Continuous-Wave Modulation (CWM)
Amplitude Modulation (AM)
Part II
1
Recall: Spectrum Symmetry of Real Functions
𝑋 −𝑓 = 𝑋
𝑋
∗
𝑓
∗
𝑓
= 𝑋 −𝑓
Conjugate Symmetry
= 𝑋 𝑓
𝜃𝑋 (−𝑓) = −𝜃𝑋 (𝑓)
Magnitude spectrum → Even function of 𝒇
Phase spectrum
→ Odd function of 𝒇
2
Spectrum Symmetry of AM Modulation
• The same information is transmitted in the
upper and lower sidebands
• Only one of the sidebands needs to be
transmitted → Bandwidth reduction
3
• Tradeoff: Cost & Complexity
Filtering the USB
What is the problem with this process?
4
Ideal vs. Practical BPF
5
Problem!
• To completely suppress one sideband, we
need a very sharp cutoff characteristics at
± 𝑓𝑐 (that is, we need an almost ideal BPF)
→ In practice, we require at least 40 dB
sideband suppression
• The filtering process can be somewhat
relaxed if the baseband signal has little low
frequency content
• One important such signal is voice (or
speech) signal
6
Power Spectral Density of Speech Signal
Subjective tests
showed no appreciable
change of intelligibility
𝟑𝟎𝟎
𝑾 = 𝟑𝟒𝟎𝟎 𝐇𝐳
𝟑𝟕𝟎𝟎
Bandwidth → 𝑊~ 3.4 − 4 KHz
7
Single Side-Band Modulation (SSB)
Frequency Discrimination (Filtering) Approach
8
Problem!
• For speech signal, the 600 Hz gap around
the carrier frequency ±𝑓𝑐 can simplify the
filtering operation provided that 𝑓𝑐 is not
too large
• However, in practice, 𝑓𝑐 ≫ 600 Hz
• Solution: A two-stage process
1. Attenuation of the sideband at a lower carrier
frequency 𝑓𝑐1
2. Frequency translation to the desired carrier
frequency 𝑓𝑐2 (to be shown later)
9
Two-stage Process
for Generating SSB Signals
𝑚(𝑡)
𝑥(𝑡)
𝐵𝑃𝐹1
𝐴1 cos(2𝜋𝑓𝑐1 𝑡)
𝑦(𝑡)
𝑠(𝑡)
𝐵𝑃𝐹2
𝑚(𝑡)
𝐴2 cos(2𝜋𝑓𝑐2 𝑡)
𝒇𝒄𝟏 ≪ 𝒇𝒄𝟐
10
𝑆𝑆𝐵
𝑿(𝝎)
−𝒇𝒄𝟏
𝒇𝒄𝟏
𝒀(𝝎)
−𝒇𝒄𝟏
𝒇𝒄𝟏
𝑺(𝝎)
−𝒇𝒄𝟐
𝒇𝒄 𝟐
𝟐𝒇𝒄𝟐
11
Single Side-Band Modulation (SSB)
Phase Discrimination Approach
• SSB is difficult to generate via filtering if the
baseband signal has significant DC and low
frequency components
• In practice, SSB suppression
▪ 20 dB → easy
▪ 30 dB → reasonable
▪ > 𝟒𝟎 𝐝𝐁 → very difficult
• This approach is based on the time-domain
representation of the SSB signal
• Used for low frequency generation of SSB and in
digital generation of SSB
12
SSB – Phase Discrimination Approach
𝑠(𝑡)
𝑆𝑆𝐵+
= 𝑚 𝑡 cos 2𝜋𝑓𝑐 𝑡 − 𝑚ℎ 𝑡 sin 2𝜋𝑓𝑐 𝑡
𝑠(𝑡)
𝑆𝑆𝐵−
= 𝑚 𝑡 cos 2𝜋𝑓𝑐 𝑡 + 𝑚ℎ 𝑡 sin 2𝜋𝑓𝑐 𝑡
𝑠(𝑡)
𝑆𝑆𝐵+
= 𝑠(𝑡)
𝑈𝑆𝐵
,
𝑠(𝑡)
𝑆𝑆𝐵−
= 𝑠(𝑡)
𝐿𝑆𝐵
𝑚ℎ 𝑡 = Hilbert Transform of 𝑚(𝑡)
13
Hilbert Transform
∞
1
1
𝑚(𝜏)
𝑚ℎ 𝑡 =
∗𝑚 𝑡 = න
𝑑𝜏
𝜋𝑡
𝜋
𝑡−𝜏
−∞
∞
1
1
𝑚ℎ (𝜏)
𝑚 𝑡 = − ∗ 𝑚ℎ 𝑡 = න
𝑑𝜏
𝜋𝑡
𝜋
𝑡−𝜏
−∞
Note: the above integrals are improper integrals because
the integrand has a singularity at 𝜏 = 𝑡. In order to avoid
the singularity, the Hilbert transform must be defined as
the Cauchy-principle value, that is
∞
𝑚ℎ 𝑡 = 𝑃 න
−∞
𝑡−𝜀
∞
𝑚(𝜏)
𝑚(𝜏)
𝑚(𝜏)
𝑑𝜏 = lim න
𝑑𝜏 + න
𝑑𝜏
𝜀→0
𝑡−𝜏
𝑡−𝜏
𝑡−𝜏
−∞
𝑡+𝜀
14
What does the Hilbert Transform do?
𝑚(𝑡)
𝑚ℎ (𝑡)
• The Hilbert transform is an ideal
• It produces a phase shift of
−𝜋
2
−𝝅
𝟐
phase shifter
for all positive
+𝜋
2
frequencies of the input signal and for all
negative frequencies
• Does not affect the amplitudes of all frequency
components in the signal
• Shown next
15
𝑚(𝑡)
𝑚ℎ (𝑡)
𝑚ℎ 𝑡 = ℎ(𝑡) ∗ 𝑚 𝑡
1
𝑚ℎ 𝑡 =
∗𝑚 𝑡
𝜋𝑡
1
→ℎ 𝑡 =
𝜋𝑡
1
𝐻 𝑓 =ℱ ℎ 𝑡 =ℱ
= −𝑗sgn(𝑓)
𝜋𝑡
−𝑗
=ቐ0
+𝑗
=
𝜋
−𝑗 2
𝑗>0
𝑒
𝑗=0=
0
𝜋
𝑗<0
𝑗
𝑒 2
𝜋
−𝑗 sgn(𝑓)
𝑒 2
𝑗>0
𝑗=0
𝑗<0
16
Important Remarks
• The ideal phase shifter defined by the
Hilbert transform is a non-causal system →
we can only be approximate it over a finite
band
• Fourier transform separates signals based
on their frequency contents
• Hilbert transform separates signals based
on their phase shifts
17
Complex Representation of SSB signal
• Pre-envelope
𝑚+ 𝑡 = 𝑚 𝑡 + 𝑗𝑚ℎ (𝑡)
• Complex envelope
𝑚
෥ 𝑡 = 𝑚+ 𝑡 𝑒 𝑗2𝜋𝑓𝑐 𝑡
𝑠(𝑡)
𝑠(𝑡)
𝑆𝑆𝐵+
𝑆𝑆𝐵−
= 𝑅𝑒 𝑚
෥ 𝑡
∗
= 𝑅𝑒 𝑚+
(𝑡)𝑒 𝑗2𝜋𝑓𝑐 𝑡
18
Practice Exercise 1
1. Show that if 𝑚 𝑡 = cos(2𝜋𝑓0 𝑡 + 𝜃), then
𝑚ℎ 𝑡 = sin 2𝜋𝑓0 𝑡 + 𝜃
2. Show that if 𝑚 𝑡 = sin(2𝜋𝑓0 𝑡 + 𝜃), then
𝑚ℎ 𝑡 = −cos 2𝜋𝑓0 𝑡 + 𝜃
3. Show that if 𝑚 𝑡 =
𝑎
𝑡 2 +𝑎2
, then 𝑚ℎ 𝑡 =
𝑡
𝑡 2 +𝑎2
4. Show that if 𝑔ℎ (𝑡) is the Hilbert transform of 𝑔(𝑡),
then the Hilbert transform of 𝑔ℎ (𝑡) is −𝑔(𝑡)
19
Practice Exercise 2
1. Show that 𝑔 𝑡 and its Hilbert transform 𝑔ℎ (𝑡) are
orthogonal
2. Show that 𝑔 𝑡 and its Hilbert transform 𝑔ℎ (𝑡)
have the same energy if 𝑔 𝑡 is an energy signal
and the same power if 𝑔 𝑡 is a power signal
20
SSB – Phase Discrimination – Modulation
(aka Hartley’s Method)
𝒎(𝒕)
𝒎𝒉 (𝒕)
Used when > 𝟒𝟎 dB attenuation of the
undesired sideband is required
21
Practice Exercise 3:
The Weaver’s Method
for Generating SSB Modulated Signals
22
• The message (modulating) signal 𝑚(𝑡) is limited to
the frequency band 𝑓𝑎 ≤ 𝑓 ≤ 𝑓𝑏 as shown
𝑀(𝑓)
𝑓
−𝑓𝑏 −𝑓𝑎 𝑓𝑎 𝑓𝑏
• The auxiliary carrier signal applied to the first pair of
𝑓𝑎 +𝑓𝑏
product modulators has a frequency 𝑓0 =
2
• The low-pass filters in the upper and lower branches
are identical, each with a cutoff frequency equal to
𝑓𝑏 −𝑓𝑎
2
• The carrier applied to the second pair of product
𝑓𝑏 −𝑓𝑎
modulators has frequency 𝑓𝑐 >
2
23
Sketch the spectra at the various points in the
modulator, and hence show that:
1. For the lower sideband, the contributions of the
upper and lower branches are of opposite polarity,
and by adding them at the modulator output, the
lower sideband is suppressed.
2. For the upper sideband, the contributions of the
upper and lower branches are of same polarity, and
by adding them at the modulator output, the upper
sideband is transmitted.
3. How would you modify the modulator shown in the
block diagram, so that only the lower sideband is
transmitted?
24
Demodulation of SSB Signals
𝑠(𝑡)
𝑆𝑆𝐵±
𝑥(𝑡)
Low-pass
Filter
𝑚(𝑡)
ෝ
𝑐 𝑡
𝑐 𝑡 = 𝐴𝑐 cos 𝜔𝑐 + ∆𝜔 𝑡 + 𝜃
→ ∆𝜔 = Frequency error
→ 𝜃 = Phase error
25
𝑠(𝑡)
𝑆𝑆𝐵±
= 𝑚 𝑡 cos 𝜔𝑐 𝑡 ∓ 𝑚ℎ 𝑡 sin 𝜔𝑐 𝑡
𝑥 𝑡 = 𝑚 𝑡 cos 𝜔𝑐 𝑡 cos 𝜔𝑐 + ∆𝜔 𝑡 + 𝜃
∓𝑚ℎ 𝑡 sin 𝜔𝑐 𝑡 cos 𝜔𝑐 + ∆𝜔 𝑡 + 𝜃
1
1
𝑚
ෝ 𝑡 = 𝑚 𝑡 cos ∆𝜔𝑡 + 𝜃 ± 𝑚ℎ 𝑡 sin ∆𝜔𝑡 + 𝜃
2
2
∆𝝎 = 𝟎 & 𝜽 = 𝟎 → Coherent Detection
1
𝑚
ෝ 𝑡 = 𝑚 𝑡
2
Any one of the coherent DSC-SC demodulators can be
used to coherently detect a SSB signal.
26
What happens to the Demodulated
SSB signal when there is only a phase error?
Phase error only → ∆𝝎 = 𝟎
1
𝑚
ෝ 𝑡 = 𝑚 𝑡 cos 𝜃 ± 𝑚ℎ 𝑡 sin 𝜃
2
1
= Re (𝑚 𝑡 ∓ 𝑚ℎ 𝑡 )𝑒 𝑗𝜃
2
෡ 𝜔 =
𝑀
1
𝑀(𝜔)𝑒 −𝑗𝜃
2
1
𝑀(𝜔)𝑒 +𝑗𝜃
2
Cross-talk
between
𝑚 𝑡 & 𝑚ℎ 𝑡
𝜔>0
Phase Distortion
𝜔<0
27
Phase Distortion & SSB
• Voice communications → Tolerable →
Human ear can interpret speech despite
phase changes → Donald Duck voice effect!
• Video & high-quality music → phase
distortion can cause some intolerable effects
• Data communications (pulse data) → Not
acceptable → Limiting the use of SSB for
such systems
28
What happens to the Demodulated
SSB signal when there is only a frequency error?
Frequency error only → 𝜽 = 𝟎
1
𝑚
ෝ 𝑡 = 𝑚 𝑡 cos ∆𝜔𝑡 ± 𝑚ℎ 𝑡 sin ∆𝜔𝑡
2
1
= Re (𝑚 𝑡 ∓ 𝑚ℎ 𝑡 )𝑒 𝑗∆𝜔𝑡
2
• Frequency errors ∆𝜔 will produce spectral shifts and phase
distortion in the demodulated signal
• If ∆𝜔 is small (~2 − 5 Hz), spectral shifts can be tolerated in
voice communications
• Otherwise, if ∆𝜔 is large, it can produce unacceptable results
(see P.E. 7 next) (this type of distortion unique to SSB)
29
Example 1
• Consider a demodulator of an SSB transmitted
signal with only a frequency error ∆𝜔
1. Show that the demodulated signal 𝑚(𝑡)
ෝ
will
be shifted outward by the amount ∆𝜔 if
• the SSB signal 𝑠(𝑡) consists of the lower sideband
only and ∆𝜔 > 0, or
• the SSB signal 𝑠(𝑡) consists of the upper sideband
only and ∆𝜔 < 0
30
Example 1 (cont.)
• Consider a demodulator of an SSB transmitted
signal with only a frequency error ∆𝜔
2. Show that the demodulated signal 𝑚(𝑡)
ෝ
will
be shifted inward by the amount ∆𝜔 if
• the SSB signal 𝑠(𝑡) consists of the lower sideband
only and ∆𝜔 < 0, or
• the SSB signal 𝑠(𝑡) consists of the upper sideband
only and ∆𝜔 > 0
31
Practice Exercise 3: A Scrambler System
• The spectrum of a voice signal 𝑚(𝑡) is zero outside
the interval 𝑓𝑎 ≤ 𝑓 ≤ 𝑓𝑏
• In order to ensure communication privacy, this
signal is applied to a scrambler that consists of the
following cascade of components: a product
modulator, a high-pass filter, a second product
modulator, and a low-pass filter.
32
• The carrier wave applied to the first product
modulator has a frequency equal to 𝑓𝑐 , where as the
carrier wave applied to the second product
modulator has a frequency equal to 𝑓𝑏 + 𝑓𝑐 ; both of
the product modulators have unit amplitudes.
• The high-pass filter have the same cutoff frequency
at 𝑓𝑐
• Assume that 𝑓𝑐 > 𝑓𝑏
a. Derive an expression for the scramble output 𝑠(𝑡)
and sketch its spectrum.
b. Show that the original signal 𝑚(𝑡) may be
recovered from 𝑠(𝑡) by using an unscrambler that is
identical to the unit described above.
33
Envelope Detection for SSB-LC
• Envelope detection can be used to recover an
SSB signal by transmitting an additional carrier
signal (see practice exercise 9)
• SSB-LC signals can be received by commercial AM
receivers with 1Τ2 the transmission Bandwidth
• However, SSB-LC requires considerably more
carrier power than AM (the power efficiency of
SSB-LC is much less than AM)
• Envelope detection is not that important for SSB
as it is for AM
34
Practice Exercise 4
• Show that if
𝑨 ≫ 𝒎(𝒕)
• The signal 𝑠(𝑡) 𝑆𝑆𝐵+𝐶 can be demodulated
correctly using an envelope detector
𝑠(𝑡)
𝑆𝑆𝐵+𝐶
= 𝑚 𝑡 cos 𝜔𝑐 𝑡 ∓ 𝑚ℎ 𝑡 sin 𝜔𝑐 𝑡
+ 𝐴 cos 𝜔𝑐 𝑡
• Recall, the condition for AM → 𝐴 ≥ 𝑚(𝑡)
35
Vestigial SideBand Modulation (VSB)
• A compromise between DSB and SSB
• Designed to reduce the problem of sideband
separation
– Impractical filtering when the message contains
significant amount of low frequency spectral
content (e.g., TV and Telegraph signals)
– Impractical phase shifting for signals with large
bandwidths
• Solution: one sideband is passed almost
completely, whereas just a trace (or vestige)
of the other sideband is also passed
36
Bandwidth Comparison
𝑚 𝑡 = 𝑊 Hz
• AM/DSB-SC = 2𝑊 Hz
• SSB = 𝑊 Hz
• VSB = 𝑊 + 𝑓𝑣 Hz
• Typically, 𝑓𝑣 = 0.25~0.3 𝑊 Hz
𝒇𝒗 = width of the vestigial band
37
Vestigial SideBand Modulation (VSB)
Filtering Approach
𝒔(𝒕)
𝒈(𝒕)
Shaping Filter
38
VSB Filter Characteristics
𝑠 𝑡 = 𝐴𝑐 𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡)
𝐴𝑐
𝑆 𝑓 =
𝑀 𝑓 + 𝑓𝑐 + 𝑀(𝑓 − 𝑓𝑐 )
2
𝑔 𝑡 =ℎ 𝑡 ∗𝑠 𝑡
↔ 𝐺 𝑓 = 𝐻 𝑓 𝑆(𝑓)
𝐴𝑐
𝐺 𝑓 =
𝐻 𝑓 𝑀 𝑓 + 𝑓𝑐 + 𝑀(𝑓 − 𝑓𝑐 )
2
39
Assume coherent detection:
𝑦 𝑡 = 𝑔(𝑡) cos(2𝜋𝑓𝑐 𝑡)
𝐴𝑐
𝑌 𝑓 =
𝐺 𝑓 + 𝑓𝑐 + 𝐺(𝑓 − 𝑓𝑐 )
2
𝐴𝑐
𝑌 𝑓 =
𝐻 𝑓 + 𝑓𝑐 + 𝐻(𝑓 − 𝑓𝑐 ) × 𝑍(𝑓)
4
𝑍 𝑓 = 𝑀 𝑓 + 2𝑓𝑐 + 𝑀 𝑓 + 𝑀 𝑓 + 𝑀(𝑓 − 2𝑓𝑐 )
= 𝑀 𝑓 + 2𝑓𝑐 + 2𝑀 𝑓 + 𝑀(𝑓 − 2𝑓𝑐 )
40
Assume a LPF with cutoff frequency < 𝑓𝑐
𝐴𝑐
෡ 𝑓 =
𝑀
𝑀(𝑓) 𝐻 𝑓 + 𝑓𝑐 + 𝐻(𝑓 − 𝑓𝑐 )
2
If
𝐻 𝑓 + 𝑓𝑐 + 𝐻 𝑓 − 𝑓𝑐 = 𝐾
𝐾𝐴𝑐
෡ 𝑓 =
֜𝑀
𝑀(𝑓)
2
𝐾 = 2𝐻 𝑓𝑐 → 𝑜𝑑𝑑 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦
Typically
𝐻 𝑓𝑐 =
1
2
41
Magnitude response of VSB filter
(only the positive-frequency portion is shown)
Odd Symmetry
H ( f − fc ) + H ( f + fc ) = 1 , − W  f  W
42
VSB Modulation
Phase Discrimination Approach
𝑠(𝑡)
𝑉𝑆𝐵±
= 𝑚𝐼 𝑡 cos(2𝜋𝑓𝑐 𝑡) ∓ 𝑚𝑄 𝑡 sin(2𝜋𝑓𝑐 𝑡)
𝐴𝑐
𝑚𝐼 𝑡 =
𝑚(𝑡)
2
𝐴𝑐
𝑚𝑄 𝑡 =
𝑚𝑣 (𝑡)
2
In-phase
component
Quadrature
component
43
𝒎(𝒕)
𝒎𝒗 (𝒕)
Frequency response of the filter for producing the
quadrature component of the upper side-band of
the VSB modulated wave
44
Example 2
(Tone VSB Modulation)
• Consider the following LSB VSB filter 𝐻(𝑓)
45
Example 2 (cont.)
• 𝐴𝑐 = 1 V
• 𝑓𝑐 = 1000 Hz
• 𝑓𝑣 = 100 Hz
• 𝑚 𝑡 = cos 80𝜋𝑡
• 𝑥 𝑡 = 𝑚(𝑡) cos 2𝜋𝑓𝑐 𝑡
• Find 𝑠(𝑡)
𝑉𝑆𝐵−
46
Application: VSB
Transmission of Analog & Digital TV
• TV channel bandwidth = 6 MHz
• The bandwidth covers the VSB-modulated
video signal and the accompanying audio
signal that modulates a carrier of its own
• Why VSB?
– Video signals has large bandwidth and
significant low-frequency content
– A need for cost effective receivers →
Envelope detection → 𝐕𝐒𝐁 − 𝐋𝐂
47
Amplitude Spectrum of a Typical Analog TV Signal
Not an exact VSB!
Note: The values presented on the frequency axis pertain to a specific TV channel
48
Important Remark
• Although we want to conserve bandwidth, in
commercial TV broadcasting the transmitted
signal is not quite VSB modulated
• Because transmitter power levels are high →
expensive process to rigidly control the
filtering of the sidebands
• Solution: a VSB filter is used in each receiver
where the power levels are low
• Penalty: some wasted power and bandwidth
49
Amplitude Response of
VSB Shaping Filter in the Receiver
50
Idealized Amplitude Spectrum
of VSB Modulated Digital TV Signal
Raised
Cosine
Pulse
54.144 MHz
True VSB Shape
Note: The values presented on the frequency axis pertain to a specific TV channel
51
Remarks on Digital TV
• Since baseband signals can be faithfully recovered
from a VSB signal with proper filtering → VSB
modulation can be used for digital signals
• The spectrum is shaped by the filter to extend
0.31 MHz below the carrier and 5.69 MHz above
the carrier
• A carrier component is added to the digital VSB
signal to simplify data detection and reduce
receiver cost
• The digital approach allows for the integration of
audio, video, and color information in one data
stream
52
Coherent Demodulation of VSB Signal
𝑔(𝑡)
𝑥(𝑡)
𝑦(𝑡)
𝑚(𝑡)
ෝ
53
Coherent Detection Techniques
Systems that can be used to generate a coherent
reference carrier from a suppressed carrier wave
for synchronous detection
𝑚(𝑡)
Modulator
𝑠(𝑡)
𝑠(𝑡)
cos(2𝜋𝑓𝑐1 𝑡 + 𝜃1 )
Modulator
𝑚(𝑡)
ෝ
cos(2𝜋𝑓𝑐2 𝑡 + 𝜃2 )
Synchronization
𝑓𝑐1 = 𝑓𝑐2
𝜃1 = 𝜃2
54
Coherent Detection Techniques (cont.)
1. Design identical high-quality crystals for the local
oscillator of both the modulator and demodulator
→ Difficult to match (e.g., quartz crystal oscillators,
as 𝑓𝑐 increases, the dimension of the quartz
decreases)
2. Transmit a pilot carrier signal at a reduced power
(≈ −20 dB) outside the passband of the modulated
signal
3. Costas Receiver
4. Squaring Loop
Only for DSB-SC
55
Costas
Receiver
In-phase Channel
𝒙𝟏 (𝒕)
𝟏
𝒚𝟏 𝒕 = 𝑨𝒄 𝐜𝐨𝐬 𝝓 𝒎(𝒕)
𝟐
𝒆
𝒚𝟐 𝒕 =
𝒙𝟐 (𝒕)
Quadrature Channel
𝟏
𝑨𝒄 𝐬𝐢𝐧 𝝓 𝒎(𝒕)
𝟐
56
Phase Discriminator
1
𝑦1 𝑡 = 𝐴𝑐 cos 𝜙 𝑚(𝑡)
2
𝒆
Narrow-Band
LPF
𝑧(𝑡)
Product
Modulator
1
𝑦2 𝑡 = 𝐴𝑐 sin 𝜙 𝑚(𝑡)
2
57
1
1
x(t ) = Ac cos  m(t )  Ac sin  m(t )
𝑧(𝑡)
2
2
2
Ac 2
=
m (t ) cos  sin 
4
2
Ac 2
=
m (t ) sin 2
8
2
=  m (t ) sin 2
58
Let
∞
𝐴 𝑓 = ℱ 𝑚2 (𝑡) = න 𝑚2 (𝑡)𝑒 −𝑗2𝜋𝑓𝑡 𝑑𝑡 = 𝑀(𝑓) ∗ 𝑀(𝑓)
−∞
∞
→ 𝐴 0 = න 𝑚2 (𝑡) 𝑑𝑡 = 𝐸𝑚
Energy of 𝑚(𝑡)
−∞
𝐴 𝑓
Arbitrary shape
𝐴 0 = 𝐸𝑚
−2𝑊
2𝑊
59
→ 𝑍 𝑓 = ℱ 𝑧 (𝑡) = ℱ 𝛼𝑚2 𝑡 sin(2𝜙)
= 𝛼sin(2𝜙)ℱ 𝑚2 𝑡
= 𝛼 sin 2𝜙 𝐴(𝑓)
Output of the narrow-band LPF
𝐴 𝑓
𝐴 0 = 𝐸𝑚
1
2𝑊
−2𝑊
∆𝒇
→ 𝒆 = 𝛼 sin 2𝜙 𝐸𝑚 ∆𝑓 = 𝐾 sin 2𝜙
𝐾 = 𝛼𝐸𝑚 ∆𝑓
60
𝑒 = 𝐾 sin 2𝜙
; 𝐾 = 𝛼𝐸𝑚 ∆𝑓
If 𝜙 is small → sin 2𝜙 ≈ 2𝜙
֜ 𝒆 ≈ 𝟐𝑲𝝓 ֜ 𝒆 ∝ 𝝓
• A DC controlled signal e is obtained that will automatically
corrects for local phase errors in the VCO
• When the LO is locked on the correct phase (i.e., exact
synchronization with the modulating signal), the output of
the receiver is the output of the in-phase channel
61
Squaring Loop
𝑠 𝑡 = 𝑚 𝑡 cos(2𝜋𝑓𝑐 𝑡 + 𝜙)
𝑠(𝑡)
𝟐
𝑦(𝑡)
Narrow-BPF
@ ± 𝟐𝒇𝒄
𝑥(𝑡)
𝒄 𝒕
2:1 Frequency 𝑧(𝑡)
Divider
Phase
Looked Loop
𝑐 𝑡 = 𝐾cos(2𝜋𝑓𝑐 𝑡 + 𝜙)
62
Phase-Locked Loop (PLL)
𝒙(𝒕)
𝒈(𝒕)
Low-Pass
Filter
𝒆(𝒕)
𝒛(𝒕)
63
Practice Exercise 5
• Show that the Costas receiver and the
squaring loop can not be used for carrier
acquisition (coherent detection) for SSB-SC
or VSB?
64
Frequency Conversion
• aka
– Frequency translation
– Frequency Mixing
– Heterodyning
65
66
Up − conversion :
 f 2 = f1 + f l 
f l = f 2 − f1
Down − conversion :
 f 2 = f1 − f l 
f l = f1 − f 2
67
Summary: Up or Down Conversion
Bandpass
Filter
𝒇𝒊𝒏
𝒇𝟐
vLO (t ) = ALO cos2 f LO t 
U − C → f LO = f 2 − f in
D − C → f LO = f in − f 2
𝒇𝑳𝑶
68
Image Signal Problem (I)
The Up-Conversion case:
• 𝑓2 = 𝑓1 − 𝑓𝑙 → 𝑓𝑙 = 𝑓1 − 𝑓2
• The signal
𝑠𝑖 𝑡 = 𝑔 𝑡 cos 2𝜋 𝑓1 − 2𝑓𝑙 𝑡
will also result in an output at 𝑓𝑙
• 𝑠𝑖 𝑡 is called the image signal of
𝑚 𝑡 cos(2π𝑓1 𝑡)
→ 𝑓1 − 2𝑓𝑙 = image frequency of 𝑓1
69
Image Signal Problem (II)
The Down-Conversion case:
• 𝑓2 = 𝑓1 + 𝑓𝑙 → 𝑓𝑙 = 𝑓2 − 𝑓1
• The signal
𝑠𝑖 𝑡 = 𝑔 𝑡 cos 2𝜋 𝑓1 + 2𝑓𝑙 𝑡
will also result in an output at 𝑓𝑙
• 𝑠𝑖 𝑡 is called the image signal of
𝑚 𝑡 cos(2π𝑓1 𝑡)
→ 𝑓1 + 2𝑓𝑙 = image frequency of 𝑓1
70
Some Remarks
• There is only one image signal or frequency
• It is always separated from the desired
signal or frequency by 2𝑓𝑙
• Why the image signal is a problem?
• Can you think of a way to solve the image
problem?
71
Example 3
• Let 𝑠(𝑡) be a narrow-band signal with bandwidth
𝑊 = 10 kHz and a mid-ban frequency 𝑓𝑐 which
may lie in the range 535 − 1605 kHz
• We want to translate this wave to a fixed
frequency band centered at 𝑓𝑜 = 455 kHz
• Find the range of tuning that must be provided in
the local oscillator of the converter
• Repeat for 𝑓𝑜 = 1700 kHz
72
Receiver Functions
In any communication systems, the receiver has
the following functions:
• Signal demodulation (main operation)
• Carrier-Frequency Tuning → To select the
desired signal (i.e., desired radio or TV station)
• Filtering → To separate the desired signal from
other modulated signals that may be picked
along the way
• Amplification → To compensate for the loss of
signal power incurred during transmission 73
Receiver Performance
Measured by its:
• Sensitivity → The ability of the receiver to
detect weak signals
• Selectivity → The ability of the receiver to
separate (or filter) closely spaced signals
74
The Superheterodyne Receiver
(superhet receiver)
• A special type of receiver that achieves all the
functions discussed in the previous slide
• Designed to overcome the difficulty of having
to build a tunable high- (and variable-) Q filter
• Has good sensitivity and selectivity
• Practically all analog radio and TV receivers
are of the superheterodyne type
75
Superheterodyne AM Receiver
𝐑𝐅 = Radio-Frequency
𝐈𝐅 = Intermediate-Frequency
Automatic
Volume Control
Super = Mixer Up-Conversion
(High-Side Tuning)
→ 𝒇𝑳𝑶 = 𝒇𝒄 + 𝒇𝑰𝑭
76
RF BPF & Amplifier
@ 𝒇𝒄
IF BPF & Amplifier
@ 𝒇𝑰𝑭
Automatic
Volume Control
𝒇𝑳𝑶
AM Demodulator
Receiver Tuning
77
The Receiver Operation
𝐑𝐅 Filter:
• Tunes to the desired carrier frequency 𝑓𝑐 by
varying the frequency of the Local Oscillator
• Main function to suppress the image signals (or
frequencies) → 𝒇𝒊𝒎𝒂𝒈𝒆 = 𝒇𝒄 ± 𝟐𝒇𝑰𝑭
• Standard AM 𝒇𝑰𝑭 = 𝟒𝟓𝟓 𝐊𝐇𝐳
• For example, an AM station operating @ 1710
KHz is said to be an image of the stations
operating at 800 KHz and 2620 KHz
78
The Receiver Operation (cont.)
Mixer + 𝐈𝐅 Filter:
• Translate 𝑓𝑐 to a fixed IF frequency 𝑓𝐼𝐹
• Standard AM 𝒇𝑰𝑭 = 𝟒𝟓𝟓 𝐊𝐇𝐳
• Up-Conversion:
𝒇𝑳𝑶 = 𝒇𝒄 + 𝒇𝑰𝑭 = 𝒇𝒄 + 𝟒𝟓𝟓 𝐊𝐇𝐳
• Amplify the signal prior to envelope detection
79
Why Mixing and Two-Stage Filtering?
• The IF stage
– Fixed IF frequency (filter need not be tunable)
– Factory tuned
– High selectivity (adjacent channel suppression) can
be achieved due to smaller IF frequency and
complex filter design
• The RF stage
– Need not be narrow-band (i.e., with high-selectivity)
– Simple design
– Tunable over a narrow range of frequencies
80
𝟓𝟒𝟎 ≤ 𝒇𝒄 ≤ 𝟏𝟔𝟎𝟎 𝐊𝐇𝐳
𝒇𝑰𝑭 = 𝟒𝟓𝟓 𝐊𝐇𝐳
𝒇𝑳𝑶 = 𝒇𝒄 + 𝟒𝟓𝟓 𝐊𝐇𝐳
81
AM Bandwidth
Bandwidth of audio signal = 4 KHz
Bandwidth of AM transmitted signal = 8 KHz
Guard Bandwidth = 2 KHz
Bandwidth of commercial AM Radio Station
= 10 KHz
→IF filter must be designed to provide good
selectivity over a 10 KHz band
→RF selectivity over a 910 KHz band!
•
•
•
•
82
Why Up-Conversion (High-Side Tuning)?
• Because it leads to much less tuning range of
the local oscillator than down-conversion (lowside tuning) → Less complex LO
• AM standard: 540 ≤ 𝑓𝑐 ≤ 1600 KHz ;
𝑓𝐼𝐹 = 455 KHz
• Up-conversion: 995 ≤ 𝑓𝐿𝑂 ≤ 2055 KHz
→ Tuning range = 2055Τ995 = 2.07 (to 1)
• Down-conversion: 85 ≤ 𝑓𝐿𝑂 ≤ 1145 KHz
→ Tuning range = 1145Τ85 = 13.47 (to 1)
83
84
Multiplexing
• An important signal processing operation
whereby a collection of independent signals can
be combined into a composite signal suitable for
transmission over a common channel
• Approaches:
– Frequency-Division Multiplexing (FDM) → Divide the
spectrum → Bandpass communications
– Time-Division Multiple Multiplexing (TDM) → Divide
the time access → Baseband communications
85
Frequency-Division Multiplexing
86
87
Time-Division Multiplexing
88
Multiple Access
• A technique that permits the sharing of the
communication resources of the channel by
multiple users to improve the overall capacity of
the system
• Channel resources to be shared:
– Spectrum
– Time
– Space
• Goal: The sharing of resources of the channel to
be achieved without causing serious interference
between users of the system
89
Multiple Access vs. Multiplexing
• Multiple access → the remote sharing of a
communication channel such as a satellite or radio
channel by users in highly dispersed locations;
Multiplexing → the sharing of a channel such as a
telephone channel by users confined to a local users
• Multiplexed system → user requirements are
usually fixed; Multiple access system → user
requirements can change dynamically with time
leading to the need for dynamic channel allocation
approaches
90
Multiple Access Techniques
• Frequency Division Multiple Access (FDMA)
• Time Division Multiple Access (TDMA)
• Code division multiple-access (CDMA)
• Orthogonal Frequency Division Multiple
Access (OFDMA)
• Space Division Multiple Access (SDMA)
91
Multiple Access Techniques (cont.)
FDMA
TDMA
CDMA
Example: Frequency hopping via
pseudo-noise (PN) sequence
92
Code-Division Multiple Access
93
Orthogonal Frequency-Division Multiple Access
Multiple Orthogonal Carriers
94
Space Division Multiple Access
(Multibeam Antenna/Array)
95
96