Chapter - 3:Digital Modulation

Aveek DuttaAssistant Professor

Department of Electrical and Computer EngineeringUniversity at Albany

Spring 2018

Images and equations adopted from:Digital Communications - John G. Proakis and Masoud Salehi 5/e. Copyright by The McGraw Hill CompaniesFundamentals of Communication Systems - Michael P. Fitz, .Copyright by The McGraw Hill Companies

3.1 Basics● Modulation with memory

○ The current k bits and the past (L — 1 )k bits are combined to the set of possible M = 2k messages. It has 2(L-1)k states.

● Memoryless modulation

○ The binary sequence is parsed into subsequences each of length k, and each sequence is mapped into one of the sm(t) waveforms (or signals or symbols)

○ Symbols are transmitted every Ts sec (signaling interval). Hence Rs = 1/Ts is the symbol rate.

○ K bits are transmitted of duration Tb in one symbol time Ts. Hence Tb = Ts/k = Ts / log2M

○ Bit rate R = Rs. (log2M). Unit is bits/sec

○ Average energy of sm(t), is . pm denotes the probability of the mth message, if equiprobable, pm = 1/M

○ If all symbols have the same energy, The average energy per bit is,

and the power transmitted by the transmitter is .

3.2-1 Pulse Amplitude Modulation (PAM / ASK)● Digital PAM is represented as

Where, p(t) is a pulse of duration T and {Am, 1≤ m ≤ M} is the set of M possible amplitudes corresponding to M = 2k possible k-bit blocks and

● p(t) is a real-valued signal pulse whose shape influences the spectrum of the transmitted signal, Hence, we can write

What is Ebavg

PAM with carrier● The PAM has a lowpass equivalent Am g(t),

● Recall that Em = ½ EG (passband E is half of lowpass)

● Basis functions for sm(t),

Where,

Therefore,

For PAM, n = 1 (it is a one dimensional signal)

√2 is for ɸn(t) to have unit energy

...DIY...

,

,Gray Coding - Adjacent signal amplitudes change by ONLY one bit - WHY?

Euclidean Distance

Slide 11 Lecture Notes 1

3.2-2 Phase Modulation (Phase Shift Keying)● Digital Phase modulation with M waveforms is given by

Where, g(t) is the signal pulse shape and θm=2 (m—1)/M, m=1, 2 ,. . . M are M possible phases of the carrier

● Since the waveforms have equal energy

● PSK is a 2-dimensional waveform

PSK Constellation● Euclidean distance for PSK

● Minimum distance for |m-n| = 1

● Substituting for Ebavg = Eb (previous slide)

● For large M, sin( /M) = /M,

● Adding additional /4 phase shift to the carrier results in better symbol synchronization and is called /4 - QPSK

DIY..

3.2-3 Quadrature Amplitude Modulation (QAM)● The bandwidth efficiency of PAM/SSB can be obtained by modulation

quadrature carriers cos2 fct and sin2 fct with two separate k-bit symbols

where Ami and Amq are the information-bearing signal amplitudes of the quadrature carriers and g(t) is the signal pulse

● Alternatively, QAM is combined phase (θm) and amplitude (rm) modulation

● QAM is a combination of M1- level PAM and M2 - phase PSK to construct an M = M1M2 combined PAM-PSK signal constellation.

○ If M1 = 2n and M2 = 2m , it transmits m + n = log2 (M1M2) binary digits at a symbol rate R/(m + n)

QAM basis functions● From the definition of sm(t),

● If M = 22k, i.e. M = 4, 16, 64, 256…. with amplitudes (±1, ±3, ±7... ±√M -1) on both directions (I and Q),

Substituting for Ebavg = Eavg / log2M

Summary● Bandpass PAM, PSK, and QAM has the general form

○ PAM : Am is real, generally equal to ±1, ±3, ... ±(M — 1)

○ M-ary PSK : Am is complex and equal to ej2 (m-1)/M

○ QAM : Am is complex number Am = Ami + jAmq

● PAM and PSK can be considered as special cases of QAM

○ In QAM signaling, both amplitude and phase carry information, whereas in PAM and PSK only amplitude or phase carries the information.

● The structure of the modulator

3.2-4 Multidimensional Signaling● Multidimensional signals can be constructed from n=1 or n=2 signals

○ By using orthogonal signals across time (T = T1/N) and frequency (F = NΔf)

○ OFDM (later in course) is an example of multidimensional signal

○ Each time-frequency slot can carry any modulated signals (symbols)

● Mathematically, sm(t) are orthogonal signals if

● The orthonormal set to represent sm(t) is

And

The signal vectors can be defined as

What is dmin and Eb

Frequency Shift Keying● Multidimensional signal is constructed by changing the frequency of fc

● The coefficient is used so that each symbol sm(t) has an energy E.

○ Can you verify the statement above?

● The transmitted symbols differ in frequency only, hence FSK

○ FSK is non-linear while QAM (ASK and PSK are special cases of QAM) is linear modulation

○ Combining two QAM symbols gives another QAM symbols but not for FSK.

● What is condition for orthogonality for sml(t) and snl(t) to be orthogonal?

For all m ≠nRecall…

⍴x,y = Re[⍴xl,yl]

Refer to Slide 12 in Lecture Notes 1

*

Orthogonality of FSK● Expand the integral

● For the inner product to be 0, Δf = k / 2T (as the angle evaluates to multiple of π and sin(kπ) = 0, for all integer k and m≠n)

● Generate orthogonal symbols is using Hadamard matrix (used in CDMA)

Normalized sinc function

sinc(x) = sin( x) / ( x)

3.4-2 PSD of modulated signals (Random process) ● A lowpass signal can be represented as , where the

basic modulated symbols are ● Therefore, in frequency domain,

● RI(k) is the autocorrelation function of the random variable IN representing the information source, hence the PSD is

Autocorrelation function of vl(t) is given by,

Rvl is periodic with symbol period T (cyclostationary, sec 2.7-2), hence Rvl expressed as an average T . Hence,

where,

sub. n = m-k

Two factors determine the shape of the power spectral density :

1] Shape of the basic pulse used for modulation.

2] The PSD of the information sequence {In }

(1)

(2)(3)

See (3.4-7)

Example 3.4-1: In a binary communication system In = ±1 with equal probability, and the In’ s are independent. This information stream linearly modulates a basic pulse

Modulation with memory L

● Xz(f) is the FT of xz(t), and Gx(f) is the Energy Spectral Density of the baseband signal

● Since XI(f) and XQ(f) has Hermitian symmetry, we can write

● This leads to

And the Energy Spectral Density is defined by

Hence,

√2 appears if the passband signal is defined as below. This is done such that the baseband and passband signal have the same energy. This is differently portrayed in the Proakis textbook, where this equality is not enforced.

Spectral Density (Deterministic signal approach)

Energy of the bandpass and the lowpass signal is same

XI(-f) = XI*(f)XQ(-f) = XQ*(f)

Example

Using the representation of Xz(f) from previous slide (conjugate of (f) is (f) itself

1.50.5

Baseband spectrum Xz(f) Passband spectrum Xc(f)