DC_Module1.pptx fundamentals of digital communication

Digital Communication (21EC51)
Dr. Nataraj Vijapur
Associate Professor, Dept. of ECE
RVITM, Bengaluru
natarajv.rvitm@rvei.edu.in

MODULE 1
Digital Modulation Techniques

Phase-Shift Keying Techniques Using Coherent Detection
COMMUNICATION CHANNELS:
In a binary PSK system, the pair of signals s1(t) and s2(t) used to represent binary symbols 1 and 0, respectively, is
defined by
…………...(3.1)
where Tb is the bit duration and Eb is the transmitted signal energy per bit. We find it convenient, although not
necessary, to assume that each transmitted bit contains an integral number of cycles of the carrier wave; that is, the
carrier frequency fc is chosen equal to nc/Tb for some fixed integer nc. A pair of sinusoidal waves that differ only in a
relative phase-shift of 180°, defined in (3.1) and (3.2), is referred to as an antipodal signal.

Signal-Space Diagram of Binary PSK Signals
From this pair of equations it is clear that, in the case of binary PSK, there is only one
basis function of unit energy:
……………(3.3)
Then, we may respectively express the transmitted signals s1(t) and s2(t) in terms of (t) as
………………(3.4)
……………….(3.5)
A binary PSK system is, therefore, characterized by having a signal space that is one-dimensional (i.e., N = 1), with a
signal constellation consisting of two message points (i.e., M = 2). The respective coordinates of the two message points
are

…………………(3.6)
………………….(3.7)
In words, the message point corresponding to s1(t) is located at and the message point corresponding to s2(t) is located
at . Figure 3.1a displays the signal-space diagram for binary PSK and Figure 3.1b shows example waveforms of
antipodal signals representing s1(t) and s2(t). Note that the binary constellation of Figure 3.1 has minimum average
energy.

Fig 3.1 : (a) Signal-space diagram for coherent binary PSK system. (b) The waveforms depicting the transmitted
signals s1(t) and s2(t), assuming nc = 2.

Error Probability of Binary PSK Using Coherent Detection
To make an optimum decision on the received signal x(t) in favor of symbol 1 or symbol 0 (i.e., estimate the original
binary sequence at the transmitter input), we assume that the receiver has access to a locally generated replica of the
basis function 1(t).
We may identify two basic components in the binary PSK receiver:
1. Correlator, which correlates the received signal x(t) with the basis function 1(t) on a bit-by-bit basis.
2. Decision device, which compares the correlator output against a zero-threshold, assuming that binary symbols 1 and 0
are equiprobable. If the threshold is exceeded, a decision is made in favor of symbol 1; if not, the decision is made in
favor of symbol 0. Equality of the correlator with the zero-threshold is decided by the toss of a fair coin (i.e., in a random
manner). With coherent detection in place, Specifically, we partition the signal space of Figure 3.1 into two regions:
 the set of points closest to message point 1 at ;and
 the set of points closest to message point 2 at - .

The decision rule is now simply to decide that signal s1(t) (i.e., binary symbol 1) was transmitted if the received signal
point falls in region Z1 and to decide that signal s2(t) (i.e., binary symbol 0) was transmitted if the received signal point
falls in region Z2. Two kinds of erroneous decisions may, however, be made:
1. Error of the first kind. Signal s2(t) is transmitted but the noise is such that the received signal point falls inside
region Z1; so the receiver decides in favor of signal s1(t).
2. Error of the second kind. Signal s1(t) is transmitted but the noise is such that the received signal point falls inside
region Z2; so the receiver decides in favor of signal s2(t).

Figure3.2 Block diagrams for (a) binary PSK transmitter and (b) coherent binary PSK receiver.
To calculate the probability of making an error of the first kind, we note from Figure 3.1 a that the decision region
associated with symbol 1 or signal s1(t) is described by
where the observable element x1 is related to the received signal x(t) by
……………..3.8

…..…....3.9
Using (3.7) in this equation yields
………….3.10
The conditional probability of the receiver deciding in favor of symbol 1, given that symbol 0 was transmitted, is
therefore
…………………..(3.11)
Putting
………………..(3.12)
and changing the variable of integration from x1 to z, we may compactly rewrite (3.11) in terms of the Q-
function:
……………..(3.13)
Using the formula of Q-function in (3.13) we get
…………………(3.14) and ………………..(3.15)

As with binary PSK, information about the message symbols in QPSK is contained in the carrier phase. In particular, the
phase of the carrier takes on one of four equally spaced values, such as /4, 3 /4, 5 /4, and 7 /4. For this set of values, we
may define the transmitted signal as
QPSK Waveforms
Figure 3.4 illustrates the sequences and waveforms involved in the generation of a QPSK signal. The input binary
sequence 01101000 is shown in Figure 3.4 a. This sequence is divided into two other sequences, consisting of odd- and
even-numbered bits of the input sequence. These two sequences are shown in the top lines of Figure 3.4 b and c. The
waveforms representing the two components of the QPSK signal, namely si1 1(t) and si2 2(t) are also shown in Figure 3.4
b and c, respectively
Quadriphase-Shift Keying

Figure 3.4 (a) Input binary sequence. (b) Odd-numbered dibits of input sequence and associated binary PSK signal. (c)
Even-numbered dibits of input sequence and associated binary PSK signal. (d) QPSK waveform defined as s(t) =si1 1
(t) + si2 2 (t).

Generation and Coherent Detection of QPSK Signals
Figure 3.5: Block diagram of (a) QPSK transmitter and (b) coherent QPSK receiver.

The functional composition of the QPSK receiver is as follows:
1. Pair of correlators, which have a common input x(t). The two correlators are supplied with a pair of locally
generated orthonormal basis functions (t) and (t), which means that the receiver is synchronized with the transmitter.
The correlator outputs, produced in response to the received signal x(t), are denoted by x1 and x2, respectively.
2. Pair of decision devices, which act on the correlator outputs x1 and x2 by comparing each one with a zero-threshold;
here, it is assumed that the symbols 1 and 0 in the original binary stream at the transmitter input are equally likely. If x1
> 0, a decision is made in favor of symbol 1 for the in-phase channel output; on the other hand, if x1 < 0, then a decision
is made in favor of symbol 0. Similar binary decisions are made for the quadrature channel.
3. Multiplexer, the function of which is to combine the two binary sequences produced by the pair of decision devices.
The resulting binary sequence so produced provides an estimate of the original binary stream at the transmitter input.

Error Probability of QPSK
In a QPSK system operating on an AWGN channel, the received signal x(t) is defined by………..(3.17)
………...(3.18)
`
………………….3.19
these two binary PSK receivers are characterized as follows:
• signal energy per bit equal to E/2, and
• noise spectral density equal to N0/2.

Hence, using (3.15) for the average probability of bit error of a coherent binary PSK receiver, we may
express the average probability of bit error in the in-phase and quadrature paths of the coherent QPSK
receiver as
…………(3.20)
………………..(3.21)
=
The average probability of symbol error for QPSK is therefore
= ……….(3.22)
In the region where (E/N0) >> 1, we may ignore the quadratic term on the right-hand side of (3.22), so the
average probability of symbol error for the QPSK receiver is approximated as

……………(3.23)
In a QPSK system, we note that since there are two bits per symbol, the transmitted signal energy per symbol is twice the
signal energy per bit, as shown by
…….(3.24)
Thus, expressing the average probability of symbol error in terms of the ratio Eb/N0, we may write
……….(3.25)
With Gray encoding used for the incoming symbols, we find from (3. 20) and (3.24) that the BER of QPSK is exactly
…………(3.26)

M-ary PSK
QPSK is a special case of the generic form of PSK commonly referred to as M-ary PSK, where the phase of the carrier
takes on one of M possible values: Өi = 2(i – 1) Π/M, where i = 1, 2,… M. Accordingly, during each signaling interval of
duration T, one of the M possible signals
…………..(3.27)
Each si(t) may be expanded in terms of the same two basis functions 1(t) and 2(t); the signal constellation of M-ary PSK
is, therefore, two-dimensional. The M message points are equally spaced on a circle of radius and center at the origin, as
illustrated in Figure 3.6a for the case of octaphase-shift-keying (i.e., M = 8).
………….(3.28)
Hence, the use of yields the average probability of symbol error for coherent M-ary PSK as
……………(3.29)

Figure 3.6 (a) Signal-space diagram for octaphase-shift keying (i.e., M = 8). The decision boundaries are shown as
dashed lines. (b) Signal-space diagram illustrating the application of the union bound for octaphase-shift keying.

M-ary Quadrature Amplitude Modulation
In an M-ary PSK system, the in-phase and quadrature components of the modulated signal are interrelated in such a way
that the envelope is constrained to remain constant. This constraint manifests itself in a circular constellation for the
message points, as illustrated in Figure 3.6a.
The QAM is a hybrid form of modulation, in that the carrier experiences amplitude as well as phase-modulation. In M-
ary PAM, the signal-space diagram is one-dimensional. M-ary QAM is a two dimensional generalization of M-ary PAM,
in that its formulation involves two orthogonal passband basis functions:
…………(3.30)

Let dmin denote the minimum distance between any two message points in the QAM constellation. Then,
the projections of the ith message point on the 1- and2-axes are respectively defined by ai dmin/2 and bi
dmin/2, where i = 1, 2, …, M. With the separation between two message points in the signal-space
diagram being proportional to the square root of energy, we may therefore set
……….(3.31)
where E0 is the energy of the message signal with the lowest amplitude. The transmitted M-ary QAM
signal for symbol k can now be defined in terms of E0:
………(3.32)
The signal sk(t) involves two phase-quadrature carriers, each one of which is modulated by a set of
discrete amplitudes; hence the terminology “quadrature amplitude modulation”.

Frequency-Shift Keying Techniques Using Coherent Detection
M-ary PSK and M-ary QAM share a common property: both of them are examples of linear modulation. In this
section, we study a nonlinear method of modulation known as FSK using coherent detection. We begin the study by
considering the simple case of binary FSK, for which M = 2.
3.2.1 Binary FSK
In binary FSK, symbols 1 and 0 are distinguished from each other by transmitting one of two sinusoidal waves that
differ in frequency by a fixed amount. A typical pair of sinusoidal waves is described by
………(3.33)
where i = 1, 2 and Eb is the transmitted signal energy per bit; the transmitted frequency is set at
…………..(3.34)

From (3.33) and (3.34), we observe directly that the signals s1(t) and s2(t) are orthogonal, but not
normalized to have unit energy. The most useful form for the set of orthonormal basis functions is
described by
………….(3.35)
where i = 1, 2. Correspondingly, the coefficient sij for where i = 1, 2 and j = 1, 2 is defined by
…………(3.36)
Carrying out the integration in (3.36), the formula for sij simplifies to
…………(3.37)

Figure 3.7: Signal-space diagram for binary FSK system.
The diagram also includes example waveforms of the two modulated signals s1(t) and s2(t).
……….(3.38)………(3.39)
The Euclidean distance |||| is equal to . Figure 3.7 also includes a couple of waveforms representative of signals s1(t) and
s2(t).

Generation and Coherent Detection of Binary FSK Signals
The block diagram of Figure 3.8a describes a scheme for generating the binary FSK signal; it consists of two
components:
1. On–off level encoder, the output of which is a constant amplitude of in response to input symbol 1 and zero in
response to input symbol 0.
2. Pair of oscillators, whose frequencies f1 and f2 differ by an integer multiple of the bit rate 1/Tb in accordance with
(3.34). The lower oscillator with frequency f2 is preceded by an inverter. When in a signalling interval, the input
symbol is 1, the upper oscillator with frequency f1 is switched on and signal s1(t) is transmitted, while the lower
oscillator is switched off. On the other hand, when the input symbol is 0, the upper oscillator is switched off, while the
lower oscillator is switched on and signal s2(t) with frequency f2 is transmitted. With phase continuity as a
requirement, the two oscillators are synchronized with each other. Alternatively, we may use a voltage-controlled
oscillator, in which case phase continuity is automatically satisfied.

Figure 3.8 Block diagram for (a) binary FSK transmitter and (b) coherent binary FSK receiver.

Error Probability of Binary FSK
The observation vector x has two elements x1 and x2 that are defined by, respectively,
………….(3.40)
and…………..(3.41)
where x(t) is the received signal, whose form depends on which symbol was transmitted. Given that symbol 1 was
transmitted, x(t) equals s1(t) + w(t), where w(t) is the sample function of a white Gaussian noise process of zero mean
and power spectral density N0/2. If, on the other hand, symbol 0 was transmitted, x(t) equals s2(t) + w(t).
To proceed further, we define a new Gaussian random variable Y whose sample value y is equal to the difference
between x1 and x2; that is,
………..(3.42)

Given that symbol 1 was sent, the Gaussian random variables X1 and X2, whose sample values are denoted by x1 and
x2, have mean values equal to and zero, respectively. Correspondingly, the conditional mean of the random variable Y
given that symbol 1 was sent is
………….(3.43)
On the other hand, given that symbol 0 was sent, the random variables X1 and X2 have mean values equal to zero and ,
respectively. Correspondingly, the conditional mean of the random variable Y given that symbol 0 was sent is
…………….(3.44)

The variance of the random variable Y is independent of which binary symbol was sent. Since the random variables
X1 and X2 are statistically independent, each with a variance equal to N0/2, it follows that
…………..(3.45)
Suppose we know that symbol 0 was sent. The conditional probability density function of the random variable Y is
then given by …………….(3.46)
Since the condition x1 > x2 or, equivalently, y > 0 corresponds to the receiver making a decision in favor of symbol
1, we deduce that the conditional probability of error given that symbol 0 was sent is
……………………..(3.47)

To put the integral in (3.47) in a standard form involving the Q-function, we set
…………….(3.48)
Then, changing the variable of integration from y to z, we may rewrite (3.47) as
…………….(3.50)
the BER for binary FSK using coherent detection is
…………………..(3.51)
Comparing (3.14) and (3.51), we see that for a binary FSK receiver to maintain the same BER as in a binary PSK
receiver, the bit energy-to-noise density ratio, Eb/N0, has to be doubled. . For a prescribed Eb, the minimum distance
dmin in binary PSK is, therefore, times that in binary FSK. Recall from (7.89) that the probability of error decreases
exponentially as ; hence the difference between (3.14) and (3.51).

Noncoherent Orthogonal Modulation Techniques
Figure 3.9 (a) Generalized binary receiver for noncoherent orthogonal modulation. (b) Quadrature receiver equivalent to
either one of the two matched filters in (a); the index i = 1, 2.

Binary Frequency-Shift Keying Using Noncoherent Detection
In binary FSK, the transmitted signal is defined in (3.33) and repeated here for convenience of presentation:
…….(3.56)
where Tb is the bit duration and the carrier frequency fi equals one of two possible values f1 and f2; to ensure
that the signals representing these two frequencies are orthogonal, we choose fi = ni/Tb, where ni is an integer.
The transmission of frequency f1 represents symbol 1 and the transmission of frequency f2 represents symbol 0.
For the noncoherentdetection of this frequency-modulated signal, the receiver consists of a pair of matched
filters followed by envelope detectors, as in Figure 3.10. The filter in the upper path of the receiver is matched to
cos(2Πf1t) and the filter in the lower path is matched to cos(2Πf2t) for the signaling interval 0 ≤ t ≤ Tb. The
resulting envelope detector outputs are sampled at t = Tb and their values are compared. The envelope samples
of the upper and lower paths in Figure 3.10

The noncoherent binary FSK described herein is a special case of noncoherent orthogonal modulation with T = Tb and E
= Eb, where Eb is the signal energy per bit. Hence, the BER for noncoherent binary FSK is
……..(3.57)
which follows directly from (3.55) as a special case of noncoherent orthogonal modulation.
Figure 3.10 Noncoherent receiver for the detection of binary FSK signals.

Differential Phase-Shift Keying
Differential encoding starts with an arbitrary first bit, serving as the reference bit; to this end, symbol 1 is used as
the reference bit. Generation of the differentially encoded sequence then proceeds in accordance with a two-part
encoding rule as follows:
1. If the new bit at the transmitter input is 1, leave the differentially encoded symbol unchanged with respect to
the current bit.
2. If, on the other hand, the input bit is 0, change the differentially encoded symbol with respect to the current bit.
The differentially encoded sequence, denoted by {dk}, is used to shift the sinusoidal carrier phase by zero and
180o, representing symbols 1 and 0, respectively. Thus, in terms of phase-shifts, the resulting DPSK signal follows
the two-part rule:
1. To send symbol 1, the phase of the DPSK signal remains unchanged.
2. To send symbol 0, the phase of the DPSK signal is shifted by 180°.

Error Probability of DPSK
let the transmitted DPSK signal befor the first-bit interval 0 ≤ t ≤ Tb, which corresponds to symbol 1. Suppose,
then, the input symbol for the second-bit interval Tb ≤ t ≤ 2Tb is also symbol 1. According to part 1 of the DPSK
encoding rule, the carrier phase remains unchanged, thereby yielding the DPSK signal.
………..(3.58)
Suppose, next, the signaling over the two-bit interval changes such that the symbol at the transmitter
input for the second-bit interval Tb ≤ t ≤ 2Tb is 0. Then, according to part 2 of the DPSK encoding rule, the
carrier phase is shifted by Π radians (i.e., 180°), thereby yielding the new DPSK signal
……….(3.59)

We now readily see from (3.58) and (3.59) that s1(t) and s2(t) are indeed orthogonal over the two-bit interval 0
≤ t ≤ 2Tb, which confirms that DPSK is indeed a special form of noncoherent orthogonal modulation with one
difference compared with the case of binary FSK: for DPSK, we have T = 2Tb and E = 2Eb. Hence, using
(3.55), we find that the BER for DPSK is given by
…………(3.60)
According to this formula, DPSK provides a gain of 3 dB over binary FSK using noncoherent detection for the
same Eb/N0.

Generation of DPSK Signal
Figure 3.11 shows the block diagram of the DPSK transmitter. To be specific, the transmitter consists of two
functional blocks:
• Logic network and one-bit delay (storage) element, which are interconnected so as to convert the raw input binary
sequence {bk} into the differentially encoded sequence {dk}.
• Binary PSK modulator, the output of which is the desired DPSK signal.
Figure 3.11 Block diagram of a DPSK transmitter.

Optimum Receiver for the Detection of DPSK
In the use of DPSK, the carrier phase Ө is unknown, which complicates the received signal x(t). To deal with the
unknown phase Ө in the differentially coherent detection of the DPSK signal in x(t), we equip the receiver with an
in-phase and a quadrature path.
Figure 3.13 Block diagram of a DPSK receiver.

This geometry of possible signals is illustrated in Figure 3.12. For the two-bit interval 0 ≤ t ≤ 2Tb, the receiver
measures the coordinates , first, at time t = Tb and then measures at time t = 2Tb. The issue to be resolved is
whether these two points map to the same signal point or different ones. Recognizing that the vectors x0 and x1,
with end points and
…………(3.61)
Figure 3.12 Signal-space diagram of received DPSK signal.

Hence, substituting this identity into (3.61), we get the equivalent test:
…………..(3.62)
where the scaling factor 1/4 is ignored. In light of this equation, the question on the binary hypothesis test for the
detection of DPSK may now be restated as follows: Given the current signal point received in the time interval 0
< t < 2Tb, is this point closer to the signal point ) or its imagereceived in the next time interval Tb < t < 2Tb?
Thus, the optimum receiver for the detection of binary DPSK is as shown in Figure 3.13, the formulation of which
follows directly from the binary hypothesis test of (3.62). This implementation is simple, in that it merely requires
that sample values be stored. The receiver of Figure 3.13 is said to be optimum for two reasons:
1. In structural terms, the receiver avoids the use of fancy delay lines that could be needed otherwise.
2. In operational terms, the receiver makes the decoding analysis straightforward to handle, in that the two signals
to be considered are orthogonal over the interval [0,2Tb] in accordance with the formula of (3.55).

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