Single Sideband Modulation
Contents
• Hilbert Transform Review
• Pre-envelope
• SSBSC Generation
• SSBSC detection
• Pilot carrier SSB
Hilbert transform
• Hilbert transform is an operation that affects the phase of a signal
H(f) |H(f)|
𝟏. 𝟎
H(f)
𝟗𝟎
𝑓
𝑓
−𝟗𝟎
Hilbert Transform Relationship
• The transfer function 𝐻 𝑓 of the Hilbert Transformer can be written as:
𝜋
−𝑗 2
𝑒 𝑓≥0
𝐻 𝑓 =ቐ 𝜋
𝑒 𝑗2 𝑓 < 0
• It can also be represented as:
𝐻 𝑓 = −𝑗𝑠𝑔𝑛 𝑓
• where 𝑠𝑔𝑛 𝑓 is the signum function defined as:
1 𝑓>0
𝑠𝑔𝑛 𝑓 = ቐ 0 𝑓 = 0
−1 𝑓 < 0
• The HT of a signal g 𝑡 is denoted by:
HT g 𝑡 = gො 𝑡
• In frequency domain the signal G 𝑓 and 𝐺 𝑓 are related as:
𝐺 𝑓 = −𝑗𝑠𝑔𝑛 𝑓 𝐺 𝑓
Simple Example
• Find the HT of the function g 𝑡 given by: Positive Negative
Frequency Frequency
g 𝑡 = 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡
• It is easier to work in frequency domain here, we know that:
1
𝐺 𝑓 = −𝑗𝑠𝑔𝑛 𝑓 𝐺 𝑓 = −𝑗𝑠𝑔𝑛 𝑓 𝛿 𝑓 − 𝑓𝑐 + 𝛿 𝑓 + 𝑓𝑐
2
𝑗 𝑗 1
= − 𝛿 𝑓 − 𝑓𝑐 + 𝛿 𝑓 + 𝑓𝑐 = 𝛿 𝑓 − 𝑓𝑐 − 𝛿 𝑓 + 𝑓𝑐
2 2 2𝑗 𝑠𝑖𝑛 2𝜋𝑓𝑐 𝑡
HT Properties
• Property 1
g 𝑡 and gො 𝑡 have the same amplitude spectrum
• Property 2
𝐻𝑇 gො 𝑡 = −g 𝑡
• Property 3
∞
g 𝑡 and gො 𝑡 are orthogonal, i.e. −∞ g 𝑡 gො 𝑡 𝑑𝑡 = 0
Problems
• Problem: Find the HT of the function g 𝑡 given by:
g 𝑡 = 𝑚 𝑡 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡
Given that 𝑚 𝑡 ՞ 𝑀 𝑓
1
• Solution: ℱ g 𝑡 =𝐺 𝑓 = 𝑀 𝑓 − 𝑓𝑐 + 𝑀 𝑓 + 𝑓𝑐
2
𝑗
𝐺 𝑓 = −𝑗𝑠𝑔𝑛 𝑓 𝐺 𝑓 = − sgn 𝑓 𝑀 𝑓 − 𝑓𝑐 + sgn 𝑓 𝑀 𝑓 + 𝑓𝑐
2
1
= 𝑀 𝑓 − 𝑓𝑐 − 𝑀 𝑓 + 𝑓𝑐
2𝑗
1 ∞
gො 𝑡 = ℱ −1 𝐺 𝑓 = 𝑀 𝑓 − 𝑓𝑐 − 𝑀 𝑓 + 𝑓𝑐 𝑒 𝑗2𝜋𝑓𝑡 𝑑𝑓 = 𝑚 𝑡 𝑠𝑖𝑛 2𝜋𝑓𝑐 𝑡
2𝑗 −∞
• Problem: Find the HT of the function g 𝑡 = 𝑚 𝑡 𝑠𝑖𝑛 2𝜋𝑓𝑐 𝑡 , given that 𝑚 𝑡 ՞ 𝑀 𝑓
Pre-envelope
• From a real-valued signal, we can extract a complex-valued signal by adding its HT as
follows:
g + 𝑡 = g 𝑡 + jොg 𝑡
• g + 𝑡 is called the pre-envelope of g 𝑡 .
g 𝑡 = 𝑅𝑒 g + 𝑡
• where 𝑅𝑒 stands for “real part of”
• ℱ g + 𝑡 = G+ 𝑓 = G 𝑓 + 𝑗 G 𝑓 = G 𝑓 + 𝑗 −𝑗𝑠𝑔𝑛 𝑓 𝐺(𝑓)
2G 𝑓 𝑓 > 0
G+ 𝑓 = ൞ G 0 𝑓 = 0
0 𝑓<0
Pre envelope
|G+(f)|
|G(f)|
−𝑓𝑐 𝑓𝑐
f 𝑓
• Fourier transform of Pre-envelope exists only for positive frequencies
• Per-envelope is complex, not a real signal.
Pre envelope contd…
g − 𝑡 = g 𝑡 − jොg 𝑡
𝑓 = G 𝑓 − 𝑗 −𝑗𝑠𝑔𝑛 𝑓 𝐺(𝑓)
• ℱ g − 𝑡 = G− 𝑓 = G 𝑓 − 𝑗G
0 𝑓>0
G− 𝑓 = ൞G 0 𝑓 = 0
2G 𝑓 𝑓 < 0
|G(𝑓)|
|G-(𝑓)|
𝑓
−𝑓𝑐 𝑓𝑐
𝑓
Pre envelope of bandpass signal
|G(f)| |G+(f)|
f f
• Fourier transform of Pre-envelope exists only for positive frequencies
• Per-envelope is complex, not a real signal.
|G+(f)|
Complex Envelope
f
• Every bandpass signal has a lowpass
equivalent or complex envelope.
• Complex and pre envelope are related
as:
g t = g + 𝑡 𝑒 −𝑗2𝜋𝑓𝑐𝑡 ෩ (f)|
|G
Or
g + (𝑡) = g 𝑡 𝑒 𝑗2𝜋𝑓𝑐𝑡
f
Single Sideband Suppressed Carrier Modulation
• Two main disadvantages of the DSBSC modulation scheme:
• Utilize twice as much bandwidth – both the upper and lower sideband
actually contains same information (redundant).
• Power requirement for two sidebands.
• Is it possible to transmit single sideband only? Benefits
• Will require half the bandwidth of DSBSC
• Considerably less power
• Sideband power makes up 100% of the total transmitted power
SSBSC
𝑚+ 𝑡 ∗ 𝑒 𝑗2𝜋𝑓𝑐 𝑡
|M(f)|
|S(f)|
Upper sideband
𝑓 −𝑓𝑐 𝑓𝑐
𝑓
−𝑗2𝜋𝑓𝑐 𝑡
𝑚− 𝑡 ∗ 𝑒
|S(f)|
Lower sideband
−𝑓𝑐 𝑓𝑐
𝑓
Mathematical Treatment
s 𝑡 = 𝑚+ 𝑡 ∗ 𝑒 𝑗2𝜋𝑓𝑐𝑡 + 𝑚− 𝑡 ∗ 𝑒 −𝑗2𝜋𝑓𝑐𝑡 (upper sideband)
= 𝑚 𝑡 + j𝑚ෝ 𝑡 ∗ 𝑒 𝑗2𝜋𝑓𝑐 𝑡 + 𝑚 𝑡 − j𝑚ෝ 𝑡 ∗ 𝑒 −𝑗2𝜋𝑓𝑐 𝑡
= 𝑒 𝑗2𝜋𝑓𝑐 𝑡 + 𝑒 −𝑗2𝜋𝑓𝑐 𝑡 ∗ 𝑚 𝑡 + j 𝑒 𝑗2𝜋𝑓𝑐 𝑡 − 𝑒 −𝑗2𝜋𝑓𝑐 𝑡 𝑚
ෝ 𝑡
= 2𝑚 𝑡 cos 2𝜋𝑓𝑐 𝑡 − 2𝑚 ෝ 𝑡 sin 2𝜋𝑓𝑐 𝑡
s 𝑡 = 𝑚+ 𝑡 ∗ 𝑒 −𝑗2𝜋𝑓𝑐𝑡 + 𝑚− 𝑡 ∗ 𝑒 𝑗2𝜋𝑓𝑐𝑡 (lower sideband)
= 2𝑚 𝑡 cos 2𝜋𝑓𝑐 𝑡 + 2𝑚ෝ 𝑡 sin 2𝜋𝑓𝑐 𝑡
SSB modulator
• Frequency-discrimination scheme for the generation of a SSB
modulated wave.
• If the message signal is limited in the frequency range 𝑓𝑎 and 𝑓𝑏
where 𝑓𝑎 > 0
SSBSC USB transmission
Baseband (message) signal
SSBSC LSB transmission
Two stage modulator (Home Assignment)
• Suppose m(t) is a voice signal in the range 0.3 Hz to 3.4 kHz f1=100 kHz and f2=10
MHz. The scheme shown below is for generation of SSB signal. Find the
specification of the band-pass filters to generate SSB signal when:
a) LSB is transmitted.
b) USM is transmitted.
c) Plot the spectrum of the signal at each point A, B, C and D for LSB and USB transmission.
m(t) s2(t)
Product Band-pass s1(t) Product Band-pass
Modulator filter 1 Modulator filter 2
A B C D
A1cos(1t) A2cos(2t)
SSB modulator 𝒎 𝒕 𝒄𝒐𝒔 𝟐𝝅𝒇𝒄 𝒕
𝒎 𝒕 𝒄𝒐𝒔 𝟐𝝅𝒇𝒄 𝒕
∓𝒎ෝ 𝒕 𝒔𝒊𝒏 𝟐𝝅𝒇𝒄 𝒕
• Phase discrimination
method for generating a
SSB-modulated wave.
ෝ 𝒕 𝒔𝒊𝒏 𝟐𝝅𝒇𝒄 𝒕
𝒎
Product Low-pass
𝑠 𝑡 𝑚′ 𝑡
SSB demodulation Modulator
d t
filter
𝐴1 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡
• Assume that the clock at the receiver is perfectly synchronized with the transmitter
clock.
d t = 𝑠 𝑡 ∗ 𝐴1 cos 2𝜋𝑓𝑐 𝑡 = 2𝑚 𝑡 cos 2𝜋𝑓𝑐 𝑡 − 2𝑚 ෝ 𝑡 sin 2𝜋𝑓𝑐 𝑡 𝐴1 cos 2𝜋𝑓𝑐 𝑡
1 + cos 2𝜋2𝑓𝑐 𝑡
= 2𝐴1 𝑚 𝑡 − 𝐴1 𝑚ෝ 𝑡 𝑠𝑖𝑛 2𝜋2𝑓𝑐 𝑡
2
• After low pass filtering (LPF)
𝑚′ 𝑡 = 𝐴1 𝑚 𝑡
• The cut-off frequency of the LPF must be less than 2𝑓𝑐 − 𝑊and greater than 𝑊 where
𝑊 is the bandwidth of the low pass signal.
SSB demodulation contd…
Product Low-pass
𝑠 𝑡 𝑚′ 𝑡
Modulator filter
d t
𝐴1 𝑐𝑜𝑠 2𝜋(𝑓𝑐 +∆𝑓)𝑡 + 𝜃(𝑡)
• Practically it is impossible to have perfect synchronization between the receiver and
transmitter clock. This will result into frequency and phase error. Assuming the LSB
is transmitted
d t = 𝑠 𝑡 ∗ 𝐴1 cos 2𝜋(𝑓𝑐 +∆𝑓)𝑡 + 𝜃(𝑡)
= 2𝑚 𝑡 cos 2𝜋𝑓𝑐 𝑡 + 2𝑚 ෝ 𝑡 sin 2𝜋𝑓𝑐 𝑡 𝐴1 cos 2𝜋(𝑓𝑐 +∆𝑓)𝑡 + 𝜃(𝑡)
cos 2𝜋∆𝑓𝑡 + 𝜃(𝑡) + cos 2𝜋(2𝑓𝑐 +∆𝑓)𝑡 + 𝜃(𝑡)
= 2𝐴1 𝑚 𝑡
2
sin 2𝜋(2𝑓𝑐 +∆𝑓)𝑡 + 𝜃(𝑡) − sin 2𝜋∆𝑓𝑡 + 𝜃(𝑡)
+ 2𝐴1 𝑚
ෝ 𝑡
2
SSB demodulation contd…
• After low pass filtering (LPF)
𝑚′ 𝑡 = 𝐴1 𝑚 𝑡 cos 2𝜋∆𝑓𝑡 + 𝜃 𝑡 − 𝐴1 𝑚
ෝ 𝑡 sin 2𝜋∆𝑓𝑡 + 𝜃(𝑡)
• We can have the following two cases
• ∆𝑓 ≠ 0 and 𝜃 𝑡 = 0
• ∆𝑓 = 0 and 𝜃(𝑡) ≠ 0 but constant.
• Case I Constant frequency error.
𝑚′ 𝑡 = 𝐴1 𝑚 𝑡 cos 2𝜋∆𝑓𝑡 − 𝐴1 𝑚
ෝ 𝑡 sin 2𝜋∆𝑓𝑡
• In case ∆𝑓 is positive and the LSB is transmitted, the spectrum of the demodulated
signal 𝑚′ 𝑡 will shift outwards by an amount ∆𝑓.
• In case ∆𝑓 is negative and the LSB is transmitted, the spectrum of the demodulated
signal 𝑚′ 𝑡 will shift inwards by an amount ∆𝑓.
• The effect will be opposite in case of USB transmission.
Pilot carrier SSB
• The major problem with the previous detection technique is to have a
coherent source at the receiver.
• If the frequency error becomes a function of time then it becomes
difficult to recover the original signal.
• Simple and effective solution is to transmit the carrier along with a
sideband but at much reduced level which may be used to
synchronise the receiver.
𝑠 𝑡 = 𝑚 𝑡 cos 2𝜋𝑓𝑐 𝑡 − 𝑚 ෝ 𝑡 sin 2𝜋𝑓𝑐 𝑡 + 𝐴𝑐 cos 2𝜋𝑓𝑐 𝑡
Pilot carrier SSB detection
• Envelop detector can be used
𝑠 𝑡 = 𝑚 𝑡 cos 2𝜋𝑓𝑐 𝑡 − 𝑚
ෝ 𝑡 sin 2𝜋𝑓𝑐 𝑡 + 𝐴𝑐 cos 2𝜋𝑓𝑐 𝑡
= 𝑚 𝑡 + 𝐴𝑐 cos 2𝜋𝑓𝑐 𝑡 − 𝑚
ෝ 𝑡 sin 2𝜋𝑓𝑐 𝑡
• The envelop detector output 𝑒 𝑡
1
2 𝑚 𝑡 +𝐴𝑐 2 cos 2𝜋𝑓𝑐 𝑡 −𝑚
ෝ 2 𝑡 sin 2𝜋𝑓𝑐 𝑡
• s 𝑡 = 𝑚 𝑡 + 𝐴𝑐 ෝ2 𝑡
+𝑚 2
1
𝑚 𝑡 +𝐴𝑐 2 +𝑚
ෝ2 𝑡 2
= e 𝑡 𝑐𝑜𝑠 2𝜋𝑓𝑐 𝑡 + 𝜃 where e 𝑡 is the envelop of the signal s 𝑡
Pilot carrier SSB detection
1
𝑚2 𝑡 𝑚(𝑡) 𝑚 ෝ2 𝑡 2
e 𝑡 = 𝐴𝑐 1 + 2 +2 𝐴 + 2
𝐴𝑐 𝑐 𝐴 𝑐
1
𝑚 𝑡 2
e 𝑡 ≅ 𝐴𝑐 1 + 2 𝐴𝑐 ≫ |𝑚 𝑡 |
𝐴𝑐
𝑚 𝑡
e 𝑡 ≅ 𝐴𝑐 1 +
𝐴𝑐
e 𝑡 ≅ 𝐴𝑐 + 𝑚 𝑡
The demodulated signal is approximately same as the message signal with a DC
offset
