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NET 222: COMMUNICATIONS AND
NETWORKS FUNDAMENTALS
(PRACTICAL PART)
Networks and
Communication
Department
Tutorial 2 : Chapter 3 Data & computer communications
Revision
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Trigonometric Functions
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Revision (Cont.)
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Trigonometric Functions
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Revision (Cont.)
4
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Revision (Cont.)
5
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Logarithms
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Revision (Cont.)
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General Sine Wave
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Revision (Cont.)
7
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Revision (Cont.)
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Revision (Cont.)
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Revision (Cont.)
10
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Revision (Cont.)
11
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Chapter 3 (Data & computer communications)
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3.1(a)
3.2
3.3
3.4
3.5
3.6
3.7
3.16
3.17
3.18
3.19
3.21
3 more question on ch3.
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Question
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3.1)
 a) For the multipoint configuration, only one
device at a time can transmit. Why?
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Answer
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
If two devices transmit at the same time, their
signals will be on the medium at the same time,
interfering with each other; i.e., their signals will
overlap and become garbled.
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Question
15
3.2)
A signal has a fundamental frequency 1000 Hz
what is it's period?
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Answer
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Period = 1/1000 = 0.001 s (×103) = 1 ms.
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Question
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3.3)
Express the following in the simplest form you can:
 a) sin(2π ft -π ) + sin(2π ft +π)
 b) sin 2π ft + sin(2π ft -π )
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Answer
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
Using:


Sin (A+B)=sin (A) cos (B) + cos (A) sin (B)
Sin (A- B) = sin (A) cos (B) - cos (A) sin (B)
a)
= sin 2ft . cos  - cos 2ft . sin  + sin 2ft . cos  + cos 2ft
. sin 
= -2 sin 2ft
OR
 Using

Sin(A+B) + Sin(A-B) = 2 sin (A) cos (B)
= 2 sin (2ft) . cos 
= -2 sin 2ft

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Answer(Cont.)
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b)
= sin 2ft + sin 2ft . cos  + cos 2ft . sin 
=0
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Question
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3.4)
Sound may be modeled as a sinusoidal function.
Compare the relative frequency and wavelength of
musical note. Use 330 m/s as the speed of sound
and the following frequencies for the musical scale
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Note
C
D
E
F
G
A
B
C
Frequency
264
297
330
352
396
440
495
528
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Answer
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Note
C
D
E
F
G
A
B
C
Frequency
264
297
330
352
390
440
495
528
Frequency
deference
Wavelength 1.25
33
33
1.11
22
1
44
0.93
44
0.83
55
0.7
5
33
0.6
7
0.6
3
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Question
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3.5)
If the solid curve in Figure 3.17 represents sin(2t),
what does the dotted curve represent? That is, the
dotted curve can he written in the form
A sin (2ft +  ); what are A, f and  ?
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Answer
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2 sin(4πt + π ); A = 2, f = 2,  = π
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Question
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3.6)
Decompose the signal (1+0.1cos 5t) cos 100t into a
linear combination of sinusoidal, function
amplitude ,frequency, and phase of each
component
hint: use the identity for cos a cos b .

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Answer
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= cos 100t + 0.1 cos 5t cos 100t.
From the trigonometric identity
cos a cos b = (1/2)[ cos(a+b) + cos(a–b) ]
, this equation can be rewritten as the linear
combination of three sinusoids
cos 100t + 0.05 cos 105t + 0.05 cos 95t
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A=1
A=0.05
A=0.05
F=15.96
F=16.72
F=15.13
=0
=0
=0
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Question
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3.7)
Find the period of the function f(t) =(10 cos t )2?
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Answer
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cos a cos b = (1/2) [ cos(a+b) + cos(a–b) ]
f(t) = 50 cos 2t + 50
The period of cos(2t) is π and therefore the period of
f(t) is π .
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Question
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3.16)
A digital signaling system is required to operate at
9600 bps.

 a-
if a signal element encodes a 4-bit word, what is
the minimum required bandwidth of the channel?
 b- Repeat part (a) for the case of 8-bit words.
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Answer
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Using Nyquist's equation: C = 2B log2M
We have C = 9600 bps
 a. log2M = 4, because a signal element encodes a
4-bit word
Therefore, C = 9600 = 2B × 4, and B = 1200 Hz
 b. 9600 = 2B × 8, and B = 600 Hz
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Question
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3.17)
What is the thermal noise level of a channel with a
bandwidth of 10 kHz carrying 1000 watts of power
operating at 50°C?
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Answer
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N = 1.38 × 10–23 × (50 + 273.15)
= 445.947× 10–23 watts/Hz
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Question
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3.18)
Given the narrow (usable) audio bandwidth of a
telephone transmission facility, a nominal SNR of
56dB (400,000), and a certain level of distortion,

 a.
What is the theoretical maximum channel capacity
(kbps) of traditional telephone lines?s
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Answer
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Question
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3.19)
Consider a channel with a 1-MHz capacity and an
SNR of 63.

 a.
What is the upper limit to the data rate that the
channel can carry?
 b. The result of part (a) is the upper limit. However,
as a practical matter, better error performance will be
achieved at a lower data rate. Assume we choose a
data rate of 2/3 the maximum theoretical limit. How
many signal levels are needed to achieve this data
rate?
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Answer
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Question
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
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3.21)
Given a channel with an intended capacity of 20
Mbps, the bandwidth of the channel is 3 MHz.
Assuming white thermal noise, what signal-tonoise ratio is required to achieve this capacity?
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Answer
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Question
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1- find the bandwidth for the signal:
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(4/π)[ sin(2πft) + (1/3)sin(2π(3f)t) + (1/5)sin(2π(5f)t) +(1/7)sin(2π(7f)t) ]
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Answer
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Bandwidth = 7f-f=6f
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Question
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
2- a signal with a bandwidth of 2000 Hz is
composed of two sine waves. the first one has a
frequency of 100 Hz with a maximum amplitude
of 20, the second one has a maximum amplitude
of 5. draw the frequency spectrum
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Answer
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B=2000
F=100
B=fh-fL
2000=fh-100=2100Hz
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Question
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3- find the DC component of the following signal.
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Answer
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DC component =13
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The End
Any Questions ?
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