pulse modulation ekt343 –principle of communication engineering

PULSE MODULATION

EKT343 –Principle of Communication Engineering

Chapter Outline

PART 1:•Basic sampling technique•Generation and recovery

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

•Advantages & Disadvantages

EKT343 –Principle of Communication Engineering 2

SamplingTo 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.


Sampling

• Sampling a signal: Analog → Digital conversion byreading the value at discrete points

• A process of taking samples of information signal at a rate of Nyquist’s sampling frequency.


• 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)


infinite bandwidth cannot be sampled.

the sampling rate must be at least 2 times the highest frequency, not the bandwidth.


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


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 & OversamplingUndersampling 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.


• 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


Aliasing effect


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


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


Natural Sampling


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.


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


Flat-top Sampling


• 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.


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)


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


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.

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


Fs >= 2 WTs =< 1/2WT « Ts =< 1/2W

2

1max f

Transmission BW of PAM Signal• Bandwidth required for transmitter of PAM

signal will be equal to maximum frequency


2

1max

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


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


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


• PWM signal output is generated by comparing summation result with reference level


Cont’d...


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


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.


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.


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


tr« Ts

rT t

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


Fs >= 2 WTs =< 1/2WT « 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


sxT

sxkHzf

T

s

ss

5

4

1025.11.0

1025.18

11

kHzB

WB

W

T

T

402

12

12

1

Example 5For 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


MHzB

tB

sxt

T

rT

r

4

2

1

1025.1100

1 7

Pulse Modulation


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.


pulse modulation ekt343 –principle of communication engineering

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