Signals : Physical Layer
Shashank Srivastava
Motilal Nehru National Institute Of Technology, Allahabad
ANALOG DATA
Analog data refers to information that is continuous.
Example: human voice. When someone speaks, an analog wave is
created in the air.
This can be captured by a microphone and converted to an analog
signal or sampled and converted to a digital signal.
DIGITAL DATA
Digital data refers to information that has discrete values.
Example: A digital clock in which the value will change suddenly
from 8.05 to 8.06.
Data are stored in computer memory in the form of 0s and 1s . It
can be converted to a digital signal when it is transferred from one
position to another inside computer or modulated into an analog
signal for transmission across a medium.
Analog and Digital Clocks
Transformation of Information
to Signals
Comparison Of Analog and Digital Signals
An analog signal has infinitely many levels of intensity
over a period of time.
Or
An analog signal can have an infinite number of values in
a range.
As the wave moves from value A to B, it passes through an
infinite number of values along its path.
A digital signal, can have only limited number of
defined values often as simple as 1 and 0 .
Comparison Of Analog and Digital Signals
 Graphical representation:
The vertical axis represents the value of strength
The horizontal axis represent the time
The curve representing the analog signal passes through an
infinite number of points.
The vertical lines of the digital signals shows the sudden
jump that the signals makes from value to value.
Analog and Digital Signals
PERIODIC AND APERIODIC (NONPERIODIC) SIGNALS
A periodic signal completes a pattern with a measurable
time frame called a period, and repeats that pattern over
subsequent identical periods.
The completion of one full pattern is called a cycle.
Both digital and analog data can be periodic or non-periodic.
A non periodic signal changes without exhibiting a pattern or
cycle that repeats over time.
Periodic Signals
Aperiodic Signals
SIMPLE AND COMPOSITE ANALOG SIGNALS
A simple periodic analog signal, a sine wave , cannot be
decomposed into simpler signals.
A composite periodic analog signal is composed of multiple sine
waves.
Sine Waves This is most fundamental form of a periodic analog
signal.
Each cycle consists of a single arc above the time axis followed by
single arc below it.
SINE WAVE
A sine wave can be represented by three parameters: peak
amplitude , the frequency and the phase.
PEAK Amplitude:
The peak amplitude of a signal is the absolute value of its
highest intensity , proportional to the energy it carries.
For electric signals, peak amplitude is measured in volts
Period refers to the amount of time , in seconds, a signal needs
to complete one cycle .
Frequency refers to the number of periods in 1 s
Frequency and periods are the inverse of each other.
F=1/T and T =1/f
 Frequency is the rate of change with respect to time. Change in
sort span of time means high frequency. Change over a long
span of time means low frequency.
 If signal does not change at all, its frequency is zero.
 If a signal changes instantaneously, its frequency is infinite.
Example
Suppose that power has a frequency of 60 Hz. The period of this
sine wave can be determined as follows:
T=1/f=1/60 = 0.0166s = 0.0166 x 10 3 ms = 16.6 ms
that the period of the power is 0.0116 s or 16.6 ms .Our eyes
are not sensitive enough to distinguish these rapid changes.
Sine Wave
PHASES
Phase describes the position of the wave form relative
to time 0
Phase is measured in degree or radians.
A phase shift of 360 0 corresponds to a shift of a complete
period.
A phase shift of 180 0corresponds to a shift of one-half of a
period.
A phase shift of 90 0corresponds to a shift of one-quarter of a
period.
Point To Be Noted
1.A sine wave with phase of 0 0 starts with a zero amplitude. The
amplitude is increasing.
2. A sine wave with a phase of 90 0 starts at time 0 with a peak
amplitude .The amplitude is decreasing.
3. A sine wave with a phase of 1800 starts at time 0 with a zero
amplitude. The amplitude is decreasing.
In terms of Shift or offset
1.A sine wave with a phase of 00 is not shifted
2.A sine wave with a phase 900 is shifted to the left by 1/4 cycle
. Note that the signal does not really exist before time 0.
3. A sine wave with a phase of 1800 is shifted to the left by 1/2
cycle. Note that the signal does not really exist before time 0.
Phases
Amplitude Change
Frequency Change
Phase Change
EXAMPLE
A sine wave is offset 1/6 cycle with respect to time 0. What
is its phase in degrees and radians?
Sol:
One complete cycle is 360 0
Therefore 1/6 cycle
=1/6 X 360=60 0
= 60 X 2 X 3.14/360 Rad
=1.046 Rad
WAVELENGTH
Wavelength : It is the distance a simple signal can travel in one period.
Wavelength binds the period or frequency of a simple sine wave to
the propagation speed of the medium.
The wavelength depends on both the frequency and the medium.
Wavelength=propagation speed x period = propagation speed
Frequency
λ= c/f
λ is wavelength, c is propagation speed, and f is frequency.
It is normally measured in micrometers(microns).
EXAMPLE
The wavelength of red light(frequency = 4X1014) in air
is
=(3X108 / 4X1014 )
=0.75X10-6m
=.75um
Wavelength is measured in micrometers.
Time and Frequency Domain
The Time domain plot shows changes in signal amplitude
with respect to time.(Amplitude versus Time)
A frequency –domain is concerned with only the peak value
and the frequency.
The advantage of the frequency domain is that we can
immediately see the values of the frequency and peak amplitude.
A complete sine wave is represented by one spike in the
frequency domain.
It is more compact and useful when we are dealing with
more than one sine wave.
The phase of a signal cannot be shown in frequency domain.
Time and Frequency Domain
Examples
COMPOSITE SIGNALS
A single frequency sine wave is not useful in data communications,
we need to send a composite signal ,a signal made of many simple
sine waves.
Example: A single sine wave can be used to send an alarm when a
burglar opens a door in our house.
But if we send a single sine wave to convey data, we would be sending
alternating 0s and 1s,which does not have any communication value.
A composite is a combination of simple sine waves with different
frequencies ,amplitude , and phases.
It can be periodic or non-periodic .
COMPOSITE SIGNAL
A periodic composite signal can be decomposed into a series of
simple sine waves with discrete frequencies-frequencies that have
integer values(1,2,3…)
A non-periodic composite signal can be decomposed into a
combination of an infinite number of simple sine waves with continuous
frequencies , frequencies that have real values.
In a time domain representation of the composite signal, there are an
infinite number of simple sine frequencies.
A medium may pass some frequencies and may block or weaken others.
So, we may not receive the same signal at the other end.
Bandwidth
The range of frequencies contained in a composite signal is its
bandwidth.
The bandwidth of a composite signals is the difference between the
highest and the lowest frequencies contained in that signal.
The range of frequencies a medium can pass is called its bandwidth.
Normally a medium cannot pass all frequencies, so bandwidth generally
refers to the range of frequencies that a medium can pass without losing
one-half of the power contained in that signal.
It is the difference between the highest and lowest frequencies that a
medium can pass.
Example: If a composite signal contains frequencies between 1000 and
5000, its bandwidth is (5000-1000) or 4000.
Bandwidth
EXAMPLE
If a periodic signal is decomposed into five sine waves
with frequencies of 100, 400, 600,900 and 1200 Hz, what is
its bandwidth? Draw the spectrum , assuming all components
have a maximum amplitude of 20V.
Solution:
Let fh be the highest frequency and fl the lowest frequency ,
and B the bandwidth .Then
B = fh-fl
=1200-100
=1100 Hz
DIGITAL SIGNALS
Data can be represented by a digital signal. Example: a 1 can be
encoded as a positive voltage and a 0 as zero voltage.
A digital signal can have more than two levels.
Example:
A digital signal has eight levels. How many bits are needed per
level ? We calculate the number of bits from the formula
Number of bits per level=log2l=log28=3
Each signal level is represented by 3 bits.
Digital Signal
Amplitude, Period, and Phase
for a Digital Signal
Bit interval and Bit Rate
Most digital signals are aperiodic, thus period or frequency is
not appropriate.
The bit interval or bit duration(instead of period) is the time
required to send one bit.
The bit rate (instead of frequency) is the number of bit
intervals per second means the number of bits sent in 1s
,expressed in bits per second (bps).
Bit interval=1/bit rate
BIT LENGTH
The bit length is the distance one bit occupies on
the transmission medium.
Bit Length=propagation speed X bit duration
The bit rate and bandwidth are proportional to each
other.
B >= n/2
or
n<=2B
n is the bit rate, B is the bandwidth.
The analog bandwidth of a medium is expressed in hertz,
the digital bandwidth in bits per second.
Digital bandwidth means the maximum bit rate that a
medium can pass.
Low-pass versus Band-pass
A low-pass channel has a bandwidth with frequencies between 0
and f.
A band-pass channel has a bandwidth with frequencies between
f1 and f2.
Digital transmission needs a low-pass channel.
Analog signal uses a band-pass channel.
In addition, bandwidth of analog signal can be shifted as long as
width of bandwidth remains same.
A band-pass channel is more available than a low-pass channel.
Data rate limits
Data rate depends on three factors:
1. The bandwidth available.
2. The levels of signals we can use.
3. The quality of channel (the level of noise).
Nyquist Bit Rate (for noiseless channel): Defines the maximum
theoretical bit rate.
BitRate = 2 * Bandwidth *
- Bandwidth is the bandwidth of the channel,
- L is the no. of signal levels used to represent data(example: if four
signal levels are used, means for each level we send two bits),
- BitRate is the bit rate in bits per second.
Shannon Capacity (for noisy channel):
Theoretical highest data rate for a noisy channel
Capacity = Bandwidth *
Bandwidth is the bandwidth of the channel,
SNR is the signal to noise ratio = power of signal / power of noise,
Capacity is the capacity of the channel in bits per second.
This formula defines the characteristics of the channel, not the method of
transmission.
For increasing data rate, either increase bandwidth or improve signal-tonoise ratio.
Example: if SNR is zero, capacity= B*0=0
means we cannot receive any data through this channel.
Using both limits: We need to use both methods in practice.
Example: we have a channel with 1 MHz bandwidth. The SNR of this
channel is 63, what is the approximate bit rate and signal level?
Solution: First we use Shannon formula to find upper limit.
C= B *
= 10^6 *
= 10^6 *
= 6 Mbps
6 Mbps is the upper limit. For better performance, we choose something
lower, for example 4 Mbps.
Now use Nyquist formula to find the number of signal levels:
4 Mbps= 2 * 1 MHz*
gives L = 4.
TRANSMISSION IMPAIRMENTS
1. Attenuation –Attenuation means loss of energy. When a signal travels
through a medium, it loses some of its energy in overcoming the
resistance of the medium. Some of the electrical energy is converted
into heat.
To compensate this loss, amplifiers are used to amplify the signal.
Decibel measures the relative strength of two signals or a signal at
two different points. Decibel is negative if a signal is attenuated and
positive if the signal is amplified.
dB = 10 log10(P1/P2)
where p1,p2 are powers of signal at points 1 and 2 respectively.
Decibel numbers can be added(or subtracted) when we are talking
about several points.
2. Distortion – Distortion means that signal changes its shape or
form.
Distortion occurs in a composite signal made of different
frequencies.
Each signal has its propagation speed and therefore its own delay
in arriving at the final destination.
Difference in delay may create a phase difference.
So signal components at the receiver have phases different from
what they had at the sender.
3. Noise –Noise like thermal noise, induced noise, cross talk,
impulse noise may corrupt the signal.
Thermal noise is the random motion of electrons in a wire which
creates an extra signal not originally sent by the transmitter.
Induced noise comes from sources and appliances. These devices
act as sending antenna and the transmission medium act as
receiving antenna.
Cross talk is the effect of one wire on the other. One wire acts as
sending antenna and other acts as receiving antenna.
Impulse noise is a spike (a signal with high energy in a very short
period of time) that comes from power lines, lightning and so
on.
PERFORMANCE-
1. Bandwidth
2. Network Throughput – Throughput is the measurement of
how fast data can pass through an entity (such as a point on
the network). Network throughput is measured in bits per
second(bps).
It is an actual measure of the how much data we can send.
3. Latency (Delay) – It is the time taken by an entire message to
completely arrive at the destination from the time the first bit is
sent out from the source machine. It is made up of four
components
-Propagation time
-Transmission time
-Queuing time
-Processing time
Latency time = Propagation time + Transmission time +
Queuing time + Processing time
Propagation time – Time taken by a bit to travel from source to
the destination. It is given by
Distance/Propagation speed
Propagation speed is the distance a bit can travel through a
medium in one second. Speed depends upon the medium and on
the frequency of the signal.
Transmission time – It is the time from the first bit until the last
bit of a message has left the transmitting node.
The packet transmission time in seconds can be obtained from
the packet size in bit and the bit rate in bit/s as:
Packet transmission time = Packet size / Bit rate
or
Transmission time = Message size/ bandwidth
What are the propagation time and the transmission time for a 2.5-kbyte message (an e-mail) if the
bandwidth of the network is 1 Gbps? Assume that the distance between the sender and the receiver is
12,000 km and that light travels at 2.4 × 108 m/s.
Solution
Queuing time – Queuing time is the time needed for each
intermediate or end device to hold the message before it can be
processed.
Queuing time is not fixed, it changes with the load of the
network. When there is heavy traffic in the network, the queuing
time increases.
Processing time – Each intermediate device processes the
message one by one. The time taken by each of the intermediate
device is the processing time.
Bandwidth Delay Product – This product defines the
number of bits that can fill the link.
Manchester Encoding –
A 0 is expressed by a low-to-high transition, a 1 by high-to-low
transition.
Transition in the middle of bit is used for both synchronization
and bit representation.
Each bit is transmitted in a fixed time.
The existence of guaranteed transitions allows the signal to be
self-clocking, and also allows the receiver to align correctly.
The receiver can identify if it is misaligned as there will no
longer always be a transition during each bit period.
Differential Manchester Encoding
A '1' bit is indicated by making the first half of the signal equal
to the last half of the previous bit's signal i.e. no transition at the
start of the bit-interval.
A '0' bit is indicated by making the first half of the signal
opposite to the last half of the previous bit's signal i.e. a zero bit
is indicated by a transition at the beginning of the bit-interval.
In the middle of the bit-interval there is always a transition,
whether from high to low, or low to high.
Uses two signal changes to represent binary 0, and one signal
change means binary 1.
Differential Manchester Encoding
Absence or presence of an additional transition at the
beginning of the interval is used to identify the bit.
Inversion at the middle of bit interval is used for
synchronization.
Transmission Modes
 A transmission mode is the manner in which data is sent over the
underlying medium.
 Transmission modes can be divided into two fundamental categories:
 Serial — one bit is sent at a time.
 Parallel — multiple bits are sent at the same time.
Categories of Transmission Modes
60
Parallel Transmission
 Parallel transmission allows transfers of multiple data bits at the same time over
separate media
 In general, parallel transmission is used with a wired medium that uses multiple,
independent wires
 Furthermore, the signals on all wires are synchronized so that a bit travels across each
of the wires at precisely the same time
 The term parallel is used to characterize the wiring
Parallel Transmission
 A parallel mode of transmission has two main advantages:
(1) High speed: it can send N bits at the same time
 a parallel interface can operate N times faster than an equivalent serial interface
(2) Match to underlying hardware: Internally, computer and
communication hardware uses parallel circuitry
 a parallel interface matches the internal hardware well
Parallel Transmission
Disadvantages:
 Cost increases as ‘n’ communication lines are required.
 Limited to short distances because of cost.
Serial Transmission
Serial Transmission
• Reduces cost roughly by a factor of ‘n’.
• Since communication within devices is parallel, so parallelto-serial interface (at sender) and serial-to-parallel interface
(at receiver) is required.
Transmission Order: Bits and Bytes
 In serial mode, when sending bits, which bit should be sent across the
medium first?
 Should a sender transmit
 the Most Significant Bit (MSB)
 or the Least Significant Bit (LSB) first?
 We use the term little-endian to describe a system that sends the LSB first
 We use the term big-endian to describe a system that sends the MSB first
 Either form can be used, but the sender and receiver must agree
Transmission
Order:
Bits
and
Bytes
 The order in which bits are transmitted does not settle the entire question of transmission
order
 Data in a computer is divided into bytes, and each byte is further divided into bits (typically 8
bits per byte)
 Thus, it is possible to choose a byte order and a bit order independently
 For example, Ethernet technology specifies that data is sent byte big-endian and bit little-endian
Timing of Serial Transmission
 Serial transmission mechanisms can be divided into three broad categories
depending on the timing of transmissions:
 Asynchronous transmission can occur at any time
 with an arbitrary delay between the transmission of two data items
 Synchronous transmission occurs continuously
 with no gap between the transmission of two data items
 Isochronous transmission occurs at regular intervals
 with a fixed gap between the transmission of two data items
Serial Communications
Asynchronous transmission
•
•
•
Data word (octet) is encapsulated between:
– start bit :(usually a 0)
– stop bits(s) :(usually 1s)
Called asynchronous because at the byte level, sender and receiver do not have to be
synchronized. Synchronization is still required but only for duration of single byte.
Receiver resynchronizes again at a start of each new word (or character) received.
Serial Communications
Asynchronous transmission
• The addition of stop and start bits and insertion of gaps into
the bit stream make asynchronous transmission slower.
• Example: connection of a keyboard to a computer. User
types characters one by one and leaves gaps between each
character.
Synchronous transmission
 Transmitter and receiver clocks synchronized.
 The bit stream is combined into frames which may contain multiple
bytes.
 No start, stop bits. Each byte is transferred without a gap between it
and the next one.
 The receiver counts the bits as they arrive and groups them in 8-bit
units.
 Advantage is speed as additional bits are removed. Used for high speed
applications such as transmission of data from one computer to
another.
Isochronous Transmission
 Isochronous transmission
 is designed to provide steady bit flow for multimedia applications
 Delivering such data at a steady rate is essential
 because variations in delay known as jitter can disrupt reception.
 Isochronous network is designed to accept and send data at a fixed rate, R.
 For example, an isochronous mechanism designed to transfer voice operates at a rate of
64,000 bits per second
 A sender must generate digitized audio continuously.
 A receiver must be able to accept and play the stream.
DIGITAL-TO-ANALOG CONVERSION
Digital-to-analog conversion is the process of changing
one of the characteristics of an analog signal based on
the information in digital data.
Example: telephone wires carry analog signals, so the data must be
converted to analog form before transmission.
There are three characteristics of a sine wave –
1. Amplitude
2. Frequency
3. Phase
By changing one characteristics of a simple signal, we can use
it to represent digital data.
Types of digital-to-analog conversion
Bit rate is the number of bits per second.
Baud rate is the number of signal units per second. A signal unit is composed of one
or more bits.
In the analog transmission of digital data, the baud rate is less than or equal to the bit
rate.
N=Sxr
Where N is the bit rate, S is the baud rate and r is the no of bits carried in one signal
and given by r = log2L
Fewer signal units required means transmission is more efficient, and bandwidth
required is less.
Example: a baud is analogous to a car, and bit is analogous to a passenger.
Example
An analog signal carries 8 bits per signal element. If
2000 signal elements are sent per second, find the bit
rate.
Solution
In this case, r = 8, S = 2000, and N is unknown. We can
find the value of N from
N = S x r = 2000 x 8 = 16000 bps
Example
An analog signal has a bit rate of 8000 bps and a baud
rate of 1000 baud. How many data elements are carried
by each signal element? How many signal elements do
we need?
Solution
In this example, S = 1000, N = 8000, and r and L are unknown.
We find first the value of r and then the value of L.
Amplitude Shift Keying (ASK)
 The amplitude (or height) of the sine wave varies to transmit
the ones and zeros.
 Both frequency and phase remain constant.
 Major disadvantage is that noise usually affects the amplitude,
and therefore ASK is most affected by noise.
ASK Illustration
1
5.80
0
0
1
Amplitude shift keying
OOK (on/off keying)—
•One of the bit values are represented by no voltage.
•Advantage is the reduction in the amount of energy required to
transmit information.
Bandwidth for ASK
Bandwidth is proportional to the signal rate (baud rate).
B = (1+d) x S
where B is bandwidth, S is baud rate, d is a factor related to modulation
process(with a min value of zero).
The value of d is between 0 and 1.
•Therefore minimum bandwidth required is equal to baud rate.
Relationship between baud rate and bandwidth-
Amplitude Shift Keying
Example
We have an available bandwidth of 100 kHz which
spans from 200 to 300 kHz. What are the carrier
frequency and the bit rate if we modulated our data by
using ASK with d = 1?
Solution
The middle of the bandwidth is located at 250 kHz. This
means that our carrier frequency can be at fc = 250 kHz.
We can use the formula for bandwidth to find the bit rate
(with d = 1 and r = 1).
Frequency Shift Keying
In FSK, the frequency of the carrier signal is varied to
represent data.
The frequency of the modulated signal is constant for
one signal element but changes for the next signal
element if the data element changes.
Both amplitude and phase remain constant for all
signal elements.
Binary frequency shift keying
Although FSK shifts between 2 carrier frequencies, it is easier to
analyze as two coexisting frequencies.
FSK spectrum is a combination of 2 ASK spectra centered on fc1 and
fc0.
Example
Find the minimum bandwidth for an FSK signal
transmitting at 2000bps. Transmission is in half-duplex
mode, and the carriers are separated by 3000 Hz.
Solution
For FSK, if fc1 and fc2 are the carrier frequencies, then
BW = baud rate + (fc1 - fc0)
Assume baud rate is same as bit rate, therefore
BW = bit rate + (fc1 – fc0)
BW = 2000 + 3000 = 5000 Hz.
Phase shift keying
In Phase shift keying, the phase of the carrier is varied to represent
two or more different signal elements
Both peak amplitude and frequency remain constant as the phase
changes.
The phase of the signal during each bit duration is constant.
Binary phase shift keying – Also called 2-PSK, because two different phases are used.
Concept of a constellation diagram (phase-state diagram)
•PSK is less susceptible to noise, therefore smaller variations in the signal can be
detected reliably by the receiver.
•Instead of using only two variations of a signal, each representing one bit, we can
use four variations and let each phase shift represent 2 bits.
The 4-PSK method (Q-PSK method)
•The pair of bits represented by each phase is called a dibit.
•We can transmit data twice as efficiently using 4-PSK as we can using 2-PSK.
Constellation diagram (phase-state diagram)
Constellation diagrams
Relationship between baud rate and bandwidth-
Find the bandwidth for a 4-PSK signal transmitting at 2000 bps.
Solution-For 4-PSK, the baud rate is one half of the bit rate.
The baud rate is therefore 1000.
A PSK signal requires bandwidth equal to its baud rate.
Therefore, the bandwidth is 1000 Hz.
QAM – Quadrature Amplitude Modulation
 Combination of phase shifting and amplitude shifting
 Because amplitude changes are susceptible to noise and require greater shift
differences than do phase changes, the number of phase shifts used by a QAM
system is always larger than the number of amplitude shifts.
4 – QAM and 8 -QAM constellations
8 – QAM in Time Domain
Bandwidth for QAM
 The minimum bandwidth required for QAM transmission is
the same as that required for ASK and PSK transmission.
Bit and baud rate comparison
Modulation
Units
Bits/Baud
Baud rate
Bit Rate
Bit
1
N
N
4-PSK, 4-QAM
Dibit
2
N
2N
8-PSK, 8-QAM
Tribit
3
N
3N
16-QAM
Quadbit
4
N
4N
32-QAM
Pentabit
5
N
5N
64-QAM
Hexabit
6
N
6N
128-QAM
Septabit
7
N
7N
256-QAM
Octabit
8
N
8N
ASK, FSK, 2-PSK
A constellation diagram consists of eight equally spaced
points on a circle. If the bit rate is 4800 bps, what is the
baud rate?
The constellation indicates 8-PSK with the points 45
degrees apart. Since 23 = 8, 3 bits are transmitted with
each signal unit. Therefore, the baud rate is
4800 / 3 = 1600 baud
Compute the bit rate for a 1000-baud 16-QAM signal.
A 16-QAM signal has 4 bits per signal unit since
log216 = 4.
Thus,
(1000)(4) = 4000 bps
Compute the baud rate for a 72,000-bps 64-QAM signal.
A 64-QAM signal has 6 bits per signal unit since
log2 64 = 6.
Thus,
72000 / 6 = 12,000 baud
ANALOG-TO-ANALOG CONVERSION
Analog-to-analog conversion (or modulation of an
analog signal) is the representation of analog
information by an analog signal.
Example: radio stations
The analog signals produced by each radio station is a
low-pass channel. To be able to listen to different
stations, the low pass signals need to be shifted, each to
a different range.
5.107
Types of analog-to-analog modulation
Amplitude modulation
Amplitude modulation (AM)
•In AM, the carrier signal is modulated so that its amplitude varies with
the changing amplitudes of the modulating signal.
•The frequency and phase of the carrier remains the same.
Amplitude modulation
AM Bandwidth -
AM Bandwidth The bandwidth of an AM is equal to twice the bandwidth of the modulating signal. And covers a
region centered around the carrier frequency.
The bandwidth of an audio signal is 5KHz. Therefore an AM radio station needs a minimum
bandwidth of 10 KHz.
AM stations are allowed carrier frequencies anywhere between 530 and 1700 KHz. However,
each station’s carrier frequency must be separated from those on either side of it by at least 10
KHz(one AM bandwidth) to avoid interference.
Frequency modulation
In FM, the frequency of the carrier signal is modulated to follow the changing amplitude of the
modulating signal.
The peak amplitude and the phase of the carrier signal remains constant.
AS the amplitude of the information signal changes, the frequency of the carrier changes
correspondingly.
Frequency modulation
FM Bandwidth
The bandwidth of an FM signal is equal to 10 times the bandwidth of the
modulating signal and covers a range centered around the carrier
frequency.
If BWm is the bandwidth of modulating signal then total bandwidth BWt
is given by
BWt = 10 x BWm
FM Bandwidth
FM Bandwidth
The bandwidth of a stereo audio signal is usually 15 KHz. Therefore an
FM station needs at least a bandwidth of 150 KHz. FCC(Federal
Communication Commission) requires the minimum bandwidth to be at
least 200 KHz.
FM stations are allowed carrier frequencies anywhere between 88 and
108 MHz. Stations must be separated by at least 200 KHz to keep their
bandwidths from overlapping.
FM band allocation
The total bandwidth required for FM can be determined from the bandwidth
of the audio signal:
BFM = 2(1 + β)B,
β has a common value of 4
Phase Modulation(PM)
In PM, the phase of the carrier signal is modulated to follow the changing
amplitude of the modulating signal.
The peak amplitude and frequency of the carrier signal remain constant.
As the amplitude of the information signal changes, the phase of the carrier
changes correspondingly.
PM Bandwidth
The total bandwidth required for PM can be determined from the
bandwidth of the audio signal:
BPM = 2(1 + β)B
β has a common value of 2