2. Modulation techniques
• Modulation :
A process through which audio, video, image or text information is added to an
electrical or optical carrier signal to be transmitted over a telecommunication or
electronic medium.
Modulation enables the transfer of information on an electrical signal to a receiving
device that demodulates the signal to extract the blended information.
• Amplitude modulation (AM)
Is defined as a carrier frequency whose amplitude is varied in proportion to the
instantaneous amplitude of a modulating voltage.
The modulating voltage is referred to as the intelligence.
The carrier frequency, or carrier of the modulating voltage, is typically much
higher than that of the modulating voltage
o Usually a radio frequency(RF) signal in the Mid-Frequency range of 300KHz to 3MHz
The frequency of the modulating voltage, is typically an audio frequency (AF) signal
o Usually in the Frequency range of 20Hz to 20KHz
4. AM cont..
To produce an AM wave requires a modulator which produces the sum and
difference products of the carrier and the modulation frequency.
A modulator is a nonlinear device capable of producing a mixing action thus resulting
in the sum and difference products. Transistors and diodes are nonlinear devices.
Combining two frequencies through a nonlinear device produces a mixing action
resulting in harmonics as well as the sum and difference frequencies of the original
signals.
5. AM cont..
• In a simple AM modulator, a summing
amplifier is used to electrically sum the
modulation and carrier frequencies.
• The diode and the resistor are used to
clip the negative half of the composite
signal to produce a pulsating DC signal
rich in harmonic content.
• When a single pulse (vo ) is fed to a
resonant tank circuit, it causes a
damped oscillation to occur at the
natural resonant frequency of the tank.
• A damped oscillation is a sinusoid
function with an exponentially
decaying envelope.
• Over time, decaying is caused by
resistive losses in the tank components.
V1
t
V2
t
vo
Vo
t
vo
t
Gnd
Ae-t
6. AM cont..
• If a parallel tank circuit is put at the diode
output, the pulsating DC signal will sustain
the oscillatory effect thereby producing
complete sinusoids whose amplitudes are
proportional to the amplitude of each pulse.
• This is known as a flywheel effect and
effectively results into an AM wave.
• The resonant tank circuit filters out the
undesirable harmonics and preserves the
AM frequency components!
• Since the diode is a passive device, it offers
no gain to the AM wave!
A transistor is a device that will add the
element of gain to the signal.
• Tank circuit
vo
t
v
7. AM cont..
Replacing the diode with a transistor.
• The amplitudes of the Carrier and
modulating frequencies are adjusted to turn
on the transistor in its nonlinear region
during the positive peaks of the cycle.
• During the negative peaks of the cycle, the
transistor turns off, thus producing the
pulsating DC signal at the tank circuit.
• The AM wave is produced at the collector
output of the transistor
• Transistor modulator circuit
em
ec
å
-
+
-
vcc
Vo
em
ec
Vo
t
8. Analysis of the AM wave
A carrier frequency, fc (angular velocity c ) is to be amplitude modulated with a sine
wave fm (angular velocity m ).
Therefore, the instantaneous voltage em of the sine wave used to modulate the carrier
frequency is given as em = Vm sin m t where m is 2fm and Vm is the peak amplitude
of the modulating signal.
The instantaneous voltage of the carrier frequency is given as ec = Vc sin c t where c
is 2fc and Vc is the peak amplitude of the carrier.
The components of an AM signal are:
• 𝑒𝐴𝑀 = 𝑉
𝑐𝑠𝑖𝑛𝑐𝑡 +
𝑚𝑉𝑐
2
cos 𝜔𝑐 + 𝜔𝑚 𝑡 −
𝑚𝑉𝑐
2
cos(𝜔𝑐 + 𝜔𝑚)𝑡
• carrier frequency Lower sideband Upper sideband
• m in the equation is the modulation index or factor given by 𝑚 =
𝑉𝑚
𝑉𝑐
• Expressed as a percentage it becomes 𝑀 =
𝑉𝑚
𝑉𝑐
× 100
9. Analysis of the AM wave cont..
Vm
Vmin
Vc Vmax
Envelope = -[Vc + Vm sin mt]
Envelope = [Vc + Vm sin mt]
10. Analysis of the AM wave cont..
• The modulation index m was given as 𝑚 =
𝑉𝑚
𝑉𝑐
• As a percentage M =
𝑉𝑚
𝑉𝐶
100
• From the AM wave, because ‘m’ is symmetrical about the carrier frequency, therefore
Vmax – Vc = Vc - Vmin
• And 𝑉
𝑚 =
𝑉𝑚𝑎𝑥−𝑉𝑚𝑖𝑛
2
and 𝑉
𝑐 =
𝑉𝑚𝑎𝑥+𝑉𝑚𝑖𝑛
2
dividing Vm by Vc gives us
• 𝑚 =
𝑉𝑚𝑎𝑥−𝑉𝑚𝑖𝑛
𝑉𝑚𝑎𝑥+𝑉𝑚𝑖𝑛
and is equal to
𝑉𝑚𝑎𝑥𝑝−𝑝−𝑉𝑚𝑖𝑛𝑝−𝑝
𝑉𝑚𝑎𝑥𝑝−𝑝+𝑉𝑚𝑖𝑛𝑝−𝑝
11. Analysis of the AM wave cont..
Envelope of the AM waveform.
• The amplitude of the AM waveform (eenv ) is the envelope (peak values of each cycle).
• The envelope outlines the peaks and troughs of the waveform and hence the envelope
can be a positive or negative signed value.
• Hence for Sine wave modulation eenv = Vc + em = Vc + Vm sin m t
• Remember 𝑚 =
𝑉𝑚
𝑉𝑐
! Therefore Vm = mVc and by substitution in the equation above
• eenv =Vc + mVc sin m t
= Vc (1 + m sin m t) the positive envelope
= -Vc (1 + m sin m t) the negative envelope
12. Worked example 1
A carrier signal with a peak voltage of 2V is amplitude modulated with a 10 KHz sine
wave. The modulation voltage has an effective value(rms) of 750mV. Determine (i)the
percent modulation index m (ii) the instantaneous voltage of the positive and negative
envelope when the 10KHz sine wave has completed 68s of its cycle (iii) draw the
resultant AM wave form.
V
t
Vc=2Vp
eenv=-1.04V
eenv =+1.04V
t=68s
Vm =750mV
13. Analysis of the AM wave cont..
• Frequency spectrum of the AM wave.
eenv represents the peak envelope of the AM waveform at any instant in time ‘t’.
The carrier frequency c as a peak value represents an AM waveform using the
envelope.
Hence eAM = eenv sin = eenv sin c t
From earlier derivations, 𝑒𝐴𝑀 = 𝑉
𝑐 1 + m sin 𝜔𝑚𝑡 sin𝜔𝑐𝑡
𝑒𝐴𝑀 = 𝑉
𝑐𝑠𝑖𝑛 𝜔𝑐𝑡 + 𝑚𝑉
𝑐𝑠𝑖𝑛𝜔𝑐𝑡𝑠𝑖𝑛 𝜔𝑚t
The trigonometric identity sin sin =
1
2
[cos 𝛼 − 𝛽 − cos 𝛼 + 𝛽 ]
By applying the same identity to the eAM equation above we get
𝑒𝐴𝑀 = 𝑉
𝑐𝑠𝑖𝑛𝜔𝑐𝑡 + 𝑚
𝑉𝑐
2
cos(𝑐 − 𝑚)𝑡 − 𝑚
𝑉𝑐
2
cos(𝑐 + 𝑚)𝑡
Carrier frequency fc
Lower sideband (LSB) fc - fm Upper sideband (USB) fc + fm
14. Analysis of the AM wave cont.
• The total composite AM signal comprised of two side bands and a carrier frequency, is
normally referred to as double sideband full carrier (DSBFC)
• The peak value terms of the LSB and the USB are equal i.e.
𝑚𝑉𝑐
2
= 𝑉𝐿𝑆𝐵 = 𝑉𝑈𝑆𝐵
• When all three frequency components are in phase they add together linearly and form
the maximum signal amplitude, Vmax i.e Vmax = Vc +VLSB + VUSB
• Vmax is also equal to the sum of Vc and Vm hence Vm = VLSB + VUSB
• Because peak sideband voltages are equal Vm = 2 VLSB = 2 VUSB
15. Analysis of the AM wave cont..
Vm
Vc
Vmax
Vmax =Vc + Vm
=Vc +VLSB + VUSB
Vm =VLSB + VUSB = 2VLSB =2VUSB
V
t
fLSB fc fUSB
𝑉𝐿𝑆𝐵 =
𝑚
2
𝑉
𝑐
𝑉𝑈𝑆𝐵 =
𝑚
2
𝑉
𝑐
Vc
Time domain
Frequency domain
16. Worked example 2
• Given a station’s peak carrier voltage of 2KV has been modulated to an index of 75% with a 2 KHz
test tone. The station’s broadcasting frequency is 810 KHz. Determine
• (i) the LSB and the USB
• (ii) the peak modulation voltage
• (iii) the peak VLSB and VUSB
• (iv) Vmax
17. Analysis of the AM wave cont..
Power distribution in the AM waveform
• The total effective(rms) power, PT in the AM wave is the sum of the effective carrier
power level Pc and the effective sideband power levels PLSB and PUSB
• Hence 𝑃𝑇 = 𝑃𝑐 + 𝑃𝐿𝑆𝐵 + 𝑃𝑈𝑆𝐵
• =
𝑉2
𝑐𝑟𝑚𝑠
𝑅
+
𝑉2
𝐿𝑆𝐵𝑟𝑚𝑠
𝑅
+
𝑉2
𝑈𝑆𝐵𝑟𝑚𝑠
𝑅
• =
(0.707𝑉𝑐)2
𝑅
+
(0.707𝑉𝐿𝑆𝐵 )2
𝑅
+
(0.707𝑉𝑈𝑆𝐵) 2
𝑅
• =
𝑉𝑐
2
2𝑅
+
𝑉𝐿𝑆𝐵
2
2𝑅
+
𝑉2
𝑈𝑆𝐵
2𝑅
Remember VLSB = VUSB =
𝑚
2
Vc hence PLSB = PUSB =
[0.707(
𝑚𝑉𝑐
2
)]2
𝑅
=
𝑚2𝑉2
𝑐
8𝑅
The total power in the carrier Pc is
𝑉𝑐
2
2𝑅
hence PLSB = PUSB =
𝑚2𝑃𝑐
4
18. Worked example 3
A spectrum analyser with an input impedance of 50Ω is used to measure the power spectrum of an AM signal at
the output of a preamplifier circuit. The AM signal has been modulated with a sine wave. The effective carrier
power, PC is 750 mW and each sideband is 120 mW. Calculate (i) The total effective power, (ii) The peak carrier
voltage, (iii) the modulation index m and the percent modulation index M, (iv) The modulation voltage (v) LSB and
USB voltages. (vi) Draw the waveform.
20. Analysis of the AM wave cont..
The total effective power PT representation
• Can be represented using the carrier power in the sidebands where PT = PC + PLSB + PUSB
• Such that 𝑃𝑇 = 𝑃𝑐 +
𝑚2𝑃𝑐
4
+
𝑚2𝑃𝑐
4
= 𝑃𝑐 +
𝑚2𝑃𝑐
2
= 𝑃𝑐(1 +
𝑚2
2
)
• Hence only knowing the carrier power and the modulation index we can compute the total
effective power of an AM wave form
Vc
21. Analysis of the AM wave cont..
Considering an AM waveform in terms of power efficiency
• The condition when 100% modulation is used m = 1 and this sets the condition for maximum
power utilisation in the sidebands.
• From 𝑃𝑇 = 𝑃𝑐 +
𝑚2𝑃𝑐
4
+
𝑚2𝑃𝑐
4
= 𝑃𝑐 +
𝑚2𝑃𝑐
2
= 𝑃𝑐 1 +
𝑚2
2
when m = 1, PT = 1.5Pc
• The equation states that at 100% modulation, the total effective power in the sidebands of an
AM waveform is half the carrier power..
• Or we can say that 2/3 of the total power is dissipated by the carrier frequency Pc = PT /1.5
22. Worked example 4
• An AM transmitter has an effective carrier power level of 50KW. If the carrier is modulated with a sine wave,
determine (i) the total effective power for a percent modulation index of 50%. (ii) The effective power in each
sideband for a modulation index of 50%. (iii) The total effective transmitted power for 100% modulation. (iv)
The effective power in each sideband for 100% modulation.
23. Analysis of the AM wave cont..
• AM waveform with 100% modulation
• 𝑚 =
𝑉𝑚𝑎𝑥𝑝−𝑝− 𝑉𝑚𝑖𝑛𝑝−𝑝
𝑉𝑚𝑎𝑥𝑝−𝑝+ 𝑉𝑚𝑖𝑛𝑝−𝑝
=
𝑉𝑚𝑎𝑥𝑝−𝑝− 0
𝑉𝑚𝑎𝑥𝑝−𝑝+ 0
= 1
• Carrier modulated by a sine wave.
• At 100% modulation there is no Vmin p-p
• Labelled diagram is the time domain of the AM
waveform with 100% modulation.
• Frequency domain of the AM wave form showing
the amount of carrier power in the sidebands and
the centre frequency
• 𝑃𝑇 = 𝑃𝑐 +
1
4
𝑃𝑐 +
1
4
𝑃
𝑐 = 𝑃𝑐 +
1
2
𝑃𝑐
Vmax p-p
Vmin p-p
fLSB fc
fUSB
t
V
f
P
1
4
𝑃𝑐
Pc
1
4
𝑃𝑐
24. Analysis of the AM wave cont..
Single sideband
There are inherent problems with AM more specifically double sideband full carrier (DSBFC).
The amount of power contained in the carrier frequency is enormous.
Modulating the carrier frequency with intelligence, or modulating voltage does not alter the amplitude or the
frequency of the carrier component itself.
It remains independent of the modulating voltage!
It is the sidebands that change in proportion to the modulating voltage and the depth of modulation.
So the intelligence resides entirely in the sidebands!
Hence the power transmitted in the AM carrier frequency is essentially wasted!
o Remember that the power in each sideband is one-fourth the power in the carrier for 100% modulation! Or we can say that
2/3 of the total power is dissipated by the carrier frequency Pc = PT /1.5!
For modulation indices less than 100% even more power is wasted by the carrier component!
LSB and USB are mirror images of each other and contain the same information with the same power levels.
There is no need to transmit both!
By suppressing one sideband, 50% of the power is saved! Thus overall for 100% modulation a saving of
83.33% is realised i.e. 2/3 for the carrier plus ½ of 1/3 for one of the sidebands
2/3 + 1/6 = 5/6 translating into 0.833333 and is 83.33%
Therefore suppressing the carrier frequency and one of the sidebands is referred to as single sideband
suppressed carrier (SSBSC) or simply SSB.
Suppressing the carrier component of the AM signal so that the transmitted wave only consists of the lower
and Upper sidebands is known as Double sideband Suppressed Carrier (DSBSC)
25. Analysis of the AM wave cont..
• Double sideband suppressed carrier (DSBSC) with
sine wave modulation in time domain.
• The carrier frequency of the AM wave is
suppressed by a balanced modulator
fc fUSB
fLSB
fc - fm fc + fm
Suppressed carrier
f
P
t
v
Suppressed carrier modulator
26. Analysis of the AM wave cont..
Advantages
• SSB requires only half the bandwidth of the AM
signal
• SSB is power efficient
Selection of transmitting of LSB or USB requires
high Q (Q = resonant frequency/bandwidth)
filters with steep attenuation characteristics
27. Analysis of the AM wave cont..
Filtering the SSB LSB or USB
• Suppose a SSB transceiver operating in the CB band has a
carrier frequency in the 7 MHz region. If the carrier
frequency is modulated with a low frequency tone of
100Hz the USB and LSB are separated by only 200 Hz.
• Selecting the USB for transmission would require the filter
to pass the USB frequency of 7.0001MHz (fc +100Hz) and
totally reject 6.9999 MHz (fc – 100Hz)
• Over a range of 200Hz selectivity requirements for the
filter would be a Q of 35,000! Remember Q = resonant
frequency/bandwidth) or (7MHz /200Hz) and for higher
carrier frequencies higher would be necessary!.
• This is highly impractical as multi stage LC filters with Qs
beyond 200 are difficult and impractical to design !
• For the same 200Hz range, if we lower the fc to say
100KHz our Q becomes 500, the USB can be selected and
up converted to the transmitting frequency (a process
known as dual-conversion).
• The fm is mixed with low carrier frequency fc1 and the LSB
and USB signals at the output of the balanced modulator
1 are at a frequency better suited for SSB filtering.
• The USB is filtered out and mixed again with a higher
carrier frequency fc2 .
• The output of balanced modulator 2 now contains
sideband frequencies that are separated by a frequency
range compatible with SSB filters
Audio
Amplifier
Balanced
Modulator
1
Carrier
Oscillator1fc1
=100KHz
Bandpass
Filter1
USB
Balanced
Modulator
2
Carrier
Oscillator2,
fc2=7MHz
Bandpass
Filter2
USB
Power
Amplifier
Antenna
Output1 Output2
100.2KHz
USB
7MHz+100.1KHz
f
fc2
7MHz
LSB
7MHz–100KHz
USB
200Hz
LSB
100.1KHz
fc1
100
KHz
99.9KHz
Output1
Output2
fm=100Hz
28. Quadrature Amplitude Modulation
(QAM)
• The technique of combining Amplitude and phase modulation is called quadrature
amplitude modulation (QAM).
• What is Phase modulation?
• See you in the next lecture!!!!!!!!!!!!!!
