PULSE MODULATIONportal.unimap.edu.my/portal/page/portal30/Lecture... · –The amplitude is held...

PULSE MODULATION

EKT358 – Communication System

Dr. Muzammil Bin Jusoh

Chapter Outline

PART 1: • Basic sampling technique • Generation and recovery

– Pulse Amplitude Modulation (PAM) – Pulse Duration Modulation (PDM) – Pulse Position Modulation (PPM)

• Advantages & Disadvantages

2



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.

4



• Nyquist’s Sampling Theorem :

5

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)



4

infinite bandwidth cannot be sampled.

the sampling rate must be at least 2 times the highest

frequency, not the bandwidth.



7

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.

Example 1



8

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.

Example 2



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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• 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 high-frequency variation

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Aliasing effect



Aliasing effect

11



12

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.

Example 3

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Recovery of a sampled sine wave for different sampling rates



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

18



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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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The amplitude of

pulses is varied in

accordance with the

information signal.

Width & position

constant.

2 types –

double polarity

single polarity

Pulse Amplitude Modulation (PAM)

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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

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Flat-top sampling – generation of PAM signals. EKT358 – Communication System


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

• If ON/OFF time of the pulse is same,

frequency of the PAM pulse is

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Fs >= 2 W

Ts =< 1/2W

T « Ts =< 1/2W

2

1max f



Transmission BW of PAM Signal

• Bandwidth required for transmitter of PAM

signal will be equal to maximum frequency

25

2

1

max

fBT



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

28 EKT358 – Communication System


• 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

• From formula above, we know that transmission BW of PPM and PDM is higher than PAM

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tr« Ts

r

Tt

B2

1



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

• If ON/OFF time of the pulse is same,

frequency of the PAM pulse is

35

Fs >= 2 W

Ts =< 1/2W

T « Ts =< 1/2W

2

1max f



Example 4

• For PAM transmission of voice signal with W = 3kHz. Calculate BT if fs = 8 kHz and τ = 0.1 Ts

• SOLUTION

36

sxT

sxkHzf

T

s

s

s

5

4

1025.11.0

1025.18

11

kHzB

WB

W

T

T

402

1

2

1

2

1



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

37

MHzB

tB

sxt

T

r

T

r

4

2

1

1025.1100

1 7



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



• PAM, PWM, PPM



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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