Transcript Chapter 4_part 1

PULSE MODULATION
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Chapter Outline
PART 1:
• Basic sampling technique
• Generation and recovery
– Pulse Amplitude Modulation (PAM)
– Pulse Duration Modulation (PDM)
– Pulse Position Modulation (PPM)
• Advantages & Disadvantages
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Sampling
 To convert a signal from continuous time to
discrete time, a process called sampling is
used. The value of the signal is measured at
certain intervals in time. Each measurement is
referred to as a sample.
 When the continuous analog signal is sampled
at a frequency F, the resulting discrete signal has
more frequency components than did the analog
signal. To be precise, the frequency components
of the analog signal are repeated at the sample
rate.
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Sampling
• Sampling a signal: Analog → Digital conversion by
reading the value at discrete points
• A process of taking samples of information signal at a rate of
Nyquist’s sampling frequency.
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• Nyquist’s Sampling Theorem :
The original information signal can be reconstructed at the receiver
with minimal distortion if the sampling rate in the pulse modulation
system equal to or greater than twice the maximum information
signal frequency.
fs >= 2fm (max)
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 infinite bandwidth cannot be sampled.
 the sampling rate must be at least 2 times the highest
frequency, not the bandwidth.
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Example 1
A complex low-pass signal has a bandwidth of
200 kHz. What is the minimum sampling rate for
this signal?
Solution:
The bandwidth of a low-pass signal is between 0 and f,
where f is the maximum frequency in the signal.
Therefore, we can sample this signal at 2 times the
highest frequency (200 kHz). The sampling rate is
therefore 400,000 samples per second.
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Example 2
A complex bandpass signal has a bandwidth of
200 kHz. What is the minimum sampling rate for
this signal?
Solution :
We cannot find the minimum sampling rate in this case
because we do not know where the bandwidth starts or
ends. We do not know the maximum frequency in the
signal.
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Undersampling & Oversampling
Undersampling is essentially sampling too
slowly, or sampling at a rate below the
Nyquist frequency for a particular signal of
interest. Undersampling leads to aliasing and
the original signal cannot be properly
reconstructed
Oversampling is sampling at a rate beyond
twice the highest frequency component of
interest in the signal and is usually desired.
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Aliasing effect
• If the required condition of the sampling theorem that fs
>= 2fmmax is not met, then errors will occur in the
reconstruction.
• When such errors arise due to undersampling, aliasing is
said to occur
• Undersampling: Sampling rate is too low to capture highfrequency variation
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Aliasing effect
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Example 3
For an intuitive example of the Nyquist theorem, let us
sample a simple sine wave at three sampling rates:
a) fs = 2f (Nyquist rate)
b) fs = 4f (2 times the Nyquist rate),
c) fs = f (one-half the Nyquist rate).
Figure shows the sampling and the subsequent recovery of
the signal.
SOLUTION:
It can be seen that sampling at the Nyquist rate can create
a good approximation of the original sine wave (part a).
Oversampling in part b can also create the same
approximation, but it is redundant and unnecessary.
Sampling below the Nyquist rate (part c) does not produce
a signal that looks like the original sine wave.
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Recovery of a sampled sine wave for different sampling rates
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Natural Sampling
• Tops of the sample pulses retain their natural
shape during the sample interval.
• Frequency spectrum of the sampled output is
different from an ideal sample.
• Amplitude of frequency components
produced from narrow, finite-width sample
pulses decreases for the higher harmonics
– Requiring the use of frequency equalizers
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Natural Sampling
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Flat-top Sampling
• Common used in PCM systems.
• Accomplish in a sample-and-hold circuit
– To periodically sample the continually changing analog
input voltage & convert to a series of constant-amplitude
PAM voltage levels.
• The input voltage is sampled with a narrow pulse
and then held relatively constant until the next
sample is taken.
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Cont’d…
• Sampling process alters the frequency
spectrum & introduces aperture error.
• The amplitude of the sampled signal changes
during the sample pulse time.
• Advantages:
– Introduces less aperture distortion
– Can operate with a slower ADC
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Flat-top Sampling
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•
•
•
•

Sampling analog information signal
Converting samples into discrete pulses
used to represent an analog signal with digital data
among the first of the pulse techniques to be utilized
Carrier signal is pulse waveform and the modulated signal
is where one of the carrier signal’s characteristic (either
amplitude, width or position) is changed according to
information signal.
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Pulse Amplitude Modulation (PAM)
 The amplitude of
pulses is varied in
accordance with the
information signal.
 Width
constant.
&
position
 2 types –
double polarity
single polarity
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Natural Sampling (PAM)
• A PAM signal is generated by using a pulse train, called the
sampling signal (or clock signal) to operate an electronic
switch or "chopper". This produces samples of the analog
message signal, as shown in Figure
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Flat Top Sampling (PAM)
• a sample-and-hold circuit is used in conjunction with the
chopper to hold the amplitude of each pulse at a constant
level during the sampling time
Flat-top sampling – generation of PAM signals.
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Cont’d
• Pulse duration (τ) supposed to be very small
compare to the period, Ts between 2 samples
• Lets max frequency of the signal, W
Fs >= 2 W
Ts =< 1/2W
T « Ts =< 1/2W
• If ON/OFF time of the pulse
is
same,
1
f max pulse

frequency of the PAM
2 is
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Transmission BW of PAM Signal
• Bandwidth required for transmitter of PAM
signal will be equal to maximum frequency
BT  f max
1

2
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Advantages & Disadvantages PAM
• Advantage:
– it allows multiplexing, i.e., the sharing of the same
transmission media by different sources (or users). This
is because a PAM signal only occurs in slots of time,
leaving the idle time for the transmission of other PAM
signals.
• Disadvantage:
– require a larger transmission bandwidth (very large
compare to its maximum frequency)
– Interference of noise is maximum
– Needed for varies transmission power
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Pulse Density Modulation (PDM)
• Sometimes called Pulse Duration Modulation/ Pulse Width
Duration (PWM).
• The width of pulses is varied in accordance to information
signal
• Amplitude & position constant.
• PDM is used in a great number of applications
Communications
• The width of the transmitted pulse corresponds to the
encoded data value
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PDM
• Immune to noise
• Power Delivery
– Reduce the total amount of power delivered to a load
• Applications: DC Motors, Light Dimmers, Anti-Lock Breaking System
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• PWM signal output is generated by comparing summation
result with reference level
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Cont’d...
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Advantages & Disadvantages PDM
• Advantage:
– Noise performance is better compare to PAM.
• Disadvantages:
– require a larger power transmission compare to
PPM
– Require very large bandwidth compare to PAM
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Pulse Position Modulation (PPM)
• Modulation in which the temporal positions of the pulses are
varied in accordance with some characteristic of the
information signal.
• Amplitude & width constant.
• The higher the amplitude of the sample, the farther to the
right the pulse is position within the prescribed time slot.
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Advantages & Disadvantages PPM
• Advantage:
– The amplitude is held constant thus less noise
interference.
– Signal and noise separation is very easy
– Due to constant pulse widths and amplitudes,
transmission power for each pulse is same.
– Require less power compare to PAM and PDM because
of short duration pulses.
• Disadvantages:
– Require very large bandwidth compare to PAM.
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Transmission BW of PDM/PPM Signal
• PPM and PDM need a sharp rise time and fall
time for pulses in order to preserve the message
information.
• Lets rise time, tr
tr« Ts
BT 
1
2t r
• From formula above, we know that transmission
BW of PPM and PDM is higher than PAM
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Transmission BW of PAM Signal
• Pulse duration (τ) supposed to be very small
compare to the period, Ts between 2 samples
• Lets max frequency of the signal, W
Fs >= 2 W
Ts =< 1/2W
T « Ts =< 1/2W
f max 
1
2
• If ON/OFF time of the pulse is same,
frequency of the PAM pulse is
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Example 4
• For PAM transmission of voice signal with W =
3kHz. Calculate BT if fs = 8 kHz and τ = 0.1 Ts
•
SOLUTION
1
1
Ts 

 1.25 x10  4 s
f s 8kHz
  0.1Ts  1.25 x10 5 s
1
 
2W
1
BT 
 W
2
1
BT 
 40 kHz
2
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Example 5
For the same information as in example 1, find
minimum transmission BW needed for PPM
and PDM. Given tr= 1% of the width of the
pulse.
SOLUTION
1
  1.25 x10  7 s
100
1
BT 
2t r
tr 
BT  4 MHz
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Pulse Modulation
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PAM
PDM
PPM
Relation with
modulating signal
Amplitude of the
pulse is
proportional to
amplitude of
modulating signal
Width of the pulse
is proportional to
amplitude of
modulating signal
Relative position of
the pulse is
proportional to
amplitude of
modulating signal
BW of the
transmission
channel
depends on width
of the pulse
Depends of rise
time of the pulse
Depends on rising
time of the pulse
Instantaneous
power
varies
varies
Remains constant
Noise interference
High
Minimum
Minimum
Complexity of the
system
Complex
Simple
simple
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• PAM, PWM, PPM
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Advantages & Drawbacks of Pulse Modulation
• Noise immunity.
• Relatively low cost digital
circuitry.
• Able to be time division
multiplexed with other pulse
modulated signal.
• Storage of digital streams.
• Error detection & correction
• Requires greater BW to transmit
& receive as compared to its
analog counterpart.
• Special encoding & decoding
methods must be used to
increased transmission rates &
more difficult to be recovered.
• Requires precise
synchronization of clocks
between Tx & Rx.
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