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Fundamentalsof Telecommunications

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208
CONCEPTS IN TRANSMISSION TRANSPORT
E
b
= -89 dBW - 10 log(2.048 × 10
6
)
= -89 dBW - 63.11 dB
= -152.11 dBW
We can now develop a formula for
E
b
/N
0
:
E
b
/N
0
= RSL
dBW
- 10 log(bit rate) - (-204 dBW + NF
dB
).
(9.14)
Simplifying, we obtain
E
b
/N
0
= RSL
dBW
- 10 log(bit rate) + 204 dBW - NF
dB
.
(9.15)
Some Notes on E
b
/N
0
and Its Use.
E
b
/N
0
, for a given BER, will be different for
different types of modulation (e.g., FSK, PSK, QAM, etc.). When working with
E
b
, we
divide RSL by the bit rate, not the symbol rate nor the baud rate. There is a theoretical
E
b
/N
0
and a practical
E
b
/N
0
. The practical is always a greater value than the theoretical,
greater by the modulation implementation loss in decibels, which compensates for system
imperfections.
Figure 9.10 is an example of where BER is related to
E
b
/N
0
. There are two curves in
the figure. The first from the left is for BPSK/QPSK (binary phase shift keying/quadrature
phase shift keying), and the second is for 8-ary PSK (an eight-level PSK modulation
scheme). The values are for coherent detection. Coherent detection means that the receiver
has a built-in phase reference as a basis to make its binary or higher level decisions.
9.2.3.5
Digital Modulation of LOS Microwave Radios
. Digital systems, typically
standard PCM as discussed in Chapter 6, are notoriously wasteful of bandwidth compared
to their analog counterparts.
6
For example, the analog voice channel is nominally of 4-
kHz bandwidth, whereas the digital voice channel requires a 64-kHz bandwidth, assuming
1-bit/Hz occupancy. This is a 16-to-1 difference in required bandwidth. Thus national
regulatory authorities, such as the U.S. FCC, require that digital systems be bandwidth
conservative. One means that is used to achieve bandwidth conservation is bit packing.
This means packing more bits into 1 Hz of bandwidth. Another driving factor for bit
packing is the need to transmit such higher bit rate formats such as SONET and SDH
(Chapter 19). Some radio systems can transmit as much as 622 Mbps using advanced
bit-packing techniques.
How Does Bit Packing Work? In the binary domain we can estimate bandwidth to
approximately equate to 1 bit/Hz. For example, if we were transmitting at 1.544 Mbps,
following this premise, we'd need 1.544 MHz of bandwidth. Suppose now that we turn
to higher levels of modulation. Quadrature phase shift keying is one example. In this case
we achieve a theoretical packing of 2 bits/Hz. Again, if we are transmitting 1.544 Mbps,
with QPSK we would need 1.544 MHz/2 or 0.772 MHz. QPSK is one of a family of
modulation schemes that are based on phase-shift keying (PSK). With binary PSK, we
might assign a binary 1 to the 0
position (i.e., no phase retardation) and a binary 0 to the
180
phase retardation point. For QPSK, the phase circle is broken up into 90
segments,
rather than 180
segments as we did with binary PSK. In this case, for every transition we
transmit 2 bits at a time. Figure 9.11 is a functional block diagram of a QPSK modulator.
It really only consists of two BPSK modulators where one is out of phase with the other
by 90
.
6
A cogent example is FDM using frequency modulation; another is single sideband modulation.

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Summary :

208 CONCEPTS IN TRANSMISSION TRANSPORT E b = -89 dBW - 10 log(2.048 × 10 6 ) = -89 dBW - 63.11 dB = -152.11 dBW We can now develop a formula for E b /N 0 : E b /N 0 = RSL dBW - 10 log(bit rate) - (-204 dBW + NF dB ). (9.14) Simplifying, we obtain E b /N 0 = RSL dBW - 10 log(bit rate) + 204 dBW - NF dB .


Tags : bandwidth,phase,modulation,binary,dbw,psk,rate,example,bit,packing,qpsk,1544,digital





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