Modern CommunicationsChapter 3. Digital Modulations

Husheng Li

Min Kao Department of Electrical Engineering and Computer ScienceUniversity of Tennessee, Knoxville

Fall, 2016

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Digital Communications: Advantages

Digital communications can betterwithstand channel noise and distortion.

Digital communications provides viabilityof regenerative repeaters.

Applicability of microprocessors.

Privacy and protection

Multiplexing

Exchanging SNR for bandwidth

More efficient storage.

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Digital Communications: Big Picture

A digital communication system consists of Information source,Baseband modulation, Digital carrier modulation, Multiplexer, et al.

At the receiver, we have demodulator, decoder and demultiplexer...

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Digital Communications: Sampling

In digital communications,both time and informationare discrete (discrete timeand bits).

If the information source isanalog, we need samplingand then quantization.

What should the samplingrate be, if the basebandbandwidth is B? (Nyquistcriterion)

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Digital Communications: Quantization

A quantizer converts a sample to adiscrete symbol and a sequence ofbits.

A quantizer always incursquantization error.

The mean square error of

quantization is given bym2

p3L2 , where L

is the number of levels, and(−mp,mp) is the range of amplitude.

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Digital Communications: NonuniformQuantization

The distribution of signal may not beuniform.

We can put more levels at theranges with more probability.

For voice, we have µ-law (northAmerica and Japan) and A-law(Europe).

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Digital Modulation: PAM

Recall the sinusoid carrier: ψ(t) = m(t) cos(wc t + θ).

Similarly to the analog communications, we can also use the amplitudeof pulses to convey the information, thus forming pulse amplitudemodulation (PAM).

If there are only two levels of power (one is zero power), it is on-offkeying (OOK).

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Gray Encoding

The constellation mapping is usually done by Gray encoding, wherethe messages associated with signal amplitudes that are adjacent toeach other differ by one bit.

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Phase Shift Keying (MPSK)

In MPSK, the information is encoded in the phase:

si(t) = Ag(t)cos[2πfc t +

2π(i − 1)M

].

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Demodulation of MPSK

The decision regions of MPSK are shown above.

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Differential Modulation

MPSK and MQAM require coherent demodulation, namely theoriginal phase needs to be known (by using pilot symbol).When channel changes very fast, it is difficult to estimate theoriginal phase (unless you use many pilot symbols, but thiscauses too much overhead). Then, we need differentialmodulation.Differential modulation conveys information in the changes.DPSK: if bit 0, keep the same phase; if bit 1, change the phaseby π. Example: 00100110→ 00πππ0π.

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Demodulation of DPSK

Differential modulation is less sensitive to a random drift in thecarrier phase.If the channel has a nonzero Doppler frequency, the signal phasecan decorrelate between symbol times.

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Frequency Modulation

The frequency modulation signal is given bysi(t) = A cos(2πfi t + φi). The frequency spacing should be0.5/Ts.

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Modulation of FSK

We can use a switch to choose the correct frequency.

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Demodulation of FSK

We can project the received signal to different possiblefrequencies and choose the closest one.

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Noncoherent Detection of FSK

For each carrier frequency fj , j = 1, ...,M, the received signal ismultiplied by a noncoherent in-phase and quadrature carrier at thatfrequency, integrated over a symbol time, sampled and then squared.

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Two Branches: The Quadrature Structure

The sinusoid carrier can be dividedinto in-phase and quadraturecomponents:

s(t) = sI(t)cos(2πfc t)−sQ(t)sin(2πfc t),

which can also be written as thecomplex baseband representation:s(t) = R

[u(t)e2πfc t].

You can consider cos and sin as twoorthogonal axes.

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Quadrature Amplitude Modulation

QAM can be considered as amplitudemodulation on the in-phase andquadrature components:

s(t) = bI cos(2πfc t) + bQ sin(2πfc t),

QAM can also be considered as acombination of PAM and PSK:

s(t) = A cos(2πfc t + φ),

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Homework 4

Deadline: Sept. 30, 2016Problem 1. Show that the minimum frequency separation for FSK suchthat the cos(2πfj t) and cos(2πfi t) are orthogonal is∆f = mini 6=j |fi − fj | = 0.5/Ts.

Problem 2. Consider 16QAM. Find an upper bound and anapproximation of the demodulation error rate if d min√

2N0= 10dB.

Problem 3. Prove that functions sin wc t and cos wc t are orthogonal toeach other.

Problem 4. In DPSK, what is the output when the input is0110010010011?

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More Examples of QAM

If we use higher order QAM, we have a faster transmission rate;meanwhile, we have less reliability.

We can make the order adaptive to the channel (adaptive modulation).

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Signal Projection

The noise should not help the detection since its projection onto thesignal space is zero.

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Decision Regions

Decision rules are represented by decision regions (decision i if thereceived signal falls r in region Zi ).

Error probability (suppose M equal probably messages)

Pe =1M

M∑m=1

P(r not in Zm|message m).

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Error Probability of MPSK

The error probability is given by

Pe = 1− 1M

M∑i=1

∫Zi−si

p(n)dn.

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Union Bound

The error probability is upper bounded by

Pe =M∑

i=1

p(mi)Pe(mi sent) ≤ 1M

M∑i=1

M∑k=1,k 6=i

Q(

dik√2N0

).

The approximation of the error can be obtained by

Pe ≈ MddimQ(

dmin

2N0

).

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Simplest Case: BPSK

Pe = 1√2π

SNR

∫∞1 e

− x22

SNR dx .

When SNR is large, the error rate decreases exponentially withSNR.

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Why Pulse Shaping?

How about simply using rectangle waveforms for the basebandsignal (before multiplied by the sinusoid)?Bad news: too much bandwidth will be used.

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Limited Bandwidth Waveform

How about using waveform with limited bandwidth?Bad news: there will be inter-symbol interference (ISI, a seriousconcern in communications).

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Limited Bandwidth without ISI

If we arrange the timing of samples correctly, we can avoid theISI while the bandwidth is limited (thus the waveform is infinite inthe time).

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Nyquist Criterion

The same Nyquist as that in signals and systems.In the Nyquist criterion, we require that the pulse wavefomr p(t)satisfy

p(t) ={

1, t = 00, t = ±nTb,

.

where Tb is the symbol period and equals 1/Rb (Rb is the symbolrate).The sinc function satisfies the Nyquist criterion.

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Nyquist Criterion in the Frequency Domain

The Nyquist criterion isequivalent to

∞∑n=−∞

P(w − nwb) = Tb.

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Nyquist Criterion in the Frequency Domain

For the above spectrum, thebandwidth is given by wb

2 + wx ,where wx is the bandwidth inexcess of the theoreticalbandwidth. We define r = 2wx/wb,which is called the roll-off factor.

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Raised Cosine

In practice, we use the raisedcosine waveform:

P(w) = cos2(

w4Rb

)rect

(w

4πRb

).

What should the time domainwaveform be?

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Eye Diagram

We can use eye diagram for visualexamination of severity of ISI, theaccuracy of timing extraction,noise immunity and otherimportant factors.

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Eye Diagram

Consider a basic oscilloscopewhich is triggered at the rate 1/Tb,and it yields a sweep lastingexactly Tb. It shows many traces oflength Tb.

We can check the impact of ISI(resulting in a partially closed eyepattern) and channel noise(tending to close the eye).

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Key Measures

Maximum opening point (noisemargin), which indicates the optimumsampling or decision-making point.

Sensitivity to timing jitter, which isdetermined by the width of theopening eye.

Level crossing jitter: The variation oflevel crossing can be seen from thewidth of the eye corners. Thezero-crossing information is used forextracting timing information about thepulse rate and sampling clock.

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