EE4512 Analog and Digital Communications
Chapter 6
Analog Modulation
and Demodulation
Chapter 6
EE4512 Analog and Digital Communications
Chapter 6
Analog Modulation
and Demodulation
• Amplitude Modulation
• Pages 306-309
Chapter 6
EE4512 Analog and Digital Communications
Chapter 6
• The analytical signal for double sideband, large carrier
amplitude modulation (DSB-LC AM) is:
sDSB-LC AM(t) = AC (c + s(t)) cos (2π fC t)
where c is the DC bias or offset and AC is the carrier
amplitude. The continuous analog signal s(t) is a baseband
signal with the information content (voice or music) to be
transmitted.
EE4512 Analog and Digital Communications
Chapter 6
• The baseband power spectral density (PSD) spectrum of
the information signal s(t) or S(f) for voice has significant
components below 500 Hz and a bandwidth of < 8 kHz:
S(f) = F(s(t))
The single-sided spectrum of the modulated signal is:
F(AC (c + s(t)) cos (2π fC t)) = S(f – fC)
500 Hz
dB
Power Spectral Density of s(t)
8 kHz
EE4512 Analog and Digital Communications
Chapter 6
• The single-sided (positive frequency axis) spectrum of the
modulated signal replicates the baseband spectrum as a
double-sided spectrum about the carrier frequency.
Double-sided spectrum
Baseband spectrum
Carrier 25 kHz
EE4512 Analog and Digital Communications
Chapter 6
• The double-sided modulated spectrum about the carrier
frequency has an lower (LSB) and upper (USB) sideband.
EE4512 Analog and Digital Communications
Chapter 6
• The modulated DSB-LC AM signal shows an outer envelope
that follows the polar baseband signal s(t).
EE4512 Analog and Digital Communications
Chapter 6
• The analytical signal for double sideband, suppressed
carrier amplitude modulation (DSB-SC AM) is:
sDSB-SC AM(t) = AC s(t) cos (2π fC t)
where AC is the carrier amplitude. The single-sided
spectrum of the modulated signal replicates the baseband
spectrum as a double-sided spectrum about the carrier
frequency but without a carrier component.
EE4512 Analog and Digital Communications
Chapter 6
• The analytical signal for double sideband, suppressed
carrier amplitude modulation (DSB-SC AM) is:
sDSB-SC AM(t) = AC s(t) cos (2π fC t)
where AC is the carrier amplitude. The modulated signal
sDSB-SC AM(t) looks similar to s(t) but has a temporal but not
spectral carrier component.
EE4512 Analog and Digital Communications
Chapter 6
• The DSB-LC AM and the DSB-SC AM modulated signals
have the same sidebands.
DSB-LC AM
Carrier 25 kHz
DSB-SC AM
No carrier
EE4512 Analog and Digital Communications
Chapter 6
• The modulated DSB-LC AM and the DSC-SC AM signals
are different.
EE4512 Analog and Digital Communications
Chapter 6
• The modulated DSB-SC AM signal has an envelope that
follows the polar baseband signal s(t) but not an outer
envelope.
EE4512 Analog and Digital Communications
Chapter 6
Analog Modulation
and Demodulation
• Coherent Demodulation
of AM Signals
• Pages 309-315
Chapter 6
EE4512 Analog and Digital Communications
Chapter 6
• The DSB-SC AM coherent receiver has a bandpass filter
centered at fC and with a bandwidth of twice the bandwidth
of s(t) because of the LSB and USB. The output of the
multiplier is lowpass filtered with a bandwidth equal to
the bandwidth of s(t).
r(t) = γ sDSB-SC(t) + n(t)
• The DSB-SC AM received signal is r(t) = γ sDSB-SC(t) + n(t).
The bandpass filter passes the modulated signal but filters
the noise:
z(t) = γ sDSB-SC(t) + no(t)
S&M Eq. 6.3
no(t) has a Gaussian distribution. The bandpass filter has a
center frequency of fC = 25 kHz and a -3 dB bandwidth of 8
kHz (25 ± 4 kHz).
EE4512 Analog and Digital Communications
Chapter 5
• The filter noise no(t) has a flat power spectral density
within the bandwidth of the bandpass filter:
no(t)
PSD
21 kHz
29 kHz
fC = 25 kHz
EE4512 Analog and Digital Communications
Chapter 6
• The filter noise no(t) can be described as a quadrature
representation:
no(t) = W(t) cos (2π fCt) + Z(t) sin (2π fCt) S&M Eq. 5.62R
In the coherent receiver the noise is processed:
no(t) cos (2π fCt) = W(t) cos2 (2π fCt) +
S&M Eq. 6.5
Z(t) cos (2π fCt) sin (2π fCt)
PSD
21 kHz
29 kHz
fC = 25 kHz
EE4512 Analog and Digital Communications
Chapter 6
• Applying the trignometric identity the filter noise no(t) is:
no(t) cos (2π fCt) = ½ W(t) + ½ W(t) cos (4π fCt) +
S&M Eq. 6.5
½ Z(t) sin (4π fCt)
After the lowpass filter in the receiver the demodulated
signal is:
sdemod(t) = ½ γ AC s(t) + ½ W(t)
PSD
21 kHz
29 kHz
fC = 25 kHz
S&M Eq. 6.7
EE4512 Analog and Digital Communications
Chapter 6
• The transmitted DSB-SC AM signal is:
sDSB-SC AM(t) = AC s(t) cos (2π fC t)
The average normalized bi-sided power of sDSB-SC(t) is
found in the spectral domain with S(f) = F (s(t)):
Ptrans = A 2 ∫
2
1
[S(f − fC ) + S(f + fC )] df
2
S&M Eq. 6.8
EE4512 Analog and Digital Communications
Chapter 6
• The dual-sided spectral do not overlap (at zero frequency)
and the cross terms are zero so that:
2
1
Ptrans = A 2 ∫ [SDSB-SC (f − fC ) + SDSB-SC (f + fC )] df
2
Ptrans
A2
=
Ps
2
S&M Eq. 6.9
where Ps is the average normalized power of s(t).
EE4512 Analog and Digital Communications
Chapter 6
• The average normalized power of s(t) is found in the
spectral domain:
2
2
Ps = ∫ S(f) df = ∫ S(f + fC ) df
S&M Eq. 6.10
In a noiseless channel the power in the demodulated
DSB-SC AM signal is:
Pdemod, noiseless
1 2 2
γ2
= γ A Ps =
Ptrans
4
2
S&M Eq. 6.11
EE4512 Analog and Digital Communications
Chapter 6
• The average normalized power of the processed noise is:
Pprocessed noise =
1
No (2 B)
4
The signal-to-noise power ratio then is:
γ2
Ptrans
2
γ
Ptrans
2
SNRcoherent DSB-SC =
=
1
No B
No (2 B)
4
2B
S&M Eq. 6.12
EE4512 Analog and Digital Communications
Chapter 6
• The DSB-SC AM coherent receiver requires a phase
and frequency synchronous reference signal. If the
reference signal has a phase error φ then:
SNRcoherent DSB-SC phase error =
γ 2 cos2 ϕ Ptrans
No B
S&M Eq. 6.17
EE4512 Analog and Digital Communications
Chapter 6
• The DSB-SC AM coherent receiver requires a phase
and frequency synchronous reference signal. If the
reference signal has a frequency error ∆f then:
Sdemod frequency error(t) = ½ γ AC s(t) cos (2π ∆f t)
+ ½ X(t) cos (2π ∆f t)
+ ½ Y(t) sin (2π ∆f t) S&M Eq. 6.18
• Although the noise component remains the same, the
amplitude of the demodulated signal varies with ∆f:
EE4512 Analog and Digital Communications
Chapter 6
Analog Modulation
and Demodulation
• Non-coherent Demodulation
of AM Signals
• Pages 315-326
Chapter 6
EE4512 Analog and Digital Communications
Chapter 6
• The non-coherent AM (DSB-LC) receiver uses an envelope
detector implemented as a semiconductor diode and a lowpass filter:
The DSB-LC AM analytical signal is:
sDSB-LC AM(t) = AC (c + s(t)) cos (2π fC t)
where c is the DC bias (offset).
EE4512 Analog and Digital Communications
Chapter 5
• The envelope detector is a half-wave rectifier and
provides a DC bias (c) to the processed DSB-LC AM
signal :
c = DC bias
EE4512 Analog and Digital Communications
Chapter 5
• The output of the half-wave diode rectifier is low-pass
filtered to remove the carrier frequency and outputs the
envelope which is the information:
EE4512 Analog and Digital Communications
Chapter 6
• The DSB-LC AM signal can be decomposed as:
sDSB-LC AM(t) = s(t) cos (2π fC t) + AC c cos (2π fC t)
S&M Eq. 6.20R
The average normalized power of the information term:
Pinfo term
A C2
=
PS
2
S&M Eq. 6.23
EE4512 Analog and Digital Communications
Chapter 6
• The average normalized transmitted power is:
Pcarrier term
Pcarrier term
1
=
T
T
∫ [A
c cos(2πfC t)] dt
2
C
0
A C2 c 2
=
2
S&M Eq. 6.24
Since s(t) + c must be >= 0 to avoid distortion in the
DSB-LC AM signal: c ≥ | min [s(t)] | or c2 ≥ s2(t) for all t.
EE4512 Analog and Digital Communications
Chapter 6
• Therefore c2 ≥ Ps and for DSB-LC AM:
Pcarrier term ≥ Pinfo term
S&M Eq. 6.28
The power efficiency η of a DSB-LC AM signal is:
Pinfo term
Pinfo term
=
η =
≤ 0.5
Pcarrier term + Pinfo term
Ptrans DSB-LC AM term
S&M Eq. 6.29
EE4512 Analog and Digital Communications
Chapter 6
• The DSB-LC AM signal wastes at least half the
transmitted power because the power in the carrier term
has no information:
Pcarrier term ≥ Pinfo term
η ≤ 0.5
The modulation index m is defined as:
m =
max [ s(t) + c ] − min [ s(t) + c ]
max [ s(t) + c ] + min [ s(t) + c ]
S&M Eq. 6.30
EE4512 Analog and Digital Communications
Chapter 6
• The modulation index m defines the power efficiency but
m must be less than 1. If m > 1 then min [s(t) + c] < 0 and
distortion occurs.
m =
max [ s(t) + c ] − min [ s(t) + c ]
max [ s(t) + c ] + min [ s(t) + c ]
S&M Eq. 6.30
EE4512 Analog and Digital Communications
Chapter 6
• The average normalized power of the demodulation
noiseless DSB-LC AM signal is:
Pdemod, noiseless = 2 γ 2 Pinfo term
S&M Eq. 6.39
Then the signal-to-noise power ratio for the DSB-LC AM
signal is:
S&M Eq. 6.40
2 γ 2 Pinfo term
γ 2 Ptrans DSB-LC
=
η
SNRnoncoherent DSB-LC =
No (2 B)
No B
2B
EE4512 Analog and Digital Communications
Chapter 6
• The non-coherent AM (DSB-LC) receiver is the crystal
radio which needs no batteries! Power for the highimpedance ceramic earphone is obtained directly from the
transmitted signal. For simplicity, the RF BPF is omitted
and the audio frequency filter is a simple RC network.
EE4512 Analog and Digital Communications
Chapter 6
Analog Modulation
and Demodulation
• Frequency Modulation and
Phase Modulation
• Pages 334-343
Chapter 6
EE4512 Analog and Digital Communications
Chapter 6
• The analytical signal for an analog phase modulated (PM)
signal is:
sPM(t) = AC cos [2π fC t + α s(t)]
S&M Eq. 6.53
where α is the phase modulation constant rad/V and AC is
the carrier amplitude. The continuous analog signal s(t) is a
baseband signal with the information content (voice or
music) to be transmitted.
EE4512 Analog and Digital Communications
Chapter 6
• The analytical signal for an analog frequency modulated
(FM) signal is:
sFM(t) = AC cos{ 2π [fC + k s(t)] t + φ]
S&M Eq. 6.53
where k is the frequency modulation constant Hz / V, AC
is the carrier amplitude and φ is the initial phase angle at
t = 0. The continuous analog signal s(t) is a baseband
signal with the information content.
EE4512 Analog and Digital Communications
Chapter 6
• The instantaneous phase of the PM signal is:
ΨPM(t) = 2π fC t + α s(t)
S&M Eq. 6.56
The instantaneous phase of the FM signal is:
ΨFM(t) = 2π [fC + k s(t)] t + φ]
S&M Eq. 6.57
The instantaneous phase is also call the angle of the signal.
The instantaneous frequency is the time rate of change of
the angle:
f(t) = (1/2π) dΨ(t) / dt
S&M Eq. 6.58
EE4512 Analog and Digital Communications
Chapter 6
• The instantaneous frequency of the unmodulated carrier
signal is:
fcarrier(t) = dΨcarrier(t) / dt = d/dt {2π fCt + φ}
S&M Eq. 6.59
The instantaneous phase is also:
t
t
Ψ(t) = ∫ f (λ) dλ = ∫ f (λ) dλ + φ
0
-∞
S&M Eq. 6.60
There are practical limits on instantaneous frequency and
instantaneous phase. To avoid ambiguity and distortion in
FM signals due to phase wrapping:
k s(t) ≤ fC for all t
S&M Eq. 6.61
EE4512 Analog and Digital Communications
Chapter 6
• To avoid ambiguity and distortion in PM signals due to
phase wrapping:
-π < α s(t) ≤ π radians for all t
S&M Eq. 6.61
Since FM and PM are both change the angle of the carrier
signal as a function of the analog information signal s(t), FM
and PM are called angle modulation.
For example, is this signal FM, PM or neither:
t
x(t) = AC cos { 2π fCt + ∫ k s(λ) dλ + φ} S&M Eq. 6.60
-∞
EE4512 Analog and Digital Communications
Chapter 6
• The instantaneous phase of the signal is:
t
Ψx(t) = 2π fCt + ∫ k s(λ) dλ + φ
S&M Eq. p. 336
-∞
which is not a linear function of s(t) so the signal is not PM.
The instantaneous frequency of the signal is:
fx(t) = (1/2π) dΨx(t) / dt = fC + k s(t) / 2π
and the frequency difference fx – fC is a linear function of s(t)
so the signal is FM.
The maximum phase deviation of a PM signal is
max | αs(t) |. The maximum frequency deviation of a FM
signal is ∆f = max | k s(t) |.
EE4512 Analog and Digital Communications
Chapter 6
• The spectrum of a PM or FM signal can be developed as
follows: S&M Eqs. 6.64 through 6.71
v(t) = A C sin(2π fC t + β sin 2π fm t)
v(t) = Re { exp(j 2π fC t + j β sin 2π fm t) }
now
exp(j 2π fC t + j β sin 2π fm t) =
cos (2π fC t + β sin 2π fm t) + j sin (2π fC t + β sin 2π fm t)
v(t) = Im { A C exp(2π fC t + jβ sin 2π fm t) }
now
exp(j β sin 2π fm t) =
∞
∑ c exp(j 2π n f
n = -∞
n
after further development
exp(j β sin 2π fm t) =
m
t)
Bessel function of the first kind
∞
∑ J (β) exp(j 2π n f
n = -∞
n
m
t)
EE4512 Analog and Digital Communications
Chapter 6
• Bessel functions of the first kind Jn(β) are tabulated for FM
with single tone fm angle modulation (S&M Table 6.1):
β
n
EE4512 Analog and Digital Communications
Chapter 6
• For single tone fm angle modulation the spectrum is periodic
and infinite in extent:
v(t) = A C
∞
∑ J (β) sin[2π (n f
n = -∞
n
m
+ fC ) t]
S&M Eq. 6.72
n
β
EE4512 Analog and Digital Communications
Chapter 6
• The complexity of the Bessel function solution for the
spectrum of a single tone angle modulation can be
simplified by the Carson’s Rule approximation for the
bandwidth B. Since β = ∆f / fm:
S&M Eq. 6.74
B = 2 (β + 1) fm = 2 (∆f + fm) Hz
n
β
EE4512 Analog and Digital Communications
Chapter 6
• Carson’s Rule for the approximate bandwidth of an angle
modulated signal was developed by John R. Carson in
1922 while he worked at AT&T. Prior to this in 1915 he
presaged the concept of bandwidth
efficiency in AM by proposing
the suppression of a sideband
(see S&M p. 326-333):
B = 2 (β + 1) fm = 2 (∆f + fm) Hz
1886-1940
EE4512 Analog and Digital Communications
Chapter 6
• The normalized power within the Carson’s Rule bandwidth
for a single tone angle modulated signals is:
Pin-band, sinusoid
A C2 β+1 2
Jn (β)
=
∑
2 n = -(β+1)
S&M Eq. 6.75
Note that J-n(β) = ± Jn(β) so that J-n2(β) = Jn2(β) and for the
normalized power calculation the sign of J(β) is not used.
Spectrum of
single tone FM
modulation
EE4512 Analog and Digital Communications
Chapter 6
• The analog FM power spectral density PSD of the voice
signal has a bandwidth predicted only by Carson’s Rule
since it is not a single tone.
Voice
PSD
EE4512 Analog and Digital Communications
Chapter 6
• Here fmax = 4 kHz, k = 25 Hz/V and ∆fmax = 40(25) = 1 kHz.
The Carson’s Rule approximate maximum bandwidth
B = 2 (∆f + fm) = 10 kHz or ± 5 kHz (but seems wrong!)
40
Voice
fC
PSD
Bandwidth
EE4512 Analog and Digital Communications
Chapter 6
• A 200 Hz single tone FM signal has a PSD with periodic
terms at fC ± n fm = 25 ± 0.2 n kHz.
fC
PSD
200 Hz
EE4512 Analog and Digital Communications
Chapter 6
• Here fm = 200 Hz, k = 25 Hz/V and ∆fmax = 40(25) = 1
kHz. The Carson’s Rule approximate maximum bandwidth
B = 2 (∆f + fm) = 2.4 kHz or ± 1.2 kHz:
fC
200 Hz
PSD
Bandwidth
EE4512 Analog and Digital Communications
Chapter 6
• Since β = ∆f / fm = 1 kHz / 0.2 kHz = 5 and the Bessel
function predicts a bandwidth of 2 n fm = 2(12)(200) =
4.8 kHz (since n = 12 for β = 5 from Table 6.1):
fC
200 Hz
PSD
Bandwidth
EE4512 Analog and Digital Communications
Chapter 6
Analog Modulation
and Demodulation
• Noise in FM and PM Systems
• Pages 347-355
Chapter 6
EE4512 Analog and Digital Communications
Chapter 6
• A general angle modulated transmitted signal, where Ψ(t)
is the instantaneous phase, is:
sangle-modulated(t) = AC cos [Ψ(t)]
S&M Eq. 6.86
The received signals is:
rangle-modulated(t) = γ AC cos [Ψ(t)] + n(t)
S&M Eq. 6.87
EE4512 Analog and Digital Communications
Chapter 6
• The analytical signal for PM is:
sPM(t) = AC cos [Ψ(t)] = AC cos [2π fC t + α s(t)]
S&M Eq. 6.53
After development the SNR for demodulated PM is:
SNRPM = (αγ AC)2 PS / (2 No fmax)
where −π < α s(t) ≤ π for all t.
S&M Eq. 6.98
EE4512 Analog and Digital Communications
Chapter 6
• The analytical signal for FM is:
sFM(t) = AC cos [Ψ(t)] = AC cos [2π fC t + ∫ k s(λ) dλ]
S&M Eq. 6.53
After development the SNR for demodulated FM is:
SNRFM = 1.5 (k γ AC /(2π) )2 PS / (No fmax3) S&M Eq. 6.98
where k s(t) ≤ fC for all t.
EE4512 Analog and Digital Communications
End of Chapter 6
Analog Modulation
and Demodulation
Chapter 6