Digital Modulation 4

Objectives
After studying this chapter, you should be able to:
( Describe the basic types of digital modulation.
( Calculate the maximum data rate for a channel with a given modulation
scheme and signal-to-noise ratio.
( Explain the use of eye diagrams and constellation diagrams.
( Explain the difference between bit rate and baud rate and calculate both
for typical digital modulation systems.
( Describe and compare FSK, PSK, and QAM and perform simple calculations
with each.
( Explain the concepts of multiplexing and multiple access using frequency
and time division.
( Describe the principles of spread-spectrum communication and distinguish
between frequency-hopping and direct-sequence systems.
( Calculate spreading gain and signal-to-noise ratio for spread-spectrum
systems.
( Describe code-division multiple access and compare with FDMA and TDMA.
112 ! CHAPTER 4

' 4.1 Introduction

Digital signals have become very important in wireless communication.
Some of the information to be transmitted is already in digital form. In
Chapter 3, we examined the ways in which analog signals such as voice can
be converted to digital form for transmission.
In order to send data by radio, it is necessary to use a higher frequency car-
rier wave, just as for analog communication. Since the high-frequency carrier is
a sine wave for digital as well as analog signals, the same three basic parameters
are available for modulation: amplitude, frequency, and phase. All three, singly
and in combination, are used in digital systems.
Often the modulator and demodulator are collectively described as a
modem. You are probably familiar with telephone modems. These modulate
digital data onto an audio-frequency carrier. Radio-frequency modems are
similar in principle though quite a bit different in construction due to the
much higher carrier frequencies used.
In a holdover from the days of Morse code (a digital, though not a bi-
nary, communication scheme), the word keying is still often used to denote
digital modulation schemes. Thus we have frequency-shift keying (FSK)
and phase-shift keying (PSK). Straightforward amplitude-shift keying
(ASK) is rare in digital communication unless we count Morse code, but
quadrature AM (QAM) is very common. In this chapter we shall look at all
of these schemes as they are used in wireless transmission.
In Chapter 1 we noted that information capacity is proportional to
bandwidth. This fundamental relationship is given by Hartley’s law, which
is repeated here for convenience:

I = ktB (4.1)
where
I = amount of information to be transmitted in bits
k = a constant that depends on the modulation scheme and the
signal-to-noise ratio
t = time in seconds
B = bandwidth in hertz

We noted in Chapter 1 that the constant k is important. In Chapter 2,

when we studied analog signals, we saw that the bandwidth required for the
same information rate varied greatly with the modulation scheme. In that
chapter we used baseband bandwidth as a rough measure of information
content. We are now in a position to discuss the value of k for different digi-
tal modulation schemes. The information rate for digital communication
will be expressed in bits per second.
 DIGITAL MODULATION ! 113

Digital signals result in discrete, rather than continuous, changes in the

modulated signal. The receiver examines the signal at specified times, and
the state of the signal at each such time is called a symbol. The eventual out-
put will be binary, but it is certainly possible to use more than two states for
the transmitted symbol. Complex schemes using several levels can send
more data in a given bandwidth.
There is a theoretical limit to the maximum data rate that can be trans-
mitted in a channel with a given bandwidth. The Shannon-Hartley theorem
states:

C = 2B log 2 M (4.2)

where
C = information capacity in bits per second
B = channel bandwidth in hertz
M = number of possible states per symbol

For instance, if an FSK modulator can transmit either of two frequencies,

M = 2. A more elaborate modulator, using four different phase angles, each of
which can have two different amplitudes, has M = 4 × 2 = 8.
The limiting effect of bandwidth on data rate is understood most easily
by looking at a low-pass, rather than a bandpass, channel. Suppose that the
channel can pass all frequencies from zero to some maximum frequency B.
Then, of course, the highest frequency that can be transmitted is B. Suppose
that a simple binary signal consisting of alternate ones and zeros is transmit-
ted through the channel. This time let a logic 1 be 1 V and a logic 0 be −1 V.
The input signal will look like Figure 4.1(a): it will be a square wave with a
frequency one-half the bit rate (since there are two bits, a one and a zero,
for each cycle). This signal, which is a square wave, has harmonics at all odd

FIGURE 4.1 Digital transmission through a low-pass channel

114 ! CHAPTER 4

multiples of its fundamental frequency, with declining amplitude as the

frequency increases. This was shown in Chapter 1, using Fourier analysis.
At very low bit rates the output signal after passage through the channel will
be similar to the input, but as the bit rate increases, the frequency of the
square wave also increases and more of its harmonics are filtered out. There-
fore, the output will become more and more distorted. Finally, for a bit rate
of 2B the frequency of the input signal will be B, and only the fundamental
of the square wave will pass through the channel, as shown in Figure 4.1(b).
The receiver will still be able to distinguish a one from a zero, and the infor-
mation will be transmitted. Thus, with binary information, the channel
capacity will be 2B, as predicted by Equation (4.2).
Often the ratio of bit rate to channel bandwidth, expressed in bits per
second per hertz, is used as a figure of merit in digital radio systems. In the
preceding paragraph, we can see that for a two-level code the theoretical
maximum bandwidth efficiency is 2 b/s/Hz.
If a random pattern of ones and zeros were transmitted using the code of
Figure 4.1 and the signal were applied to an oscilloscope sweeping at the bit
rate, we would see the patterns shown in Figure 4.2. These are called “eye
diagrams.” The reason should be obvious from Figure 4.2(b). Because some
sweeps of the scope take place during a high- and some during a low-signal
level, both levels appear on the scope. A diagram like that in Figure 4.2(a)
indicates a data rate much lower than can be carried by the channel. Fig-
ure 4.2(b) indicates an optimal system. If an attempt were made to transmit
data too quickly, the eye would begin to close due to intersymbol interfer-
ence, as shown in Figure 4.2(c).
Now suppose that four voltage levels, each corresponding to a two-bit se-
quence, are used, as in Figure 4.3. The bandwidth required for the funda-
mental of this signal is the same as before. We have, it seems, managed to
transmit twice as much information in the same bandwidth. Again, this
agrees with Equation (4.2).
Unfortunately, the information capacity of a channel cannot be in-
creased without limit by increasing the number of states because noise
makes it difficult to distinguish between signal states. The ultimate limit is
called the Shannon limit:

C = B log 2 (1 + S/N) (4.3)

where
C = information capacity in bits per second
B = bandwidth in hertz
S/N = signal-to-noise ratio (as a power ratio, not in decibels)
 DIGITAL MODULATION ! 115

FIGURE 4.2 Eye diagrams

FIGURE 4.3
Four-level code
116 ! CHAPTER 4

By the way, it is quite easy to find logs to the base 2, even if your calcula-
tor lacks this function. Simply find the log to the base 10 (the common log)
of the given number and divide by the log of 2, that is:

log 10 N
log 2 N = (4.4)
log 10 2

The effect of noise on a signal can be seen in the eye diagram of Fig-
ure 4.4. The noise causes successive oscilloscope traces to be at different
amplitudes. If the noise is severe enough, the eye closes and data recovery is
unreliable.

FIGURE 4.4
Eye diagram showing
inter-symbol interference
and noise

EXAMPLE 4.1 Y
A radio channel has a bandwith of 10 kHz and a signal-to-noise ratio of
15 dB. What is the maximum data rate than can be transmitted:
(a) Using any system?
(b) Using a code with four possible states?

SOLUTION
(a) We can find the theoretical maximum data rate for this channel from
Equation (4.3). First, though, we need the signal-to-noise ratio as a
power ratio. We can convert the given decibel value as follows:

S  15 
= log −1  
N  10 
= 316
.
 DIGITAL MODULATION ! 117

Now we can use Equation (4.3):

C = B log2(1 + S/N)
= 10 × 103 log2(1 + 31.6)
= 10 × 103 × 5.03
= 50.3 kb/s

(b) We can use Equation (4.2) to find the maximum possible bit rate given
the specified code and bandwidth. We will then have to compare this
answer with that of part (a). From Equation (4.2),

C = 2B log2 M
= 2 × 10 × 103 × log24
= 2 × 10 × 103 × 2
= 40 kb/s

Since this is less than the maximum possible for this channel, it should
be possible to transmit over this channel, with a four-level scheme, at
40 kb/s. A more elaborate modulation scheme would be required to attain
the maximum data rate of 50.3 kb/s for the channel.
X

At this point we should distinguish between bit rate and baud rate. The
bit rate is simply the number of bits transmitted per second (C in the preced-
ing two equations), while the baud rate is the number of symbols per second.
Therefore, if we let the baud rate be S (for symbols, since B is already being
used for bandwidth), then:

C = S log 2 M (4.5)
where
C = capacity in bits per second
S = baud rate in symbols per second
M = number of possible states per symbol

EXAMPLE 4.2 Y
A modulator transmits symbols, each of which has 64 different possible
states, 10,000 times per second. Calculate the baud rate and bit rate.

SOLUTION
The baud rate is simply the symbol rate, or 10 kbaud. The bit rate is given by
Equation (4.5):
118 ! CHAPTER 4

C = S log2 M
= 10 × 103 × log264
= 60 kb/s

' 4.2 Frequency-Shift Keying

Probably the simplest digital modulation scheme in current use is fre-
quency-shift keying (FSK). In its simplest form two frequencies are transmit-
ted, one corresponding to binary one, the other to zero. In digital
communication systems, a one is often denoted by the term mark, and a
zero is called a space. This is another holdover from telegraphy in which
Morse code messages were often recorded as marks on a paper tape. FSK is a
robust scheme; that is, like analog FM, it tends to be reliable in the presence
of noise. Its disadvantage is: since each symbol has only two possible states,
it is not very efficient in terms of bandwidth. It tends to be used for
low-data-rate applications, such as pagers, and for transmitting bursts of
data over systems that are mainly analog.
FSK is also used extensively in high-frequency radio systems for radio-
teletype transmission. High-frequency radio channels tend to be very noisy,
and phase shifts induced into the signal by travel through the ionosphere
make the use of any scheme requiring accurate phase information impracti-
cal. Data rates for HF communication are very low (on the order of 100 b/s)
and frequency shifts between mark and space vary from 170 Hz to 850 Hz.
Usually HF FSK is actually transmitted and received using SSBSC AM
equipment. Two different audio tones, one for mark and one for space, with
the required frequency difference are chosen. These are applied one at a time
to the microphone input of the transmitter. This results in the transmission
of RF frequencies that differ by the same amount as the audio frequencies. At
the receiver two audio tones differing by the required frequency shift will be
produced. Figure 4.5 shows the spectrum of a typical HF FSK transmission, as
well as a typical pair of audio tones used to modulate the transmitter.
A variant of FSK is AFSK (audio frequency-shift keying), in which two
different audio frequencies, corresponding to mark and space, are used to
modulate a carrier by any of the analog methods discussed in Chapter 2.
AFSK is often combined with FM modulation for the audio tones. Figure 4.6
shows how this can work using a VHF amateur radio link as an example.
There is an extensive amateur packet-radio network, using AFSK transmis-
sion with a frequency shift of 1 kHz between mark and space, operating at a
data rate of 1200 bits per second.
 DIGITAL MODULATION ! 119

FIGURE 4.5 Spectrum of HF FSK transmission

FIGURE 4.6 AFSK using FM equipment

120 ! CHAPTER 4

It is possible to build an FSK system with more than two different fre-
quencies in order to increase the number of bits per symbol, but it is usually
more efficient to move to a system using phase shifts, or a combination of
amplitude and phase shifts, when this is required.

Gaussian A special case of FSK called Gaussian minimum-shift keying (GMSK) is

Minimum-Shift used in the GSM cellular radio and PCS systems to be described later. In a
Keying minimum-shift system, the mark and space frequencies are separated by half
the bit rate, that is:

ƒ m − ƒ s = 0.5 ƒ b (4.6)
where
ƒ m = frequency transmitted for mark (binary 1)
ƒ s = frequency transmitted for space (binary 0)
ƒ b = bit rate
If we use the conventional FM terminology from Chapter 2, we see that
GMSK has a deviation each way from the center (carrier) frequency, of

δ = 0.25 ƒ b
which corresponds to a modulation index of
δ
mƒ =
ƒm

0.25ƒ b
=
ƒb

= 0.25
The word Gaussian refers to the shape of a filter that is used before the
modulator to reduce the transmitted bandwidth of the signal. GMSK uses
less bandwidth than conventional FSK, because the filter causes the trans-
mitted frequency to move gradually between the mark and space frequen-
cies. With conventional FSK the frequency transition is theoretically
instantaneous, and in practice as rapid as the hardware allows, producing
sidebands far from the carrier frequency.

EXAMPLE 4.3 Y
The GSM cellular radio system uses GMSK in a 200-kHz channel, with a
channel data rate of 270.833 kb/s. Calculate:
(a) the frequency shift between mark and space
 DIGITAL MODULATION ! 121

(b) the transmitted frequencies if the carrier (center) frequency is exactly

880 MHz
(c) the bandwidth efficiency of the scheme in b/s/Hz

SOLUTION
(a) The frequency shift is

ƒm − ƒs = 0.5 ƒb = 0.5 × 270.833 kHz = 135.4165 kHz

(b) The shift each way from the carrier frequency is half that found in (a) so
the maximum frequency is

ƒmax = ƒc + 0.25 ƒb = 880 MHz + 0.25 × 270.833 kHz = 880.0677 MHz

and the minimum frequency is

ƒmin = ƒc − 0.25 ƒb = 880 MHz − 0.25 × 270.833 kHz = 879.93229 MHz

(c) The GSM system has a bandwidth efficiency of 270.833/200 = 1.35

b/s/Hz, comfortably under the theoretical maximum of 2 b/s/Hz for a
two-level code.
X

' 4.3 Phase-Shift Keying

When somewhat higher data rates are required in a band-limited channel
than can be achieved with FSK, phase-shift keying (PSK) is often used. Mea-
suring phase requires a reference phase, which would be hard to maintain
accurately. Usually, the phase of each symbol is compared with that of the
previous symbol, rather than with a constant reference. This type of PSK is
more completely described as delta phase-shift keying (DPSK). Most DPSK
modems use a four-phase system called quadrature phase-shift keying
(QPSK or DQPSK). In QPSK, each symbol represents two bits and the bit rate
is twice the baud rate. This is called a dibit system. Such a system can carry
twice as much data in the same bandwidth as can a single-bit system like
FSK, provided the signal-to-noise ratio is high enough.
Figure 4.7 is a vector diagram that represents a typical DQPSK system.
Phase shifts are given with respect to the phase of the previous symbol. Each
of the four possible phase shifts (including no shift at all) is associated with a
two-bit sequence, as shown in Table 4.1.
122 ! CHAPTER 4

FIGURE 4.7
Delta quadrature
phase-shift keying

TABLE 4.1 DQPSK Coding

Phase Shift (degrees) Symbol

0 00
+90 01
−90 10
180 11

π/4 Delta The system shown in Figure 4.7 and Table 4.1 requires a 180 degree transi-
Phase-Shift tion for the symbol 11. The transmitted signal has to go to zero amplitude
Keying momentarily as it makes this transition. Accurate transmission of this signal
therefore requires a linear amplifier, unlike the case for FSK. In fact, the am-
plifier should be linear all the way down to zero output. This is quite possi-
ble, of course, but linear amplifiers are markedly less efficient than
nonlinear amplifiers. The need for linearity can be reduced, though not
eliminated, by changing to a system called π/4 DQPSK. Here the allowable
transitions from the previous phase angle are ±45° and ±135°. Neither of
these requires the signal amplitude to go through zero, relaxing the linearity
requirements somewhat. A typical π/4 DQPSK system has the state table
shown in Table 4.2, and a vector diagram showing the possible transitions

TABLE 4.2 π/4 DQPSK Coding

Phase Shift (degrees) Symbol

45 00
135 01
−45 10
−135 11
 DIGITAL MODULATION ! 123

can be found in Figure 4.8. This system is used for the North American
TDMA cell phone and PCS systems.

FIGURE 4.8
π/4 DQPSK

EXAMPLE 4.4 Y
The North American TDMA digital cell phone standard transmits at
24.3 kilobaud using DQPSK. What is the channel data rate?

SOLUTION
Since this is a dibit system, the symbol rate, also known as the baud rate, is
half the bit rate. Therefore the data rate is 48.6 kb/s.
X

' 4.4 Quadrature Amplitude Modulation

The only way to achieve high data rates with a narrowband channel is
to increase the number of bits per symbol, and the most reliable way to do
that is to use a combination of amplitude and phase modulation known as
quadrature amplitude modulation (QAM). For a given system, there is a finite
number of allowable amplitude-phase combinations. Figure 4.9(a) is a con-
stellation diagram that shows the possibilities for a hypothetical system
with sixteen amplitude/phase combinations. Thus each transmitted symbol
represents four bits. This diagram is similar to the previous figure except that
the vectors are not drawn. Each dot represents a possible amplitude/phase
combination or state. With a noiseless channel, the number of combina-
tions could be increased indefinitely, but a practical limit is reached when
the difference between adjacent states becomes too small to be detected
reliably in the presence of noise and distortion. If a QAM signal is applied to
124 ! CHAPTER 4

FIGURE 4.9 Quadrature AM

the oscilloscope, the noise can be seen as a blurring of the points in the con-
stellation, as shown in Figure 4.9(b).
In fixed terrestrial microwave systems QAM is used with quite a large
number of states—up to 1024 in some cases. This requires a very high sig-
nal-to-noise ratio however, and portable and mobile systems are much more
limited.
QAM is more efficient in terms of bandwidth than either FSK or QPSK,
but it is also more susceptible to noise. Another disadvantage compared to
FSK is that QAM signals, like analog AM signals, vary in amplitude. This
means that transmitter amplifiers must be linear.

EXAMPLE 4.5 Y
A modem uses 16 different phase angles and 4 different amplitudes. How
many bits does it transmit for each symbol?

SOLUTION
The number of possible states per symbol is 16 × 4 = 64
The number of bits per symbol is log264 = 6
X

' 4.5 Multiplexing and Multiple Access

Multiplexing allows several signals to be carried on a single transmission
channel or medium. When the signals to be multiplexed originate at differ-
ent locations, the system is often described by using the term multiple
access. Both are used in wireless communication.
 DIGITAL MODULATION ! 125

Frequency-Division The simplest multiple access scheme is the one used by radio and television
Multiplexing and broadcasting stations. Each signal is assigned a portion of the available fre-
Multiple Access quency spectrum on a full-time basis. This is called frequency-division
multiplexing (FDM) or frequency-division multiple access (FDMA) de-
pending on the situation. For instance, over-the-air broadcasts are FDMA
while a cable-television system, where all the signals are assigned slots on
the same cable by the headend equipment, is an example of FDM. Frequency
division can be and is used with both analog and digital signals.

Time-Division Time-division multiplexing (TDM) is used mainly for digital communica-

Multiplexing and tion. In TDM, each information signal is allowed to use all the available
Multiple Access bandwidth, but only for part of the time. From Hartley’s Law (Equation 4.1),
it can be seen that the amount of information transmitted is proportional to
both bandwidth and time. Therefore, at least in theory, it is equally possible
to divide the bandwidth or the time among the users of a channel. Continu-
ously varying signals, such as analog audio, are not well adapted to TDM,
because the signal is present at all times. On the other hand, sampled au-
dio is very suitable for TDM, as it is possible to transmit one sample or bit
from each of several sources sequentially, then send the next sample or
bit from each source, and so on. As already mentioned, sampling itself does
not imply digital transmission, but in practice sampling and digitizing usu-
ally go together.
Many signals can be sent on one channel by sending either a sample
from each signal, or a fixed number of bits from each signal, in rotation.
Time-division multiplexing requires that the total bit rate be multiplied
by the number of channels multiplexed. This means that the bandwidth
requirement is also multiplied by the number of signals.

TDM in TDM is used extensively in digital telephony. The simplest North American
Telephony standard is known as the DS-1 signal, which consists of 24 PCM voice chan-
nels, multiplexed using TDM. Each channel is sampled at 8 kHz, with 8 bits
per sample, as previously described. This gives a bit rate of 8 kb/s × 8 = 64 kb/s
for each voice channel.
The DS-1 signal consists of frames, each of which contains the bits repre-
senting one sample from each of the 24 channels. One extra bit, called the
framing bit, is added to each frame to help synchronize the transmitter and
receiver. Each frame contains 24 × 8 + 1 = 193 bits.
The samples must be transmitted at the same rate as they were obtained
in order for the signal to be reconstructed at the receiver without delay. This
requires the multiplexed signal to be sent at a rate of 8000 frames per second.
126 ! CHAPTER 4

Thus the bit rate is 193 × 8000 b/s = 1.544 Mb/s. See Figure 4.10 for an illus-
tration of a frame of a DS-1 signal.

FIGURE 4.10
DS-1 signal

Time-Division Time-division multiple access (TDMA) is like TDM except that it involves
Multiple Access signals originating at different points. The telephone system observed in the
previous section uses TDM because all the signals are combined at one point.
An example of TDMA is a digital cellular radio system where several signals
from mobile units are combined on one channel by assigning each a time
slot. TDMA systems are very similar in principle to TDM, but they tend to be
more complex to design. One complicating feature in TDMA radio systems
is: the propagation time for the signal from a mobile unit to a base station
varies with its distance to the base. We will look at the details later when we
discuss specific systems.

' 4.6 Spread-Spectrum Systems

As radio communication systems proliferate and traffic increases, interfer-
ence problems become more severe. Interference is nothing new, of course,
but it has been managed reasonably successfully in the past by careful regu-
latory control of transmitter locations, frequencies, and power levels. There
are some exceptions to this, however. Two examples are CB radio and cord-
less telephones. In fact, wherever government regulation of frequency use is
informal or nonexistent, interference is likely to become a serious problem.
Widespread use of such systems as cordless phones, wireless local-area net-
works, and wireless modems by millions of people obviously precludes the
tight regulation associated with services such as broadcasting, making both-
ersome interference almost inevitable. One approach to the problem, used
by cellular radio systems, is to employ a complex system of frequency reuse,
with computers choosing the best channel at any given time. However, this
system too implies strong central control, if not by government then by one
or more service providers, each having exclusive rights to certain radio chan-
nels. That is, in fact, the current situation with respect to cellular telephony,
 DIGITAL MODULATION ! 127

but it can cause problems where several widely different services use the
same frequency range. The 49-MHz band, for instance, is currently used by
cordless phones, baby monitors, remote controlled models, and various
other users in an almost completely unregulated way. Similarly, the 2.4-GHz
band is shared by wireless LANs, wireless modems, cordless phones—and
even microwave ovens!
Another problem with channelized communication, even when tightly
controlled, is that the number of channels is strictly limited. If all available
channels are in use in a given cell of a cellular phone system, the next at-
tempt to complete a call will be blocked, that is, the call will not go through.
Service does not degrade gracefully as traffic increases; rather, it continues as
normal until the traffic density reaches the limits of the system and then
ceases altogether for new calls.
There is a way to reduce interference that does not require strong central
control. That technique, known as spread-spectrum communication, has
been used for some time in military applications where interference often
consists of deliberate jamming of signals. This interference, of course, is
not under the control of the communicator, nor is it subject to government
regulation.
Military communication systems need to avoid unauthorized eavesdrop-
ping on confidential transmissions, a problem alleviated by the use of spread-
spectrum techniques. Privacy is also a concern for personal communication
systems, but many current analog systems, such as cordless and cellular tele-
phone systems, have nonexistent or very poor protection of privacy.
For these reasons, and because the availability of large-scale integrated
circuits has reduced the costs involved, there has recently been a great deal
of interest in the use of spread-spectrum technology in personal communi-
cation systems for both voice and data.
The basic idea in spread-spectrum systems is, as the name implies, to
spread the signal over a much wider portion of the spectrum than usual. A
simple audio signal that would normally occupy only a few kilohertz of spec-
trum can be expanded to cover many megahertz. Thus only a small portion
of the signal is likely to be masked by any interfering signal. Of course, the
average power density, expressed in watts per hertz of bandwidth, is also re-
duced, and this often results in a signal-to-noise ratio of less than one (that
is, the signal power in any given frequency range is less than the noise power
in the same bandwidth).
It may seem at first glance that this would make the signal almost impos-
sible to detect, which is true unless special techniques are used to
“de-spread” the signal while at the same time spreading the energy from in-
terfering signals. In fact, the low average power density of spread-spectrum
signals is responsible for their relative immunity from both interference and
eavesdropping.
128 ! CHAPTER 4

EXAMPLE 4.6 Y
A voice transmission occupies a channel 30 kHz wide. Suppose a
spread-spectrum system is used to increase its bandwidth to 10 MHz. If the
signal has a total signal power of −110 dBm at the receiver input and the sys-
tem noise temperature referred to the same point is 300 K, calculate the
signal-to-noise ratio for both systems.

SOLUTION
Recall from Chapter 1 that thermal noise power is given by

PN = kTB
where
PN = noise power in watts
k = Boltzmann’s constant: 1.38 × 10−23 joules/kelvin (J/K)
T = absolute temperature in kelvins
B = noise power bandwidth in hertz
In general, the noise power bandwidth for a system will be approxi-
mately equal to the receiver bandwidth. For the signal with a bandwidth of
30 kHz and a noise temperature of 300 K,

PN (30 kHz) = 1.38 × 10−23 J/K × 300 K × 30 × 103 Hz

= 124 × 10−18 W
= −129 dBm

When the signal bandwidth increases to 10 MHz, the signal is spread

over a much wider region of the spectrum, and a receiver designed to receive
the whole signal bandwidth would need a bandwidth of 10 MHz as well. It
would receive a noise power equal to

PN(10 MHz) = 1.38 × 10−23 J/K × 300 K × 10 × 106 Hz

= 41.4 × 10−15 W
= −104 dBm

With both signal and noise in dBm, we can subtract to get the signal-to-noise
ratio.
For the 30 kHz bandwidth,

S/N = −110 dBm − (−129 dBm) = 19 dB

For the 10 MHz bandwidth,

S/N = −110 dBm − (−104 dBm) = −6 dB

X
 DIGITAL MODULATION ! 129

Spread-spectrum communication is especially effective in a portable or

mobile environment. Cancellation due to the reflection of signals often
causes very deep fades, called Rayleigh fading, over a narrow frequency
range. When one of these fades happens to coincide with the frequency in
use, the signal can be lost completely. With spread spectrum, only a small
part of the communication will be lost, and this can usually be made up, at
least in digital schemes, by the use of error-correcting codes.
There are two important types of spread-spectrum systems. They are
known as frequency hopping and direct sequence.

Frequency- Frequency hopping is the simpler of the two spread-spectrum techniques.

Hopping Systems A frequency synthesizer is used to generate a carrier in the ordinary way.
There is one difference, however: instead of operating at a fixed frequency,
the synthesizer changes frequency many times per second according to
a preprogrammed sequence of channels. This sequence is known as a
pseudo-random noise (PN) sequence because, to an outside observer who
has not been given the sequence, the transmitted frequency appears to hop
about in a completely random and unpredictable fashion. In reality, the
sequence is not random at all, and a receiver which has been programmed
with the same sequence can easily follow the transmitter as it hops and the
message can be decoded normally.
Since the frequency-hopping signal typically spends only a few millisec-
onds or less on each channel, any interference to it from a signal on that fre-
quency will be of short duration. If an analog modulation scheme is used
for voice, the interference will appear as a click and may pass unnoticed. If
the spread-spectrum signal is modulated using digital techniques, an error-
correcting code can be employed that will allow these brief interruptions in
the received signal to be ignored, and the user will probably not experience
any signal degradation at all. Thus reliable communication can be achieved
in spite of interference.

EXAMPLE 4.7 Y
A frequency-hopping spread-spectrum system hops to each of 100 frequen-
cies every ten seconds. How long does it spend on each frequency?

SOLUTION
The amount of time spent on each frequency is

t = 10 seconds/100 hops
= 0.1 second per hop
X
130 ! CHAPTER 4

If the frequency band used by the spread-spectrum system contains

known sources of interference, such as carriers from other types of service,
the frequency-hopping scheme can be designed to avoid these frequencies
entirely. Otherwise, the communication system will degrade gracefully as
the number of interfering signals increases, since each new signal will sim-
ply increase the noise level slightly.

Direct-Sequence The direct-sequence form of spread-spectrum communication is commonly

Systems used with digital modulation schemes. The idea is to modulate the transmit-
ter with a bit stream consisting of pseudo-random noise (PN) that has a
much higher rate than the actual data to be communicated. The term
pseudo-random means that the bit stream appears at first glance to be a ran-
dom sequence of zeros and ones but is actually generated in such a way as to
repeat exactly from time to time. The data to be transmitted is combined
with the PN. One common technique is to invert all the bits of the PN stream
during the time the “real” data is represented by a one and to leave the PN bit
stream unchanged when a data zero is to be transmitted. The extra bits trans-
mitted in this way are called chips, and the resulting bit rate is known as the
chipping rate. Most direct-sequence spread-spectrum systems use a chip-
ping rate at least ten times as great as the bit rate of the actual information to
be transmitted.
The use of the high-speed PN sequence results in an increase in the band-
width of the signal, regardless of what modulation scheme is used to encode
the bits into the signal. Recall from Hartley’s Law (Equation 4.1) that for any
given modulation scheme, the bandwidth is proportional to the bit rate. It
follows from Hartley’s Law that a direct-sequence system transmitting a to-
tal of ten bits for each information bit will use ten times as much bandwidth
as a narrowband signal with the same type of modulation and the same in-
formation rate. That is, the sidebands will extend ten times as far from the
carrier, as illustrated in Figure 4.11. Direct-sequence spread-spectrum
schemes typically use some form of phase-shift keying (PSK).

FIGURE 4.11 Direct-sequence spread-spectrum signal

 DIGITAL MODULATION ! 131

EXAMPLE 4.8 Y
A digital communication scheme uses DQPSK. It is to transmit a compressed
PCM audio signal which has a bit rate of 16 kb/s. The chipping rate is 10
to 1. Calculate the number of signal changes (symbols) which must be trans-
mitted each second.

SOLUTION
The total bit rate, including the chips, is 10 times the data rate, or 160 kb/s.
Since there are four signal states, each state represents two bits. Therefore
the symbol rate is

160/2 = 80 kilobaud
X
Expanding the bandwidth by a factor of ten while keeping the transmit-
ted power constant will decrease the received signal-to-noise ratio by
the same factor. As before, the pseudo-random sequence is known to the re-
ceiver, which has to separate the information signal from the chips. A pro-
cessing gain, also called spreading gain, can be defined equal to the
bandwidth expansion (which, for direct-sequence spread spectrum, is also
equal to the ratio of chips to information bits):
BRF
Gp = (4.7)
BBB

where
G p = processing gain
B RF = RF (transmitted) bandwidth
B BB = baseband (before spreading) bandwidth
The processing gain also describes the amount by which the signal-
to-noise ratio of the signal is reduced by the spreading process during trans-
mission. Of course, this reduction is reversed at the receiver. Since
signal-to-noise ratio is generally given in decibels, it would make sense to ex-
press the processing gain that way too, that is:
G p (dB) = 10 log G p
G p (dB) = (S/N) i (dB) − (S/N) o (dB) (4.8)
where
G p (dB) = processing gain in decibels
(S/N) i (dB) = signal-to-noise ratio in dB before spreading
(S/N) o (dB) = signal-to-noise ratio in dB after spreading
132 ! CHAPTER 4

EXAMPLE 4.9 Y
A signal would have a bandwidth of 200 kHz and a signal-to-noise ratio of
20 dB if transmitted without spreading. It is spread using a chipping rate
of 50:1. Calculate its bandwidth and signal-to-noise ratio after spreading.

SOLUTION
The bandwidth after spreading can be found from Equation (4.7):
BRF
Gp =
BBB
BRF = G p BBB
= 50 × 200 kHz
= 10 MHz

The processing gain in decibels is

Gp (dB) = 10 log Gp
= 10 log 50
= 17 dB

The signal-to-noise ratio after spreading is given by Equation (4.8):

Gp (dB) = (S/N)i (dB) − (S/N)o (dB)
(S/N)o (dB) = (S/N)i (dB) − Gp (dB)
= 20 dB − 17 dB
= 3 dB
X

Reception of The type of receiver required for spread-spectrum reception depends on how
Spread-Spectrum the signal is generated. For frequency-hopped transmissions, what is needed is
Signals a relatively conventional narrowband receiver that hops in the same way as
and is synchronized with the transmitter. This requires that the receiver be
given the frequency-hopping sequence, and there be some form of synchroniz-
ing signal (such as the signal usually sent at the start of a data frame in digital
communication) to keep the transmitter and receiver synchronized. Some
means must also be provided to allow the receiver to detect the start of a trans-
mission, since, if this is left to chance, the transmitter and receiver will most
likely be on different frequencies when a transmission begins.
One way to synchronize the transmitter and receiver is to have the trans-
mitter send a tone on a prearranged channel at the start of each transmis-
sion, before it begins hopping. The receiver can synchronize by detecting
 DIGITAL MODULATION ! 133

the end of the tone and then begin hopping according to the prearranged PN
sequence. Of course, this method fails if there happens to be an interfering
signal on the designated synchronizing channel at the time synchronization
is attempted.
A more reliable method of synchronizing frequency-hopping systems is
for the transmitter to visit several channels in a prearranged order before be-
ginning a normal transmission. The receiver can monitor all of these chan-
nels sequentially, and once it detects the transmission, it can sample the
next channel in the sequence for verification and synchronization.
Direct-sequence spread-spectrum transmissions require different recep-
tion techniques. Narrowband receivers will not work with these signals,
which occupy a wide bandwidth on a continuous basis. A wideband receiver
is required, but a conventional wideband receiver would output only noise.
In order to distinguish the desired signal from noise and interfering signals,
which over the bandwidth of the receiver are much stronger than the desired
signal, a technique called autocorrelation is used. Essentially this involves
multiplying the received signal by a signal generated at the receiver from the
PN code. When the input signal corresponds to the PN code, the output from
the autocorrelator will be large; at other times this output will be very small.
Of course, once again the transmitter and receiver will probably not be syn-
chronized at the start of a transmission, so the transmitter sends a preamble
signal, which is a prearranged sequence of ones and zeros, to let the receiver
synchronize with the transmitter.

Code-Division Traditionally, most analog communication systems have used frequency-

Multiple Access division multiplexing, and digital systems have employed time-division
(CDMA) multiplexing, in order to combine many information signals into a single
transmission channel. When the signals originate from different sources,
these two methods (which have already been studied) become frequency-
division multiple access (FDMA) and time-division multiple access (TDMA),
respectively.
Spread-spectrum communication allows a third method for multi-
plexing signals from different sources—code-division multiple access
(CDMA). All that is required is for each transmitter to be assigned a different
pseudo-noise (PN) sequence. If possible, orthogonal sequences should
be chosen; that is, the transmitters should never be in the same place at
the same time. The PN sequence for the transmitter is given only to the re-
ceiver that is to operate with that transmitter. This receiver will then receive
only the correct transmissions, and all other receivers will ignore these sig-
nals. This technique, which is applicable to both frequency-hopping and
direct-sequence transmissions, allows many transmissions to share the same
spread-spectrum channel.
134 ! CHAPTER 4

It is not necessary for all transmitters to have orthogonal PN sequences

in order to use CDMA. If some of the sequences are not orthogonal, there
will be some mutual interference between users. However, the system can
still operate until there are so many users that the signal-to-noise ratio
becomes unacceptably low.
In order that all transmissions other than the desired one blend together
at the receiver as noise, it is highly desirable in a CDMA system that all re-
ceived signals have about the same strength at the receiver. Obviously this is
not always possible, such as in an unlicensed environment like that of cord-
less phones. But where it is possible, the benefits of close control over trans-
mitter power levels are great. We will see this idea applied later when we look
at CDMA PCS systems.

' Summary The main points to remember from this chapter are:

( Digital wireless transmission uses frequency, phase, and amplitude varia-

tions, just as does analog transmission.
( The maximum data rate for a channel is a function of bandwidth, modu-
lation scheme, and signal-to-noise ratio.
( In general, more complex modulation schemes can achieve higher data
rates, but only when the signal-to-noise ratio is high.
( Frequency-shift keying (FSK) uses two (and occasionally more than two)
transmitted frequencies to achieve modest data rates with good perfor-
mance in noisy channels.
( Gaussian minimum-shift keying is a special case of FSK that achieves the
minimum bandwidth possible for a two-frequency FSK system at a given
data rate.
( Most phase-shift keying (PSK) systems use four phase angles for some-
what higher data rates than are achievable with FSK.
( Quadrature amplitude modulation (QAM) achieves higher data rates than
FSK or PSK by using a combination of amplitude and phase modulation.
QAM requires a relatively noise-free channel to realize its advantages.
( Frequency-division multiplexing and multiple-access schemes divide
available bandwidth among channels, with each operating full-time.
( Time-division multiplexing and multiple-access schemes divide the avail-
able time among channels, with each using the full bandwidth.
( Spread-spectrum systems reduce interference by spreading each signal
over a large bandwidth.
 DIGITAL MODULATION ! 135

( Code-division multiple-access (CDMA) schemes assign all channels

full-time to the full available bandwidth. Channels are separated by
spreading and despreading codes.

( Equation List
I = ktB (4.1)

C = 2B log2 M (4.2)

C = B log2(1 + S/N) (4.3)

C = S log2 M (4.5)

ƒm − ƒs = 0.5 ƒb (4.6)

BRF
Gp = (4.7)
BBB

Gp (dB) = (S/N)i (dB) − (S/N)o (dB) (4.8)

( Key Terms
amplitude-shift keying (ASK) data transmission by varying the
amplitude of the transmitted signal
audio frequency-shift keying (AFSK) use of an audio tone of two or more
different frequencies to modulate a conventional analog transmitter
for data transmission
baud rate speed at which symbols are transmitted in a digital
communication system
bit rate speed at which data is transmitted in a digital communication
system
chips extra bits used to spread the signal in a direct-sequence
spread-spectrum system
code-division multiple access (CDMA) system to allow multiple users to
use the same frequency using separate PN codes and a
spread-spectrum modulation scheme
136 ! CHAPTER 4

constellation diagram in digital communication, a pattern showing all

the possible combinations of amplitude and phase for a signal
delta phase-shift keying (DPSK) digital modulation scheme that
represents a bit pattern by a change in phase from the previous state
dibit system any digital modulation scheme that codes two bits of
information per transmitted symbol
direct-sequence spread spectrum technique for increasing the
bandwidth of a transmitted signal by combining it with a
pseudo-random noise signal with a higher bit rate
frequency hopping form of spread-spectrum communication in which
the RF carrier continually moves from one frequency to another
according to a prearranged pseudo-random pattern
frequency-division multiple access (FDMA) sharing of a communication
channel among multiple users by assigning each a different carrier
frequency
frequency-division multiplexing (FDM) combining of several signals into
one communication channel by assigning each a different carrier
frequency
frequency-shift keying (FSK) digital modulation scheme using two or
more different output frequencies
Gaussian minimum-shift keying (GMSK) variant of FSK which uses the
minimum possible frequency shift for a given bit rate
mark in digital communication, a logic one
modem acronym for modulator-demodulator; device to enable data to
be transmitted via an analog channel
multiple access use of a single channel by more than one transmitter
multiplexing use of a single channel by more than one signal
phase-shift keying (PSK) digital modulation scheme in which the phase
of the transmitted signal is varied in accordance with the baseband
data signal
processing gain improvement in interference rejection due to spreading
in a spread-spectrum system
pseudo-random noise (PN) sequence a transmitted series of ones and
zeros that repeats after a set time, and which appears random if the
sequence is not known to the receiver
quadrature AM (QAM) modulation scheme in which both the amplitude
and phase of the transmitted signal are varied by the baseband signal
 DIGITAL MODULATION ! 137

quadrature phase-shift keying (QPSK or DQPSK) digital modulation

scheme using four different transmitted phase angles
Rayleigh fading variation in received signal strength due to multipath
propagation
space binary zero
spreading gain improvement in interference rejection due to spreading
in a spread-spectrum system
symbol in digital communication, the state of the signal at a sampling
time
time-division multiple access (TDMA) system to allow several
transmissions to use a single channel by assigning time slots to each
time-division multiplexing (TDM) system to combine several data
streams onto a single channel by assigning time slots to each

( Questions
1. What is the meaning of the term modem?
2. What parameters of a sine-wave carrier can be modulated?
3. Name the three most common basic types of digital modulation.
4. Which type of modulation is likely to be used for:
(a) low data rates
(b) moderate data rates
(c) high data rates
5. What signal parameters are varied with QAM?
6. What factors limit the maximum data rate for a channel?
7. What is an eye diagram?
8. Explain the difference between the terms bit rate and baud rate.
9. Explain the origin and meaning of the terms mark and space.
10. What is the difference between FSK, AFSK, and GMSK?
11. Why is delta phase-shift keying the most common form of PSK?
12. What is the advantage of π/4 DQPSK?
13. What is represented by the dots in a constellation diagram for a QAM
system?
14. Compare the modulation schemes studied in this chapter, listing as
many advantages and disadvantages for each as you can.
138 ! CHAPTER 4

15. List and describe the three multiple-access systems in common use.
16. What is a DS-1 signal?
17. Compare frequency-hopping and direct-sequence spread-spectrum
systems.
18. What happens when a call is blocked?
19. How do spread-spectrum systems reduce the effect of fading?
20. Briefly describe what is meant by orthogonal spread-spectrum signals.

( Problems
1. The North American analog cellular radio system uses FM with channels
30 kHz wide. Suppose such a channel were used for digital communica-
tion. If the available signal-to-noise ratio is 20 dB, calculate the maxi-
mum theoretical bit rate and the corresponding baud rate using:
(a) a two-level code
(b) a four-level code
2. How much bandwidth would be required to transmit a DS-1 signal
(1.544 Mb/s) using a four-level code:
(a) assuming a noiseless channel?
(b) with a signal-to-noise ratio of 15 dB?
3. The AFSK system described in the text operates at 1200 bits per second
using an FM signal modulated by tones at 1200 and 2200 Hz, with a fre-
quency deviation of 5 kHz. Calculate the efficiency of this system in
bits per second per hertz of bandwidth by using Carson’s rule (see Chap-
ter 2) to calculate the approximate radio frequency bandwidth for this
system. Is this system bandwidth-efficient?
4. A typical HF radioteletype system uses 170 Hz shift between mark and
space frequencies and a bit rate of 45 bits per second. What would be the
bit rate if GMSK were used for this system?
5. Consider a QPSK system that will transmit three bits of information per
symbol.
(a) How many phase angles are needed?
(b) Draw a vector diagram for such a system.
(c) Would this system have any advantages compared with the dibit
systems described in the text? Any disadvantages?
 DIGITAL MODULATION ! 139

6. A constellation diagram for a modem is shown in Figure 4.12.

FIGURE 4.12 (
(
(
(
(((( ((((
(
(
(
(

(a) What type of modulation is this?

(b) If the transmitted bit rate is 9600 b/s, what will be the baud rate us-
ing this modem?
7. A microwave radio system uses 256-QAM, that is, there are 256 possible
amplitude and phase combinations.
(a) How many bits per symbol does it use?
(b) If it has a channel with 40-MHz bandwidth, what is its maximum
data rate, ignoring noise?
8. Draw a constellation pattern for a modem that uses eight equally-
spaced phase angles and four equally-spaced amplitude levels. If the
modem operates at 4800 baud, what is its bit rate?
9. An ordinary broadcast television channel has a bandwidth of 6 MHz.
How many FM radio stations (with a bandwidth of 200 kHz each) could
be accommodated in one television channel using FDM?
10. Suppose the TV channel described in the previous problem is used for
voice-quality transmission using SSBSC AM with a bandwidth of 4 kHz
per channel. How many voice channels can be carried?
11. Suppose the TV channel described in the previous problem is used for
digital audio communication. Assume the RF system can support
2 b/s/Hz of RF bandwidth.
(a) What is the total available bit rate?
(b) How many telephone voice signals, at a bit rate of 64 kb/s each,
could be accommodated on this channel using TDM?
(c) How many high-quality stereo audio signals could be accommo-
dated? Assume 16-bit samples, with a sampling rate of 44.1 kSa/s,
and ignore error correction and data compression.
140 ! CHAPTER 4

12. Suppose that a voice signal normally occupies 30 kHz of bandwidth and
has a signal-to-noise ratio of 20 dB. Spread-spectrum techniques are
used to increase its bandwidth to 2 MHz.
(a) What is the signal-to-noise ratio of the spread signal?
(b) What is the processing gain, in decibels?
13. Suppose a frequency-hopping system hops among 500 channels. How
many orthogonal PN sequences are possible?
14. Suppose there is a narrowband analog signal on one of the channels vis-
ited by a frequency-hopping system.
(a) What is the effect on the narrowband signal of the spread-spectrum
signals?
(b) What is the effect on the spread-spectrum signals of the narrow-
band signal?
15. A direct-sequence spread-spectrum system uses FSK with a chipping rate
of 20 to 1. The signal-to-noise ratio for the spread signal is −5 dB (that is,
the signal is 5 dB weaker than the noise in the same bandwidth). If the
data is transmitted at 50 kb/s, calculate:
(a) the chipping rate
(b) the bandwidth occupied by the spread signal if the modulation
scheme used allows 1.5 bits/s/Hz
(c) the signal-to-noise ratio for the despread signal at the receiver
16. A signal has a bit rate of 20 kb/s. Find the baud rate if the signal is trans-
mitted using:
(a) FSK with two frequencies
(b) QPSK with four phase angles
(c) QAM with four phase angles and four amplitudes
17. Ten voice signals are to be multiplexed and transmitted. The analog sig-
nal occupies 4 kHz of bandwidth and can be digitized using a vocoder at
12 kb/s. Calculate the required bandwidth for each of the following pos-
sibilities.
(a) FDMA using analog FM with 12 kHz deviation (use Carson’s rule to
find the bandwidth for one signal). Ignore guard bands between
channels.
(b) FDMA using SSBSC AM. Ignore guard bands.
(c) TDM using GMSK. Assume a noiseless channel.
(d) TDMA using QPSK. Assume a noiseless channel.
(e) CDMA using frequency-hopping with 10 available channels. Use
GMSK and assume a noiseless channel.
(f) CDMA using direct-sequence, QPSK with a chipping rate of 10:1.