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Net 222: Communications and networks fundamentals ( Practical Part). Tutorial 2 : Chapter 3 Data & computer communications . Revision . Trigonometric Functions. Revision (Cont.). Trigonometric Functions. Revision (Cont.). Revision (Cont.). Logarithms. Revision (Cont.).
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Net 222: Communications and networks fundamentals (Practical Part) Tutorial 2 : Chapter 3 Data &computer communications Networks and Communication Department
Revision • Trigonometric Functions Networks and Communication Department
Revision (Cont.) • Trigonometric Functions Networks and Communication Department
Revision (Cont.) Networks and Communication Department
Revision (Cont.) • Logarithms Networks and Communication Department
Revision (Cont.) • General Sine Wave Networks and Communication Department
Revision (Cont.) Networks and Communication Department
Revision (Cont.) Networks and Communication Department
Revision (Cont.) Networks and Communication Department
Revision (Cont.) Networks and Communication Department
Revision (Cont.) Networks and Communication Department
Chapter 3 (Data & computer communications) • 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. Networks and Communication Department
Question 3.1) • a) For the multipoint configuration, only one device at a time can transmit. Why? Networks and Communication Department
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
Question • 3.2) A signal has a fundamental frequency 1000 Hz what is it's period? Networks and Communication Department
Answer • Period = 1/1000 = 0.001 s (×103) = 1 ms. Networks and Communication Department
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
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 2ft . cos - cos 2ft . sin + sin 2ft . cos + cos 2ft . sin = -2 sin 2ft OR • Using • Sin(A+B) + Sin(A-B) = 2 sin (A) cos (B) = 2 sin(2ft) . cos = -2 sin 2ft Networks and Communication Department
Answer(Cont.) • b) = sin 2ft + sin 2ft . cos + cos 2ft . sin = 0 Networks and Communication Department
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 Networks and Communication Department
Answer Networks and Communication Department
Question • 3.5) If the solid curve in Figure 3.17 represents sin(2t), what does the dotted curve represent? That is, the dotted curve can he written in the form A sin (2ft + ); what are A, f and ? Networks and Communication Department
Answer • 2 sin(4πt + π ); A = 2, f = 2, = π Networks and Communication Department
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
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 + 0.05 cos 105t + 0.05 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
Question • 3.7) Find the period of the function f(t) =(10 cos t )2? Networks and Communication Department
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
Question • 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. Networks and Communication Department
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. 9600 = 2B × 8, and B = 600 Hz Networks and Communication Department
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
Answer • N = 1.38 × 10–23 × (50 + 273.15) = 445.947× 10–23 watts/Hz Networks and Communication Department
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
Answer Networks and Communication Department
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
Answer Networks and Communication Department
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
Answer Networks and Communication Department
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
Answer • Bandwidth = 7f-f=6f Networks and Communication Department
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
Answer • B=2000 • F=100 • B=fh-fL • 2000=fh-100=2100Hz Networks and Communication Department
Question • 3- find the DC component of the following signal. Networks and Communication Department
Answer • DC component =13 Networks and Communication Department
The End Any Questions ? Networks and Communication Department