Data Communication &Network
Tutorial 2
DKT224/3
Date : 28/1/2010
1. a. Suppose that a digitized TV picture is to be transmitted from a source that uses a
matrix of 480 X 500 picture elements (pixels) , where each pixel can take on one
of 32 intensity values. Assume that 30 pictures are sent per second. Find the
source rate R (bps).
b. Assume that the TV picture is to be transmitted over a channel with 4.5 MHz
bandwidth and a 35 dB SNR. Find the capacity of the channel (bps).
c. Discuss how the parameters given in part (a) could be modified to allow
transmission of color TV signals without increasing the required value for R.
a. (30 pictures/s) (480 × 500 pixels/picture) = 7.2 × 106 pixels/s
Each pixel can take on one of 32 values and can therefore be represented by 5 bits:
R = 7.2 × 106 pixels/s × 5 bits/pixel = 36 Mbps
b. We use the formula: C = B log2 (1 + SNR)
B = 4.5 × 106 MHz = bandwidth, and
SNRdB = 35 = 10 log10 (SNR), hence
SNR = 1035/10
= 103.5 , and therefore
C = 4.5 × 106 log2 (1 + 103.5 ) = 4.5 × 106 × log2 (3163)
C = (4.5 × 106 × 11.63) = 52.335 × 106 bps
c. Allow each pixel to have one of ten intensity levels and let each pixel be one of
three colors (red, blue, green) for a total of 10 × 3 = 30 levels for each pixel
element.
2. Suppose a file of 10,000 bytes is to be sent over a line at 2400 bps.
a. Calculate the overhead in bits and time in using asynchronous communication.
Assume one start bit and a stop element of length one bit, and a 8 bits to send the byte
itself for each character. The 8 bits character consists of all data bits, with no parity
bit.
b. Calculate the overhead in bits and time in using synchronous communication.
Assume that the data are sent in frames. Each frame consists of 1000 characters =
8000 bits and an overhead of 48 control bits per frame.
c. What would the answers to part (a) and (b) be for a file of 100,000 characters.
a. Each character has 25% overhead. For 10,000 characters, there are 20,000 extra
bits. This would take an extra 20,000/2400 = 8.33 seconds.
b. The file takes 10 frames or 480 additional bits. The transmission time for the
additional bits is 480/2400 = 0.2 seconds.
c. Ten times as many extra bits and ten times as long for both.
Tutorial 1
Pn. Shahadah Ahmad
Data Communication &Network
DKT224/3
3. For P = 110011 and M = 11100011, find the CRC.
4.
A CRC (Cyclic Redundancy Check) is constructed to generate a 4-bit FCS
(Frame Check Sequence) for an 11-bit message. The generator polynomial is X4
+ X3 + 1.
i) Encode the data bit sequence 10011011100 (leftmost bit is the least
significant) using the generator polynomial and give the codeword.
ii) Now assume that bit 7 (counting from LSB) in the codeword is in error and
show that the detection algorithm detects the error.
Tutorial 1
Pn. Shahadah Ahmad
Data Communication &Network
5.
DKT224/3
Calculate the Hamming pairwise distances among the following codewords:
a. 00000, 10101, 01010
b. 000000, 010101, 101010, 110110
6. List three differences techniques in serial transmission and explain the differences.
Three differences techniques in serial transmission are synchronous,
asynchronous, and isochronous. In both synchronous and asynchronous
transmissions, a bit stream is divided into independent frames. In
synchronous transmission, the bytes inside each frame are synchronized; in
asynchronous transmission, the bytes inside each frame are also
independent. In isochronous transmission, there is no independency at all.
All bits in the whole stream must be synchronized.
Tutorial 1
Pn. Shahadah Ahmad
Data Communication &Network
7.
DKT224/3
Distinguish between synchronous and statistical TDM.
In synchronous TDM, each input has a reserved slot in the output frame. This can
be inefficient if some input lines have no data to send. In statistical TDM, slots
are dynamically allocated to improve bandwidth efficiency. Only when an input
line has a slot’s worth of data to send is it given a slot in the output frame.
8.
We need to use synchronous TDM and combine 20 digital sources, each of
100Kbps.Each output slot carries 1 bit from each digital source, but one extra bit
is added to each frame for synchronization. Answer the following questions.
a.
b.
c.
d.
e.
What is the size of an output frame in bits?
What is the output frame rate?
What is the duration of an output frame?
What is the output data rate?
What is the efficiency of the system?
a. Each output frame carries 1 bit from each source plus one extra bit for
synchronization. Frame size = 20 × 1 + 1 = 21 bits.
b. Each frame carries 1 bit from each source. Frame rate = 100,000 frames/s.
c. Frame duration = 1 /(frame rate) = 1 /100,000 = 10 μs.
d. Data rate = (100,000 frames/s) × (21 bits/frame) = 2.1 Mbps
e. In each frame 20 bits out of 21 are useful. Efficiency = 20/21= 95%
9.
Ten sources, six with a bit rate of 200 kbps and four with a bit rate of 400 kbps
are to be combined using multilevel TDM with no synchronizing bits. Answer
the following questions about the final stage of multiplexing:
a.
b.
c.
d.
What is the size of a frame in bits?
What is the frame rate?
What is the duration of a frame?
What is the data rate?
We combine six 200-kbps sources into three 400-kbps. Now we have seven 400kbps channel.
a. Each output frame carries 1 bit from each of the seven 400-kbps line. Frame
size = 7 × 1 = 7 bits.
b. Each frame carries 1 bit from each 400-kbps source. Frame rate = 400,000
frames/s.
c. Frame duration = 1 /(frame rate) = 1 /400,000 = 2.5 μs.
d. Output data rate = (400,000 frames/s) × (7 bits/frame) = 2.8 Mbps. We can also
calculate the output data rate as the sum of input data rate because there is no
synchronizing bits. Output data rate = 6 × 200 + 4 × 400 = 2.8 Mbps.
Tutorial 1
Pn. Shahadah Ahmad
Data Communication &Network
10.
DKT224/3
Two channels, one with a bit rate of 190 kbps and another with a bit rate of
180 kbps, are to be multiplexed using pulse stuffing TDM with no
synchronization bits. Answer the following questions:
a. What is the size of a frame in bits?
b. What is the frame rate?
c. What is the duration of a frame?
d. What is the data rate?
We need to add extra bits to the second source to make both rates = 190 kbps.
we have two sources, each of 190 Kbps.
a. The frame carries 1 bit from each source. Frame size = 1 + 1 = 2 bits.
b. Each frame carries 1 bit from each 190-kbps source. Frame rate = 190,000
frames/s.
c. Frame duration = 1 /(frame rate) = 1 /190,000 = 5.3 μs.
d. Output data rate = (190,000 frames/s) × (2 bits/frame) = 380 kbps. Here the
output bit rate is greater than the sum of the input rates (370 kbps) because of
extra bits added to the second source.
Chapter 7
11.
You are setting up a network for a small company, which has seven computers,
two printers, and one scanner that need to be connected. All equipment is located
on one floor in a relatively small area. Costs must be kept at a minimum, and the
network does not need to be especially fast. As an engineer you need to
recommend which transmission medium should be used. State few reasons for
your choice.
�Given that costs must be kept at minimum and network does not need to be
Especially fast the most appropriate medium is Twisted Pair cable
�Copper wires are shielded and then twisted together. This minimizes both EM
radiation and susceptibility to radiation
�Typical bit rates are in the order of 1-100 Mbps over short distances
�TP is cheap and easy to install, hence the most widely used cable in networking
apps
�Suggestion for radio waves (with appropriate description) would also be accepted
12.
In considering the design of data transmission systems, key concerns are data rate
and distance , the greater the data rate and distance the better. A number of design
factors relating to the transmission medium and the signal determine the data rate
and distance . Briefly explain the following factors :
i)
ii)
iii)
iv)
Tutorial 1
Bandwidth
Transmission impairment
Interference
Number of receivers
Pn. Shahadah Ahmad
Data Communication &Network
DKT224/3
A number of design factors relating to the transmission medium and the signal
determine the data rate and distance:
• Bandwidth: All other factors remaining constant, the greater the bandwidth of a
signal, the higher the data rate that can be achieved.
• Transmission impairments: Impairments, such as attenuation, limit the distance.
For guided media, twisted pair generally suffers more impairment than coaxial cable,
which in turn suffers more than optical fiber.
• Interference: Interference from competing signals in overlapping frequency
bands can distort or wipe out a signal. Interference is of particular concern for
unguided media, but is also a problem with guided media. For guided media,
interference can be caused by emanations from nearby cables. For example, twisted
pairs are often bundled together and conduits often carry multiple cables. Interference
can also be experienced from unguided transmissions. Proper shielding of a guided
medium can minimize this problem.
• Number of receivers: A guided medium can be used to construct a point-to-point
link or a shared link with multiple attachments. In the latter case, each attachment
introduces some attenuation and distortion on the line, limiting distance and/or data
rate.
Tutorial 1
Pn. Shahadah Ahmad