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1.introduction› Signal space representation of signals.› Signal space model of digital communication system.› Different modulation schemes.› Points of comparison bet each mod scheme.

2-Binary modulation techniques 3-quadrature modulation techniques 4-M-ary modulation techniques 5-comparing different scheme

› Power spectra.› Probability of error.› Bandwidth efficiency.

6-bit vs. symbol error probabilities 7-applications

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Why modulate?• Antenna’s length• Multiplexing

Main idea: shifting(keying) the amplitude/frequency/phase of a high frequency carrier.

Modulator:› In digital communication :

modulating wave :binary data/M-ary encoded version of it Carrier: usually sinusoidal wave

› Hybrid modulation: Both amplitude and phase of carrier are combined to produce APK

Demodulator-receiver:› Coherent vs. non coherent detection

Coherent :› exact replicas of possible arriving signals are available at receiver ( exact knowledge of the

carrier’s wave reference) › “receiver is phase locked to the transmitter› autocorrelator

Non coherent:› cross correlating the received signal with each one of he replicas and then make a decision

made on comparisons with preselected threshold’s› Complexity increase while Pe decreases.

Trade offs:› Multitude of modulation/detection schemes› Trade off bet. Primary communication resources

transmission power channel bandwidth

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max data rate min Pe min Pt min BW max resistance to noise min circuit complexity

But Conflictions may occur :1-power and bandwidth2-data rate and probability of error

Bit vs symbol:› M_ary signaling: modulator produces one of available set

of M=2^m distinct signals in response to m bits of source data.

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1- Signal space representation of signalsConcept of basis:

› The set of basis vectors {e1, e2, …, en} of a space are› chosen such that:

Should be complete or span the vector space: any vector a can be expressed as a linear combination of these vectors. Each basis vector should be orthogonal to all others Each basis vector should be normalized: energy=1

A set of basis vectors satisfying these properties is also said to be a complete orthonormal basis In an n-dim space, we can have at most n basis vectors Basic idea:

› representation of M energy signals {s(t)} as a linear combinations of N _orthonormal basis functions

› signals used in communications can be expressed and visualized graphically› Let φ1(t), φ2(t),…., φn(t) be n signals› Consider a signal x(t) and suppose that

If every signal can be written as linear combination of (φ1(t), φ2(t),…., φn(t)) basis functions then we have a n-dim signal space

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If method

If inspection does not work we can use analytical method called Gram_shmidt method

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Vector representation:

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Transmitter: Message source: emits one symbol every t seconds Vector transmitter: each signal si(t) corresponds to a symbol

and using previous theorem each signal can be represented as a linear combination of the basis of the system .

i.e. si=[si1 ; si2;… SiN ]

 

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Modulator : The modulator target is to multiply the vector corresponding to

the signal to be transmitted with the basis of the signal and transmits the signal Si(t)through the channel.

AWGN is added to the signal and it can be viewed as adding vector W to the signal vector S producing the received signal vector X=S+W

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Its task will be observing the received message x(t) for T seconds and make the best estimate of si(t) or equivalently the symbol mi

This occurs in 2 stages: 1-detector: takes x(t) and produces a vector of random Variables X 2-vector receiver : by using an observation vector x(which is a sample

value of X),prior knowledge of si(t) the vector receiver produces an estimate of mi

*Think of this

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In a binary FSK system, 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

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Thus symbol 1 is represented by s1(t), and symbol 0 by s2(t),the signals s1(t) and s2(t) are orthogonal.

(a) binary FSK transmitter when we have symbol 0 at the input, the oscillator in the upper channel is switched off, and the oscillator in the lower channel is switched on, with the result that frequency f2 is transmitted. The two frequencies f1 and f2 are chosen to equal integer multiples of the bit rate 1 /Tb.

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(b) coherent binary FSK receiver.• In order to detect the original binary sequence given the noisy received wave x(t), we may use the receiver shown . • It consists of two correlators with a common input, which are supplied with locally generated coherent reference signals φ1(t) and φ2(t). • The correlator outputs are then subtracted, one from the other, and the resulting difference, l, is compared with a threshold of zero volts. If l > 0, the receiver decides in favor of 1. On the other hand, if l < 0, it decides in favor of 0. The average probability of symbol error for coherent binary FSK is

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• In a coherent binary PSK system, the pair of signals, s1(t) and s2(t), used to represent binary symbols 1 and 0, respectively, are defined by

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• In order to ensure that each transmitted bit contains an integral number of cycles of the carrier wave, the carrier frequency fc = nc/Tb for some fixed integer nc. • A pair of sinusoidal waves that differ only in a relative phase-shift of 180 degrees.

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• To detect the original binary sequence of 1s and 0s, we apply the noisy PSK wave x ( t ) (at the channel output) to a correlator, which is also supplied with a locally generated coherent reference signal φ1(t), as in Figure 4.2b. • The correlator output, x1, is compared with a threshold of zero volts. If x1 > 0, the receiver decides in favor of symbol 1. If x1 < 0, it decides in favor of symbol 0.• The average probability of symbol error for coherent binary PSK equals

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• As with binary PSK the information carried by the transmitted wave is contained in the phase. • In quadriphase-shift keying (QPSK), the phase of the carrier takes on one of four equally spaced values, such as π /4, 3π/4, 5π/4 and 7π/4 as shown by

Using a well-known trigonometric identity, we may rewrite Eq.

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In a QPSK system, we note that there are two bits per symbol. This means that the transmitted signal energy per symbol is twice the signal energy per bit :

• So The average probability of symbol error for coherent binary QPSK equals

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• This binary wave is divided by means of a de-multiplexer into two separate binary waves consisting of the odd- and even numbered input bits.•Example: if (01) is the input so the 1st bit (odd) is {0} and the 2nd bit (even) is {1}

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• The QPSK receiver consists of a pair of correlators with a common input and supplied with a locally generated pair of coherent reference signals φ1(t) and φ2(t) as in Figure . The correlator outputs, x1 and x2, are each compared with a threshold of zero volts. • If x1 > 0, a decision is made in favor of symbol 1 for the upper or in-phase channel output, but if x1 < 0 a decision is made in favor of symbol 0. • If x2 > 0, a decision is made in favor of symbol 1 for the lower or quadrature channel output, but if x2 < 0, a decision is made in favor of symbol 0. • Finally, these two binary sequences at the in-phase and quadrature channel outputs are combined in a multiplexer to reproduce the original binary sequence at the transmitter input with the minimum probability of symbol error.

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A digital communication receiver with no provision make for carrier phase recovery is said to be non-coherent.

Non-coherent Orthogonal Modulation: Scheme. For a binary signaling scheme that involves the use of two signals which are orthogonal with equal energy. 

the received signal x(t) can be expressed as follows

The receiver tries to discriminate between s1(t) and s2(t), regardless of the carrier phase. This goal can be achieved by the following receiver structure:

A non-coherent matched filter may be viewed as being equivalent to a quadrature receiver, as illustrated below. The quadrature receiver itself has two channel

Let φ1(t) and φ 2(t) be the orthonormal set of s1(t) and s2(t) and Ǿ(t) be the version of φ(t) that results from shifting the carrier phase by -90 degrees(Hilbert Transform)

The average probability of error for the non-coherent receiver

  where E is the signal energy per

symbol and N0/2 is the noise spectral density.

For the binary FSK case, the transmitted signal is

Thus the non-coherent binary FSK is a special case of non-coherent orthogonal modulation with T=Tb and E=Eb, where Tb is the bit duration and Eb is the signal energy per bit.

Transmitter’s two operations:(1) differential encoding of the input

binary sequence and(2) phase-shift keying Generation of DPSK:

For an input binary sequence {bk},a differential encoded sequence{dk} is made as follows:

DPSK is another example of non-coherent orthogonal modulation, when it is considered over two bit intervals.

Where T = 2Tb and E = 2Eb.

Note that DPSK is 3dB poorer than BPSK, however it is much easier to implement as it doesn’t require phase synchronization.

we send any one of M possible signals S1(t), S2(t), ………, SM(t) during each signaling interval of duration T› The requirement is to conserve bandwidth at the expense of

both increased power and increased system complexity› When the bandwidth of the channel is less than the required

value, we resort to an M-ary modulation scheme for maximum bandwidth conservation

M-ary Phase-Shift Keying› If we take blocks of m bits to produce a symbol and use an M-

ary PSK scheme with M=2n and symbol duration T=nTb

› The bandwidth required is proportional to 1/(nTb)

› The use of M-ary PSK provides a reduction in transmission bandwidth by a factor by a factor n=log2M

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The phase of the carrier takes one of M possible values, namely,

A M-ary signal set is represented as , for

where T is the symbol duration and E is the signal energy per symbol. The carrier frequency where nc is a fixed integer.

each signal si(t) can be expanded in terms of the following two basis

,  the signal constellation of M-ary PSK is two-dimensional. The M messages

are equally spaced on a circle of radius and center at the origin

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The decision boundaries are shown as dashed lines.

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w1 & w2 are samples of two independent Gaussian random variables of zero mean and common variance

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The optimum receiver includes a pair of correlators with reference signals in phase quadrature.

The two correlator outputs XI and XQ are fed into a phase discriminator that computes the phase estimated

Phase discriminator selects from the set of the signal whose phase is closet to the estimatedThe decision region is bounded by the threshold Probability of correct reception is therefore

Where is the probability density function of the random variable

The probability of symbol error is

For large M and high values of

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In an M-aryFSK scheme, the transmitted signals are defined by

the carrier frequency where nc is a fixed integerThe transmitted signals are of equal duration T and have equal energy

The individual signal frequencies are separated by 1/2T hertzThe signals are orthogonal For coherent M_ary FSK ,the optimum receiver consists of a bank of M

correlators or matched filters.At the sampling time the receiver makes decisions on the largest

matched filter output.

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In this modulation scheme the carrier experiences amplitude and phase modulation.

signal constellation for M=16

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Transmitter

Receiver

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The transmitted energy in Mary QAM is variable , as its instantaneous value depends on the particular symbol transmitted. Therefore the probability of error is expressed in terms of average value of the transmitted energy.

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PSKThe symbol duration of M_ary PSK is defined by T=Tb log2M

The base band power spectral density of M_ary PSK is given by:

SB(f)=2Esinc2(Tf)

SB(f)=2Eb log2M sinc2(Tb log2M f)

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Our main objective is to maximize the bandwidth efficiency at a minimum practical expenditure of average signal power.

With the data rate Rb and the channel bandwidth B, we may express the bandwidth efficiency, as :

Bandwidth Efficiency of M-ary PSK Signals: The bandwidth of M-ary PSK (B) is defined as the null-to-null

bandwidth (contains most of the signal power).

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The bandwidth efficiency of M-ary PSK signals is given by: [Bit/s/Hz]

The table gives the values of calculated from previous equationfor varying M.

Bandwidth efficiency of M-ary PSK signals

We observe that as M inc. , the bandwidth efficiency increases.

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M 2 4 8 16 32 64

0.5 1 1.5 2 2.5 3

M-ary FSK signals consist of an orthogonal set of M frequency-shifted signalsThe adjacent signals are separated by 1/2TWe may define We may express the bandwidth efficiency of M-ary FSK signals as

The table gives the values of calculated from previous equation for varying M

Bandwidth efficiency of M-ary FSK signals

We can see that as M inc. , bandwidth efficiency decreases.

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M 2 4 8 16 32 64

1 1 0.75 0.5 0.31 0.18

For M-ary PSK System: Consider first a coherent M-ary PSK system that employs a

nonorthogonal set of M phase-shifted signals for transmission of binary data.

Using the definition of Null-to-null bandwidth, we may Express bandwidth efficiency (channel capacity) as

Each symbol corresponds to an average probability of

symbol Error Pe= We observe that as M inc. ,the bandwidth Efficiency inc ,

but the ratio of symbol energy to noise power spectral density (Eь/Nо) moves away from shannon limit.

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For M-ary FSK System: Consider a coherent M-ary FSK that uses an orthogonal set

of M frequency shifted signals for the transmission of binary data, with the separation between adjacent signal frequency set at 1/2T.

The bandwidth efficiency is defined as

Each symbol has average probability of error Pe= We observe that as M inc. , bandwidth requirement inc. ,

the operating point moves closer to the Shannon limit.

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For Orthogonal SignalsFor an M-ary orthogonal signal

In the limit as k increases, we get

A simple example of octal message set will make this equation more acceptable.The message symbols are transmitted on orthogonal waveforms such as FSK To calculate symbol error probability (Pe)For the transmitted symbol o11An error might occur in anyone of the other Symbols with equal probability

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For non orthogonal signaling such as M-ary PSKEx: 8-ary signal:In order to illustrate decision space of 8-ary signal we use

Binarycode or Gray code

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If symbol (011) is transmitted, it is clear that should an error occur, the transmitted signal will most likely be mistaken for one of its closest neighbors.

Using Binary code:(011) could be transmitted as (010) or (100)some symbol errors will usually result in 2 or more bit errors,

evenwith large (S/N)Using Gray code:(011) could be transmitted as 001 or 010one bit will be in errorutilizing the Gray code assignment, it can be shown that

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