BIF 58.21 1.2/2 34.9 MHz
From Example 14.6 it will be seen that if the satellite transponder has a bandwidth of 36 MHz, and an EIRP that results in a [C/N0] of 87.3 dBHz at the receiving ground station, the system is rear optimum in that the bandwidth is almost fully occupied.
14.7.12 Comparison of uplink power requirements for FDMA and TDMA
With FDMA, the modulated carriers at the input to the satellite are retransmitted from the satellite as a combined frequency-divisionmultiplexed signal. Each carrier retains its modulation, which may be analog or digital. For this comparison, digital modulation will be assumed. The modulation bit rate for each carrier is equal to the input bit rate [adjusted as necessary for forward error correction (FEC)]. The situation is illustrated in Fig. 14.24a, where for simplicity, the input bit rate Rb is assumed to be the same for each earth station. The [EIRP] is also assumed to be the same for each earth station. With TDMA, the uplink bursts that are displaced in time from one another are retransmitted from the satellite as a combined time-divisionmultiplexed signal. The uplink bit rate is equal to the downlink bit rate in this case, as illustrated in Fig. 14.24b. As described in Sec. 14.7, compression buffers are needed in order to convert the input bit rate Rb to the transmitted bit rate RTDMA. Because the TDMA earth stations have to transmit at a higher bit rate compared with FDMA, a higher [EIRP] is required, as can be deduced from Eq. (10.24). Equation (10.24) states that c C d N0 c Eb N0 d [R]
where [R] is equal to [Rb] for an FDMA uplink and [RTDMA] for a TDMA uplink. For a given BER the [Eb/N0] ratio is fixed as shown by Fig. 10.17. Hence, assuming that [Eb/N0] is the same for the TDMA and the FDMA uplinks, an increase in [R] requires a corresponding increase in [C/N0]. Assuming that the TDMA and FDMA uplinks operate with the same [LOSSES] and satellite [G/T], Eq. (12.39) shows that the increase in [C/N0] can be achieved only through an increase in the earth station [EIRP], and therefore [EIRP]TDMA [EIRP]FDMA [RTDMA] [Rb] (14.31)
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