3.

1 Introduction

(3.1)

 Referring to Equation (3.1), if the information signal is digital and

the amplitude (V) of the carrier is varied proportional to the
information signal, a digitally modulated signal called amplitude
shift keying (ASK) is produced.

 If the frequency (f) is varied proportional to the information signal,

frequency shift keying (FSK) is produced, and if the phase of the
carrier (Ɵ) is varied proportional to the information signal, phase
shift keying (PSK) is produced.

 If both the amplitude and the phase are varied proportional to the
information signal, quadrature amplitude modulation (QAM)
results. ASK, FSK, PSK, and QAM are all forms of digital
modulation

FIGURE 3-1
Figure 3-1 shows a simplified block diagram for a digital modulation
system.

1
  In the transmitter, the precoder performs level conversion and
then encodes the incoming data into groups of bits that
modulate an analog carrier.

 The modulated carrier is shaped (filtered), amplified, and then

transmitted through the transmission medium to the receiver.

 The transmission medium can be a metallic cable, optical fiber

cable, Earth's atmosphere, or a combination of two or more
types of transmission systems.

 In the receiver, the incoming signals are filtered, amplified, and

then applied to the demodulator and decoder circuits, which
extracts the original source information from the modulated
carrier.

 The clock and carrier recovery circuits recover the analog

carrier and digital timing (clock) signals from the incoming
modulated wave since they are necessary to perform the de-
modulation process.

3.2.1 Information Capacity, Bits, and Bit Rate

I  B×t (3.2)

where I= information capacity (bits per second)

B = bandwidth (hertz)
t = transmission time (seconds)

 From Equation 3-2, it can be seen that information capacity is a

linear function, and is directly proportional to both bandwidth
and transmission time.

2
 If either the bandwidth or the transmission time changes, a
directly proportional change occurs in the information capacity.

 In 1948, Shannon stated in his paper, the information capacity is

related to bandwidth and signal-to-noise ratio (SNR), The higher
the (SNR), the better the performance and the higher the
information capacity.

 Mathematically stated, the Shannon limit for information

capacity is

(3.3)
or

(3.4)

Where, I = information capacity (bps)

B = bandwidth (hertz)
S
= signal-to-noise power ratio (unitless)
N

 For a standard telephone circuit with a signal-to-noise power

ratio of 1000 (30 dB) and a bandwidth of 2.7 kHz, the Shannon
limit for information capacity is

I = (3.32)(2700) log 10 (1 + 1000) = 26.9 kbps

Shannon's formula is often misunderstood. The results of the

preceding example indicate that 26.9 kbps can be propagated
through a 2.7-kHz communications channel. This may be true, but

3
it cannot be done with a binary system. To achieve an information
transmission rate of 26.9 kbps through a 2.7-kHz channel, each
symbol transmitted must contain more than one bit.

3.2.2 M-ary Encoding

 M-ary is a term derived from the word binary.

 M simply represents a digit that corresponds to the number of

conditions, levels, or combinations possible for a given
number of binary variables.

 For example, a digital signal with four possible conditions

(voltage levels, frequencies, phases, and so on) is an M-ary
system where M = 4. If there are eight possible conditions, M
= 8 and so forth.

The number of bits necessary to produce a given number of

conditions is expressed mathematically as

N  log2 M (3.5)

Where, N = number of bits necessary

M = number of conditions, levels, or combinations
possible with N bits

Equation 3.5 can be simplified and rearranged to express the

number of conditions possible with N bits as

2 N =M (3.6)
For example, with one bit, only 2 1 = 2 conditions are possible.
With two bits, 2 2 = 4 conditions are possible, with three bits, 2 3 = 8
conditions are possible, and so on.

4
3.2.3 Baud and Minimum Bandwidth

 Baud refers to the rate of change of a signal on the transmission

medium after encoding and modulation have occurred.

 Hence, baud is a unit of transmission rate, modulation rate, or

symbol rate and, therefore, the terms symbols per second and baud
are often used interchangeably. (bit rate is a rate of change of
information)

 Mathematically, baud is the reciprocal of the time of one output

signaling element, and a signaling element may represent several
information bits. Baud is expressed as

1
baud = t (3.7)
s

where baud = symbol rate (baud per second)

ts = time of one signaling element (seconds)

 Signaling element is sometimes called a symbol and could be

encoded as a change in the amplitude, frequency or phase.

 Baud rate represent more than one bit and usually less than bit rate
(QPSK, 8-PSK). For binary FSK, PSK, baud rate =bit rate.

 According to Nyquist, the binary digital signals can be propagated

through an ideal noiseless transmission medium at a rate equal to
two times the bandwidth.

 Thus, fb = 2B, where fb is the bit rate in bps and B is the ideal
Nyquist bandwidth.

 The relationship between bandwidth and bit rate also applies to the
opposite situation. For a given bandwidth (B), the highest
theoretical bit rate is 2B.

5
 For example, a standard telephone circuit has a bandwidth of
approximately 2700 Hz, which has the capacity to propagate 5400
bps through it. However, if more than two levels are used for
signaling (higher-than-binary encoding), more than one bit may be
transmitted at a time, and it is possible to propagate a bit rate that
exceeds 2B.

 For multilevel signaling, the Nyquist formulation for channel

capacity is
fb = B log2 M (3.8)

Where, fb = channel capacity (bps)

B = minimum Nyquist bandwidth (hertz)
M = number of discrete signal or voltage levels

Equation 3.8 can be rearranged to solve for the minimum

bandwidth necessary to pass M-ary digitally modulated carriers
 fb 
B =   (3.9)
 log 2 M 

If N is substituted for log2 M, Equation 3.9 reduces to

 fb 
B = N  (3.10)
 
 Where N is the number of bits encoded into each signaling
element.

 In addition, since baud is the encoded rate of change, it also

equals the bit rate divided by the number of bits encoded into
one signaling element. Thus,

 f 
baud =  N 
b
(3.11)
 

6
 By comparing Equation 3.10 with Equation 3.11 the baud and the
ideal minimum Nyquist bandwidth have the same value and are
equal to the bit rate divided by the number of bits encoded.

3-3 AMPLITUDE-SHIFT KEYING (ASK)

 Simplest digital modulation technique, where a binary
information signal directly modulates the amplitude of an
analog carrier.

 ASK is similar to standard amplitude modulation (AM) except

there are only two output amplitudes possible. Amplitude-shift
keying is sometimes called digital amplitude modulation
(DAM).
 Mathematically, ASK is

(3.12)
Where,
vask(t) = amplitude-shift keying wave
vm(t) = digital information (modulating) signal (volts)
A/2 = unmodulated carrier amplitude (volts)
ωc = analog carrier radian frequency (radians per second, 2πfc)

 In Equation 3.12, the modulating signal vm(t) is a normalized

binary waveform, where + 1 V = logic 1 and -1 V = logic 0.
 for a logic 1, vm(t) = + 1 V, Equation 3.12 reduces to

 for a logic 0 , vm(t) = -1 V, Equation 3.12 reduces to

=0

7
 Thus, the modulated wave vask(t), is either A cos(ωc t) or 0. Hence,
the carrier is either "on"or "off," which is why amplitude-shift
keying is sometimes referred to as on-off keying (OOK).

 Figure 3-2 shows the input and output waveforms from an ASK
modulator.

FIGURE 3-2 Digital amplitude modulation: (a) input binary; (b)

output DAM waveform

 for every change in the input binary data there is one change in
the ASK waveform, and the time of one bit (t b) equals the time of
one analog signaling element (t,).
Substituting N=1 in 3.10 and in 3.11

B = fb /1 = fb baud = fb /1 = fb

 baud = B = bit rate

Example 3-1

Determine the baud and minimum bandwidth necessary to pass a 10 kbps

binary signal using amplitude shift keying.
Solution

8
For ASK, N = 1, and the baud and minimum bandwidth are
determined from Equations 3.11 and 3.10, respectively:

B = 10,000 / 1 = 10,000 Hz

baud = 10, 000 /1 = 10,000

The use of amplitude-modulated analog carriers to transport digital

information is a relatively low-quality, low-cost type of digital
modulation and, therefore, is seldom used except for very low speed
telemetry circuits.

3-4 FREQUENCY-SHIFT KEYING

 FSK is a form of constant-amplitude angle modulation similar
to standard FM except the modulating signal is a binary signal
that varies between two discrete voltage levels rather than a
continuously changing analog waveform.

 Consequently, FSK is sometimes called binary FSK (BFSK).

The general expression for FSK is

(3.13)
Where,
vfsk(t) = binary FSK waveform
Vc = peak analog carrier amplitude (volts)
fc = analog carrier center frequency (hertz)
Δf = peak change (shift) in the analog carrier frequency (hertz)
vm(t) = binary input (modulating) signal (volts)

 peak shift in the carrier frequency (Δf) is proportional to the

binary input signal (vm[t]), and the direction of the shift is
determined by the polarity.

9
 for a logic l input, vm(t) = + 1, Equation 3.13 can be rewritten as

 For a logic 0 input, vm(t) = -1, Equation 3.13 becomes

 With binary FSK, the carrier center frequency (fc) is shifted by the
binary input signal as shown in Figure 3-3.

FIGURE 3-3 FSK in the frequency domain

 As the binary input signal changes from a logic 0 to a logic 1 and

vice versa, the output frequency shifts between two frequencies: a
mark, or logic 1 frequency (fm), and a space, or logic 0 frequency
(fs).

 The mark and space frequencies are separated from the carrier
frequency by the peak frequency deviation (Δf) and from each
other by 2 Δf.

 Frequency deviation is mathematically expressed as

Δf = |fm – fs| / 2 (3.14)

10
 where Δf = frequency deviation (hertz)
|fm – fs| = absolute difference between the mark and space
frequencies (hertz)

FIGURE 3-4 FSK in the time domain: (a) waveform: (b) truth table

 mark frequency is the higher frequency (fc + Δf)

 space frequency is the lower frequency (fc - Δf)
 The truth table shows the input and output possibilities for a
given digital modulation scheme.

3-4-1 FSK Bit Rate, Baud, and Bandwidth

 In Figure 3-4, the time of one bit (tb) is the same as the time the FSK
output is a mark or space frequency (ts).
 bit rate equals the baud.

The baud for binary FSK can also be determined by substituting

N = 1 in Equation 3.11,
baud = fb / 1 = fb
FSK is the exception to the rule for digital modulation, the
minimum bandwidth not determined from 3.10, thus

B = |(fs – fb) – (fm – fb)|

11
 = |(fs– fm)| + 2fb

 |(fs– fm)| = 2Δf, the minimum bandwidth can be approximated as

B = 2(Δ f + f b ) (3.15)

where

B = minimum Nyquist bandwidth (hertz)

Δ f = frequency deviation |(fm– fs)| (hertz)
fb = input bit rate (bps)

Example 3-2

Determine (a) the peak frequency deviation, (b) minimum bandwidth,

and (c) baud for a binary FSK signal with a mark frequency of 49 kHz, a
space frequency of 51 kHz, and an input bit rate of 2 kbps.

Solution

a. The peak frequency deviation is determined from Equation 3.14:

Δf = |49 kHz - 51 kHz| / 2 =1 kHz

b. The minimum bandwidth is determined from Equation 3.15:

B = 2(1000 + 2000)
= 6 kHz
c. For FSK, N = 1, and the baud is determined from Equation 3.11 as

baud = 2000 / 1 = 2000

Bessel functions can also be used to determine the approximate

bandwidth for an FSK wave.

12
3-4-2 FSK Transmitter

 Figure 3-5 shows a simplified binary FSK modulator,

which is very similar to a conventional FM modulator and
is very often a voltage-controlled oscillator (VCO).

 The center frequency (fc) is chosen such that it falls halfway

between the mark and space frequencies.

FIGURE 3-5 FSK modulator

 A logic 1 input shifts the VCO output to the mark frequency,

and a logic 0 input shifts the VCO output to the space
frequency.

 A VCO-FSK modulator can be operated in the sweep mode

where the peak frequency deviation is simply the product of the
binary input voltage and the deviation sensitivity of the VCO.

Frequency deviation is expressed mathematically as

Δf = vm(t) kl (3-19)

vm(t) = peak binary modulating-signal voltage (volts)

kl = deviation sensitivity (hertz per volt).

13
3-4-3 FSK Receiver

Noncoherent FSK demodulator - is quite simple with a circuit such as

the one shown in Figure 3-6.

FIGURE 3-6 Noncoherent FSK demodulator

The FSK input signal is simultaneously applied to the inputs of both

bandpass filters (BPFs) through a power splitter.

 The respective filter passes only the mark or only the space fre-
quency on to its respective envelope detector.

 The envelope detectors, in turn, indicate the total power in each

passband, and the comparator responds to the largest of the two
powers.

Coherent FSK demodulator

 The incoming FSK signal is multiplied by a recovered carrier signal

that has the exact frequency and phase as the transmitter reference.

 However, the two transmitted frequencies (the mark and space

frequencies) are not generally continuous; it is not practical to
reproduce a local reference that is coherent with both of them.
Consequently, coherent FSK detection is seldom used.

14
FIGURE 3-7 Coherent FSK demodulator

PLL-FSK demodulator
 The most common circuit used for demodulating binary FSK.

FIGURE 3-8 PLL-FSK demodulator

 Input to the PLL shifts between the mark and space frequencies,
produces the dc error voltage.

 Mark and space frequencies shifts the error voltages to only two
levels, one represents a logic 1 and the other a logic 0.

 Binary FSK has a poorer error performance than PSK or QAM and,
consequently, is seldom used for high-performance digital radio
systems.

 Its use is restricted to low-performance, low-cost, asynchronous data

modems that are used for data communications over analog, voice-
band telephone lines.

15
3-4-4 Continuous-Phase Frequency-Shift Keying (CP-FSK)

 Mark and space frequencies are synchronized with the input binary
bit rate.

 Mark and space frequencies are selected such that they are separated
from the center frequency by an exact multiple of one-half the bit
rate (fm and fs = n[f b / 2]), where n = any integer).

This ensures a smooth phase transition in the analog output signal when
it changes from a mark to a space frequency or vice versa.

Figure 3-10 shows a noncontinuous FSK waveform.

FIGURE 3-10 Noncontinuous FSK waveform

FIGURE 3-11 Continuous-phase MSK waveform

 CP-FSK has a better bit-error performance than conventional

binary FSK for a given signal-to-noise ratio.

 The disadvantage of CP-FSK is that it requires synchro-

nization circuits and is, therefore, more expensive to implement.

16
3-5 PHASE-SHIFT KEYING

Phase-shift keying (PSK) is another form of angle-modulated,

constant-amplitude digital modulation.

3-5-1 Binary Phase-Shift Keying (BPSK)

 In (BPSK), N = 1 and M = 2.

 Therefore, with BPSK, two phases (2 1 = 2) are possible for the

carrier.

 As the input digital signal changes state, the phase of the output
carrier shifts between two angles that are separated by 180°.

 other names for BPSK are phase reversal keying (PRK) and
biphase modulation.

 BPSK is a form of square-wave modulation of a continuous

wave (CW) signal.

3-5-1-1 BPSK transmitter.

FIGURE 3-12 BPSK transmitter

17
  The balanced modulator acts as a phase reversing switch.
 Depending on the input it changes the phase of the carrier, in
phase or 180° out of phase with the
reference.

FIGURE 3-13 (a) Balanced ring modulator; (b) logic 1; (c) logic 0

18
FIGURE 3-14 BPSK modulator: (a) truth table; (b) phasor diagram;
(c) constellation diagram

3-5-1-2 Bandwidth considerations: BPSK

 In a BPSK modulator, each time the input logic condition changes,

the output phase changes.
 The fundamental frequency (fa) of an alternative 1/0 bit sequence is
equal to (fb/2)
Mathematically, the output of a BPSK modulator is proportional to

BPSK output = [sin (2πfat)] x [sin (2πfct)] (3.20)

where
fa = maximum fundamental frequency of binary input (hertz)
fc = reference carrier frequency (hertz)

19
 Using trig identity,
½ cos [2π (fc – fa)t] – ½ cos [2π(fc + fa)t]
Thus, the minimum double-sided Nyquist bandwidth (B) is

B= (fc + fa) - (fc - fa) = 2fa

and because fa = fb / 2, where fb = input bit rate,

B= 2fb/2 = fb

where B is the minimum double-sided Nyquist bandwidth.

Figure 3-15 shows the output phase-versus-time relationship for a

BPSK waveform.

FIGURE 3-15

 Logic 1 produces an analog output signal with a 0° phase angle,

and a logic 0 produces an output signal with a 180° phase angle.

 BPSK signaling element (ts) is equal to the time of one

information bit (tb), which indicates that the bit rate equals the
baud.

20
Example 3-4

For a BPSK modulator with a carrier frequency of 70 MHz and an input

bit rate of 10 Mbps, determine the maximum and minimum upper and
lower side frequencies, draw the output spectrum, determine the minimum
Nyquist bandwidth, and calculate the baud.
Solution
Substituting into Equation 3-20 yields
output = [sin (2πfat)] x [sin (2πfct)] ; fa = fb / 2 = 5 MHz

= [sin 2π(5MHz)t)] x [sin 2π(70MHz)t)]

= 0.5cos [2π(70MHz – 5MHz)t] – 0.5cos[2π(70MHz + 5MHz)t]

lower side frequency upper side frequency

Minimum lower side frequency (LSF):

LSF=70MHz - 5MHz = 65MHz

Maximum upper side frequency (USF):

USF = 70 MHz + 5 MHz = 75 MHz

Therefore, the output spectrum for the worst-case binary input conditions
is as follows: The minimum Nyquist bandwidth (B) is

B = 75 MHz - 65 MHz = 10 MHz

and the baud = fb or 10 megabaud.

21
3-5-1-3 BPSK receiver.

FIGURE 3-16 Block diagram of a BPSK receiver

 The input signal will be either + sin ω ct or - sin ωct .
 The coherent carrier recovery circuit detects and regenerates a
carrier signal that is both frequency and phase coherent with the
original transmit carrier.
 The balanced modulator output is the product of the two inputs (the
BPSK signal and the recovered carrier).
 The low-pass filter (LPF) operates the recovered binary data from
the complex demodulated signal.

Mathematically, the demodulation process is as follows.

 For a BPSK input signal of + sin ωct (logic 1), the output of the
balanced modulator is

output = (sin ω ct )(sin ωct) = sin 2ωct (3.21)

or

sin2ωct = 0.5(1 – cos 2ωct) = 0.5 - 0.5cos 2ωct

filtered out

22
leaving
output = + 0.5 V = logic 1
 The LPF has a cutoff frequency much lower than 2 ωct, and, thus,
blocks the second harmonic of the carrier and passes only the
positive constant component.

 For a BPSK input signal of -sin ωct (logic 0), the output of the
balanced modulator is

output = (-sin ω ct )(sin ωct) = sin 2ωct

or

sin2ωct = -0.5(1 – cos 2ωct) = -0.5 + 0.5cos 2ω ct

filtered out

leaving
output = - 0.5 V = logic 0

3.5.2 Quaternary Phase Shift Keying (QPSK)

 M-ary encoding scheme with N=2 and M=4, hence the name
“quaternary” meaning “4”.
 Four possible output phase changes (-135º, -45º, +135º, +45º,)
corresponds to input dibit (two bits) 00, 01, 10, 11.

 For each dibit clocked into the modulator, a single output

change occurs.
 Rate of change at the output is equal to one half of the rate of
change of the input.

23
 3.5.2.1 QPSK Transmitter

FIGURE 3-17 QPSK modulator

 Two bits (a dibit) are clocked into the bit splitter. After both bits
have been serially inputted, they are simultaneously parallel
outputted.

 The I bit modulates a carrier that is in phase with the reference

oscillator (hence the name "I" for "in phase" channel), and the Q
bit modulate, a carrier that is 90° out of phase.

 For a logic 1 = + 1 V and a logic 0= - 1 V, two phases are

possible at the output of the I balanced modulator (+sin ωct and -
sin ωct), and two phases are possible at the output of the Q
balanced modulator (+cos ωct), and (-cos ωct).

 When the linear summer combines the two quadrature (90° out
of phase) signals, there are four possible resultant phasors given
by these expressions: + sin ωct + cos ωct, + sin ωct - cos ωct, -
sin ωct + cos ωct, and -sin ωct - cos ωct.

24
Truth Table, Phasor Diagram, And Constellation Diagram

 For a binary data input of Q = 0 and I= 0, the two inputs to the I

balanced modulator are -1 and sin ωct, and the two inputs to the Q
balanced modulator are -1 and cos ωct.

 Consequently, the outputs are

I balanced modulator = (-1)(sin ωct) = -1 sin ωct
Q balanced modulator = (-1)(cos ωct) = -1 cos ωct

 Output of the linear summer is

-1 cos ωct - 1 sin ωct = 1.414 sin(ωct - 135°)

For the remaining dibit codes (01, 10, and 11), the procedure is the
same. The results are shown in Figure 3-18a.

FIGURE 3-18 QPSK modulator: (a) truth table; (b) phasor diagram;
(c) constellation diagram

25
  The angular separation between any two adjacent phasors in
QPSK is 90°.

 Undergo almost a+45° or -45° shift in phase during

transmission and still retain the correct encoded information
when demodulated at the receiver.

FIGURE 3-19 Output phase-versus-time relationship for a PSK

modulator.

3-5-2-2 Bandwidth considerations of QPSK

 Since, the input data are divided into two channels, the bit rate in
either the I or the Q channel is equal to one-half of the input data
rate (fb/2)

 The highest fundamental frequency present in the balanced

modulator is (one-half of fb/2 = fb/4).

FIGURE 3-20 Bandwidth considerations of a QPSK modulator

26
 The worse-case input condition to the I or Q balanced
modulator is an alternative 1/0 pattern, which occurs when the
binary input data have a 1100 repetitive pattern. One cycle of
the fastest binary transition (a 1/0 sequence in the I or Q
channel takes the same time as four input data bits.

 The output of the balanced modulators can be expressed

mathematically as

(2.22)
where

 The output frequency spectrum extends from f'c + fb / 4 to f'c -

fb / 4 and the minimum bandwidth (fN) is

27
Example 3-5
For a QPSK modulator with an input data rate (fb) equal to 10 Mbps and
a carrier frequency 70 MHz, determine the minimum double-sided
Nyquist bandwidth (fN) and the baud. Also, compare the results with
those achieved with the BPSK modulator in Example 3-4. Use the
QPSK block diagram shown in Figure 3-17 as the modulator model.

Solution
The bit rate in both the I and Q channels is equal to one-half of the
transmission bit rate, or
fbQ = fb1 = fb / 2 = 10 Mbps / 2 = 5 Mbps

The highest fundamental frequency presented to either balanced

modulator is
fa= fbQ / 2 = 5 Mbps / 2 = 2.5 MHz

The output wave from each balanced modulator is

(sin 2πfat)(sin 2πfct)

0.5 cos 2π(fc – fa)t – 0.5 cos 2π(fc + fa)t

0.5 cos 2π[(70 – 2.5)MHz]t – 0.5 cos 2π[(70 – 2.5)MHz]t

0.5 cos 2π(67.5MHz)t - 0.5 cos 2π(72.5MHz)t

The minimum Nyquist bandwidth is

B=(72.5-67.5)MHz = 5MHz

The symbol rate equals the bandwidth: thus,

symbol rate = 5 megabaud

28
The output spectrum is as follows:

Comparing with BPSK modulator in Example 3-4 for the same input bit
rate the minimum bandwidth required for the QPSK modulator is equal
to one-half.
The minimum bandwidth for the QPSK system described in
Example 3-5 can also be determined by simply substituting into
Equation 3-10:

B = 10 Mbps / 2 = 5 MHz

3-5-2-3 (QPSK receiver)

FIGURE 3-21 QPSK receiver

The power splitter directs the input QPSK signal to the I and Q
product detectors and the carrier recovery circuit.

The carrier recovery circuit reproduces the original transmit carrier

oscillator signal. The recovered carrier must be frequency and

29
phase coherent with the transmit reference carrier.

The QPSK signal is demodulated in the I and Q product detectors,

which generate the original I and Q data bits. The outputs of the
product detectors are fed to the bit combining circuit, where they
are converted from parallel I and Q data channels to a single binary
output data stream.

 Let the incoming QPSK signal be -sin ωct + cos ωct.

Mathematically, the demodulation process is as follows.

 The inputs to the I product detector are (-sin ωct + cos ωct) and
(sin ωct). The output of the I product detector is

(2.23)

 The inputs to the Q product detector are (-sin ωct + cos ωct) and
(cos ωct). The output of the Q product detector is

30
 (2.24)

The demodulated I and Q bits (0 and 1, respectively) correspond to

the constellation diagram and truth table for the QPSK modulator

31