linear modulation techniques

Chapter 3Linear Modulation Techniques

Contents

3.1 Linear Modulation . . . . . . . . . . . . . . . . . . 3-33.1.1 Double-Sideband Modulation (DSB) . . . . 3-33.1.2 Amplitude Modulation . . . . . . . . . . . . 3-83.1.3 Single-Sideband Modulation . . . . . . . . . 3-213.1.4 Vestigial-Sideband Modulation . . . . . . . . 3-353.1.5 Frequency Translation and Mixing . . . . . . 3-38

3.2 Interference . . . . . . . . . . . . . . . . . . . . . . 3-463.2.1 Interference in Linear Modulation . . . . . . 3-46

3.3 Sampling Theory . . . . . . . . . . . . . . . . . . . 3-493.4 Analog Pulse Modulation . . . . . . . . . . . . . . . 3-54

3.4.1 Pulse-Amplitude Modulation (PAM) . . . . . 3-543.4.2 Pulse-Width Modulation (PWM) . . . . . . . 3-563.4.3 Pulse-Position Modulation . . . . . . . . . . 3-56

3.5 Delta Modulation and PCM . . . . . . . . . . . . . . 3-573.5.1 Delta Modulation (DM) . . . . . . . . . . . 3-573.5.2 Pulse-Code Modulation (PCM) . . . . . . . 3-60

3.6 Multiplexing . . . . . . . . . . . . . . . . . . . . . 3-63

3-1

CONTENTS

3.6.1 Time-Division Multiplexing (TDM) . . . . . 3-64

3-2 ECE 5625 Communication Systems I

3.1. LINEAR MODULATION

We are typically interested in locating a message signal to somenew frequency location, where it can be efficiently transmitted

The carrier of the message signal is usually sinusoidal

A modulated carrier can be represented as

xc.t/ D A.t/ cos2fct C .t/

where A.t/ is linear modulation, fc the carrier frequency, and.t/ is phase modulation

3.1 Linear Modulation

For linear modulation schemes, we may set .t/ D 0 withoutloss of generality

xc.t/ D A.t/ cos.2fct /

with A.t/ placed in one-to-one correspondence with the mes-sage signal

3.1.1 Double-Sideband Modulation (DSB)

Let A.t/ / m.t/, the message signal, thus

xc.t/ D Acm.t/ cos.2fct /

From the modulation theorem it follows that

Xc.f / D1

2AcM.f fc/C

1

2AcM.f C fc/

ECE 5625 Communication Systems I 3-3

CONTENTS

t t

m(t) xc(t) carrier filled envelope

DSB time domain waveforms

M(0)M(f)

Xc(f)

AcM(0)1

2

fc

-fc

f

f

LSB USB

DSB spectra

Coherent Demodulation

The received signal is multiplied by the signal 2 cos.2fct /,which is synchronous with the transmitter carrier

LPF

m(t) xc(t) x

r(t) d(t) y

D(t)

Accos[2πf

ct] 2cos[2πf

ct]

DemodulatorModulator Channel



For an ideal channel xr.t/ D xc.t/, so

d.t/ DAcm.t/ cos.2fct /

2 cos.2fct /

D Acm.t/C Acm.t/ cos.2.2fc/t/

where we have used the trig identity 2 cos2 x D 1C cos 2x

The waveform and spectra of d.t/ is shown below (assumingm.t/ has a triangular spectrum in D.f /)

D(f)

AcM(0)

2fc

-2fc

f

AcM(0)1

2A

cM(0)1

2

W-W

Lowpass modulation recovery filter

d(t)

t

Lowpass filtering will remove the double frequency carrier term

Waveform and spectrum of d.t/

Typically the carrier frequency is much greater than the mes-sage bandwidth W , so m.t/ can be recovered via lowpass fil-tering

The scale factor Ac can be dealt with in downstream signalprocessing, e.g., an automatic gain control (AGC) amplifier


CONTENTS

Assuming an ideal lowpass filter, the only requirement is thatthe cutoff frequency be greater than W and less than 2fc W

The difficulty with this demodulator is the need for a coherentcarrier reference

To see how critical this is to demodulation ofm.t/ suppose thatthe reference signal is of the form

c.t/ D 2 cosŒ2fct C .t/

where .t/ is a time-varying phase error

With the imperfect carrier reference signal

d.t/ D Acm.t/ cos .t/C Acm.t/ cosŒ2.2fc/t C .t/yD.t/ D m.t/ cos .t/

Suppose that .t/ is a constant or slowly varying, then thecos .t/ appears as a fixed or time varying attenuation factor

Even a slowly varying attenuation can be very detrimental froma distortion standpoint

– If say .t/ D 2f t and m.t/ D cos.2fmt /, then

yD.t/ D1

2ŒcosŒ2.fm f /tC cosŒ2.fm Cf /t

which is the sum of two tones

Being able to generate a coherent local reference is also a prac-tical manner



One scheme is to simply square the received DSB signal

x2r .t/ D A2cm

2.t/ cos2.2fct /

D1

2A2cm

2.t/C1

2A2cm

2.t/ cosŒ2.2fc/t

( )2 BPF

LPF

divideby 2

xr(t) y

D(t)

xr(t)2

Acos2πfct

very narrow (tracking) band-pass filter

Carrier recovery concept using signal squaring

Assuming that m2.t/ has a nonzero DC value, then the doublefrequency term will have a spectral line at 2fc which can bedivided by two following filtering by a narrowband bandpassfilter, i.e., Ffm2.t/g D kı.f /C

k

2fc

fSpec

trum

of m

2 (t) Filter this component

for coherent demod

Note that unless m.t/ has a DC component, Xc.f / will notcontain a carrier term (read ı.f ˙ fc), thus DSB is also calleda suppressed carrier scheme


CONTENTS

Consider transmitting a small amount of unmodulated carrier

k

m(t) xc(t)

Accos2πf

ct

ffc

-fc

AcM(0)/2

use a narrowband filter (phase-locked loop) to extract the carrier in the demod.

k << 1

3.1.2 Amplitude Modulation

Amplitude modulation (AM) can be created by simply addinga DC bias to the message signal

xc.t/ DACm.t/

A0c cos.2fct /

D Ac1C amn.t/

cos.2fct /

where Ac D AA0c, mn.t/ is the normalized message such thatminmn.t/ D 1,

mn.t/ Dm.t/

jminm.t/j

and a is the modulation index

a Djminm.t/j

A



t

xc(t)

a < 1

m(t)

A

A + m(t)

Accos[2πf

ct]

xc(t)

Bias term

Note that the enve-lope does not cross zero in the case of AM having a < 1

0A

c(1 - a)

A + max m(t) A + min m(t)

Generation of AM and a sample wavefrom

Note that ifm.t/ is symmetrical about zero and we define d1 asthe peak-to-peak value of xc.t/ and d2 as the valley-to-valleyvalue of xc.t/, it follows that

a Dd1 d2

d1 C d2

proof: maxm.t/ D minm.t/ D jminm.t/j, so

d1 d2

d1 C d2D2Œ.AC jminm.t/j/ .A jminm.t/j/2Œ.AC jminm.t/j/C .A jminm.t/j/

Djminm.t/j

AD a


CONTENTS

The message signal can be recovered from xc.t/ using a tech-nique known as envelope detection

A diode, resistor, and capacitor is all that is needed to constructand envelope detector

eo(t)x

r(t) C R

t

Recovered envelope with proper RC selection

eo(t)

0

The carrier is removed if 1/fc << RC << 1/W

Envelope detector

The circuit shown above is actually a combination of a nonlin-earity and filter (system with memory)

A detailed analysis of this circuit is more difficult than youmight think

A SPICE circuit simulation is relatively straight forward, but itcan be time consuming if W fc



The simple envelope detector fails if AcŒ1C amn.t/ < 0

– In the circuit shown above, the diode is not ideal andhence there is a turn-on voltage which further limits themaximum value of a

The RC time constant cutoff frequency must lie between bothW and fc, hence good operation also requires that fc W


CONTENTS

Digital signal processing based envelope detectors are also pos-sible

Historically the envelope detector has provided a very low-costmeans to recover the message signal on AM carrier

The spectrum of an AM signal is

Xc.f / DAc

2

ı.f fc/C ı.f C fc/

„ ƒ‚ …

pure carrier spectrum

CaAc

2

Mn.f fc/CMn.f C fc/

„ ƒ‚ …

DSB spectrum

AM Power Efficiency

Low-cost and easy to implement demodulators is a plus forAM, but what is the downside?

Adding the bias term to m.t/ means that a fraction of the totaltransmitted power is dedicated to a pure carrier

The total power in xc.t/ is can be written in terms of the timeaverage operator introduced in Chapter 2

hx2c .t/i D hA2cŒ1C amn.t/

2 cos2.2fct /i

DA2c2hŒ1C 2amn.t/C a

2m2n.t/Œ1C cos.2.2fc/t i

If m.t/ is slowly varying with respect to cos.2fct /, i.e.,

hm.t/ cos!cti ' 0;



then

hx2c .t/i DA2c2

1C 2ahmn.t/i C a

2hm2

n.t/i

DA2c2

1C a2hm2.t/i

D

A2c2„ƒ‚…

Pcarrier

Ca2A2c2hm2

n.t/i„ ƒ‚ …Psidebands

where the last line resulted from the assumption hm.t/i D 0

(the DC or average value of m.t/ is zero)

Definition: AM Efficiency

EffD

a2hm2n.t/i

1C a2hm2n.t/i

alsoD

hm2.t/i

A2 C hm2.t/i

Example 3.1: Single Sinusoid AM

An AM signal of the form

xc.t/ D AcŒ1C a cos.2fmt C =3/ cos.2fct /

contains a total power of 1000 W

The modulation index is 0.8

Find the power contained in the carrier and the sidebands, alsofind the efficiency

The total power is

1000 D hx2c .t/i DA2c2Ca2A2c2 hm2

n.t/i


CONTENTS

It should be clear that in this problemmn.t/ D cos.2fmt /, sohm2

n.t/i D 1=2 and

1000 D A2c

1

2C1

40:64

D33

50A2c

Thus we see that

A2c D 1000 50

33D 1515:15

and

Pcarrier D1

2A2c D

1515

2D 757:6 W

and thus

Psidebands D 1000 Pc D 242:4 W

The efficiency is

Eff D242:4

1000D 0:242 or 24.2%

The magnitude and phase spectra can be plotted by first ex-panding out xc.t/

xc.t/ D Ac cos.2fct /C aAc cos.2fmt C =3/ cos.2fct /D Ac cos.2fct /

CaAc

2cosŒ2.fc C fm/t C =3

CaAc

2cosŒ2.fc fm/t =3



f

f

0

0

fc+f

mfc-fm

fc

-fc

Ac/2

0.8Ac/4

|Xc(f)|

Xc(f)

π/3

-π/3

Amplitude and phase spectra for one tone AM

Example 3.2: Pulse Train with DC Offset

t

2

-1

m(t)

Tm/3 T

m

Find mn.t/ and the efficiency E

From the definition of mn.t/

mn.t/ Dm.t/

jminm.t/jDm.t/

j 1jD m.t/

The efficiency is

E Da2hm2

n.t/i

1C a2hm2n.t/i


CONTENTS

To obtain hm2n.t/i we form the time average

hm2n.t/i D

1

Tm

"Z Tm=3

0

.2/2 dt C

Z Tm

Tm=3

.1/2 dt

#D

1

Tm

Tm

3 4C

2Tm

3 1

D4

3C2

3D6

3D 2

thus

E D2a2

1C 2a2

The best AM efficiency we can achieve with this waveform iswhen a D 1

Eff

ˇaD1D2

3D 0:67 or 67%

Suppose that the message signal is m.t/ as given here

Now minm.t/ D 2 and mn.t/ D m.t/=2 and

hm2n.t/i D

1

3 .1/2 C

2

3 .1=2/2 D

1

2

The efficiency in this case is

Eff D.1=2/a2

1C .1=2/a2D

a2

2C a2

Now when a D 1 we have Eff D 1=3 or just 33.3%

Note that for 50% duty cycle squarewave the efficiency maxi-mum is just 50%



Example 3.3: Multiple Sinusoids

Suppose that m.t/ is a sum of multiple sinusoids (multi-toneAM)

m.t/ D

MXkD1

Ak cos.2fkt C k/

where M is the number of sinusoids, fk values might be con-strained over some band of frequencies W , e.g., fk W , andthe phase values k can be any value on Œ0; 2

To find mn.t/ we need to find minm.t/

A lower bound on minm.t/ is PM

kD1Ak; why?

The worst case value may not occur in practice depending uponthe phase and frequency values, so we may have to resort to anumerical search or a plot of the waveform

Suppose that M D 3 with fk D f65; 100; 35g Hz, Ak Df2; 3:5; 4:2g, and k D f0; =3;=4g rad.

>> [m,t] = M_sinusoids(1000,[65 100 35],[2 3.5 4.2],...[0 pi/3 -pi/4], 20000);>> plot(t,m)

>> min(m)

ans = -7.2462e+00

>> -sum([2 3.5 4.2]) % worst case minimum value

ans = -9.7000e+00

>> subplot(311)>> plot(t,(1 + 0.25*m/abs(min(m))).*cos(2*pi*1000*t))>> hold


CONTENTS

Current plot held>> plot(t,1 + 0.25*m/abs(min(m)),’r’)>> subplot(312)>> plot(t,(1 + 0.5*m/abs(min(m))).*cos(2*pi*1000*t))>> holdCurrent plot held>> plot(t,1 + 0.5*m/abs(min(m)),’r’)>> subplot(313)>> plot(t,(1 + 1.0*m/abs(min(m))).*cos(2*pi*1000*t))>> holdCurrent plot held>> plot(t,1 + 1.0*m/abs(min(m)),’r’)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−8

−6

−4

−2

0

2

4

6

8

Time (s)

m(t

) A

mpl

itude

min m(t)

Finding minm.t/ graphically

The normalization factor is approximately given by 7.246, thatis

mn.t/ Dm.t/

7:246

Shown below are plots of xc.t/ for a D 0:25; 0:5 and 1 usingfc D 1000 Hz



0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−2

0

2

x c(t),

a =

0.2

5

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−2

0

2

x c(t),

a =

0.5

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−2

0

2

x c(t),

a =

1.0

Time (s)

Modulation index comparison (fc D 1000 Hz)

To obtain the efficiency of multi-tone AM we first calculatehm2

n.t/i assuming unique frequencies

hm2n.t/i D

MXkD1

A2k2jminm.t/j2

D22 C 3:52 C 4:22

2 7:2462D 0:3227

The maximum efficiency is just

Eff

ˇaD1D

0:3227

1C 0:3227D 0:244 or 24.4%


CONTENTS

A remaining interest is the spectrum of xc.t/

Xc.f / DAc

2

ı.f fc/C ı.f C fc/

CaAc

4

MXkD1

Ak

hejkı.f .fc C fk//

C ejkı.f C .fc C fk//i

(USB terms)

CaAc

4

MXkD1

Ak

hejkı.f .fc fk//

C ejkı.f C .fc fk//i

(LSB terms)

ï1000 ï800 ï600 ï400 ï200 0 200 400 600 800 10000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Frequency (Hz)

Ampl

itude

Spe

ctra

(|X c(f)

|)

SymmetricalSidebands for

a = 0.5

Carrier with Ac = 1

Amplitude spectra



3.1.3 Single-Sideband Modulation

In the study of DSB it was observed that the USB and LSBspectra are related, that is the magnitude spectra about fc haseven symmetry and phase spectra about fc has odd symmetry

The information is redundant, meaning thatm.t/ can be recon-structed from one or the other sidebands

Transmitting just the USB or LSB results in single-sideband(SSB)

For m.t/ having lowpass bandwidth of W the bandwidth re-quired for DSB, centered on fc is 2W

Since SSB operates by transmitting just one sideband, the trans-mission bandwidth is reduced to just W

M(f)

f

fc - W

fc - W f

c+W

fc+Wf

c

fc

fc

XDSB

(f)

f

XSSB

(f)XSSB

(f)

f

LSB USB

W

USBremoved

LSBremoved

DSB to two forms of SSB: USSB and LSSB

The filtering required to obtain an SSB signal is best explainedwith the aid of the Hilbert transform, so we divert from text


CONTENTS

Chapter 3 back to Chapter 2 to briefly study the basic proper-ties of this transform

Hilbert Transform

The Hilbert transform is nothing more than a filter that shiftsthe phase of all frequency components by =2, i.e.,

H.f / D j sgn.f /

where

sgn.f / D

8<:1; f > 0

0; f D 0

1; f < 0

The Hilbert transform of signal x.t/ can be written in terms ofthe Fourier transform and inverse Fourier transform

Ox.t/ D F1 j sgn.f /X.f /

D h.t/ x.t/

where h.t/ D F1fH.f /g

We can find the impulse response h.t/ using the duality theo-rem and the differentiation theorem

d

dfH.f /

F ! .j 2t/h.t /

where here H.f / D j sgn.f /, so

d

dfH.f / D 2jı.f /



Clearly,F1f2jı.f /g D 2j

so

h.t/ D2j

j 2tD

1

t

and1

t

F ! j sgn.f /

In the time domain the Hilbert transform is the convolutionintegral

Ox.t/ D

Z1

1

x./

.t /d D

Z1

1

x.t /

d

Note that since the Hilbert transform of x.t/ is a =2 phaseshift, the Hilbert transform of Ox.t/ is

OOx.t/ D x.t/

why? observe that .j sgn.f //2 D 1

Example 3.4: x.t/ D cos!0t

By definition

OX.f / D j sgn.f / 1

2

ı.f f0/C ı.f C f0/

D j

1

2ı.f f0/C j

1

2ı.f C f0/


CONTENTS

so from ej!0tF) ı.f f0/

Ox.t/ D j1

2ej!0t C j

1

2ej!0t

Dej!0t ej!0t

2jD sin!0t

or2cos!0t D sin!0t

It also follows that

2sin!0t D22cos!0t D cos!0t

since OOx.t/ D x.t/

Hilbert Transform Properties

1. The energy (power) in x.t/ and Ox.t/ are equal

The proof follows from the fact that jY.f /j2 D jH.f /j2jX.f /j2

and jj sgn.f /j2 D 1

2. x.t/ and Ox.t/ are orthogonal, that isZ1

1

x.t/ Ox.t/ dt D 0 (energy signal)

limT!1

1

2T

Z T

T

x.t/ Ox.t/ dt D 0 (power signal)



The proof follows for the case of energy signals by generaliz-ing Parseval’s theoremZ

1

1

x.t/ Ox.t/ dt D

Z1

1

X.f / OX.f / df

D

Z1

1

.j sgn.f //„ ƒ‚ …odd

jX.f /j2„ ƒ‚ …even

df D 0

3. Given signals m.t/ and c.t/ such that the corresponding spec-tra are

M.f / D 0 for jf j > W (a lowpass signal)C.f / D 0 for jf j < W (c.t/ a highpass signal)

then3m.t/c.t/ D m.t/ Oc.t/

Example 3.5: c.t/ D cos!0t

Suppose that M.f / D 0 for jf j > W and f0 > W then

5m.t/ cos!0t D m.t/2cos!0tD m.t/ sin!0t

Analytic Signals

Define analytic signal z.t/ as

z.t/ D x.t/C j Ox.t/

where x.t/ is a real signal


CONTENTS

The envelope of z.t/ is jz.t/j and is related to the envelopediscussed with DSB and AM signals

The spectrum of an analytic signal has single-sideband charac-teristics

In particular for zp.t/ D x.t/C j Ox.t/

Zp.f / D X.f /C j˚ j sgn.f /X.f /

D X.f /

1C sgn.f /

D

(2X.f /; f > 0

0; f < 0

Note: Only positive frequencies present

Similarly for zn.t/ D x.t/ j Ox.t/

Zn.f / D X.f /1 sgn.f /

D

(0; f > 0

2X.f /; f < 0



W-W

W-W

W-W

f

f

f

X(f)

Zp(f)

Zn(f)

2

2

1

The spectra of analytic signals can suppress positive or negativefrequencies

Return to SSB Development

SidebandFilter

Accosω

ct

m(t)

xDSB

(t)x

SSB(t)

LSB or USB

Basic SSB signal generation

In simple terms, we create an SSB signal from a DSB signalusing a sideband filter

The mathematical representation of LSSB and USSB signalsmakes use of Hilbert transform concepts and analytic signals


CONTENTS

DSB Signal Starting Point

Formation of HL(f)

-fc

fc

-fc

fc

f

f

f

f

sgn(f + fc)/2

-sgn(f - fc)/2

+1/2

+1/2

-1/2

-1/2

HL(f) = [sgn(f + f

c) - sgn(f - f

c)]/21

An ideal LSSB filter

From the frequency domain expression for the LSSB, we canultimately obtain an expression for the LSSB signal, xcLSSB.t/,in the time domain

Start with XDSB.f / and the filter HL.f /

XcLSSB.f / D1

2AcM.f C fc/CM.f fc/

1

2

sgn.f C fc/ sgn.f fc/



XcLSSB.f /rewrite 1D

1

4AcM.f C fc/sgn.f C fc/

M.f fc/sgn.f fc/

1

4AcM.f C fc/sgn.f fc/

M.f fc/sgn.f C fc/

rewrite 2D

1

4AcM.f C fc/sgn.f C fc/

M.f fc/sgn.f fc/

C1


The inverse Fourier transform of the second term is DSB, i.e.,

1

2Acm.t/ cos!ct

F !

1


The first term can be inverse transformed using the Hilbert

transform definition

Om.t/F ! j sgn.f / M.f /

soF1

˚M.f ˙ fc/sgn.f ˙ fc/

D j Om.t/ej!ct

since m.t/e˙j!ctF !M.f ˙ fc/

Thus

1

4AcF1

˚M.f Cfc/sgn.f Cfc/M.f fc/sgn.f fc/

D

1

4Acj Om.t/ej!ct j Om.t/ej!ct

D

1

2Om.t/ sin!ct


CONTENTS

Finally,

xcLSSB.t/ D1

2Acm.t/ cos!ct C

1

2Ac Om.t/ sin!ct

Similarly for USSB it can be shown that

xcUSSB.t/ D1

2Acm.t/ cos!ct

1

2Ac Om.t/ sin!ct

The direct implementation of SSB is very difficult due to therequirements of the filter

By moving the phase shift frequency from fc down to DC (0Hz) the implementation is much more reasonable (this appliesto a DSP implementation as well)

The phase shift is not perfect at low frequencies, so the modu-lation must not contain critical information at these frequencies

H(f) = -jsgn(f)

0o

-90o

m(t)

xc(t)

sinωct

cosωct

cosωct

Carrier Osc.+

+-LSBUSB

Phase shift modulator for SSB



Demodulation

The coherent demodulator first discussed for DSB, also worksfor SSB

LPFxr(t)

d(t) yD(t)

4cos[2πfct + θ(t)]

1/Ac scale factor

included

Coherent demod for SSB

Carrying out the analysis to d.t/, first we have

d.t/ D1

2Acm.t/ cos!ct ˙ Om.t/ sin!ct

4 cos.!ct C .t//

D Acm.t/ cos .t/C Acm.t/ cosŒ2!ct C .t/ Ac Om.t/ sin .t/˙ Ac Om.t/ sinŒ2!ct C .t/

so

yD.t/ D m.t/ cos .t/ Om.t/ sin .t/.t/ small' m.t/ Om.t/.t/

– The Om.t/ sin .t/ term represents crosstalk

Another approach to demodulation is to use carrier reinsertionand envelope detection

EnvelopeDetector

xr(t)

e(t)yD(t)

Kcosωct


CONTENTS

e.t/ D xr.t/CK cos!ct

D

1

2Acm.t/CK

cos!ct ˙

1

2Ac Om.t/ sin!ct

To proceed with the analysis we must find the envelope of e.t/,which will be the final output yD.t/

Finding the envelope is a more general problem which will beuseful in future problem solving, so first consider the envelopeof

x.t/ D a.t/„ƒ‚…inphase

cos!ct b.t/„ƒ‚…quadrature

sin!ct

D Re˚a.t/ej!ct C jb.t/ej!ct

D Re

˚Œa.t/C jb.t/„ ƒ‚ …QR.t/Dcomplex envelope

ej!ct

In a phasor diagram x.t/ consists of an inphase or direct com-ponent and a quadrature component

R(t)

θ(t)

a(t)

b(t)

In-phase - I

Quadrature - Q

Note: R(t) =

R(t)ejθ(t) = a(t) + jb(t)

~



where the resultant R.t/ is such that

a.t/ D R.t/ cos .t/b.t/ D R.t/ sin .t/

which implies that

x.t/ D R.t/

cos .t/ cos!ct sin .t/ sin!ct

D R.t/ cos!ct C .t/

where .t/ D tan1Œb.t/=a.t/

The signal envelope is thus given by

R.t/ Dpa2.t/C b2.t/

The output of an envelope detector will be R.t/ if a.t/ andb.t/ are slowly varying with respect to cos!ct

In the SSB demodulator

yD.t/ D

s1

2Acm.t/CK

2C

1

2Ac Om.t/

2 If we choose K such that .Acm.t/=2CK/2 .Ac Om.t/=2/

2,then

yD.t/ '1

2Acm.t/CK

Note:

– The above analysis assumed a phase coherent reference

– In speech systems the frequency and phase can be ad-justed to obtain intelligibility, but not so in data systems


CONTENTS

– The approximation relies on the binomial expansion

.1C x/1=2 ' 1C1

2x for jxj 1

Example 3.6: Noncoherent Carrier Reinsertion

Let m.t/ D cos!mt , !m !c and the reinserted carrier beK cosŒ.!c C!/t

Following carrier reinsertion we have

e.t/ D1

2Ac cos!mt cos!ct

1

2Ac sin!mt sin!ct CK cos

.!c C!/t

D1

2Ac cos

.!c ˙ !m/t

CK cos

.!c C!/t

We can write e.t/ as the real part of a complex envelope times

a carrier at either !c or !c C!

In this case, since K will be large compared to Ac=2, we write

e.t/ D1

2AcRe

ne˙j!mtej!ct

oCKRe

n1 ej.!cC!/t

oD Re

n 12Ace

j.˙!m!/t CK

„ ƒ‚ …complex envelope QR.t/

ej.!cC!/to



Finally expanding the complex envelope into the real and imag-inary parts we can find the real envelope R.t/

yD.t/ Dhn12Ac cosŒ˙!m C!/tCK

o2C

n12Ac sinŒ.˙!m C!/t

o2i1=2'1

2Ac cosŒ.!m !/tCK

where the last line follows for K Ac

Note that the frequency error ! causes the recovered mes-sage signal to shift up or down in frequency by !, but notboth at the same time as in DSB, thus the recovered speechsignal is more intelligible

3.1.4 Vestigial-Sideband Modulation

Vestigial sideband (VSB) is derived by filtering DSB such thatone sideband is passed completely while only a vestige remainsof the other

Why VSB?

1. Simplifies the filter design

2. Improves the low-frequency response and allows DC topass undistorted

3. Has bandwidth efficiency advantages over DSB or AM,similar to that of SSB


CONTENTS

A primary application of VSB is the video portion of analogtelevision (note HDTV replaces this in the US with 8VSB1)

The generation of VSB starts with DSB followed by a filterthat has a 2ˇ transition band, e.g.,

jH.f /j D

8<:0; f < Fc ˇf .fcˇ/

2ˇ; fc ˇ f fc C ˇ

1; f > fc C ˇ

ffcf

c - β f

c + β

|H(f)|

1

Ideal VSB transmitter filter amplitude response

VSB can be demodulated using a coherent demod or using car-rier reinsertion and envelope detection

ff - f

1f + f

1f - f

2f + f

2fc

B/2A(1 - ε)/2

Aε/2

Transmitted Two-Tone Spectrum(only single-sided shown)

0

Two-tone VSB signal1http://www.tek.com/document/primer/fundamentals-8vsb



Suppose the message signal consists of two tones

m.t/ D A cos!1t C B cos!2t

Following the DSB modulation and VSB shaping,

xc.t/ D1

2A cos.!c !1/t

C1

2A.1 / cos.!c C !1/t C

1

2B cos.!c C !2/t

A coherent demod multiplies the received signal by 4 cos!ctto produce

e.t/ D A cos!1t C A.1 / cos!1t C B cos!2tD A cos!1t C B cos!2t

which is the original message signal

The symmetry of the VSB shaping filter has made this possible

In the case of broadcast TV the carrier in included at the trans-mitter to insure phase coherency and easy demodulation at theTV receiver (VSB + Carrier)

– Very large video carrier power was required for typicalTV station, i.e., greater than 100,000 W

– To make matters easier still, the precise VSB filtering isnot performed at the transmitter due to the high powerrequirements, instead the TV receiver did this

– Further study is needed on today’s 8VSB


CONTENTS

0

0

-0.75

-0.75 0.75

-1.75 4.0

4.0

4.5

4.75

4.75(f - f

cv) MHz

(f - fcv

) MHz

Video CarrierAudio Carrier

TransmitterOutput

ReceiverShapingFilter

1

2β interval

Broadcast TV transmitter spectrum and receiver shaping filter

3.1.5 Frequency Translation and Mixing

Used to translate baseband or bandpass signals to some newcenter frequency

BPFatf2 ff

f1 f

2

Local oscillator of the form

e(t)

m(t)cosω1t

1 2 LO2cos[( )t cos2 )] ( tw w w± =

Frequency translation system

Assuming the input signal is DSB of bandwidth 2W the mixer(multiplier) output is

e.t/ D m.t/ cos.!1t /

local osc (LO)‚ …„ ƒ2 cos.!1 ˙ !2/t

D m.t/ cos.!2t /Cm.t/ cosŒ.2!1 ˙ !2/t



The bandpass filter bandwidth needs to be at least 2W Hz wide

Note that if an input of the form k.t/ cosŒ.!1˙2!2/t is presentit will be converted to !2 also, i.e.,

e.t/ D k.t/ cos.!2t /C k.t/ cosŒ.2!1 ˙ 3!2/t ;

and the bandpass filter output is k.t/ cos.!2t /

The frequencies !1˙2!2 are the image frequencies of !1 withrespect to !LO D !1 ˙ !2

Example 3.7: AM Broadcast Superheterodyne Receiver

TunableRF-Amp

IF Filt/Amp

LocalOsc.

EnvDet

AudioAmp

Joint tuning

Automatic gaincontrol

fIF

AM Broadcast Specs: fc = 540 to 1600 kHz on 10 kHz spacings

carrier stability Modulated audio flat 100 Hz to 5 kHz Typical f

IF = 455 kHz

For AM BT = 2W

Classical AM superheterodyne receiver

We have two choices for the local oscillator, high-side or low-side tuning


CONTENTS

– Low-side: 540455 fLO 1600455 or 85 fLO

1145, all frequencies in kHz

– High-side: 540 C 455 fLO 1600 C 455 or 995 fLO 2055, all frequencies in kHz

The high-side option is advantageous since the tunable oscil-lator or frequency synthesizer has the smallest frequency ratiofLO,max=fLO,min D 2055=995 D 2:15

Suppose the desired station is at 560 kHz, then with high-sidetuning we have fLO D 560C 455 D 1015 kHz

The image frequency is at fimage D fc C 2fIF D 560 C 2

455 D 1470 kHz (note this is another AM radio station centerfrequency

f (kHz)

f (kHz)

f (kHz)

f (kHz)0

Input

fLO

MixerOutput

ImageOut ofmixer

455 560 1470

1470

1575(560+1015)

2485(1470+1015)

1015(560+455)

455

Desired Potential Image

This is removedwith RF BPF

IF BPF

1470-1015

1015-560

fIF

fIF

BRF

BIF

Receiver frequency plan including images



Example 3.8: A Double-Conversion Receiver

TunableRF-Amp

10.7 MHzIF BPF

455 kHzIF BPF

1stLO

2ndLO

FMDemod

fc = 162.475 MHz

(WX #4)

fLO1

= 173.175 MHz fLO2

= 11.155 MHz

10.7 MHz &335.65 MHz

455 kHz &21.855 MHz

Double-conversion superheterodyne receiver (Lab 4)

Consider a frequency modulation (FM) receiver that uses double-conversion to receive a signal on carrier frequency 162.475MHz (weather channel #4 here in Colorado Springs)

– Frequency modulation will be discussed in the next sec-tion

The dual-conversion allows good image rejection by using a10.7 MHz first IF and then can provide good selectivity byusing a second IF at 455 kHz; why?

– The ratio of bandwidth to center frequency can only be sosmall in a low loss RF filter

– The second IF filter can thus have a much narrower band-width by virtue of the center frequency being much lower

A higher first IF center frequency moves the image signal fur-ther away from the desired signal


CONTENTS

– For high-side tuning we have fimage D fc C 2fIF D fc C

21:4 MHz

Double-conversion receivers are more complex to implement

Mixers

The multiplier that is used to implement frequency translationis often referred to as a mixer

In the world of RF circuit design the term mixer is more ap-propriate, as an ideal multiplier is rarely available

Instead active and passive circuits that approximate signal mul-tiplication are utilized

The notion of mixing comes about from passing the sum of twosignals through a nonlinearity, e.g.,

y.t/ D Œa1x1.t/C a2x2.t/2C other terms

D a21x21.t/C 2a1a2x1.t/x2.t/C a

22x

22.t/

In this mixing application we are most interested in the centerterm

ydesired.t/ D 2a1a2x1.t/ x2.t/

Clearly this mixer produces unwanted terms (first and third),

and in general many other terms, since the nonlinearity willhave more than just a square-law input/output characteristic



A diode or active device can be used to form mixing productsas described above, consider the dual-gate MEtal Semiconduc-tor FET (MESFET) mixer shown below

+5V

R2

C4

L3

47pF

L4

C8C7Q1NE25139

IF270nH270nH

82pF

C3

42pF

0.01uF

10Ω

C8R1 C5

47pF0.01uF

270Ω

L15 turns, 28 AWG.050 I.D.

C10.5pFLO

5 turns, 28 AWG.050 I.D.

C2

L2

0.5pF

RF

G1

G2D

S

Dual-Gate MESFET Active Mixer

VRF

VOUT

VIN

VLO

zL

Nonlinear Device

Mixer concept

The double-balanced mixer (DBM), which can be constructedusing a diode ring, provides better isolation between the RF,LO, and IF ports

When properly balanced the DBM also allows even harmonicsto be suppressed in the mixing operation


CONTENTS

A basic transformer coupled DBM, employing a diode ring, isshown below, followed by an active version

The DBM is suitable for use as a phase detector in phase-locked loop applications

02 91 81 71

DNG

+FI

-FI

DNG

61DNG

13

12

11

14

15

GND

GND

GND

LO2LO2

LO1LO1

4

3

2

1

RFBIAS

TAP

RFRFIN

5GND

LO SELECT

GND

6 7 8 9

VCC VCCD NG

D NG

LE SOL

01

MAX9982

IF OUT

4:1 (200:50)TRANSFORMER

41

25V

3C9

C10

C7

C6

C11C8

R3L1

R4L2

T1 6

5V5V

C5C4

R1

C2

C1

C3

825 MHz to 915 MHz SiGe High-Linearity Active DBM

vo(t)

mixer

D3 D4

D1D2inputLO

R G

LO source

vp(t)

inputRF

R G

RF source

R L

IF load

vi (t)

IFout

Passive Double-Balanced Mixer (DBM)



Example 3.9: Single Diode Mixer

TRANSIENT

PspecTranf

ap_dio_1N4148_19930601RAM_ModTunedVtSine

VtSine

R

R

PspecTran

Tran

RR

VtSine

Splitter_Resistive_6dB

I_Probe

D1R3

R6MOD1SRC4

SRC5

R8

PspecTran1

Tran1

R1R2

SRC2

X3

I_Probe1R=50 Ohm

R=50 OhmModIndex=0.5Fnom=560 kHzRout=50 Ohm

Vdc=0 VAmplitude=0.5 VFreq=560 kHz

Vdc=0 VAmplitude=1.0 VFreq=10 kHz

R=50 Ohm

PspecTran1=pspec_tran(V_out, 0,I_Probe1.i,5kHz,200) MaxTimeStep=10.0 nsecStopTime=400.0 usec

R=50 OhmR=50 Ohm

Freq=1.015 MHzAmplitude=2 VVdc=0 V

V_out

420 440 460 480 500 520 540 560 580400 600

-40

-30

-20

-50

-10

freq, KHz

Pow

er (d

Bm)

455 kHz IF Output Input Signal (RF)

AM Signal at 560 kHz, 10 kHz Tone Message Signal Sum Mixing Diode

VI Probe for PSD

10 kHzsidebands

LO Mix @ 1015 kHz

ADS single diode mixer simulation: 560 kHz! 455 kHz


CONTENTS

3.2 Interference

Interference is a fact of life in communication systems. A throughunderstanding of interference requires a background in random sig-nals analysis (Chapter 6 of the text), but some basic concepts canbe obtained by considering a single interference at fc C fi that liesclose to the carrier fc

3.2.1 Interference in Linear Modulation

ffc - f

mfc + f

mfc + f

ifc

Xr(f)

Sing

le-S

ided

Spe

ctru

m

12A

m

12A

m

Ac

Ai

AM carrier with single tone interference

If a single tone carrier falls within the IF passband of the re-ceiver what problems does it cause?

Coherent Demodulator

xr.t/ DAc cos!ct C Am cos!mt cos!ct

C Ai cos.!c C !i/t

– We multiply xr.t/ by 2 cos!ct and lowpass filter

yD.t/ D Am cos!mt C Ai cos!i t„ ƒ‚ …interference

Envelope Detection: Here we need to find the received enve-lope relative to the strongest signal present


3.2. INTERFERENCE

– Case Ac Ai

– We will expand xr.t/ in complex envelope form by firstnoting that

Ai cos.!cC!i/t D Ai cos!i t cos!ctAi sin!i t sin!ct

now,

xr.t/ D Re˚Ac C Am cos!mt C Ai cos!i t

jAi sin!i tej!ct

D Re

˚QR.t/ej!ct

so

R.t/ D j QR.t/j

D

h.Ac C Am cos!mt C Ai cos!i t /2

C .Ai sin!i t /2i1=2

' Ac C Am cos!mt C Ai cos!i t

assuming that Ac Ai

– Finally,

yD.t/ ' Am cos!mt C Ai cos!i t„ ƒ‚ …interference

– Case Ai >> Ac

– Now the interfering term looks like the carrier and the re-maining terms look like sidebands, LSSB sidebands rela-tive to fc C fi to be specific


CONTENTS

– From SSB envelope detector analysis we expect

yD.t/ '1

2Am cos.!i C !m/t C Ac cos!i t

C1

2Am cos.!i !m/t

and we conclude that the message signal is lost!


3.3. SAMPLING THEORY

3.3 Sampling Theory

We now return to text Chapter 2, Section 8, for an introduc-tion/review of sampling theory

Consider the representation of continuous-time signal x.t/ bythe sampled waveform

xı.t/ D x.t/

"1X

nD1

ı.t nTs/

#D

1XnD1

x.nTs/ı.t nTs/

x(t) xδ(t)

tt0 0 T

s-T

s2T

s3T

s4T

s5T

s

Sampling

How is Ts selected so that x.t/ can be recovered from xı.t/?

Uniform Sampling Theorem for Lowpass Signals

GivenFfx.t/g D X.f / D 0; for f > W

then choose

Ts <1

2Wor fs > 2W .fs D 1=Ts/

to reconstruct x.t/ from xı.t/ and pass xı.t/ through an idealLPF with cutoff frequency W < B < fs W

2W D Nyquist frequencyfs=2 D folding frequency


CONTENTS

proof:

Xı.f / D X.f /

"fs

1XnD1

ı.f nfs/

#but X.f / ı.f nfs/ D X.f nfs/, so

Xı.f / D fs

1XnD1

X.f nfs/

-W W

-W W

0

0

f

f

X(f)X0

Xδ(f)

-fs

fs

X0fs

fs-W

Guard band= f

s - 2W

Lowpass reconstruction filter

. . . . . .

f-2f

sfs

-fs

2fs

X0fs

. . . . . .

Aliasingfs < 2W

fs > 2W

Spectra before and after sampling at rate fs

As long as fs W > W or fs > 2W there is no aliasing(spectral overlap)



To recover x.t/ from xı.t/ all we need to do is lowpass filterthe sampled signal with an ideal lowpass filter having cutofffrequency W < fcutoff < fs W

In simple terms we set the lowpass bandwidth to the foldingfrequency, fs=2

Suppose the reconstruction filter is of the form

H.f / D H0…

f

2B

ej 2f t0

we then choose W < B < fs W

For input Xı.f /, the output spectrum is

Y.f / D fsH0X.f /ej 2f t0

and in the time domain

y.t/ D fsH0x.t t0/

If the reconstruction filter is not ideal we then have to designthe filter in such a way that minimal desired signal energy is re-moved, yet also minimizing the contributions from the spectraltranslates either side of the n D 0 translate

The reconstruction operation can also be viewed as interpolat-ing signal values between the available sample values

Suppose that the reconstruction filter has impulse response h.t/,


CONTENTS

then

y.t/ D

1XnD1

x.nTs/h.t nTs/

D 2BH0

1XnD1

x.nTs/sincŒ2B.t t0 nTs/

where in the last lines we invoked the ideal filter describedearlier

Uniform Sampling Theorem for Bandpass Signals

If x.t/ has a single-sided bandwidth of W Hz and

Ffx.t/g D 0 for f > fu

then we may choose

fs D2fu

m

where

m D

fu

W

;

which is the greatest integer less than or equal to fu=W

Example 3.10: Bandpass signal sampling

4 2 2 4

0.20.40.60.81

f

X(f)

Input signal spectrum



In the above signal spectrum we see that

W D 2; fu D 4 fu=W D 2 ) m D 2

sofs D

2.4/

2D 4

will work

The sampled signal spectrum is

Xı.f / D 4

1XnD1

X.f nfs/

15 10 5 5 10 15

1234

f

Xδ(f)

Recover withbandpass filter

fs

2fs

3fs

-fs

-2fs

-3fs

Spectrum after sampling


CONTENTS

3.4 Analog Pulse Modulation

The message signal m.t/ is sampled at rate fs D 1=Ts

A characteristic of the transmitted pulse is made to vary in aone-to-one correspondence with samples of the message signal

A digital variation is to allow the pulse attribute to take onvalues from a finite set of allowable values

3.4.1 Pulse-Amplitude Modulation (PAM)

PAM produces a sequence of flat-topped pulses whose ampli-tude varies in proportion to samples of the message signal

Start with a message signal, m.t/, that has been uniformlysampled

mı.t/ D

1XnD1

m.nTs/ı.t nTs/

The PAM signal is

mc.t/ D

1XnD1

m.nTs/…

t .nTs C =2/

m(t)

mc(t)

τ

0 τ Ts

2Ts

3Ts

4Ts

t

PAM waveform


3.4. ANALOG PULSE MODULATION

It is possible to create mc.t/ directly from mı.t/ using a zero-order hold filter, which has impulse response

h.t/ D …

t =2

and frequency response

H.f / D sinc.f /ejf

h(t)mδ(t) mc(t)

How does h.t/ change the recovery operation from the case ofideal sampling?

– If Ts we can get by with just a lowpass reconstruc-tion filter having cutoff frequency at fs=2 D 2=Ts

– In general, there may be a need for equalization if is onthe order of Ts=4 to Ts=2

f-W W f

s-fs

sinc() function envelope

Lowpass reconstruction filter

Lowpassmc(t) m(t)

Recovery of m.t/ from mc.t/


CONTENTS

3.4.2 Pulse-Width Modulation (PWM)

A PWM waveform consists of pulses with width proportionalto the sampled analog waveform

For bipolar m.t/ signals we may choose a pulse width of Ts=2to correspond to m.t/ D 0

The biggest application for PWM is in motor control

It is also used in class D audio power amplifiers

A lowpass filter applied to a PWM waveform recovers themodulation m.t/

20 10 10 20

1

0.5

0.5

1PWM Signal

Analog input m(t)

t

Example PWM signal

3.4.3 Pulse-Position Modulation

With PPM the displacement in time of each pulse, with re-spect to a reference time, is proportional to the sampled analogwaveform

The time axis may be slotted into a discrete number of pulsepositions, then m.t/ would be quantized

Digital modulation that employs M slots, using nonoverlap-ping pulses, is a form of M -ary orthogonal communications


3.5. DELTA MODULATION AND PCM

– PPM of this type is finding application in ultra-widebandcommunications

20 10 10 20

1

0.5

0.5

1PPMSignal

Analog input m(t)

t

Example PPM signal

3.5 Delta Modulation and PCM

This section considers two pure digital pulse modulation schemes

Pure digital means that the output of the modulator is a binarywaveform taking on only discrete values

3.5.1 Delta Modulation (DM)

The message signal m.t/ is encoded into a binary sequencewhich corresponds to changes in m.t/ relative to referencewaveform ms.t/

DM gets its name from the fact that only the difference fromsample-to-sample is encoded

The sampling rate in combination with the step size are the twoprimary controlling modulator design parameters


CONTENTS

1

-1

+

−

m(t)

ms(t) =

xc(t)∆(t)

Pulse Modulator

d(t)

δ0

Control the step size

Delta modulator with step size parameter ı0

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

−1

−0.5

0

0.5

1

Time (s)

m(t

) an

d m

s(t)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05

−1

−0.5

0

0.5

1

Time (s)

x c(t)

m(t) (blue)m

s(t) (red)

δ0 = 0.15Slope

overload

Start-up transient

Delta modulator waveforms

The maximum slope that can be followed is ı0=Ts



A MATLAB DM simulation function is given below

function [t_o,x,ms] = DeltaMod(m,fs,delta_0,L)% [t,x,ms] = DeltaMod(m,fs,delta_0,L)%% Mark Wickert, April 2006

n = 0:(L*length(m))-1;t_o = n/(L*fs);ms = zeros(size(m));x = zeros(size(m));ms_old = 0; % zero initial conditionfor k=1:length(m)

x(k) = sign(m(k) - ms_old);ms(k) = ms_old + x(k)*delta_0;ms_old = ms(k);

end

x = [x; zeros(L-1,length(m))];x = reshape(x,1,L*length(m));

The message m.t/ can be recovered from xc.t/ by integratingand then lowpass filtering to remove the stair step edges (low-pass filtering directly is a simplification)

Slope overload can be dealt with through an adaptive scheme

– If m.t/ is nearly constant keep the step size ı0 small

– If m.t/ has large variations, a larger step size is needed

With adaptive DM the step size is controlled via a variable gainamplifier, where the gain is controlled by square-law detectingthe output of a lowpass filter acting on xc.t/


CONTENTS

1

-1

+

−

m(t)

ms(t)

xc(t)∆(t)

Pulse Modulator

d(t)

LPF( )2

VGA

Means to obtain a variable step size DM

3.5.2 Pulse-Code Modulation (PCM)

Each sample of m.t/ is mapped to a binary word by

1. Sampling

2. Quantizing

3. Encoding

Sampler Quantizer Encoder

Sample&

Hold

Analog toDigital

Converter

Parallelto Serial

Converter

PCM Output

Serial Data

m(t)

m(t)

EquivalentViews

n



EncodedOutput

111110101100011010001000

Quant.Level

76543210 t

0 Ts

2Ts

3Ts

4Ts

5Ts

6Ts

7Ts

Quantizer Bits: n = 3, q = 2n = 8

Encoded Serial PCM Data: 001 100 110 111 110 100 010 010 ...

m(t)

3-Bit PCM encoding

Assume that m.t/ has bandwidth W Hz, then

– Choose fs > 2W

– Choose n bits per sample (q D 2n quantization levels)

– ) 2nW binary digits per second must be transmitted

Each pulse has width no more than

./max D1

2nW;

so using the fact that the lowpass bandwidth of a single pulseis about 1=.2/ Hz, we have that the lowpass transmissionbandwidth for PCM is approximately

B ' kW n;

where k is a proportionality constant

When located on a carrier the required bandwidth is doubled


CONTENTS

Binary phase-shift keying (BPSK), mentioned earlier, is a pop-ular scheme for transmitting PCM using an RF carrier

Many other digital modulation schemes are possible

The number of quantization levels, q D log2 n, controls thequantization error, assumingm.t/ lies within the full-scale rangeof the quantizer

Increasing q reduces the quantization error, but also increasesthe transmission bandwidth

The error betweenm.kTs/ and the quantized valueQŒm.kTs/,denoted e.n/, is the quantization error

If n D 16, for example, the ratio of signal power in the samplesofm.t/, to noise power in e.n/, is about 95 dB (assumingm.t/stays within the quantizer dynamic range)

Example 3.11: Compact Disk Digital Audio

CD audio quality audio is obtained by sampling a stereo sourceat 44.1 kHz

PCM digitizing produces 16 bits per sample per L/R audiochannel


3.6. MULTIPLEXING

Synch(27 bits)

SubCode Data (96 bits)Data (96 bits)

Parity(32 bits)

Parity(32 bits)

0.163 mm

One Frame of 12 Audio Samples

(8 bits)

CD recoding frame format

The source bit rate is thus 2 16 44:1ksps D 1:4112 Msps

Data framing and error protection bits are added to bring thetotal bit count per frame to 588 bits and a serial bit rate of4.3218 Mbps

3.6 Multiplexing

It is quite common to have multiple information sources lo-cated at the same point within a communication system

To simultaneously transmit these signals we need to use someform of multiplexing


CONTENTS

There is more than one form of multiplexing available to thecommunications engineer

In this chapter we consider time-division multiplexing, whilein Chapter 4 frequency division multiplexing is described

3.6.1 Time-Division Multiplexing (TDM)

Time division multiplexing can be applied to sampled analogsignals directly or accomplished at the bit level

We assume that all sources are sample at or above the Nyquistrate

Both schemes are similar in that the bandwidth or data rate ofthe sources being combined needs to be taken into account toproperly maintain real-time information flow from the sourceto user

For message sources with harmonically related bandwidths wecan interleave samples such that the wideband sources are sam-pled more often

To begin with consider equal bandwidth sources


3.6. MULTIPLEXING

Info. Source 1

Info. Source 2

Info. Source N

...

Channel

Info.User 1

Info. User 2

Info. User N

...

Commutators

For equal bandwidth: s1s

2s

3 s

1s

2s

3 s

1s

2s

3 s

1s

2s

3 s

1s

2s

3 s

1s

2s

3 s

1s

2s

3 ....

SynchronizationRequired

Analog TDM (equal bandwidth sources)

Suppose thatm1.t/ has bandwidth 3W and sourcesm2.t/,m3.t/,andm4.t/ each have bandwidthW , we could send the samplesas

s1s2s1s3s1s4s1s2s1 : : :

with the commutator rate being fs > 2W Hz

The equivalent transmission bandwidth for multiplexed signalscan be obtained as follows

– Each channel requires greater than 2Wi samples/s

– The total number of samples, ns, over N channels in T sis thus

ns D

NXiD1

2WiT

– An equivalent signal channel of bandwidth B would pro-duce 2BT D ns samples in T s, thus the equivalent base-


CONTENTS

band signal bandwidth is

B D

NXiD1

Wi Hz

which is the same minimum bandwidth required for FDMusing SSB

– Pure digital multiplexing behaves similarly to analog mul-tiplexing, except now the number of bits per sample, whichtakes into account the sample precision, must be included

– In the earlier PCM example for CD audio this was takeninto account when we said that left and right audio chan-nels each sampled at 44.1 ksps with 16-bit quantizers,multiplex up to

2 16 44; 100 D 1:4112 Msps


3.6. MULTIPLEXING

Example 3.12: Digital Telephone System

The North American digital TDM hierarchy is based on a sin-gle voice signal sampled at 8000 samples per second using a7-bit quantizer plus one signaling bit

The serial bit-rate per voice channel is 64 kbps

North American Digital TDM Hierarchy)

Digital No. of 64 kbpsSignal Bit Rate PCM VF Transmission

Sys. Number R (Mb/s) Channels Media UsedDS-0 0.064 1 Wire pairs

T1 DS-1 1.544 24 Wire pairsT1C DS-1C 3.152 48 Wire pairsT2 DS-2 6.312 96 Wire pairsT3 DS-3 44.736 672 Coax, radio, fiber

DS-3C 90.254 1344 Radio, fiberDS-4E 139.264 2016 Radio, fiber, coax

T4 DS-4 274.176 4032 Coax, fiberDS-432 432.00 6048 Fiber

T5 DS-5 560.160 8064 Coax, fiber

Consider the T1 channel which contains 24 voice signals

Eight total bits are sent per voice channel at a sampling rate of8000 Hz

The 24 channels are multiplexed into a T1 frame with an extrabit for frame synchronization, thus there are 24 8C 1 D 193bits per frame


CONTENTS

Frame period is 1=8000 D 0:125 ms, so the serial bit rate is193 8000 D 1:544 Mbps

Four T1 channels are multiplexed into a T2 channel (96 voicechannels)

Seven T2 channels are multiplexed into a T3 channel (672voice channels)

Six T3 channels are multiplexed into a T4 channel (4032 voicechannels)


linear modulation techniques

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