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Net 222: Communications and networks fundamentals (Practical Part)

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Presentation on theme: "Net 222: Communications and networks fundamentals (Practical Part)"— Presentation transcript:

1 Net 222: Communications and networks fundamentals (Practical Part)
Networks and Communication Department Tutorial 2 : Chapter 3 Data & computer communications

2 Revision Trigonometric Functions Networks and Communication Department

3 Revision (Cont.) Trigonometric Functions
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4 Revision (Cont.) Networks and Communication Department

5 Revision (Cont.) Logarithms Networks and Communication Department

6 Revision (Cont.) General Sine Wave
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7 Revision (Cont.) Networks and Communication Department

8 Revision (Cont.) Networks and Communication Department

9 Revision (Cont.) Networks and Communication Department

10 Revision (Cont.) Networks and Communication Department

11 Revision (Cont.) Networks and Communication Department

12 Chapter 3 (Data & computer communications)
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. Networks and Communication Department

13 Question 3.1) a) For the multipoint configuration, only one device at a time can transmit. Why? Networks and Communication Department

14 Answer 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. Networks and Communication Department

15 Question 3.2) A signal has a fundamental frequency 1000 Hz what is it's period? Networks and Communication Department

16 Answer Period = 1/1000 = 0.001 s (×103) = 1 ms.
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17 Question 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 ) Networks and Communication Department

18 Answer 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  Networks and Communication Department

19 Answer(Cont.) b) = sin 2ft + sin 2ft . cos  + cos 2ft . sin  = 0
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20 Question 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 C D E F G A B Frequency 264 297 330 352 396 440 495 528 Networks and Communication Department

21 Answer Note Frequency Frequency deference Wavelength C D E F G A B 264
297 330 352 390 440 495 528 Frequency deference 33 22 44 55 Wavelength 1.25 1.11 1 0.93 0.83 0.75 0.67 0.63 Networks and Communication Department

22 Question 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  ? Networks and Communication Department

23 Answer 2 sin(4πt + π ); A = 2, f = 2,  = π
Networks and Communication Department

24 Question 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 . Networks and Communication Department

25 Answer = 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 cos 105t cos 95t A=1 F=15.96 =0 A=0.05 F=16.72 =0 A=0.05 F=15.13 =0 Networks and Communication Department

26 Question 3.7) Find the period of the function f(t) =(10 cos t )2?
 3.7) Find the period of the function f(t) =(10 cos t )2? Networks and Communication Department

27 Answer 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 . Networks and Communication Department

28 Question 3.16) A digital signaling system is required to operate at 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. Networks and Communication Department

29 Answer 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 = 2B × 8, and B = 600 Hz Networks and Communication Department

30 Question 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? Networks and Communication Department

31 Answer N = 1.38 × 10–23 × (50 + 273.15) = 445.947× 10–23 watts/Hz
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32 Question 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 Networks and Communication Department

33 Answer Networks and Communication Department

34 Question 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? Networks and Communication Department

35 Answer Networks and Communication Department

36 Question 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? Networks and Communication Department

37 Answer Networks and Communication Department

38 Question 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) ] Networks and Communication Department

39 Answer Bandwidth = 7f-f=6f Networks and Communication Department

40 Question 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 Networks and Communication Department

41 Answer B=2000 F=100 B=fh-fL 2000=fh-100=2100Hz
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42 Question 3- find the DC component of the following signal.
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43 Answer DC component =13 Networks and Communication Department

44 The End Any Questions ? Networks and Communication Department


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