Data and signals

4.1
Chapter 4
Data and Signals
 Computer Networks
 Al-Mustansiryah University
 Elec. Eng. Department College of Engineering
Fourth Year Class

4.2
Figure 4.1 Two digital signals: one with two signal levels and the other
with four signal levels

4.4
Assume we need to download text documents at the rate
of 100 pages per sec. What is the required bit rate of the
channel (page is an average of 24 lines with 80 characters
in each line)?
Solution
If we assume that one character requires 8 bits (ascii), the
bit rate is
Example 4.1

4.5
A digitized voice channel is made by digitizing a 4-kHz
bandwidth analog voice signal. We need to sample the
signal at twice the highest frequency (two samples per
hertz). We assume that each sample requires 8 bits. What
is the required bit rate?
Solution
The bit rate can be calculated as
Example 4.2

4.6
What is the bit rate for high-definition TV (HDTV)?
Solution
HDTV uses digital signals to broadcast high quality video
signals. The HDTV screen is normally a ratio of 16 : 9.
There are 1920 by 1080 pixels per screen, and the screen
is renewed 30 times per second. Twenty-four bits
represents one color pixel.
Example 4.3
The TV stations reduce this rate to 20 to 40 Mbps through
compression.

4.7
4-2 PERFORMANCE
One important issue in networking is the performance of
the network—how good is it? We discuss quality of
service, an overall measurement of network performance,

4.8
In networking, we use the term
bandwidth in two contexts.
 The first, bandwidth in hertz, refers to the
range of frequencies in a composite signal or
the range of frequencies that a channel can
pass.
 The second, bandwidth in bits per second,
refers to the speed of bit transmission in a
channel or link. Often referred to as Capacity.
Note

4.9
A network with bandwidth of 10 Mbps can pass only an
average of 12,000 frames per minute with each frame
carrying an average of 10,000 bits. What is the throughput
of this network?
Solution
We can calculate the throughput as
Example 4.6
The throughput is almost one-fifth of the bandwidth in
this case.

4.10
Propagation & Transmission delay
 Propagation speed - speed at which a bit travels
though the medium from source to destination.
 Transmission speed - the speed at which all the
bits in a message arrive at the destination.
(difference in arrival time of first and last bit)

4.11
Propagation and Transmission Delay
 Propagation Delay = Distance/Propagation speed
 Transmission Delay = Message size/bandwidth bps
 Latency = Propagation delay + Transmission delay +
Queueing time + Processing time

4.12
What is the propagation time if the distance between the
two points is 12,000 km? Assume the propagation speed
to be 2.4 × 108 m/s in cable.
Solution
We can calculate the propagation time as
Example 4.7
The example shows that a bit can go over the Atlantic
Ocean in only 50 ms if there is a direct cable between the
source and the destination.

4.13
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
We can calculate the propagation and transmission time as
shown on the next slide:
Example 4.8

4.14
Note that in this case, because the message is short and the
bandwidth is high, the dominant factor is the propagation
time, not the transmission time. The transmission time can
be ignored.
Example 4.8 (continued)

4.15
What are the propagation time and the transmission time
for a 5-Mbyte message (an image) if the bandwidth of the
network is 1 Mbps? 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
We can calculate the propagation and transmission times
as shown on the next slide.
Example 4.9

4.16
Note that in this case, because the message is very long
and the bandwidth is not very high, the dominant factor is
the transmission time, not the propagation time. The
propagation time can be ignored.
Example 4.10 (continued)

4.17
Figure 4.6 Filling the link with bits for case 1

4.18
We can think about the link between two points as a pipe.
The cross section of the pipe represents the bandwidth,
and the length of the pipe represents the delay. We can say
the volume of the pipe defines the bandwidth-delay
product, as shown in Figure 4.7.
Example 4.11
The bandwidth-delay product defines
the number of bits that can fill the link.

4.19
Figure 4.7 Filling the link with bits in case 2

4.20
A network link is transferring a data file in two minutes period. The figure shown
below shows the throughput vs. time. Use this figure to calculate the following:
1) Average data transfer rate for this whole time.
2) Amount of data transferred between second 70 and second 120 ?
Example 4.11
P Kbps
500
t sec30 70 120

4.21
Size of data=0.5*30*500+40*500+0.5*50*500= 40000 KB
1) Average data transfer =S/T=40000/120= 334.3333 Kbps
2) Amount of data transferred =0.5*50*500= 12500 bit
Solution

4.22
Example 4.12
We are sending a 30 Mbit MP3 file from a source host to a destination host.
All links in the path between source and destination have a transmission rate
of 10 Mbps. Assume that the propagation speed is 2 * 10
8
meters/sec, and the
distance of the each link is 10,000 km. the processing time of the router is
0.01 sec
a) Suppose there is only one link between source and destination. Find the
latency?
b)Suppose there is three links between source and destination. The router is
found between each two links Find the latency?
Sol:
a)Latency = Transmission delay +Propagation delay
=30* 10
6
/10
7
+10
7
/(2 * 10
8
) = 4.05 sec
b)Latency = 3*Transmission delay +3*Propagation delay + 2* Processing
time(waiting time)
=3*( 30* 10
6
/10
7
)+ 3*(10
7
/(2 * 10
8
))+2*0.01=3*4.05+2*0.01=9.17

Data and signals

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