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NET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS (PRACTICAL PART)

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.)

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Revision (Cont.)

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

Logarithms

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Revision (Cont.)

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

General Sine Wave

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Revision (Cont.)

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Revision (Cont.)

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Revision (Cont.)

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Revision (Cont.)

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Revision (Cont.)

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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

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

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 (×10 3 ) = 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  . sin  = -2 sin 2  ft

OR

- cos 2  ft . sin  + sin 2  ft . cos  

Using



Sin(A+B) + Sin(A-B) = 2 sin (A) cos (B)

= 2 sin (2  ft) . cos  = -2 sin 2  ft + cos 2  ft Networks and Communication Department

Answer(Cont.)

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

b)

= sin 2  ft + sin 2  ft . cos  + cos 2  ft . sin  = 0 Networks and Communication Department

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 Note Frequency C 264 D 297 E 330 F 352 G 396 A 440 B 495 C 528

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Answer

Note Frequency Frequency deference C 264 33 D 297 33 E 330 22 F 352 44 G 390 44 A 440 55 B 495 33 C 528 Wavelength 1.25

1.11

1 0.93

0.83

0.7

5 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 dotted curve can he written in the form



t), what does the dotted curve represent? That is, the 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 Networks and Communication Department

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

A=1 A=0.05

A=0.05

F=15.96



=0 F=16.72



=0 F=15.13



=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 log 2 M We have C = 9600 bps 

a.

log 2 M = 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 Networks and Communication Department

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 Networks and Communication Department

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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

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-to noise 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: (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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