- Some advantages of using digital systems:
- improved noise immunity
- more efficient transmission and better quality of service - error
correction codes, bandwidth efficient modulation schemes, efficient
source coding, joint source and channel coding
- can easily perform data encryption and other security provisions
- flexibility (variable bit-rate sources can be accommodated easily,
newer coding schemes can be easily incorporated)
- lower power consumption (unless lots of processing needed)
- can easily allow for different Quality of Service guarantees - can
specify priorities
- easier to network, compatible with current wireline standards
- easier to use signal processing
- easily implementable in VLSI (low-cost)
- software receivers which can be upgraded
- possibility of using various multiple access techniques (CDMA/TDMA)
and TDD
- needed for efficient data transmission
Some disadvantages of using digital systems:
- intersymbol interference if high data rates - need for equalization
- analog systems already in place; replacement by digital systems
involves new investments
- voice quality may not be good - quality degradation of compressed
voice in the presence of channel errors is perceptually worse than for
analog
- complex algorithms, control procedures and protocols may need to be
used
- need for synchronization
- Some techniques which can be used to increase the spectral
efficiency:
- cellular structure
- better receivers which tolerate lower SIR
- bandwidth efficient modulation
- decreased protocol overhead
- use of speech compression
- antenna sectorization
- diversity and power control
- interference suppression techniques
- interference-robust schemes (e.g., spread spectrum)
- digital transmission
- dynamic channel allocation
- multilevel modulation techniques
- From equation (3.40), we have
If 
(as it usually is), we use the expansion
to obtain
The accuracy of this approximation can be assessed by noting that
is approximated by 1+x/2 to within 0.1% for x=0.1.
- The average signal power at d=10 m is 35 dB below its level at
m, i.e., its is equal to -35 dBm (0 dBm = 1 mW). Since with
probability 10% the signal level exceeds the average by 10 dB or more, we
must have 
, i.e., 
and
dB. - Taking into account that
and 
, we have
On a logarithmic plot, these curves are straight lines with different slopes.
- A SNR of 20 dB is provided with probability 0.95 in the presence of
log-normal shadowing with
dB if its average, SNR 
, is at least
where 
, i.e., 
and
SNR = 33 dB.
The thermal
noise power at the receiver is computed as 
dBm, which becomes
-121.2 dBm when the noise figure of 8 dB is taken into account.
The radiated power is 53.76 dBm (15 W = 41.76 dBm). The value of SNR
at km is given by the free-space expression and is about 80 dB.
In order to guarantee the required performance, the additional loss
due to distance must not exceed 80-33=47 dB. The maximum distance d
is then found from 
, i.e., 
km. - For channel 1, we have
For channel 2, we have
The maximum data rate which can be supported with adequate performance
is 
, i.e., about 4 Mbps in channel 1 and about 60
kbps in channel 2.
- Note: signal level 10 dB below the rms value corresponds to
(not 0.1!). A fade duration of 1 ms corresponds to a Doppler frequency of
132 Hz, i.e., a mobile speed of 44 m/s at 900 MHz. The mobile therefore
travels 440 m. With 
, we have a fading rate of
fades per second, and therefore the
number of crossings in 10 s is about 1220. - (a) time difference is
s and speed of light is 
m/s, so that length difference is 2.4 km.
(b) 
s and 
s.
(c) using the estimate 
we have a coherence bandwidth
of about 101 kHz.
(d) IS-54 uses a channel bandwidth of 30 kHz, and experiences flat fading in
this case. GSM has a channel bandwidth of 200 kHz and experiences some
frequency selective fading. IS-95 uses channels with 1.25 MHz bandwidth and
therefore the channel is frequency selective.
