linear modulation techniques - uccs college of · pdf file · 2018-02-273.1. linear...
TRANSCRIPT
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
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
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3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-5
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/ cosŒ2fct 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 f 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
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3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-7
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
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3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-9
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
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3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-11
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;
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3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-13
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
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3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-15
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%
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3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-17
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
3-18 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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%
ECE 5625 Communication Systems I 3-19
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
3-20 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-21
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 /
3-22 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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 fc/C j
1
2ı.f C f0/
ECE 5625 Communication Systems I 3-23
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)
3-24 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-25
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
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3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-27
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/
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3.1. LINEAR MODULATION
XcLSSB.f / D1
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/
D1
4AcM.f C fc/CM.f fc/
C1
4AcM.f C fc/sgn.f C fc/
M.f fc/sgn.f fc/
The inverse Fourier transform of the second term is DSB, i.e.,
1
2Acm.t/ cos!ct
F !
1
4AcM.f C fc/CM.f fc/
The first term can be inverse transformed using
Om.t/F ! j sgn.f / M.f /
soF1
˚M.f C fc/sgn.f C 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
ECE 5625 Communication Systems I 3-29
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
3-30 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-31
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)
~
3-32 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-33
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
3-34 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-35
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
3-36 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-37
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
3-38 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-39
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
3-40 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-41
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
3-42 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-43
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)
3-44 ECE 5625 Communication Systems I
3.1. LINEAR MODULATION
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
ECE 5625 Communication Systems I 3-45
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
3-46 ECE 5625 Communication Systems I
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
ECE 5625 Communication Systems I 3-47
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!
3-48 ECE 5625 Communication Systems I
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
ECE 5625 Communication Systems I 3-49
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)
3-50 ECE 5625 Communication Systems I
3.3. SAMPLING THEORY
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/,
ECE 5625 Communication Systems I 3-51
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
3-52 ECE 5625 Communication Systems I
3.3. SAMPLING THEORY
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
ECE 5625 Communication Systems I 3-53
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
3-54 ECE 5625 Communication Systems I
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/
ECE 5625 Communication Systems I 3-55
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
3-56 ECE 5625 Communication Systems I
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
ECE 5625 Communication Systems I 3-57
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
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3.5. DELTA MODULATION AND PCM
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/
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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
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3.5. DELTA MODULATION AND PCM
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
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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
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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
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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
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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-
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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
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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
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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)
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