Communications 1

Hebat-Allah M. MouradProfessor

Cairo UniversityFaculty of Engineering

Electronics & Communications Department

Text Book• Modern Digital and Analog Communication

Systems

Third edition

B.P.Lathi

- Chapters: 4 – 5 – 6 (1st term) 7 -11 (2nd term)

• Communication systems, S. Haykin, John Wiley and Sons inc., 4th edition

Course Contents-Introduction to modulation

- Different analog modulation techniques: Amplitude, Frequency and Phase

-Transformation from analog to digital:

Sampling, Quatization, PCM, DM, ADM

-Introduction to satellite and mobile communications

Why Modulation?Mainly for two reasons

1 - Practical antenna dimensions light velocity

Wavelength frequency

Dimension is in the order of a quarter wavelength

fc

λ

Baseband Power Spectrum

Why Modulation?2- Multiplexing

Better utilization of the available frequency band

Spectrum

Frequency

M1(f) M2(f)M3(f) MN(f)

Basic Modulation Types • s (t) = A (t) cos [Ө (t) ]

• Ө (t) = ω t + φ (t)

• ‘ Ac cos ωc t ‘ is called un-modulated carrier

• Analog Modulation

• Digital Modulation

Modulator

m(t)

Acos(2πfct+φ)

modulated signal: s(t)

Un-modulated carrier

Modulation Types (Analog Modulation)

Modulation Types (Digital Modulation)

Analog Modulation

- Different analog modulation techniques

- For each type:-

- Mathematical presentation

* Bandwidth * transmitted power

- Modulators

- Demodulators

- Applications

1 -Amplitude Modulation (A.M)Conventional Amplitude Modulation

• Consider a sinusoidal Carrier wave c(t) the un-modulated carrier.

c(t) = Ac cos 2 π fc t

= Ac cos ωc t

Ac = carrier amplitude

fc = carrier frequency

• A.M is the process in which the amplitude of the carrier wave c(t) is varied about a mean value, linearly with the base band signal

1-Amplitude Modulation• s(t) = Ac cos 2 π fc t + m(t) cos 2 π fc t

= [Ac + m(t) ] cos 2 π fc t

= modulated signal

S(ω) = π Ac [ δ(ω + ω c) + δ(ω - ω c) ]

+ (1/2) [ M(ω + ω c) + M(ω - ω c) ]

1 -Amplitude Modulation

• s(t)= un-modulated carrier + upper side band (U.S.B) + lower sideband (L.S.B.)

• Bandwidth (B.W.) = 2W

| M(ω) || S(ω) |

0 0W- W - ω c + ω c

• Condition : Ac + m(t) > 0 for all t

Ac ≥ m(t)min (absolute –ve peak amp.)

• Define: μ = modulation index

= m(t)min / Ac

• 0 ≤ μ ≤ 1

since there is no upper bound on Ac and Ac ≥ m(t)min

• The envelope has the same shape of m(t)

• Over modulation : μ > 1

• μ x 100 = percentage modulation

1 -Amplitude Modulation• Example:

Let

m(t) = Am cos 2 π fm t single tone signalThen

s (t) = Ac cos 2 π fc t + Am cos 2 π fm t cos 2 π fc t

s (t) = Ac [1 + μ cos 2 π fm t ] cos 2 π fc t

μ = Am / Ac

= modulation factor= modulation index

1 -Amplitude Modulation

| M(ω) |π

ωω m- ω m 0 -ωc

ωω c+ ω m fc

| S(ω) |

πμAc/2

ω c-ω m-ω c+ ω m-ω c-ω m

πAc

-s (t) = Ac [1 + μ cos 2 π fm t ] cos 2 π fc t

-B.W. = 2 fm

-To get the power or energy : divide S(ω) by 2π, square the modulus and add all terms

1 -Amplitude Modulation• S(ω) = πAc [ (ω - ω c) + (ω - ω c) ]

+ (π Acμ/2) [ (ω +(ω c + ω m)) ]

+ (π Acμ/2) [ (ω -(ω c + ω m)) ]

+ (π Acμ/2) [ (ω -(ω c - ω m)) ]

+ (π Acμ/2) [ (ω +(ω c - ω m)) ]

1 -Amplitude Modulation• Un-modulated Power = carrier power= Ac

2 / 2

• U.S.B power = L.S.B. power = μ2 Ac2 / 8

• Total side bands power = μ2 Ac2 / 4

• Pt = Pc + Ps = (Ac2 / 2) + (μ2 Ac

2 / 4)= Pc [ 1+ (μ2 /2) ]

η = Side band power = μ2 total power 2+ μ2

AM ModulatorsSwitching Modulator

• 1- Switching Modulators

By multiplying the signal by any periodic waveform whose fundamental is ωc .

This periodic function can be expressed using F.S. as:

n

n

-Ψ(t) is a periodic pulse train

-Ex: ring modulator,bridge modulators (different electronic circuits performing the switching operation)

For c >> m(t) , the diode acts as a switch controlled mainly by the value of ‘c’

Vbb’(t) = c cosωct + m(t) for c > 0

Vbb’(t) = 0 for c < 0

A.M.demodulator1-Non coherent/

Asynchronous/Envelope Detector

(1/W) > RC > Tc

Non-coherent demodulator

Amplitude Modulation• Virtues, limitations and modifications of

A.M.• Advantages: Ease of Modulation and

demodulation ( cheap to build the system)

• Disadvantages- Waste of power.- Waste of B.W.

• We trade off system complexity to overcome these limitations- DSB-SC - DSB-QAM- SSB - VSB

2 -Double Side bands Suppressed Carrier (DSB-SC)

• The carrier component doesn’t appear in the modulated wave.

• s(t) = m(t) cos ( 2 ω ct)

• S(ω) = (1/2) [ M(ω - ω c) + M(ω - ω c) ]

• Total power = side bands power = (1/2) m2(t)

• B.W = 2W

DSB-SC Modulators

• 1- Switching Modulators

By multiplying the signal by any periodic waveform whose fundamental is ωc .

This periodic function can be expressed using F.S. as:

n

n

-Ψ(t) is a periodic pulse train

-Ex: ring modulator,bridge modulators (different electronic circuits performing the switching operation)

DSB-SC Modulators• 2- Multiplier modulators

• x(t)= m(t) cos 2 π fc t

• X(f) = (½)[ M(f+fc) + M(f-fc)]

X

m(t)

cos 2 πfct

x(t)

DSB-SC Modulators• 3- Non linear modulators

DSB-SC Demodulators• Coherent (synchronous ) detection

• Main problem : the carrier at the receiver must be synchronized in frequency and phase with the one at the transmitter.

• Otherwise we can have serious problems in the demodulation as will be seen.

DSB-SC Demodulators

X L.P.F

~

m(t) cosωct

cos[ (ωc+ ∆ ω )t + φ]

Frequency shift Phase shift

eo(t)

DSB-SC Demodulators• eo(t) = m(t) cos ωct cos[ (ωc+ ∆ ω )t + φ]

= (1/2) m(t) cos [ (∆ ω t) + φ]

+ cos [ (2 ωc +∆ ω) t + φ]

• Second term will be suppressed by the L.P.F.

• If ∆ ω =0 and φ = 0

eo(t) = (1/2) m(t) ( no frequency or phase error )

DSB-SC Demodulators• If ∆ω=0

eo(t) = (1/2) m(t) cos φ

- If φ = constant, eo(t) is proportional to m(t)

- Problems for φ either varying with time or equals to ± (π/2)

- The phase error may cause attenuation of the output signal without causing distortion as long as it is constant.

DSB-SC Demodulators• If φ = 0 , ∆ω ≠ 0

eo(t) = (1/2) m(t) cos ∆ω t (Donald Duck)

- The output is multiplied by a low frequency sinusoid, this causes attenuation and distortion of the output signal.

- This could be solved by using detectors by square law device, or phase locked loops (PLL) (ex: Costas loop)

Carrier Acquisition in DSB-SC• Signal squaring method

S.L.D N.B.F(2fc) Freq div. by 2m(t) cos ωct

x(t)

DSB-SC Demodulators• Costas Receiver

Product modulator

Product modulator

-90 phase shift

L.P.F

V.C.O

L.P.F

Phase Discriminator

cos(2πfct+φ)

Ac cos(2πfct)m(t)

)1/2(Accos φ m(t)

)1/2(Acsin φ m(t)

K sin 2 φ

DSB-SC Demodulators• The L.O. freq. is adjusted to be the same as fc

• Suppose that the L.O. phase is the same as the carrier wave:

‘I’ o/p = m(t) ‘Q’ o/p = 0

• Suppose that the L.O. phase drifts from its proper value by a small angle ‘φ’ radians.

• ‘I’ o/p will be the same and ‘Q’ o/p will be proportional to sin φ ~ φ (for small φ)

DSB-SC Demodulators• This ‘Q’ channel o/p will have the same polarity

as the ‘I’ channel for one direction of L.O. phase drift and opposite polarity for the opposite direction of L.O. phase drift.

• By combining the ‘I’ and ‘Q’ channel ouputs in a phase discriminator (multiplier followed by L.P.F.)

• A d.c. control signal is obtained that automatically corrects for local phase error in the V.C.O.

3 -DSB quadrature carrier multiplexing (QAM)

• It means quadrature amplitude modulation.

• We make use of the orthogonality of the sines and cosines to transmit and receive two different signals simultaneously on the same carrier frequency

3 -DSB quadruture carrier multiplexing (QAM)

X

X

X

X

π/2π/2

~~

+

+

L.P.F

L.P.F

m1(t)

m2(t)

cosωct s(t)cosωct

x1

x2

x3

x4

3 -DSB quadruture carrier multiplexing (QAM)

s(t) =m1(t) cos ωct + m2(t) sin ωct

x1(t) = s(t) cos ωct

= (1/2) m1(t)[1 + cos 2ωct ]

+ (1/2) m2(t) sin 2ωct

x3(t) = (1/2) m1(t)

x2(t) = s(t) sin ωct

= (1/2) m2(t)[1 - cos 2ωct ]

+ (1/2) m1(t) sin 2ωct

x4(t) = (1/2) m2(t)

4 -Single Side Band (SSB)• Can be generated by:

- Selective Filtering Method

- Phase Shift Method

Filter Method

X S.B. Filter

~

m(t)

s(t)

S.S.B.

4 -Single Side Band (SSB)• Phase shift method

– Hilbert transform.

hilbert transform of s(t)(t)ss(t)

h(t)s(t)

|H(ω)|

ω

ω

ӨH(ω)

π/2

-π/2

(t)s

4 -Single Side Band (SSB)

It can be shown that:-

s(t)S.S.B= m(t) cos ωct ± sin ωct

(t)m

X

-π/2-π/2

cos ωct

X

m(t)

(t)m

(t)m sin ωct

m(t) cos ωct

+

±

sS.S.B

Detection of SSB• Coherent Detection

y(t) = [ m(t) cos ωct ± sin ωct ] cos ωct

= (1/2)m(t) [1+ cos 2ωct ] ± (1/2) sin 2 ωct

After the L.P.F the output equals (1/2) m(t)

• Envelope Detection

sS.S.B = R(t) cos [ωct + Ө(t)]

(t)m

(t)m

Detection of SSB

• R(t) =

• Ө(t) = tan-1

• The envelope didn’t express the signal but if we have large

carrier in phase with the transmitted signal the non-coherent

could be used.

sS.S.B(t) + A cos ωct = R(t) cos [ωct + Ө(t)]

(t)2m(t)2m

m(t)

(t)m

Detection of SSB

m(t)A

] A

2m(t)[1A 1/2]

A

2m(t)A[1 R(t)

ignored be will2A

(t)2mand

2A

(t)2m termsThe

(t)mA hen tm(t) A If

1/2]2A

(t)2m2A

(t)2m

A

2m(t)A[1

(t)2m2A)(m(t)R(t)

Detection of SSB• In case of large carrier, the envelope of

the S.S.B. signal has the form of m(t) (the

base band signal), so the signal can be

demodulated by the envelope detector

with the condition that

A >>> І m(t) І

5 -Vestigial Side Band (V.S.B)• Vestige means trace.

Since S.S.B. is difficult to realize, a compromise between S.S.B. and D.S.B. in spectrum can be obtained using V.S.B. [especially when we have important components at low frequency and for large bandwidth base band signals like T.V.]

• Most of one side band is passed along with a vestige of the other side band. (B.W ~ 1.125 W)

(Reproduced by filters)

5 -Vestigial Side Band (V.S.B)

• What is the shape of Hv(f)?

X Hv(f) X L.P.FA

B C D E

cos ωctcos ωct

5 -Vestigial Side Band (V.S.B)• Spectrum at A prop. to: M(f)• Spectrum at B prop. to: (1/2)[M(f+fc) + M(f-fc)]

• Spectrum at C prop. to: Hv(f)[M(f+fc) + M(f-fc)]• Spectrum at D prop. to :

(1/2) Hv(f+fc) [ M(f+2fc) + M(f) ]

+(1/2) Hv(f-fc) [ M(f) + M(f-2fc) ]• The central lobe of the spectrum at D must

equal M(f)

(1/2) [ Hv(f+fc) + Hv(f-fc) ] M(f)

f

ffc-fc

5 -Vestigial Side Band (V.S.B)• What are the conditions on the vestigial filter?

1 – It must have an odd symmetry around fc

2 – 50% response level at fc

[ Hv(f+fc) + Hv(f-fc) ] = constant in ‘W’Hv(f)

1

1/2

- fc fc

f

Comparison of Various AM signals• AM and AM_SC

- Detectors required for AM are simpler

- AM signals are easier to generate

- SC requires less power at the transmitter for the same information ( cheaper transmitter)

- Effect of fading is must serious in AM because the carrier must maintain a certain strength in relation to the side bands.

Comparison of Various AM signals• DSB and SSB

- SSB needs only half the bandwidth

- Fading disturbs the relationship of the two sidebands and causes more serious distortion than in the case of SSB,

AM Applications• 1- Frequency Division Multiplexing (FDM)

If we have ‘N’ signals to be transmitted using AM modulation. All of these are band limited to ‘W’.

A1

B1

C1

Mod

Mod

Mod

1

2

3

+

f

1

2

3

Dem

Dem

Dem

1

2

3

A2

B2

C2

1

2

3

1

2

3

f1

f2

f3

Examples: radio, TV, telephone backbone, satellite, …

AM ApplicationsY(f)

ff2f1

Carrier Demod.

2W at fN

2W at f2

2W at f1 Detector

~

B.P.F

Detector

Detector

m1

m2

mN

AM Applications

• Example: Telephone channel multiplexing

First

Level

Mux

1

2

12

Basic Group

secondLevelMux

1

2

5

Super Group

ThirdLevelMux

1

2

10

Master Group

North American FDM hierarchy

AM Applications• All long-haul telephone channels are

multiplexed by FDM using SSB.

• Basic Group consists of 12 FDM SSB voice signals each of BW = 4 KHz. It uses LSB spectra and occupies 60 to 108 KHz. [ alternate group configuration of 12 USB occupies 148 to 196 KHz]

Basic group A (LSB)

AM Applications• Basic Super group

consists of 60 channels. It is formed by multiplexing ‘5’ basic groups and it occupies a band of 312 to 552 KHz. [alternate 60 KHz to 300 KHz]

Super group 1 (LSB)

AM Applications• Basic master group

consiste of 600 channels. It is formed by multiplexing 10 supergroups.

AM Applications• 2- Super heterodyne receiver

The receiver not only has the task of demodulation but it is also required to perform some other tasks such as:-

- Tuning : select the desired signal

- Filtering : separate the desired signal from other modulated signals.

- Amplification: to compensate for the loss in the signal power

Super heterodyne receiver fulfils efficiently all the three functions

• Basic elements of an AM receiver of the superheterodyne type

Audio Amplifier

E

MixerR.F.

SectionI.F.

Section

Envelop. detector

A B C D

~Local oscillator (fL.O)

.Common tuning

• Spectrum at different points

R.F. response

at ‘A’at ‘C’at ‘E’

fcf

I.F. response

A.F. response

fI.F 0

• at ‘A’: [A + m(t)] cos ωct

• at ‘B’: [A1+a1m(t)] cos ωct cos (ωc + ωI.F)t

[A2+a2m(t)] {cos (2ωc + ωI.F)t +cos ωI.Ft }

• at ‘C’: [A3+a3m(t)] cos ωI.Ft

• at ‘D’: a4m(t)

• The R.F cannot provide adequate selectivity since fc is high. It rejects a lot of adjacent channel interference and amplifies the signal.

• This is why we translate it to IF frequency to obtain good selectivity . All selectivity is realized in the IF section.

• The main role of RF section is image frequency rejection.

What is the image frequency?

• If fc = 1000 KHz , fL.O. = 1000+455 =1455KHz the image frequency = fc+2fI.F.= 1910 KHz. Will also be picked up.

• Stations that are 2fI.F apart are called image stations and are rejected by RF filter.

• Up-conversion: fL.O. = fc+fI.F

• Down conversion: fL.O= fc-fI.F