Lecture 2 data communication physical layer

Data Communication
Lecture 2
Physical Layer: Signals & Digital Transmission

Physical Layer Topics to Cover
Signals
Digital Transmission
Analog Transmission
Multiplexing
Transmission Media

Analog & Digital Data
 Data can be analog or digital. The term
analog data refers to information that is
continuous; digital data refers to information
that has discrete states. Analog data take on
continuous values. Digital data take on
discrete values.

To be transmitted, data must be
transformed to electromagnetic signals.
Note

Signals can be analog or digital.
Analog signals can have an infinite
number of values in a range; digital
signals can have only a limited
number of values.
Note

Cont’
 Three parameters to describe a sine wave
1. Peak amplitude
2. Frequency and time period
3. Phase

The bandwidth of a composite signal is
the difference between the
highest and the lowest frequencies
contained in that signal.
Note

Digital Signals
 In addition to being represented by an analog
signal, information can also be represented
by a digital signal. For 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. In this case, we can send
more than 1 bit for each level.

Bit Rate & Bit Interval (contd.)

Bit Interval and Bit Rate
Example
Example
A digital signal has a bit rate of 2000 bps. What is the
duration of each bit (bit interval)
Solution
Solution
The bit interval is the inverse of the bit rate.
Bit interval = 1/ 2000 s = 0.000500 s
= 0.000500 x 106
s = 500 s

The bit rate and the bandwidth are
The bit rate and the bandwidth are
proportional to each other.
proportional to each other.
Note

Data Rate Limits
 A very important consideration in data
communications is how fast we can send data, in bits
per second, over a channel. Data rate depends on
three factors:
1. The bandwidth available
2. The level of the signals we use
3. The quality of the channel (the level of noise)

Noiseless Channel: Nyquist Bit Rate
 Defines theoretical maximum bit rate for
Noiseless Channel:
 Bit Rate=2 X Bandwidth X log2L
 Such that L is he signal level

Example
Example
Consider a noiseless channel with a bandwidth of 3000
Hz transmitting a signal with two signal levels. The
maximum bit rate can be calculated as
Bit
Bit Rate = 2
Rate = 2 
 3000
3000 
 log
log2
2 2 = 6000 bps
2 = 6000 bps

Increasing the levels of a signal may
reduce the reliability of the system.
Note

Noisy Channel: Shannon Capacity
 Defines theoretical maximum bit rate for
Noisy Channel:
 it represents the maximum rate at which
information can be reliably transmitted over a
communication channel
 Capacity=Bandwidth X log2(1+SNR)
 SNR stands for "Signal-to-Noise Ratio." It is a measure used to
quantify the ratio of the strength of a signal.
 A higher SNR indicates a stronger signal relative to the noise,
which generally results in better signal quality and more reliable
communication.

Example
Example
Consider an extremely noisy channel in which the value
of the signal-to-noise ratio is almost zero. In other words,
the noise is so strong that the signal is faint. For this
channel the capacity is calculated as
C = B log
C = B log2
2 (1 + SNR) = B log
(1 + SNR) = B log2
2 (1 + 0)
(1 + 0)
= B log
= B log2
2 (1) = B
(1) = B 
 0 = 0
0 = 0

Example
Example
We can calculate the theoretical highest bit rate of a
regular telephone line. A telephone line normally has a
bandwidth of 3000. The signal-to-noise ratio is usually
3162. For this channel the capacity is calculated as
C = B log
C = B log2
2 (1 + SNR) = 3000 log
(1 + SNR) = 3000 log2
2 (1 + 3162)
(1 + 3162)
= 3000 log
= 3000 log2
2 (3163)
(3163)
C = 3000
C = 3000 
 11.62 = 34,860 bps
11.62 = 34,860 bps

Example
Example
We have a channel with a 1 MHz bandwidth. The SNR
for this channel is 63; what is the appropriate bit rate and
signal level?
Solution
Solution
C = B log
C = B log2
2 (1 + SNR) = 10
(1 + SNR) = 106
6
log
log2
2 (1 + 63) = 10
(1 + 63) = 106
6
log
log2
2 (64) = 6 Mbps
(64) = 6 Mbps
Then we use the Nyquist formula to find the
number of signal levels.
Bit Rate=2 X Bandwidth X log2L
6
6 Mbps = 2
Mbps = 2 
 1 MHz
1 MHz 
 log
log2
2 L
L 
 L = 8
L = 8
First, we use the Shannon formula to find our upper
First, we use the Shannon formula to find our upper
limit.
limit.

Transmission Impairments
 Signals travel through transmission media,
which are not perfect. The imperfection
causes signal impairment. This means that
the signal at the beginning of the medium is
not the same as the signal at the end of the
medium. What is sent is not what is received.
Three causes of impairment are attenuation
attenuation,
,
distortion
distortion,
, and noise
noise.
.

Decibel
 Used to signal gained or lost strength
 dB = 10 log P2/P1 (Power)
 dB Voltage = 20 log (V1/V2)
 dB Current = 20 log (I1/I2)
The dB is not an absolute quantity, it is always a RATIO of
two quantities. The unit can be used to express power
gain (P2>P1), or power loss (P2<P1) -- in the latter case
the result will be a negative number.

Example
 Suppose a signal travels through
transmission medium and its power is
reduced to one half…. Calculate attenuation
loss?

Signal Distortion
attenuation
distortion
noise

Performance
 One important issue in networking is the
performance of the network—how good is it?

Performance
 Bandwidth
 Throughput
 Latency (Delay)
 Bandwidth-Delay Product

Throughput
Throughput is a measure of the actual data transfer rate in
a communication system or network, representing the
amount of data successfully transmitted from the source to
the destination over a given period of time. Throughput is
often expressed in bits per second (bps), kilobits per second
(Kbps), megabits per second (Mbps), or gigabits per second
(Gbps), depending on the scale of the communication
system.

Latency
 Latency = propagation time
+
transmission time
+
queuing time
+
processing time

 In computer networks, propagation delay is
the amount of time it takes for the head of the
signal to travel from the sender to the
receiver. It can be computed as the ratio
between the link length and the propagation
speed over the specific medium.
 Propagation delay is equal to d / s where d is
the distance and s is the wave propagation
speed. In wireless communication, s=c, i.e.
the speed of light. In copper wire, the speed s
generally ranges from .59c to .77c

In data communications, bandwidth-delay
product refers to the product of a data link's
capacity (in bits per second) and its end-to-
end delay (in seconds). The result, an
amount of data measured in bits (or bytes),
is equivalent to the maximum amount of
data on the network circuit at any given
time, i.e. data that has been transmitted but
not yet received. Sometimes it is calculated
as the data link's capacity multiplied by its
round trip time

According to the Nyquist theorem, the
sampling rate must be
at least 2 times the highest frequency
contained in the signal.
Note

Lecture 2 data communication physical layer

Lecture 2 data communication physical layer

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