By: Dr. Uri Mahlab

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Simulation in Digital Communication
By Dr. Uri Mahlab
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chapter 3 Analog Modulation
By Dr. Uri Mahlab
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Introduction and Goals Each member of the class
of analog modulation systems is characterized by
five basic properties 1. Time-domain
representation of the modulated signal 2.
Frequency-domain representation of the modulated
signal 3. Bandwidth of the modulated
signal 4.Power content of the modulated
signal 5. Signal to noise ratio (SNR) after
demodulation.
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Amplitude Modulation(AM)
Amplitude modulation(AM) ,which is frequently
referred to as linear modulation, is the family
of modulation schemes in which the sinusoidal
carrier is changed as a function of the
modulating signal
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DSB-AM
In DSB-AM, the amplitude of the modulated signal
is proportional to the message signal. This means
that the time-domain representation of the
modulated signal is given by
where
Is the carrier and m(t) is the message signal.
The frequency-domain representation of DSB-AM
signal is obtained by taking the Fourier
transform of u(t) and results in
Where M(f) is the Fourier transform of m(t). The
transmission bandwidth denoted by BT is twice the
bandwidth of message signal
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M(f)
A
f
W
W-
fc
fc-
Figure3.1 Spectrum of DSB-AM modulated signal
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The power content of the modulated signal is
given by
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The SNR for DSB-AM system is equal to the
baseband SNR it is given by
Where PR is the received power (the power in the
modulated (signal at the receiver), N0/2 is the
noise power spectral density (assuming white
noise),and W is the message bandwidth.
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Example 3.1DSB-AM modulation The message signal
m(t) is defined as
This message DSB-AM modulates the carrier
c(t)cos2?fct, and the resulting modulated signal
is denoted by u(t).it is assumed that t00.15 s
and fc250Hz. 1. Obtain the expression for
u(t) 2. Derive the spectra of m(t) and
u(t). 3.assuming that the message signal is
periodic T0t0, determine the power in the
modulated signal. 4. If a noise is added to the
modulated signal in part(3) such that the
resulting SNR is 10 dB, find the noise power.
Answer DSB1.m
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Example 3.2DSB-AM modulation for an almost
bandlimited signal The message signal m(t) is
given by
Where t0 0.1. This message modulates the carrier
c(t)cos(2?fct), where fc250Hz. 1. Determine the
modulated signal u(t). 2.Determine the spectra of
m(t) and u(t) 3. If the message signal is
periodic with period T00.2,s determine the power
in the modulated signal. 4. If gaussian noise is
added to the modulated signal such that the
resulting SNR is 10 dB, find the noise power.
Answer DSB2.m
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Conventional AM
A is the index modulation and
The bandwidth is equal to the bandwidth of DSB-AM
and is given by
Typical frequency-domain plots of the message and
the corresponding conventional AM signal are
shown in figure 3.4
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M(f)
A
f
W
W-
fc
fc-
figure 3.4
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The power content of the modulated signal,
assuming that the message signal is a zero mean
signal is given by
Modulation efficiency and
The signal-to-noise ratio is given by
Where ? is the modulation efficiency
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Example 3.3conventional AM The message signal
Modulates the carrier c(t) cos(2?fct) using a
conventional AM scheme. It is assumed that
fc250Hz and t00.15 the modulation index is a
0.85 1. Derive an expression for the modulated
signal 2. Determine the spectra of the message
and the signals. 3.if the message signal is
periodic with a periodic with a period equal
t0, determine the power in the modulated signal
and the modulation efficiency 4.if a noise signal
is added to the message signal such that the SNR
at the output of the demodulator is 10 dB, find
the power content of the noise signal
Answer am.m
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SSB-AM
SSB-AM is derived from DSB-AM by eliminating one
of the sidebands.the time representation for
these signals is given by
Where the minus sign corresponds to USSB-AM and
the plus sign corresponds to LSSB-AM. The signal
denoted m(t) is the Hilbert transform of m(t),
defined by m(t)m(t) 1/?t, or, in the
frequency domain, by M(f)-j sgn(f) M(f).
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In other words, the Hilbert transform of a signal
represents a ?/2 phase shift in all signal
components. In the frequency domain, we
have end
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The bandwidth of SSB signal is half the bandwidth
of DSB and conventional AM and so is equal to the
bandwidth of message signal I.e., The power in
the SSB signal is given by
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M(f)
A
f
W
W-
fc
fc-
Figure 3.7 Spectra of the message and the
USSB-AM signal.
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Example 3.4 single-sideband example The
message signal
Modulates the carrier c(t) cos(2?fct) using an
LSSB-AM scheme. It is assumed that t00.15 s and
fc250Hz. 1.Plot the Hilbert transform of the
message signal and the modulated signal
u(t) 2.Find the spectrum of the modulated
signal. 3. Assuming the message signal periodic
with period t0, determine the power in the
modulated signal . 4.If a noise is added to the
modulated signal such that the SNR after
demodulation is 10 dB, determine the power in the
noise
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Demodulation of AM signal
DSB-AM Demodulation
In the DSB case the modulated signal is given by
Acm(t) cos(2?fct), which when multiplied by
cos(2?fct) (or mixed with cos(2?fct)) results in
Where y(t) denotes the mixer output, and its
Fourier transform is given by
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t0.15 signal
duration ts0.001
sampling interval fc250
carrier frequency snr10
SNR in dB (logarithmic) fs1/ts
sampling
frequency df0.25
desired freq. resolution t0tst0
time vector snr_lin10(snr/10)
SNR the message
vector mones(1,t0/(3ts)),-2ones(1,t0/(3ts)),z
eros(1,t0/(3ts)1) ccos(2pifc.t)
carrier vector udsbm.c
DSB modulated signal UDSB,udssb,
df1fftseq(udsb,ts,df) Fourier
transform UDSBUDSB/fs
scaling f0df1df1(length(udssb)-1)-fs/2
frequency vector n2ceil(fc/df1)
location of carrier in freq. vector
remove the upper sideband from DSB UDSB(n2length(
UDSB)-n2)zeros(size(UDSB(n2length(UDSB)-n2))) U
LSSBUDSB generate
LSSB-AM spectrum M,m,df1fftseq(m,ts,df)
Fourier transform MM/fs
scaling ureal(ifft(ULSSB))fs
generate LSSB signal from spectrum
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Figure 3.9 Demodulation of DSB-AM signal.
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Example 3.5DSB-AM demodulation The message
signal m(t) is defined
This message DSB-AM modulates the carrier c(t)
cos(2?fct), and the resulting modulated signal
is denoted by u(t). It is that t00.15 s and f
c250 Hz. 1.Obtain the expression for u(t). 2.
Derive the spectra of m(t) and u(t). 3.demodulate
the modulated signal u(t) and recover m(t). Plot
the results in the time and frequency.
Answer dsb_dem.m
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Example 3.6Effect of phase error on DSB-AM
demodulation In the demodulation of DSB-AM
signals we assumed that the phase of the local
oscillator is equal to the phase of carrier. If
that is not case-I.e., if there exist a phase
shift ? between the local oscillator and the
carrier- how would the demodulation process
change?
Answer lssb_dem.m
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The demodulation process of SSB-AM signals in
basically the same as the demodulation process
for DSB-AM signals, I,e., mixing followed by
lowpass filtering. In this case where the minus
sign corresponds to the USSB and the plus sign
corresponds to the LSSB. Mixing u(t) with the
local oscillator output, we obtain.
Which contains bandpass components at 2fc and a
lowpass component proportional to the message
signal. The lowpass component can be filtered out
using a lowpass filter to recover the message
signal. This process for the USSB-AM case is
depicted in Figure 3.13.
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Figure 3.13Demodulation of USSB-AM signals
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Example 3.7LSSB-AM example In USSB-AM
modulation system, if the message signal is
With t0 0.15 s, and the carrier has a frequency
of 259 Hz, find U(f) and Y(f) and compare the
demodulated signal with the message signal.
Answer lssb_dem.m
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Example 3.8 Effect of phase error on SSB-AM
what is the effect error in the SSB-AM?
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Conventional AM Demodulation
In envelope detection the envelope of the
modulated signal is detected via a simple circuit
consisting of diode, a resistor, and capacitor,as
shown in Figure 3.17.
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Figure 3.17 A simple envelope detector
Mathematically, the envelope detector generates
the envelope of the conventional AM signal, which
is Because 1mn(t)gt0, we conclude that
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Where mn(t) is proportional to the message signal
,m(t) and 1 corresponds to the carrier component
that can be separated by a dc-block circuit.
There is no need for knowledge of ?,the phase of
the carrier signal. That is why such a
demodulation scheme is called noncoherent,or
asynchronous, demodulation. Recall from chapter
1 that the envelope of a bandpass signal. Thus if
u(t) is the bandpass signal with central
frequency fc and the lowpass equivalent to u(t)
is denoted by ul(t), than the envelope of u(t),
denoted by V(t),can be expressed as
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• Where uc(t) and us(t) represent the in-phase and
  the quadrate components of the bandpass signal
  u(t).
• Therefore, in order to obtain the envelope, it is
  enough to obtain the lowpass equivalent of
  bandpass signal.
• The envelope is simply the magnitude of the
  lowpass equivalent of the bandpass signal.

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Example 3.9Envelope detection The message
signal
Modulates the carrier c(t)cos(2 fct)using a
conventional AM modulation scheme. It is assumed
that fc250Hz and t00.15 s, and the modulation
index is a0.85 1. Using envelope detection,
demodulate the message signal. 2. If the message
signal is periodic with a period to t0 and if an
AWGN process is added to the modulated signal,
use an envelope demodulator to demodulate the
received signal. Compare this case with the case
where is no noise present.
Answer am_dem.m
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Angle-modulation

• Angle-modulation schemes, which include frequency
  modulation(FM) and phase modulation(PM), belong
  to the class of nonlinear modulation schemes.

• The time-domain representation of angle-modulated
  signals, when the carrier is c(t) Accos(2pifct)
  and the message signal is m(t) is given by

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Where kf and kp represent the deviation constants
of FM and PM, respectively. The modulated signal
is of the from where
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And p and f are the modulation indices PM
and FM, respectively. In general, for a non
sinusoidal m(t) the modulation indices are
defined as
Where W is the bandwidth of the message signal
m(t).
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In case of a sinusoidal message signal the
modulated signal can be represented by
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Example 3.10Frequency modulation The message
signal
Modulates the carrier c(t)-cos(2 ) using a
frequency-modulation scheme. It is assumed that
fc200Hz and t0 0.15s the deviation constant is
kf 50 1.plot the modulated signal. 2. Determine
the spectra of the message and the modulated
signals
Answer fm1.m
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t0.15 signal
duration ts0.0005
sampling interval fc200
carrier frequency kf50
Modulation index fs1/ts
sampling
frequency t0tst0
time vector df0.25
required frequency resolution message
signal mones(1,t0/(3ts)),-2ones(1,t0/(3ts)),z
eros(1,t0/(3ts)1) int_m(1)0 for
i1length(t)-1 Integral of
m int_m(i1)int_m(i)m(i)ts end M,m,df1fft
seq(m,ts,df) Fourier transform
MM/fs
scaling f0df1df1(length(m)-1)-fs/2
frequency vector ucos(2pifct2pikfint_m)
modulated signal U,u,df1fftseq(u,ts,df)
Fourier transform UU/fs
scaling
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Example 3.11Frequency modulation The message
signal be
Where t00.1. This message modulates the carrier
where fc250 Hz. The deviation constant is
kf100. 1. Plot the modulated signal in the time
and frequency domain 2. Compare the demodulator
output and the original message signal.
Answer fm2.m
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t0.2 signal
duration ts0.001
sampling interval fc250
carrier frequency snr20
SNR in dB (logarithmic) fs1/ts
sampling
frequency df0.3
required freq. resolution t-t0/2tst0/2
time vector kf100
deviation constant df0.25
required frequency
resolution msinc(100t)
the message signal int_m(1)0 for
i1length(t)-1 Integral of
m int_m(i1)int_m(i)m(i)ts end M,m,df1fft
seq(m,ts,df) Fourier transform
MM/fs
scaling f0df1df1(length(m)-1)-fs/2
frequency vector ucos(2pifct2pikfint_m)
modulated signal U,u,df1fftseq(u,ts,df)
Fourier transform UU/fs
scaling v,phaseenv_phas(
u,ts,250) demodulation, find phase of
u phiunwrap(phase) restore
original phase dem(1/(2pikf))(diff(phi)/ts)
demodulator output, differentiate and scale
phase pause Press any key to see a plot of the
message and the modulated signal