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NET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS Tutorial 2 : Chapter 3 Lecture 4+ Lecture 5 Networks and communications Department Spring 2014

Tutorial 2 : Chapter 3 - · PDF fileNET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS Tutorial 2 : Chapter 3 Lecture 4+ Lecture 5 Networks and communications Department Spring 2014

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Page 1: Tutorial 2 : Chapter 3 - · PDF fileNET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS Tutorial 2 : Chapter 3 Lecture 4+ Lecture 5 Networks and communications Department Spring 2014

NET 222: COMMUNICATIONS AND

NETWORKS FUNDAMENTALS

Tutorial 2 : Chapter 3

Lecture 4+ Lecture 5

Networks and communications Department Spring 2014

Page 2: Tutorial 2 : Chapter 3 - · PDF fileNET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS Tutorial 2 : Chapter 3 Lecture 4+ Lecture 5 Networks and communications Department Spring 2014

Assignment Problem(s)

Answer these Question, plus exercises (3.2) (3.4) (3.5) (3.16 a,b) (3.18 a)

(3.19 a,b) (3.21) Chapter 3 Problems page 122 from (Data and

Computer Communications book)

1- find the spectrum and bandwidth for the signal:

s(t)=(4/π)[sin(400πt)+(1/3)sin(600πt)+(1/5)sin(6000πt)]

2- find the bandwidth and draw the spectrum following:

A- If aperiodic signal is decomposed into five sine waves. With

frequencies of 100,300,500,700,900 Hz, What is its bandwidth?

Draw the spectrum ,assuming all components have a maximum

amplitude of 10 V.

Page 3: Tutorial 2 : Chapter 3 - · PDF fileNET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS Tutorial 2 : Chapter 3 Lecture 4+ Lecture 5 Networks and communications Department Spring 2014

B- 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

B=2000

F=100

B=fh-fL

2000=fh-100=2100Hz

3- A periodic signal has a bandwidth of 20 Hz. The highest frequency

is 60 Hz. What is the lowest frequency?

Page 4: Tutorial 2 : Chapter 3 - · PDF fileNET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS Tutorial 2 : Chapter 3 Lecture 4+ Lecture 5 Networks and communications Department Spring 2014

4- find the bandwidth for the signal:

A- (4/π)[ sin(2πft) + (1/3)sin(2π(3f)t) + (1/5)sin(2π(5f)t)

+(1/7)sin(2π(7f)t) ]

Bandwidth = 7f-f=6f

B- S(t)=sin(2πft)+sin(2π(3f)t)

C- a signal that ranges from 40 KHz to 4 MHz?

40KHz=40*10−3=0.04MHz

BW=4-0.04=3.96

5- Write the mathematical equation corresponding to the sine wave

having a maximum amplitude of 8, a period of 62.5ms, and a phase

shift of 135 degrees with respect to time 0. S(t)=A+sin (2 π f t+ϕ )

A=8,T=62.5,f=1/62.5

R=135*π/180=3π/4

S(t)=8+sin(2 π1/62.5t+3π/4)

3.2)A signal has a fundamental frequency 1000 Hz what is it's period?

) = 1 ms.310×Period = 1/1000 = 0.001 s (

Page 5: Tutorial 2 : Chapter 3 - · PDF fileNET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS Tutorial 2 : Chapter 3 Lecture 4+ Lecture 5 Networks and communications Department Spring 2014

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 C

Frequency 264 297 330 352 396 440 495 528

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 ?

2 sin(4πt + π); A = 2, f = 2, = π

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.

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

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,

Note C D E F G A B C

Frequency 264 297 330 352 390 440 495 528

Frequency

deference 33 33 22 44 44 55 33

Wavelength 1.25 1.11 1 0.93 0.83 0.75 0.67 0.63

Page 6: Tutorial 2 : Chapter 3 - · PDF fileNET 222: COMMUNICATIONS AND NETWORKS FUNDAMENTALS Tutorial 2 : Chapter 3 Lecture 4+ Lecture 5 Networks and communications Department Spring 2014

a. What is the theoretical maximum channel capacity (kbps) of

traditional telephone lines?s

3.19) Consider a channel with a 1-Mbps 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?

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?

Mbps

Mbps