1. The digital signal X(t) given below.

X(t)  1                        0                      1 2 3 4 5 6 7 8 t (msec)

   

a. If the carrier is sin (2000 π t), plot Amplitude Shift Keying (ASK) Modulated signal.

b. If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented by sin (4000 π t), plot Frequency Shift Keying (FSK) Modulated signal.

c. If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented by cos (2000 π t), plot Phase Shift Keying (PSK) Modulated signal.

2. The analog signal is given as x(t) = 5 [1 + 0.5 sin (2000π t) - cos (8000π t) ].

a. Find the minimum number of samples per second needed to recover the signal without loosing information.

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Maximum frequency = 4 KHz, i.e. sampling frequency = 8 KHz

The minimum number of samples per second needed to recover the signal without loosing information = 8000

b. If one sample is represented by 256 levels, find the number of bits needed to transmit one sample.

256 levels mean that each sample is represented by 8 bits since 256 = 2 8

The number of bits needed to transmit one sample = 8 bits

c. Find the mimimum rate in Kbps at which this signal is transmitted.

The mimimum rate in Kbps at which this signal is transmitted

= 8000 samples / sec x 8 bits / sample = 64 Kbps

d. What happens if you transmit this signal at a rate higher than the minimum rate found in 2.c. above

The signal will be recovered without loss of information, however, rate more than necessary will be utilized.

e. What happens if you transmit this signal at a rate lower than the minimum rate found in 2.c. above

The signal will be recovered with loss of information which is not desired.

3. Assume a character is coded by 8 bits. Assuming no control bits or other bit redundancy is involved in the communication link.

a. How many characters can be downloaded in a minute when standard 56 Kbps modem is used (assume full rate can be utilized) ?

56 Kbps = 56000 bits / sec = ( 56000 bits / sec ) / (8 bits / character)

= 7000 characters / sec

No. of characters that can be downloaded in a minute = ( 7000 characters / sec ) x ( 60 sec / minute) = 4.2 x 10 5 characters / minute

b. How many characters can be downloaded in a minute when Dense Wavelength Division Multiplexing (DWDM) system is used which has 15000 separate wavelengths to transmit the information and each wavelength is modulated at 10 Gbps ?

For one wavelength,

10 Gbps = 10 10 bits / sec = (10 10 bits / sec) / (8 bits / character) = 1.25 x 10 9 characters / sec

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No. of characters that can be downloaded in a minute = (1.25 x 10 9 characters / sec ) x ( 60 sec / minute) = 7.5 x 10 10 characters / minuteFor 15000 wavelengths,

No. of characters that can be downloaded in a minute = No. of characters that can be downloaded in a minute for one wavelength x 15000

= 7.5 x 10 10 characters / minute x 15000 = 1.125 x 10 15 characters / minutec. How many characters can be downloaded in a minute when an xDSL system with

the highest possible rate is used.

xDSL system with the highest possible rate is provided by VDSL at 52 Mbps

52 Mbps = 52 x 10 6 bits / sec = (52 x 10 6 bits / sec) / (8 bits / character) = 6.5 x 10 6 characters / sec

No. of characters that can be downloaded in a minute = (6.5 x 10 6 characters / sec ) x ( 60 sec / minute) = 3.9 x 10 8 characters / minute

d. Compare and comment on the results you have found in 3.a, 3.b and 3.c above.

56 Kbps modem is the slowest,

VDSL is 3.9 x 10 8 / 4.2 x 10 5 ≈ 905 times faster than 56 Kbps modem,

DWDM in 3.b is 1.125 x 10 15 / 4.2 x 10 5 ≈ 2678 billion times faster than 56 Kbps modem.

4. A 10 mile link operates at 10 GHz . Both transmitting and receiving antenna gains are 28.3 dBi each and cabling loss both at the transmitter and at the receiver are 5 dB each. Output power of the transmitter is 10 dBm.

a. Find the Unfaded Received Signal Level.

FSL = 96.6+20 log D+20 logF = 96.6+20 log10+20 log10 = 96.6+20+20 = 136.6 dB

Po - Lctx + Gatx - Lcrx + Gatx - FSL = RSL

RSL = 10 dBm - 5 dB + 28.3 dBi - 5 dB + 28.3 dBi - 136.6 dB = - 80 dBm

b. If a Fade Margin of 20 dB is used in the design, find the Receiver Sensitivity Threshold required.

Fade Margin = Unfaded Receive Signal Level - Receiver Sensitivity Threshold

Receiver Sensitivity Threshold = - 80 dBm - 20 dB = -100 dBm

c. Changing the operating frequency of the link to 1 GHz and keeping all the other link parameters the same, find the Unfaded Received Signal Level.

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FSL = 96.6+20 log D+20 logF = 96.6+20 log10+20 log1 = 96.6+20+0 = 116.6 dB

Po - Lctx + Gatx - Lcrx + Gatx - FSL = RSL

RSL = 10 dBm - 5 dB + 28.3 dBi - 5 dB + 28.3 dBi - 116.6 dB = - 60 dBm

d. If for the 1 GHz link, the same receiver is used as in 4.b above, find the Fade Margin.

Fade Margin = Unfaded Receive Signal Level - Receiver Sensitivity Threshold

= - 60 dBm - ( -100 dBm ) = 40 dB

e. Which is a better design, 4.b or 4.d above ? Explain.

If the fade margin of 40 dB is needed due to atmospheric conditions of the microwave link in 4.d, then 4.d is a better design. If the atmospheric conditions of the microwave link in 4.d do not require 40 dB fade margin, but can still perform with 20 dB fade margin, then 4.b is a better design.

5. Based on E-Carrier European (CEPT) hierarchies, you own 4 different types of multiplexers, E-1, E-2, E-3 and E-4.

a. Which of these multiplexers would you prefer to send one digital video channel ?

E-4

b. Which of these multiplexers would you prefer to send 150 digital voice channels ?

E-3

c. How efficient is your choice in 5.b above ? What can happen if you use statistical multiplexer instead ? Explain.

It is not efficient because only 150 digital voice channels are used whereas there are 480 digital voice channels available in E-3.

If statistical mux of the right size is used instead, the traffic flow would be more efficient.

5. An analog signal has time variation f(t) = 3 + 0.2 cos (8000π t) - 0.3 sin (4000π t) .

a. Minimum how many samples should be taken to satisfy Nyquist requirement?

Maximum frequency = 4 KHz, i.e. sampling frequency = 8 KHz

The minimum number of samples per second needed to satisfy Nyquist requirement = 8000

b. 256 levels is used to represent one sample. How many bits are required to transmit one sample value?

256 levels mean that each sample is represented by 8 bits since 256 = 2 8

The number of bits needed to transmit one sample = 8 bits

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c. What is the mimimum transmission rate of this signal?

The mimimum rate in Kbps at which this signal is transmitted

= 8000 samples / sec x 8 bits / sample = 64 Kbps

d. Would you allocate a 256 kbps channel to transmit this signal? Why?

No because more than 64 kbps rate will be unnecessary in the recovery of the original signal.

e. Would you allocate a 32 kbps channel to transmit this signal? Why?

No because with a rate less than 64 kbps, the original signal will only be recovered with loss of information which is not desired.

6. The first microwave link (LINK-1) operating at 10 GHz has a link distance of 1 mile. The second microwave link (LINK-2) operates at 1 GHz. In both of the links, both transmitting and receiving antenna gains are 20 dBi each and cabling loss both at the transmitter and at the receiver are 2 dB each.

a. What should be the link distance in the second link (LINK-2) so that both links (LINK-1 and LINK-2) have the same free space loss?

For LINK-1: FSL = 96.6+20log1+20log10 = 96.6+0+20 = 116.6 dB For LINK-2: FSL = = 116.6 dB = 96.6+20logD+20log1 = 96.6+20logD +0 116.6 dB = 96.6+20logD ,i.e., logD=1 , D=10 miles b. Find the Received Signal Level in LINK-1 if the output power of the transmitter in

LINK-1 is 10.6 dBm.

Po - Lctx + Gatx - Lcrx + Gatx - FSL = RSL

RSL = 10.6 dBm - 2 dB + 20 dBi - 2 dB + 20 dBi - 116.6 dB = - 70 dBm

c. Find the output power of the transmitter in LINK-2 if the Received Signal Level in LINK-2 is -72.6 dBm.

Po = RSL + Lctx - Gatx + Lcrx - Gatx + FSL

Po = -72.6 dBm + 2 dB - 20 dBi + 2 dB - 20 dBi + 116.6 dB = 8 dBm

d. The Receiver Sensitivity Threshold (Rx) for LINK-1 is - 90 dBm and the Receiver Sensitivity Threshold for LINK-2 is - 60 dBm. Can LINK-1 and LINK-2 operate? Why?

For LINK-1, RSL = - 70 dBm > Rx = - 90 dBm. i.e., LINK-1 can operate. For LINK-2, RSL = -72.6 dBm < Rx = - 60 dBm. i.e., LINK-2 can not operate.e. For a given microwave transmitter and receiver system, you have made an

unsuccessful link design. What can you do to make this link operate?

For the given microwave transmitter and receiver system, to make the link operate, link distance should be reduced.

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7. a. For a carrier of sin (2000 π t), the Amplitude Shift Keying (ASK) Modulated signal is given below. Plot the digital information signal x(t).

X(t)  1                        0                      1 2 3 4 5 6 7 8 t (msec)

   

b. If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented by sin (4000 π t), plot the Frequency Shift Keying (FSK) Modulated signal for the

digital information signal x(t) found in part a.

c. If digital level “1” is represented by sin (2000 π t) and digital level “0” is represented by cos (2000 π t), plot the Phase Shift Keying (PSK) Modulated signal for the

digital information signal x(t) found in part a.

d. If the carrier in part a becomes sin (4000 π t), re-plot the Amplitude Shift Keying (ASK) Modulated signal given in part a.

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e. Find the rate of the digital information signal x(t) found in part a.

Rate of x(t) found in part a = 1 bit / msec = 1 bit / 10 -3 sec = 10 3 bits / sec = 1 kbps.

8. 10000000 =10 7 books will be downloaded. Each book has average 100 pages, each page has average 100 words, each word has average 4 letters, each letter is encoded by 8 bits. Dense Wavelength Division Multiplexing (DWDM) system with 100 separate wavelengths (channels) is used to download the information. Each wavelength is modulated at 10 Gbps. Assuming no control bits or other bit redundancy is involved in the communication link:

a. What is the total number of bits that will present the total information content in 10 7

books?

The total number of bits that will present the total information content in 10 7 books = 10 7 x 100 x 100 x 4 x 8 = 3200 x 10 9 bits = 3.2 Tbits

b. What is the time required to download the total information content in the 10 7 books with the given DWDM system?

The time required to download the total information content in the 10 7 books with

the given DWDM system 3.2 Tbits / (10 Gbps x 100) = 3.2 sec.

c. What is the number of DWDM channels required so that the same total information content in 10 7 books is downloaded in 32 milliseconds?

The number of DWDM channels required so that the same total information content in 10 7 books is downloaded in 32 milliseconds

⇒ 3.2 Tbits / (10 Gbps x (no.of channels)) = 32 x 10-3 sec

⇒ no.of channels = 3.2 Tbits / (10 Gbps x 32 x 10 -3 sec) = 10000 channels

d. Find the number of years required to download the total information content in 10 7

books when a standard 56 Kbps modem is used (assuming full rate is utilized).

The number of years required to download the total information content in 10 7

books when a standard 56 Kbps modem is used (assuming full rate is utilized) = 3.2 Tbits/ 56 Kbps = 57.143.000 sec = 57.143.000 sec/ (365 days / year x 24 hrs/day x 60 min./hr x 60 sec / min ) = 57.143.000 sec/ (365 days / year x 86400 sec / day ) = 1.812 years

e. xDSL technology is used to download the same total information content in the 10 7

books. If the download takes 17.094 hours, find the rate of the download. Specify the type of xDSL used.

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Rate of the download = 3.2 Tbits / (17.094 hours x 60 min./hr x 60 sec / min) = 52 Mbps

The type of xDSL used is VDSL.

9. For a single TV channel, the bandwidth can be taken as 6 MHz.a. What is the maximum number of analog voice channels that can be transmitted in

two TV channels? One analog voice channel bandwidth is 4 KHz = 4 x 10 3 Hz.

In 6 MHz band, there are (6 x 10 6 Hz) / (4 x 10 3 Hz) = 1500 times 4 KHz. So, maximum number of analog voice channels that can be transmitted in two TV

channels = 3000. b. What is the required minimum bit rate to transmit one TV channel digitally, if one

sample value is represented by 10 bits? Sampling by twice the maximum frequency ⇒ 6 MHz x 2 = 12 M samples per second

To represent the sample with 10 bits means 10 bits / sample Minimum Bit Rate= 12 M samples per second x 10 bits / sample = 120 Mbps

c. What is the required minimum bit rate to transmit one TV channel digitally, if one sample value is represented by 256 levels?

Sampling by twice the maximum frequency ⇒ 6 MHz x 2 = 12 M samples per second To represent the sample with 256 levels means 8 bits / sample

Minimum Bit Rate= 12 M samples per second x 8 bits / sample = 96 Mbps d. Maximum how many digital voice channels can be transmitted in one digital TV

channel given in part c? One digital voice channel at the same number of bits representing one sample is 64 Kbps In 96 Mbps, there are 96 M / 64 K = 1500 .

So, the maximum number of digital voice channels that can be transmitted in one digital channel is 1500.

e. For the transmission of the bit rate you found in part b above, which E-Carrier European (CEPT) level do you need?

Fourth level (E-4) 139.264 Mb/s (1920 Ch.)

10. Microwave links; M1 operates at milimeter wave, M2 operates at C-Band and M3 operates at L-Band. Transmitter power, atmospheric conditions, receiver sensitivity and all the other system parameters are the same for all these 3 links.

a. For transmitting the highest information bandwidth, which one of these links would you choose? Why?

Highest carrier frequency has the possibility to carry the highest information bandwidth. Thus M1 will transmit the highest information bandwidth.

b. For the longest possible link, which one of these links would you choose? Why? Longest possible link is achieved by the lowest frequency link which is M3.c. Find the Received Signal Level in dBm if 1 mile microwave link operating at 1 GHz

is used whose output power is 1 dBm. The antenna gains are 20 dBi each, cabling loss is zero for the transmitter and the receiver.

Free Space Loss FSL=96.6+20 log D+20 log F=96.6+20 log1+20 log1 = 96.6 dB Po - Lctx + Gatx - Lcrx + Gatx - FSL = RSL Received Signal Level, RSL=1 dBm–0 dB+20 dBi –0 dB+20 dBi –96.6 dB = - 55.6

dBm

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RSL = - 55.6 dBm d. Find the minimum Receiver Sensitivity Threshold that will operate the link given in

part c Minimum Receiver Sensitivity Threshold that will operate the link given in

part c is equal to RSL= dBm e. For the link given in part c, would you prefer to use a receiver with Receiver

Sensitivity Threshold of –70 dBm or –90 dBm? Why? Depends on the application.

11. Sixteen bits of information is sent in the following modulated signal where time axis is in microseconds:

a. Write the type of modulation used. Why?

Solution: ASK (Amplitude Shift Keying) because digits “1” and “0” are differentiated with different amplitudes.

b. Find the carrier frequency. Solution: For one bit, duration is 0.5 sec. and the number of cycles=1 Thus,the carrier frequency = 1 cycle in 0.5 sec., i.e., 2 x 10 6 cycles/sec = 2 MHz.

c. Find the rate of the information signal. Solution: One bit has duration of 0.5 sec. Thus , the rate of the signal = 2 x 10 6 bits/sec = 2 Mbps.

d. Plot the information signal if “1” is represented by no signal, and “0” is represented by 0.5 mV and no carrier.

Solution:

e. Is this information signal convenient to carry 1 digital voice channel ? Why? Is this information signal convenient to carry 1 digital video channel ? Why?

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Solution: This information signal is convenient to carry 1 digital voice channel because the information signal has a rate of 2 Mbps and 1 digital voice channel needs only 64 kbps.

However, this information signal is not convenient to carry 1 digital video channel because the information signal has a rate of 2 Mbps and 1 digital video channel needs (6 x 10 6 x 2) samples / sec x 8 bits / sample = 96 Mbps.

12. a. Find the number of seconds needed to download 1 Petabits of information by using a DWDM system which has 10000 separate wavelengths (channels) and at each wavelength, 10 Gbps information can be carried.

Solution: Such DWDM system will be able to download 10 Gbps / wavelength x 10000 wavelengths = 10 x 10 9 x 10 4 bps = 10 14 bps To download 1 Petabits = 10 15 bits of information, 10 15 bits / 10 14 bps = 10 seconds are needed.

b. Find the number of years needed to download the same amount of information given in part a by using a 56 kbps modem (assuming full rate can be utilized).

Solution: Such modem will be able to download 10 15 bits / 56 kbps = 10 15 bits / 56 x 10 3 bps = (10 12 / 56) seconds = (10 12 / 56) seconds In one year there are 365 x 24 x 60 x 60 seconds Thus we need (10 12 / 56) / (365 x 24 x 60 x 60) years = 566.25 years c. Would you prefer to use the DWDM system as described in part a or the 56 kbps

modem as described in part b? Solution: Preference will depend on the application.

13. a. Assuming that there are 2 billion telephone subscribers in the world and each subscriber is connected to the telephone exchange with twisted pair cable at an average distance of 4 km. If the cost of the twisted pair cable is 0.5 YTL /meter, find the total value (in YTL) of the twisted pair cable installed in such infrastructure.

Solution: The total value (in YTL) of the twisted pair cable installed in such infrastructure is (2 billion telephone subscribers) x (3 km / subscriber) x (1000 m / km) x (0.1 YTL /meter)

= 2 x 10 9 x 4000 x 0.5 YTL = 4 x 10 12 YTL = 4000 x 10 9 YTL ≈ 3000 bilion $

b. Based on the result obtained in part a, what can you comment on the feasibility of fiber optics and DSL technology applications used in the local loop part of the telecommunication network? Explain.

Solution: Currently, fiber optics connection to all the local loop subscribers is not feasible since very big investment will be needed to replace the existing twisted pair installations. Under the present conditions, DSL technologies that utilize the existing twisted pair infrastructure seem much more feasible.

14. Among E-1, E-2, E-3 and E-4 multiplexers belonging to E-Carrier European (CEPT) hierarchy,

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a. Which one do you choose to send 90 digital voice channels?

E-2

b. How efficient is your choice in part a ? Explain.

It is not very efficient because only 90 digital voice channels are used whereas there are 120 digital voice channels available in E-2.

c. What can happen if you use statistical multiplexer instead of your choice in part a ? Explain.

If statistical mux of the right size is used instead, the traffic flow would be more efficient.

15. A character is coded by 10 bits. No control or other redundant bits are involved in the communication link.

a. Find the number of characters that can be downloaded in a minute when standard 56 kbps modem is used (assume full rate can be utilized).

56 kbps = 56000 bits / sec = ( 56000 bits / sec ) / (10 bits / character)

= 5600 characters / sec

No. of characters that can be downloaded in a minute = (5600 characters / sec ) x ( 60 sec / minute) = 3.36 x 10 5 characters / minute

b. Find the number of characters that can be downloaded in a minute when a DSL system with the highest possible rate is used.

DSL system with the highest possible rate is provided by VDSL at 52 Mbps

52 Mbps = 52 x 10 6 bits / sec = (52 x 10 6 bits / sec) / (10 bits / character) = 5.2 x 10 6 characters / sec

No. of characters that can be downloaded in a minute = (5.2 x 10 6 characters / sec ) x ( 60 sec / minute) = 3.12 x 10 8 characters / minute

c. Find the number of characters that can be downloaded in a minute when Dense Wavelength Division Multiplexing (DWDM) system is used which has 10000 separate wavelengths to transmit the information and each wavelength is modulated at 10 Gbps.

For one wavelength,

10 Gbps = 10 10 bits / sec = (10 10 bits / sec) / (10 bits / character) = 1 x 10 9 characters / sec

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No. of characters that can be downloaded in a minute = (1 x 10 9 characters / sec ) x ( 60 sec / minute) = 6 x 10 10 characters / minuteFor 10000 wavelengths,

No. of characters that can be downloaded in a minute = No. of characters that can be downloaded in a minute for one wavelength x 10000

= 6 x 10 10 characters / minute x 10000 = 6 x 10 14 characters / minuted. Compare and comment on the rates you have found in parts a, b and c.

56 Kbps modem in part a is the slowest,

VDSL in part b is 3.12 x 10 8 / 3.36 x 10 5 ≈ 928 times faster than 56 Kbps modem,

DWDM in patc c is 6 x 10 14 / 3.36 x 10 5 ≈ 1.79 billion times faster than 56 Kbps modem.

e. If one minute is needed to download a certain number of characters by using the DWDM system given in part c, find the number of years needed to download the same number of characters when the modem given in part a is used.

With the DWDM given in part c, downloading rate is 6 x 10 14 characters / minute, so in one minute 6 x 10 14 characters are downloaded

With the DWDM given in part a, downloading rate is 3.36 x 10 5 characters / minute

So, the number of years needed to download the same number of characters when the modem = 6 x 10 14 characters / 3.36 x 10 5 characters / minute = 1.7857 x 10 9

minutes = 1.7857 x 10 9 / (60 x 24 x 365) years = 3397.5 years

16. a. What is multiplexing ? Why do you need multiplexing in telecommunications ?

Multiplexing is a technique in which many signals belonging to different telecommunication channels are combined. In telecommunications, multiplexing is needed to transmit more than one channel simultaneously in one transmission link.

b. Compare TDM and FDM.

In TDM, the signals in each channel are multiplexed by using different time slots for each channel. However, in FDM, the signals in each channel are multiplexed by using different frequencies for each channel.

17. The digital information signal X(t) is given below which contains bits up to 8 microseconds:

X(t) 

1               0                    

12

  1 2 3 4 5 6 7 8 t (microsec)

a. If level “1” is presented by and level “0” is presented by , plot the Frequency Shift Keying (FSK) Modulated signal. Clearly indicate the numbers on the time axis.

b. If level “1” is represented by and level “0” is represented by

, plot the Phase Shift Keying (PSK) Modulated signal. Clearly indicate the numbers on the time axis.

c. If the carrier is , plot the Amplitude Shift Keying (ASK) Modulated signal. Clearly indicate the numbers on the time axis.

d. Can you use the same ASK you have shown in part c to transmit the digital information signal Y(t) given below which contains bits up to 8 microseconds? Explain.

Y(t)

13

 1           

0                

  1 2 3 4 5 6 7 8 t (microsec) 

Yes, the same ASK shown in part c can be used to transmit the digital information

signal Y(t), only the definition of “1” and “0” will change.

e. Find the rate of the digital information signal X(t).

Rate is 1 bit in 1 microsecond, i.e., 1 bit / 10 -6 sec = 10 6 bits / sec = 1 Mbps

18. is the analog information signal.

a. How many samples are needed in a second in order to recover the signal without loosing information.

Maximum frequency = 2 MHz, i.e. sampling frequency = 4 MHz

The minimum number of samples per second needed to recover the signal without loosing information = 4000000 = 4 million

b. If one sample is presented by 1024 levels, what is the number of bits needed to transmit one sample?

1024 levels mean that each sample is presented by 10 bits since 1024 = 2 10

The number of bits needed to transmit one sample = 10 bits

c. Find the mimimum rate in Mbps at which this signal can be transmitted without changing the understandibility of the information signal.

The mimimum rate in Mbps at which this signal is transmitted

= 4000000 samples / sec x 10 bits / sample = 40000000 bps = 40 Mbps

d. What happens if you transmit this signal at a rate higher than the minimum rate found in part c? Explain.

The signal will be recovered without loss of information, however, rate more than necessary will be utilized.

e. What happens if you transmit this signal at a rate lower than the minimum rate found in part c? Explain.

The signal will be recovered with loss of information which is not desired.

19. An image is represented by 1 million bits.

a. If 56 kbps modem is used, how many images can be downloaded in 20 hours?

Answer:14

56 kbps = 56000 bits/sec There are (60 minutes/hour) x (60 seconds/minute) = 3600 seconds in an hourThere are 3600 x 20 = 72000 seconds in 20 hoursSo, 72000 seconds in 10 hours x 56000 bits/sec = 4032 x 106 bits in 10 hours No. of images that can be downloaded in 10 hours = 4032 x 106 bits / 106 bits = 4032 images

b. If an ADSL with download rate of 1 Mbps is used, how many images can be downloaded in an hour?

Answer:1 Mbps = 106 bits/sec There are (60 minutes/hour) x (60 seconds/minute) = 3600 seconds in an hourSo, 3600 seconds in 1 hour x 106 bits/sec = 3600 x 106 bits in 1 hour No. of images that can be downloaded in 1 hour = 3600 x 106 bits / 106 bits= 3600 images

c. If a Dense Wavelength Division Multiplexing (DWDM) system is used which has 15000 separate wavelengths with each wavelength modulated at 10 Gbps, how many images can be downloaded in a millisecond?

Answer:For one wavelength, 10 Gbps = 10 10 bits/sec = 10 7 bits/msecFor 15000 wavelengths, 15000 x 10 7 bits/msec No. of images that can be downloaded in 1 msec = 15000 x 10 7 / 106 bits= 150000 images

d. Find the ratio of the rates of the systems in parts b and c to the rate of the system in part a.

Answer:Rate of the system in part a is 56 Kbps,Rate of the system in part b is 1 Mbps,Rate of the system in part c is 10 10 bits/sec per wavelength x 15000 wavelengths = 15 x 10 13 bits/sec,

So, the ratio of the rate of the system in part b to the rate of the system in part a = 1 Mbps / 56 Kbps = 17.86The ratio of the rate of the system in part c to the rate of the system in part a = 15 x 10 13 bits/sec / 56 Kbps = 2.68 billion

e. Among the systems described in parts a, b and c, which one would you prefer to use?

Answer:Depends on what is required. For example, if the requirement is high rate, then part c is preferable.

20. The digital information signal f (t) is given below which contains bits up to 8 nanoseconds:

f (t) 

1                 

0                      1 2 3 4 5 6 7 8 t (nanosec)

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a. If level “0” is presented by and level “1” is presented by , plot the Frequency Shift Keying (FSK) Modulated signal. Clearly indicate the numbers on the time axis.

Answer:

b. If the carrier is , plot the Amplitude Shift Keying (ASK) Modulated signal if level “1” is represented by an amplitude of 2 and level “0” is represented by an amplitude of 0.5. Clearly indicate the numbers on the time axis.

Answer:

c. If level “1” is represented by and level “0” is represented by

, plot the Phase Shift Keying (PSK) Modulated signal. Clearly indicate the numbers on the time axis.

Answer:

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d. What is the carrier frequency and the rate of the digital information signal f (t) in part b?

Answer:

Carrier frequency is 2 GHz, rate is 1 bit in 1 nanosecond, i.e., 1 bit / 10 -9 sec = 10 9 bits / sec = 1 Gbps

e. What is the carrier frequency and the rate of the digital information signal f (t) in part c?

Answer: Carrier frequency is 2 GHz, rate is 1 bit in 1 nanosecond, i.e., 1 bit / 10 -9 sec = 10 9 bits /

sec = 1 Gbps

21. In a microwave link, free space loss (FSL) is 136.6 dB and frequency of operation is 10 GHz.

a. Find the link distance.

Answer:FSL = 96.6+20 log D+20 logF = 96.6+20 log D +20 log10=96.6 +20 logD+20 =136.6 dB

20 log D +116.6 dB = 136.6 dB 20 log D = 20 dB → D = 10 miles

b. Cable loss of the transmitter cable is 6 dB, cable loss of the receiver cable is 4 dB, transmitter output power is 20 dBm and the received signal level is – 66.6 dBm. Find the antenna gain if the transmitter and the receiver antenna gains are the same. Answer: RSL = 20 dBm – 6 dB + G dBi – 136.6 dB + G dBi – 4 dB = – 66.6 dBm 20 dBm + G dBi + G dBi = (– 66.6 +146.6) dBm = 80 dBm 2G = 80 –20 → G = 30 dBi

c. If receiver sensitivity threshold is –96.6 dBm, find the fade margin. Answer: Fade Margin = Received Signal Level – Receiver Sensitivity Threshold

Fade Margin = – 66.6 – (– 96.6) = 30 dB

d. Keeping all the other link parameters the same, what operation frequency should be chosen to have 50 dB fade margin? Answer: Fade Margin = Received Signal Level – Receiver Sensitivity Threshold 50 dB = Received Signal Level – (– 96.6) dBm → Received Signal Level = 50 dB + (– 96.6) dBm = –46.6 dBm

RSL = 20 dBm – 6 dB + 30 dBi – FSL dB + 30 dBi – 4 dB = –46.6 dBm 70 dBm – FSL= –46.6 dBm → FSL = 116.6 dB FSL = 96.6+20 log 10 +20 log F = 96.6 +20 +20 log F =116.6 dB F = 1 GHz.

e. Write 5 parameters that will effect the design of a microwave link. Keeping all the other parameters the same, for each individual parameter, indicate whether an increase or decrease is needed in order to increase the received signal level. Answer:

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Frequency, link length, transmitter (or receiver) cable loss need to be decreased, transmitter output power, transmitter (or receiver) antenna gain need to be increased.

22. a. To digitize an analog signal, if minimum 12 million samples per second are needed to be taken from an analog signal, what would be the minimum and maximum frequency involved in that signal?

Answer:

Maximum frequency = 6 MHz, we cannot say anything about the minimum frequency b. If we want to transmit a signal of 5 kHz maximum frequency in a 100 kbps channel, how

many levels do we need to represent one sample.

Answer: 5 kHz x 2 = 10 k samples per second are needed. To have 100 kbps channel we need 10

digits to represent one sample, i.e., wee need 210 = 1024 levels.

c. Explain what happens to the rate of transmitted signal in multiplexing and in inverse multiplexing?

Answer:In multiplexing rate increases but in inverse multiplexing rate decreases

d. Write five different points related to the E-hierarchy structure. Answer:

In the existing infrastructure, the smallest hierarchy is E-1, the largest is E-7 Unit channel is 64 kbps Each upper hierarchy is formed by 4 of the lower hierchy They operate on TDM basis This hierarchy is used in Türkiye and Europe e. In each of the following items (i, ii, iii, iv, v) 5 different concepts are given. For each item,

write the concept which is unrelated to the other 4 concepts.

i. Multiplexing, FSK, E-hierarchy, J-hierarchy, TDM. Answer: FSKii. FM, PSK, Modulation, Statistical Multiplexing, Carrier Frequency. Answer: Statistical Multiplexing iii. Coaxial Cable, Microwave, FDM, Transmission Systems, Twisted Pair Cable . Answer: FDM iv. ADSL, Twisted Pair, VDSL, DWDM, Access Network. Answer: DWDM

v. Bandwidth, Frequency Content, Rate, Virtual Reality Systems, Information Content. Answer: Virtual Reality Systems

23. a. How many times more voice channels can be transmitted in E-6 hierarchy as compared to E-2 hierarchy?

Answer:

4 x 4 x 4 x 4 = 256 b. In each of the following items (i, ii, iii, iv, v) 5 different concepts are given. For each item,

write the concept which is unrelated to the other 4 concepts.

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i. FSK, ASK, PSK, FDM, Modulation Answer: FDM ii. Information Signal, Carrier Frequency, Modulation, SDSL, FSK. Answer: SDSLiii. Fiber, Radio Link, Coaxial Cable, UTP, STP. Answer: Radio Link iv. Radio Link, Satellite, Fiber, FSO, Microwave. Answer: Fiber

v. 3-D holography, Rate, Virtual Reality Systems, Virtual Museums, Multiplexing. Answer: Multiplexing

c. Write 5 different multiplexer types.

Answer: TDM, FDM, DWDM, Statistical multiplexing, WDM

d. Write five telecommunication applications that require high rate. Answer:

Virtual reality, holographic TV, Human Senses Added in Telecommunications, Machine-to-machine communication, Intelligent home and office utilities

e. Write whether the following statements are correct or wrong: i. Carrier wavelengths used in satellite communications are lower than the carrier

wavelengths used in fiber optics communications. Answer: Wrong ii. Carrier frequency is always larger than the maximum frequency involved in the information signal. Answer: Correctiii. Coaxial cable transmission can support higher frequencies as compared to twisted pair transmission. Answer: Correct iv. STP brings more security to the communication but interference is worse as compared

to UTP. Answer: Wrong

v. 40 Gbps transmission can only be done by using fiber optics. Answer: Correct

24. A video signal is composed of 18 million pictures and each picture is made up of 28 million

bits.

a. How many hours do you need to download this video signal using an SDSL operating at 2 Mbps rate?

Answer:Total bits in the video signal = 18 x 106 pictures x 28 x 106 bits / picture = 504 x 1012 bitsTo find the no. of seconds to download this video signal → 504 x 1012 bits / 2 x 106 bits / sec = 252 x 106 secTo find the no. of hours to download this video signal → 252 x 106 sec / 3600 sec / hour = 7 x 104 hours

b. How many hours do you need to upload this video signal using the same system as in part a?

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Answer: Since SDSL is a symmetrical system, download and upload rates are equal. Thus, for the

same amount of information of upload, the same duration as in download will be needed, i.e., 7 x 104 hours

c. Using Dense Wavelength Division Multiplexing (DWDM) system having 100 separate wavelengths and each wavelength modulated at 40 Gbps, how many seconds do you need to download this video signal?

Answer:Rate of DWDM = 100 wavelengths x 40 Gbps / wavelength = 10 2 x 40 x 10 9 bps= 40 x 10 11 bpsTo find the no. of seconds to download this video signal → 504 x 1012 bits / 40 x 10 11

bits/sec = 126 sec

d. How many pictures of this video can you download in 500 seconds by using a 56 kbps modem?

Answer:With 56 kbps modem, in 500 seconds, total 56 x 10 3 bps x 500 sec = 28 x 10 6 bits are downloadedOne picture is composed of 28 x 10 6 bitsThus only 1 picture of this video can be downloaded in 500 seconds by using a 56 kbps

e. Among parts a, b, c and d, which one is an asymmetrical application? Assume that two way connection is available in all the parts.

Answer:None

25. A voice signal has a band of frequencies in between a minimum frequency of 50 Hz and a maximum frequency of 6 kHz. This signal is digitized according to Nyquist rate and a digital signal is obtained at a rate of 108 kbps.

a. Find the number of levels used to represent one sample.

Answer: According to Nyquist, sampling rate is 6 kHz x 2 = 12000 samples / sec Since the rate is 108 kbps, each sample is represented by 108 kbps / 12000 samples / sec = 108000 bits / 12000 samples = 9 bits / sample which corresponds to 29 = 512 levels

b. Repeat part a, if the minimum frequency of this voice signal becomes 1 kHz.

Answer: Since Nyquist rate is based on the maximum frequency, the result found in part a will not

change, i.e., 512 levels will be needed.

c. Repeat part a, if the minimum frequency of this voice signal is also 6 kHz.

Answer: In this case there is only one frequency in the signal which is 6 kHz which is also the

maximum frequency so result found in part a will not change, i.e., 512 levels will be needed.

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d. If a video signal with a band of frequencies in between a minimum frequency of 100 Hz and a maximum frequency of 7 Mhz is added to this audio signal, find the sampling rate required according to Nyquist.

Answer:

When the described video signal is added to this voice signal, the maximum frequency will still be the maximum frequency of the video, i.e., 7 MhzThus, the sampling rate required according to Nyquist = 7 Mhz x 2 = 14 x 106 samples / sec

e. What is the rate of the digital signal obtained if 256 levels are used to digitize the signal in part c?

Answer: 256 levels correspond to 8 bits / sample so the rate of digital signal is 6000 x 2 samples / sec x 8 bits / sample = 96 bits / sec

26. Below modulated signal has 8 bits.

a. Write the type of modulation and the forms of the signals representing digit “0” and digit “1” knowing that both “0” and “1” are sinusoidals.

Answer:

PSK and level “0” is represented by , level “1” is represented by

b. Draw the digital information signal which has the corresponding modulated form given. Clearly indicate the numbers on the time axis and be consistent with the definitions of “0” and “1” in your answer to part a.

Answer:

 1      

           

0                  t(microsec)  1 2 3 4 5 6 7 8

c. What is the carrier frequency of the modulated signal? Answer:

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so Hz or 2 MHz.

d. If you use another type of modulation instead of the given modulation type, write the type of the modulation you use and write the forms of the signals representing digit “0” and digit “1” knowing that both “0” and “1” are sinusoidals.

Answer:

ASK and level “0” is represented by , level “1” is represented by 0.

e. What is the rate of the information signal?

Answer: 1 bit in 1 micro sec, thus the rate of the information signal is 106 bps or 1 Mbps.

27. A modulated bit sequence is shown below where the duration of one bit is 1 microsecond. Both level “0” and “1” are represented by sinusoidal signals, and level “1” has frequency higher than level “0”.

a. Write the modulation type and the bit sequence in “1”s and “0”s.

Answer: Frequency Shift Keying (FSK). 11110001.

b. Write the forms of the signals representing level “1” and level “0”. Answer: Level “1” is , level “0” is .

c. Find the bit rate of this sequence. Answer: 1 bit in 1 microsecond, i.e 106 bits/sec = 1 Mbps.

d. Keeping the same signal representation of level “1”, write the signal representations of level “0” if the other two types of modulations are used. Also indicate which signal representation of level “0” belongs to which modulation type. Answ:If ASK, level “0” is represented by 0. If PSK, level “0” is represented by .

e. What are the bit rates for the other 2 types of modulation?

Answer: For both modulations, bit rate does not change, i.e., for both 1 Mbps.

28. The fade margin in a microwave link is 20 dB. Link distance is 1 mile, free space loss (FSL) is 116.6 dB, receiver sensitivity threshold is – 96.6 dBm. Cable loss of the transmitter cable is 4 dB, cable loss of the receiver cable is 6 dB, transmitter antenna gain is 25 dBi and the receiver antenna gain is equal to the transmitter antenna gain.

a. Find the frequency of operation.

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Answer:FSL = 96.6+20 log D+20 logF = 96.6+20 log 1 +20 logF=96.6 +0+20 logF =116.6 dB

20 log F+96.6 dB = 116.6 dB 20 log F = 20 dB → F = 10 GHz

b. Find the received signal level (RSL). Answer: Fade Margin = RSL – receiver sensitivity threshold 20 dB = RSL – (– 96.6 dBm) RSL = – 96.6 dBm + 20 dB = – 76.6 dBm

c. Find the output power of the transmitter. Answer: RSL = P0 – 4 dB + 25 dBi – 116.6 dB + 25 dBi – 6 dB = – 76.6 dBm P0 = 4 dB – 25 dBi + 116.6 dB – 25 dBi + 6 dB – 76.6 dBm = 0 dBm

d. Keeping all the other parameters in the link the same, if we change the transmitter antenna gain to 15 dBi and the receiver antenna gain to 10 dBi, find whether the link will operate or not. Answer: RSL = 0 dBm – 4 dB + 15 dBi – 116.6 dB + 10 dBi – 6 dB = – 101.6 dBm Receiver sensitivity threshold is – 96.6 dBm. Since RSL < Receiver sensitivity threshold the link will not operate.

e. In a microwave link which does not operate, what changes will you make in the parameters of the link so that the link will become operational? Write 5 changes. Answer: Decrease the frequency Decrease the link distance Increase the output power of the transmitter Increase the antenna gains Decrease cable losses

29. A library has 360 million pages and each page is expressed by 96 thousand bits.

a. If the whole library is downloaded by using a 1 Mbps SDSL, find the number of days needed for this download.

Answer:Total bits = 360 x 106 pages x 96 x 103 bits / page = 360 x 96 x 109 bits To find the no. of seconds to download the whole library → 360 x 96 x 109 bits / 1 x 106 bits / sec = 360 x 96 x 103 secTo find the no. of hours to download the whole library → 360 x 96 x 103 sec / 3600 sec / hour = 96 x 102 hoursTo find the no. of days to download the whole library → 96 x 102 hours / 24 hours / day = 4 x 102 days = 400 days.

b. Find the number of days needed to upload this whole library if the same DSL system given in part a is used?

Answer:

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Since SDSL is a symmetrical system, download and upload rates are equal. Thus, for the same amount of information of upload, the same duration as in download will be needed, i.e., 400 days.

c. If Dense Wavelength Division Multiplexing (DWDM) system is used which has 96 separate wavelengths and each wavelength is modulated at 10 Gbps, how many seconds will be needed to download this whole library?

Answer:Rate of DWDM = 96 wavelengths x 10 Gbps / wavelength = 960 x 10 9 bps= 96 x 10 10 bpsTo find the no. of seconds to download this video signal → 360 x 96 x 109 bits / 96 x 10 10

bits/sec = 36 sec

d. If a 56 kbps modem is used, find the number of pages that can be downloaded in 96 seconds.

Answer:With 56 kbps modem, in 96 seconds, total 56 x 10 3 bps x 96 sec = 56 x 10 3 x 96 bits are downloaded. One page is composed of 96 x 10 3 bitsThus the number of pages that can be downloaded in 96 seconds by using a 56 kbps= 56 x 10 3 x 96 bits / 96 x 10 3 bits = 56 pages.

e. Among the systems given in parts a, part c and d, which one would you prefer to use? Why?

Answer:Depends on what is required. If rate is very important, then the system given in part c will be preferred. If low price is important, then probably the system given in part a or part d will be preferred.

30. a. In each of the following items (i, ii, iii, iv, v) 5 different concepts are given. For each item,

write the concept which is unrelated to the other 4 concepts.

i. Virtual Reality Systems, Trunk, 3-D holography, Remote handling of hazardous materials, Virtual Museums. Answer: Trunk

ii. Wireless, Digital Modulation, ASK, FSK, PSK.

Answer: Wireless

iii. SDSL, HDSL, ISDN BRI, ADSL, ISDN PRI.

Answer: ADSL

iv. UHF, ADSL, VHF, EHF, Ka Band. Answer: ADSL

v. Shielding, UTP, Coaxial Cable, STP, Radio Link. Answer: Radio Link

b. If the number of required voice channels to be transmitted is 64 times the number of voice channels in E-2, what E hierarchy will you need? What is the rate of this E hierarchy?

Answer:

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4 x 4 x 4 = 64, i.e., E-5 will be needed. The rate is 565 Mbps.

c. Write 5 different multiplexer types.

Answer: TDM, FDM, DWDM, Statistical multiplexing, WDM

d. An analog signal is composed of frequencies from 20 kHz to 3 MHz. According to Nyquist, what is the number of samples required in a minute?

Answer: According to Nyquist, sampling rate is 3 MHz x 2 = 6 million samples / sec Thus in a minute we require 6 million samples / sec x 60 sec / minute = 360 million samples / minute.

e. If the digital sequence formed in part d has 60 million bits in two seconds, what is the number of levels to represent the sample?

Answer: 6 million samples / sec, i.e., 12 million samples in 2 seconds 60 million bits / 12 million samples = 5 bits / sample which means 25 = 32 levels

31. a. The rate of digital data is 120 Mbps. This digital data is obtained by converting an analog signal sampled at the rate of 12 million samples per second. With how many bits is one sample represented? How many levels does this correspond to?

Answer: 120 million bits/sec / 12 million samples = 10 bits / sample which means 210 = 1024 levelsb. Based on Nyquist rate of sampling how many samples per second are required to properly

digitize an analog signal which has minimum frequency of 100 Hz and maximum frequency of 15 kHz?

Answer: According to Nyquist, sampling rate is 15 kHz x 2 = 30 thousand samples / sec c. Do the following multiplexers use division of time or frequency?

i. WDM Answer: Frequency division

ii. FDM Answer: Frequency division

iii. Statistical multiplexing Answer: Time division

iv. DWDM Answer: Frequency division

v. TDM Answer: Time division

d. We want to transmit 16 times more than the voice channels in E1. What E hierarchy of multiplexing do we need? Write the rate of this E hierarchy you found.

Answer:

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We need 16 times more, i.e., 4 x 4 more, i.e., we need E3. The rate is 34 Mbps.

e. For the following items (i, ii, iii, iv, v) 4 different concepts are given. Add one more concept which is closely related to the given 4 concepts.

i. EHF, VHF, Ka Band, UHF. Answer: Ku Band

ii. ASK, Digital Modulation, Carrier, FSK.

Answer: PSK.

iii. Copper Wires, UTP, Shielding, STP. Answer: Twisted Pair Cable

iv. Virtual Museums, 3-D holography, Virtual Reality Systems, Remote handling of hazardous materials,

Answer: Machine to machine comunications

v. VDSL, ISDN BRI, SDSL, ISDN PRI. Answer: ADSL

32. In a radio link system, the following parameters are given: Carrier frequency is 10 GHz, receiver signal level (RSL) is –86.6 dBm, fade margin is 10 dB, transmitter antenna gain is 10 dBi, receiver antenna gain is 15 dBi, free space loss (FSL) is 136.6 dB, cable loss of the transmitter cable is 2 dB and cable loss of the receiver cable is 3 dB.

a. What is the link distance?

FSL = 96.6+20 log D+20 logF = 96.6+20 log D +20 logF=96.6 +20 log D +20 log10 =96.6 +20 log D +20 =136.6 dB

20 log D+96.6 dB = 116.6 dB 20 log D = 20 dB → D = 10 miles

b. What is the output power of the transmitter?

Answer: RSL = P0 – 2 dB + 10 dBi – 136.6 dB + 15 dBi – 3 dB = – 86.6 dBm P0 = 2 dB – 10 dBi + 136.6 dB – 15 dBi + 3 dB – 86.6 dBm = 30 dBm

c. Find the maximum receiver sensitivity threshold (Rx) of the receiver that can be used in this link. Answer: Fade Margin = RSL – receiver sensitivity threshold 10 dB = – 86.6 dBm – Rx

Rx = – 86.6 dBm – 10 dB = – 96.6 dBm

d. What happens if you choose a receiver whose receiver sensitivity threshold (Rx) is 10 dBm larger than the Rx found in part c? Answer: In this case RSL will be equal to the receiver sensitivity threshold, i.e., Fade Margin will be 0 which means there is no safety margin for the link availability.

e. If a radio link is designed for zero fade margin. What 3 parameters can you decrease and 2 parameters can you increase in order to have a positive fade margin in this link. Answer: 3 parameters to be decrease are frequency, link distance and cable losses 2 parameters to be increase are the antenna gains and the output power of the transmitter

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33. In the following modulated bit sequence, level “0” is represented by sinusoidal signal of amplitude 1 and level “1” is represented by zero amplitude signal. Duration of 1 bit is 1 nanosecond.

a. What is the the bit rate of this sequence? Answer: 1 bit in 1 nanosecond, i.e 109 bits/sec = 1 Gbps. b. What are the frequencies of level “1” and level “0” signals? Answer: Frequency of level “1” signal is 0, frequency of level “0” is 3 cycles in 1

nanosecond, i.e., 3 x 109 cycles in one second, i.e., 3 GHz.

c. What type of modulation is used in this bit sequence? Answer: Amplitude Shift Keying (ASK).

d. Write the bit sequence in “1”s and “0”s. Answer: 00001110

e. Draw the same sequence in another modulation type. Write the modulation type you used and clearly define your level “1” and level “0” signals.

Answ:

Level “1” is , level “0” is .

34. A picture is represented by 9600 bits and we want to download 180 million such pictures.

a. How many days are needed to download the total 180 million pictures if an ADSL having a downstream rate of 2 Mbps is used?

Answer:Total bits = 180 x 106 pictures x 9600 bits / picture = 180 x 9.6 x 109 bits To find the no. of seconds to download total pictures → 180 x 9.6 x 109 bits / 2 x 106 bits / sec = 90 x 9.6 x 103 secTo find the no. of hours to download total pictures → 90 x 9.6 x 103 sec / 3600 sec / hour = 240 hours

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To find the no. of days to download the whole library → 240 hours / 24 hours / day = 10 days.

b. If the same amount of data in part a is uploaded do we need less, more or the same number of days as found in part a? Why?

Answer: ADSL is an asymmetrical system in which download rates are higher than the upload rates.

Thus, for the same amount of data of upload, we need more number of days as found in part a.

c. The same total 180 million pictures are downloaded using a Dense Wavelength Division Multiplexing (DWDM) system that has 48 separate wavelengths and each wavelength can carry 10 Gbps. Find the number of seconds required to download the total 180 million pictures.

Answer:Rate of DWDM = 48 wavelengths x 10 Gbps / wavelength = 480 x 10 9 bps= 48 x 10 10 bpsTo find the no. of seconds to download this video signal → 180 x 9.6 x 109 bits / 48 x 10 10

bits/sec = 3.6 sec

d. How many times faster is the DWDM given in part c as compared to the ADSL system given in part a?

Answer:Rate of DWDM given in part c = 48 wavelengths x 10 Gbps / wavelength = 480 x 10 9 bpsRate of ADSL given in part a = 2 MbpsSo, DWDM given in part c is 480 x 10 9 bps / 2 Mbps = 240 x 10 3 = 240,000 times faster as compared to the ADSL system given in part a.

e. How many pictures can be downloaded in one minute by a 56 kbps modem?

Answer:With 56 kbps modem, in 1 minute, total 56 x 10 3 bps x 60 secs are downloaded. One picture is composed of 9600 bits Thus the number of pictures that can be downloaded in 1 second by a 56 kbps modem= 56 x 60 x 10 3 bits / 9600 bits = 350 pictures.

35. Below plot shows the digital information signal g(t) that has 5 bits and each bit has 1 nanosecond duration. Indicating which signal denotes level “1” and level “-1” and using the 3

types of carrier signals which are , and ,

a. Plot Amplitude Shift Keying.

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Answer: Level “1” is and level “-1” is 0.

b. Plot Frequency Shift Keying.

Answer: Level “1” is and level “-1” is .

c. Plot Phase Shift Keying.

Answer: Level “1” is and level “-1” is .

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d. Write the carrier frequencies of levels “1” and “-1” in parts (a), (b) and (c).

Answer:

Carrier frequencies of levels “1” and “-1” in part (a) are 3 GHz and 0. Carrier frequencies of levels “1” and “-1” in part (b) are 3 GHz and 4 GHz. Carrier frequencies of levels “1” and “-1” in part (c) are 3 GHz and 3 GHz.

e. Find the rate of the digital information signal g(t).

Answer: 1 bit has 1 nanosecond duration, i.e., 1 bit / 10 -9 sec = 10 9 bits / sec = 1 Gbps

36. In a 10 mile microwave link, 10 dB fade margin is used. In this link, a receiver is used which has a receiver sensitivity threshold, Rx = – 106.6 dBm and the free space loss FSL = 116.6 dB. Cable losses are 1 dB at the transmitter and 2 dB at the receiver. Antenna gains are 20 dBi at the transmitter and 30 dBi at the receiver.

a. At which carrier frequency does this link operate?

Answer: FSL = 96.6+20 log D+20 logF = 96.6+20 log 10 +20 logF=96.6 +20+20 logF =116.6 dB

20 log F+116.6 dB = 116.6 dB 20 log F = 0 dB → F = 1 GHz

b. What is the received signal level (RSL)? Answer: Fade Margin = RSL – receiver sensitivity threshold 10 dB = RSL – (– 106.6 dBm) RSL = – 106.6 dBm + 10 dB = – 96.6 dBm

c. What is the output power of the transmitter, P0 in dBm? Answer: Po - Lctx + Gatx - FSL - Lcrx + Gatx = RSL RSL = P0 – 1 dB + 20 dBi – 116.6 dB + 30 dBi – 2 dB = – 96.6 dBm P0 = 1 dB – 20 dBi + 116.6 dB – 30 dBi + 2 dB – 96.6 dBm = -27 dBm

d. Will this link operate if all the other parameters of the link are kept the same except the free space loss (FSL) becomes 130 dB? Why? Answer: When FSL = 116.6 dB, the link has a fade margin of 10 dB. When FSL = 130 dB, there will bean extra loss of 130 dB – 116.6 bB = 13.4 dB which is larger than the fade margin of 10 dB. Thus, this link will not operate.

e. Write 8 parameters that will effect the operation of a microwave link.

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Answer: Output power of the transmitter, transmitter cable loss, transmitter antenna gain, frequency of operation, link distance, receiver antenna gain, receiver cable loss, receiver sensitivity threshold

37. We want to download a digital image which is composed of 20 thousand bits.

a. An ADSL which has a download rate of 8 Mbps is used. Find the number of images that can be downloaded in one minute.

Answer:8 Mbps = 8 x106 bits/sec There are 60 seconds/minuteSo, 60 seconds in 1 minute x 8 x 106 bits/sec = 480 x 106 bits in 1 minute No. of images that can be downloaded in 1 minute = 480 x 106 bits / 20 x 103 bits= 24000 images

b. A Dense Wavelength Division Multiplexing (DWDM) which has 100 separate wavelengths with each wavelength modulated at 10 Gbps is used. Find the number of images that can be downloaded in one minute.

Answer:For one wavelength, 10 Gbps = 10 10 bits/sec For 100 wavelengths, 100 x 10 10 bits/sec = 10 12 bits/secIn 1 minute, i.e., in 60 sec, 60 x 10 12 bits are downloaded. No. of images that can be downloaded in 1 minute = 60 x 10 12 bits / (20 x 103) bits= 3 x 10 9 images = 3 billion images.

c. A modem which has a download rate of 56 kps is used. Find the number of images that can be downloaded in one minute.

Answer:56 kbps = 56000 bits/sec There are 60 seconds/minuteSo, 60 seconds in 1 minute x 56000 bits/sec = 336 x 104 bits in 1 minute = 3.36 x 106 bits in 1 minute No. of images that can be downloaded in 1 minute = 3.36 x 106 bits / (20 x 103) bits= 168 images

d. How many times faster can you download the same volume of data by using the DWDM in part (b) as compared to the modem in part (c)?

Answer:Rate of the modem in part c is 56 x 103 bps,Rate of the system in part b is 10 12 bps,

So, the same volume of data can be downloaded by using the DWDM in part (b) as compared to the modem in part (c) by

10 12 / (56 x 103)bps ≈ 0,0178571428571429 x 109 times ≈ 17857143 times

e. Is ADSL described in part (a) or DWDM described in part (b) or modem described in part (c) preferable?

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Answer: Depends on what is required. For example, if the requirement is high rate, then part b is preferable.

38. a. An analog video signal is converted to a digital video signal. The data rate of the video signal is 100 Mbps. Find the maximum frequency of this analog video signal if the number of levels used in digitalization is 1024.

Answer: Since the number of levels used in digitalization is 1024, one sample is represented by 10 bits. This means there are 100 Mbps / 10 bits = 10 million samples / sec. Applying the Nyquist rate, the maximum frequency of this analog video signal is 10 million / 2 = 5 MHz.

b. Write 5 types of multiplexing. Answer: TDM, FDM, Statistical multiplexing, WDM, DWDM.

c. If a video channel demands the same rate requirement as 100 telephone voice channels, what level of E-hierarchy would you use to transmit 2 of these video channels? Answer: 2 of these video channels will require the rate equivalent to transmit 2 x 100 = 200 telephone voice. This number of telephone channels can be handled by E-3 which has the capacity to carry 480 telephone voice channels.

d. i) Write 2 techniques used in separating the channels of the communication from the base station to the multi users and the communication from the multi users to the base station in a wireless system. Answer: TDD, FDD

ii) Write 2 techniques used in separating the channels of the multi users in a wireless system. Answer: TDMA, FDMA

iii) Write the technique used in separating the channels of the multi users in a wireless system by assigning different codes to each user. Answer: CDMA

e. Among Unshielded Twisted Pair (UTP), Shielded Twisted Pair (STP) and Coaxial Cable, which one can be used in

i) 1 GHz transmission? Answer: Coaxial Cable ii) Transmission with very small interference? Answer: Shielded Twisted Pair (STP) or Coaxial Cable.

iii) DSL? Answer: Shielded Twisted Pair (STP)

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iv) Transmission of many TV channels? Answer: Coaxial Cable.

v) The local loop to the major extent? Answer: Unshielded Twisted Pair (UTP) or Shielded Twisted Pair (STP)

39. a. What rate of E-hierarchy do you need to transmit 200 standart telephone voice channels? Answer: 200 telephone voice channels can be handled by E-3 which has the capacity to carry 480 telephone voice channels. Rate of E-3 is 34 Mbps.

b. We have Coaxial Cable, Unshielded Twisted Pair (UTP) and Shielded Twisted Pair (STP). Which one of these cables do you prefer to use in

i) ADSL? Answer: Shielded Twisted Pair (STP) ii) Transmission with almost no interference? Answer: Shielded Twisted Pair (STP) or Coaxial Cable. iii) The local loop? Answer: Unshielded Twisted Pair (UTP) or Shielded Twisted Pair (STP) iv) 900 MHz transmission? Answer: Coaxial Cable v) Transmission of cable TV signal? Answer: Coaxial Cable.

c. i) Which type of multiplexing gives the largest data rate? Answer: DWDM ii) In which type of multiplexing, time is divided into slots? Answer: TDM iii) In which type of multiplexing, all the channels are sent at the same time duration? Answer: FDM iv) In which type of multiplexing, the samples from the channels are not necessarily placed in the same channel ordering? Answer: Statistical multiplexing v) How many times more telephone channels can be carried in E-7 as compared to E-5? Answer: 4 x 4 = 16

d. Do the following abbreviations indicate multiplexing or modulation? i) TDM Answer: Multiplexing ii) DWDM Answer: Multiplexing iii) FSK Answer: Modulation iv) FDM Answer: Multiplexing v) PSK Answer: Modulation

e. After an analog audio signal is converted to a digital signal, data rate of 128 kbps is obtained. The number of levels used in digitilization is 1024. What is the maximum frequency of this analog audio signal?

Answer:

Since the number of levels used in digitalization is 1024, one sample is represented by 10 bits. This means there are 128 kbps / 10 bits = 12.8 x 10 3 samples / sec. Applying the

Nyquist rate, the maximum frequency of this analog video signal is 12.8 x 10 3 / 2 = 6.4 kHz.

40. We want to download text pages and each text page consists of 10 thousand bits.

a. How many text pages can be downloaded in one minute if a modem having a download rate of 56 kps is used?

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Answer:56 kbps = 56000 bits/sec There are 60 seconds/minuteSo, 60 seconds in 1 minute x 56000 bits/sec = 336 x 104 bits in 1 minute = 3.36 x 106 bits in 1 minute No. of text pages that can be downloaded in 1 minute = 3.36 x 106 bits / (10 x 103) bits= 336 text pages

b. How many text pages can be downloaded in one minute if an ADSL having download rate of 2 Mbps is used?

Answer:2 Mbps = 2 x106 bits/sec There are 60 seconds/minuteSo, 60 seconds in 1 minute x 2 x 106 bits/sec = 120 x 106 bits in 1 minute No. of text pages that can be downloaded in 1 minute = 120 x 106 bits / 10 x 103 bits= 12000 text pages

c. How many text pages can be downloaded in one minute if a Dense Wavelength Division Multiplexing (DWDM) which has 50 separate wavelengths with each wavelength modulated at 2.5 Gbps is used?

Answer:For one wavelength, 2.5 Gbps = 2.5 x 10 9 bits/sec For 50 wavelengths, 50 x 2.5 x 10 9 bits/sec = 125 x 10 9 bits/secIn 1 minute, i.e., in 60 sec, 60 x 125 x 10 9 bits are downloaded. No. of text pages that can be downloaded in 1 minute = 60 x 125 x 10 9 bits / (10 x 103) bits= 750 x 10 6 text pages = 750 billion text pages.

d. How many times faster can you download the same volume of data by using the DWDM in part (c) as compared to the modem in part (a)?

Answer:Rate of the modem in part a is 56 x 103 bps,Rate of the system in part c is 125 x 10 9 bits/sec bps,

So, the same volume of data can be downloaded by using the DWDM in part (c) as compared to the modem in part (a) by

125 x 10 9 bits/sec / (56 x 103) bps ≈ 2232143 times

e. Do you prefer to use the modem in part (a), ADSL in part (b) or DWDM in part (c)? Why?

Answer:

Depends on what is required. For example, if the requirement is high rate, then part b is preferable.

41. A microwave link (LINK-1) has a link distance of 1 mile and the fade margin is 5 dB. The receiver sensitivity threshold of this link is Rx = – 101.6 dBm and the free space loss is FSL = 116.6 dB. Cable losses are 2 dB at the transmitter and 1 dB at the receiver. Antenna gains are 25 dBi at the transmitter and 15 dBi at the receiver.

a. Find received signal level (RSL).

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Answer:

Fade Margin = RSL – receiver sensitivity threshold 5 dB = RSL – (– 101.6 dBm) RSL = – 101.6 dBm + 5 dB = – 96.6 dBm

b. Find the output power of the transmitter, P0 in dBm.

Answer:

Po - Lctx + Gatx - FSL - Lcrx + Gatx = RSL RSL = P0 – 2 dB + 25 dBi – 116.6 dB + 15 dBi – 1 dB = – 96.6 dBm P0 = 2 dB – 25 dBi + 116.6 dB – 15 dBi + 1 dB – 96.6 dBm = -17 dBm

c. Find the carrier frequency of LINK-1.

Answer:

FSL = 96.6+20 log D+20 logF = 96.6+20 log 1 +20 logF=96.6 +0+20 logF =116.6 dB

20 log F+96.6 dB = 116.6 dB 20 log F = 20 dB → F = 10 GHz

d. A new link (LINK-2) is designed in which all the other parameters of the LINK-1 are kept the same except the carrier frequency of 1 GHz is used. Will this link (LINK-2) operate? Why? Answer: When F = 1 GHz, FSL = 96.6+20 log D+20 logF = 96.6 +0+20 log1 = 96.6 dB i.e., FSL in LINK-2 is 20 dB less compared to LINK-1, i.e., RSL in LINK-2 is 20 dBm more in LINK-2 as compared to LINK-1. Thus LINK-2 will operate.

e. If a link is not operational, write 5 points that you can do in order to make the link operational. Answer: i) Increase the output power of the transmitter, ii) Decrease the transmitter cable loss, iii) Increase the transmitter antenna gain, iv) Decrease the carrier frequency, v) Decrease the link distance.

42. In f (t) which is shown below, there are 5 bits.

The duration of each bit is 1 microsec. You have 3 types of signals which are , and . Using these 3 signals,

a. Plot Frequency Shift Keying. 35

Answer: Level “1” is and level “-1” is .

b. Plot Phase Shift Keying.

Answer: Level “1” is and level “-1” is .

c. Plot Amplitude Shift Keying.

Answer: Level “1” is and level “-1” is 0.

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d. What is the rate of the digital information signal f (t).

Answer: 1 bit has 1 microsecond duration, i.e., 1 bit / 10 -6 sec = 10 6 bits / sec = 1 Mbps

e. What are the carrier frequencies used for the levels “1” and “-1” in parts (a), (b) and (c).

Answer:

Carrier frequencies of levels “1” and “-1” in part (a) are 3 MHz and 2 MHz. Carrier frequencies of levels “1” and “-1” in part (b) are 2 MHz and 2 MHz. Carrier frequencies of levels “1” and “-1” in part (c) are 3 MHz and 0 Hz.

43. In SDH system, mapping of data packets on an STM-1 frame is done by placing the data packets in the STM-1 payload area where three columns of bytes is secured as Path Overhead (POH) in the STM-1 payload area. RSOH, AU Pointer and MSOH occupy their known columns of bytes. Data packets to be transported are composed of 53 bytes each where 5 bytes are overhead and 48 bytes are data. Assuming that the packets can be split between the two consecutive rows in the STM-1 payload area:

a. Find the maximum integer number of packets that an STM-1 frame can carry.

In the STM-1 frame, first 9 columns of bytes are reserved for RSOH, AU Pointer and MSOH, leaving 270-9 = 261 columns of bytes for the payload area.

Three column of bytes are secured for the path overhead (POH) in the payload area leaving 261 – 3 = 258 colums of bytes for the payload area.

One column is 9 bytes. Thus there are total space for 258 x 9 = 2322 bytes in the payload area.

Packet size is fixed and 53 bytes

Since packets can be split between the two consecutive rows in the payload area, the maximum integer number of ATM cells that an STM-1 frame can carry = 2322 / 53 = 43 packets.

b. If each of the packets in 1.a above has 47 bytes data, find the number of data bits (carrying information) in one STM-1 frame.

In one packet, there are 47 bytes of data, i.e., 47 payload bytes.

Thus, in one packet, there are 47 bytes x 8 bits / byte = 376 bits of data in one packet, i.e. 376 payload bits.

In 1.a above 43 packets are found in one STM-1 frame.

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İ.e., there are 376 x 43 = 16,168 bits of data (carrying information) in one STM-1 frame

c. What percent of the total STM-1 capacity is used by the information found in 1.b above.

STM-1 frame carries a total of (headers + information) 9 rows X 270 columns = 2,430 bytes = 2,430 bytes x 8 bits / byte) = 19,440 bits

Thus 16,168 bits / 19,440 bits x 100 = 83.17 % of the total STM-1 capacity is used by the actual information as found in 1.b above.

d. Find the rate of the information found in 1.b above.

One STM-1 frame is transmitted every 0.000125 seconds (1/8000th of a second)

In 1.b above it is found that the information in one STM-1 frame is 16,168 bits

Thus rate of the information found in 1.b above is 16,168 bits in 0.000125 seconds. İ.e 16,168 bits x (1/0.000125 sec) = 129.344 Mbps

e. Find the rate (in bits / sec) of transport of the packets (overhead+data) found in 1.a above.

In one STM-1 frame, 43 packets are found in 1.a above.

43 packets are 43 packets x 53 bytes/packet x 8 bites / byte =18,232 bites in one STM-1 frame.

Thus rate of packets found in 1.a above is 18,232 bits in 0.000125 seconds, i.e., 18,232 bits x (1/0.000125 sec) = 145.856 Mbps

44. A basic block diagram of a fiber optics communication system is given below:

Link values are given in the below table:

LINK ELEMENT VALUELaser output power 10 dBmLaser to optical fiber connector loss 1 dBOptical fiber attenuation 0.3 dB / kmOptical fiber to receiver connector loss 1 dBReceiver Sensitivity -60 dBmLength of the optical fiber 100 km

a. Find the optical power delivered at the optical receiver.

Optical power delivered at the optical receiver

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= 10 dBm - 1 dB - (0.3 dB / km) x 100 km - 1 dB = - 22 dBm

b. Find the power margin in the link design of 2.a above.

Power margin = - 22 dBm - (-60 dBm) = 38 dB

c. We have the same link values as given by the above table, except the length of the optical fiber is changed. Assuming that a power margin of 14 dBm is reasonable in the link design, find the maximum length of the optical fiber that can be used in the link design.

Power margin = 14 dBm = Optical power delivered at the optical receiver - (-60 dBm)

i.e., Optical power delivered at the optical receiver = 14 dBm - 60 dBm = -46 dBm

-46 dBm = 10 dBm - 1 dB - (0.3 dB / km) x (the maximum length of the optical fiber that can be used) - 1 dB

(0.3 dB / km) x (the maximum length of the optical fiber that can be used)= 8 - (- 46 ) = 54 dB

The maximum length of the optical fiber that can be used = 54 dB / 0.3 dB / km = 180 km

d. What happens if the optical fiber used in 2. c above is 300 km?

The link will not function

e. What happens if the optical fiber used in 2. c above is 50 km?

The link will function, however the laser and / or the optical fiber and / or the receiver chosen will have unnecessarily better specifications than required. Thus the link design will be unnecessarily expensive.

45. In each of the below items (i, ii, iii, iv, v) 5 systems are named. For each item, write the name of the system which is unrelated to the other 4 systems.

i. E-1, T-3, STM-4, 140 Mbps/565 MBps Multiplexer, PDH STM-4

ii. MMDS, DWDM, Microwave Systems, LMDS, Satellite Communications DWDM

iii. SDH, FSO System, xDSL, Optical Fiber Communication System, STM-4 XDSL

iv. PSTN, Local Exchange, WLL, PBX, Packet Switched Network Packet Switched Network

v. Backbone Network, xDSL, Access Network, BRI, Local Loop Backbone Network

b. Write five points which you think would be the most important in telecommunication networks in the year 2020. Extremely high rates in the backbone networks and very high rates in the access

networks,

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All optical networks, Almost all digital networks, Networks being able to follow multiprotocol, i.e., one huge network worldwide, Intelligent programmable network: From anywhere in the world, any service or feature

will be accessible irrespective of the connected network provider or network platform.

46. In a city there are 5 local exchanges and one tandem exchange. All the local exchanges are connected to the tandem exchange and the connection of each local exchange to the tandem exchange is made by 4 Mbps trunks. Do not consider the trunk connections to the toll exchange and to the international gateway. a. How many 4 Mbps trunk connections are installed in this city?

1 from each local exhange to the tandem exchange, thus five 4 Mbps trunk connections are installed.

b. If the tandem exchange is not used in 4.a above and all the local exchanges are connected to each other by 2 Mbps trunk connections, how many 2 Mbps trunk connections would be installed in this city?

Four 2 Mbps trunk connections from the first local exhange to the other 4 local exchanges; three 2 Mbps trunk connections from the second local exhange to the other 3 local exchanges; two 2 Mbps trunk connections from the third local exhange to the other 2 local exchanges; one 2 Mbps trunk connection from the fourth local exhange to the last local exchange. Thus 4 + 3 + 2 + 1 = Ten 2 Mbps trunk connections are installed.

c. Do you prefer the network design in 4.a or 4.b above if both designs can handle the traffic properly? Why? 4.a above is preferred because it needs smaller number of connections and smaller distances of installation.

47. a. Write 2 basic similarities and 3 basic differences between Fiber Optic and FSO Systems. Basic similarities:

Both operate at optical carrier wavelengths, Both can handle very high information rates

Basic differences: Fiber Optics is used both in backbone and access networks, FSO is used mainly in the

access networks, Repeaterless link distance in Fiber Optics can be hundred (or even thousand) kilometers

where in FSO repeaterless link distance is maximum 4 – 5 km, Fiber Optics uses transmission medium of optical fiber cable whereas FSO uses

atmospheric transmission.

b. Write 2 basic similarities and 3 basic differences between Microwave and Satellite. Systems. Basic similarities:

Both operate at microwave frequencies, Both are wireless systems

Basic differences: Repeaterless transmission distance in Microwave Systems is much shorter, Satellites can transmit to, and receive from, a large area (foot print or coverage), thus

advantageous in point-to-multipoint and broadcast applications, Propagation delay is quite important in satellite communications, not so important in

microwave.

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c. Write 1 basic similarity and 4 basic differences between the Twisted Pair and the Coaxial Systems. Basic similarity:

Both use cable transmission,

Basic differences: Twisted Pair is not secure, coaxial is relatively more secure, Twisted Pair is usually for slow rate transmission and very high rate transmission can be

realized for very short distances. Coaxial is higher bandwidth for longer distances. Coaxial Systems are usually used mainly in CATV networks whereas Twisted Pairs are

employed in the classical analog local loop, xDSL and LAN applications,

Twisted Pair is usually less expensive than coax and easier to install and reconfigure.

d. Among the Twisted Pair, FSO, Microwave, Satellite and Optical Fiber Communication Systems, which one will fit the best for the following telecommunication applications: i. Heavy telephone traffic from İstanbul to İzmir. Optical Fiber Communication

ii. ADSL. Twisted Pair

iii. Broadcast of Turkish TV channels to Germany. Satellite

iv. Multimedia transmission from Mersin to Konya. Microwave

v. 1 Gbps access network in Ulus, Ankara.

FSO

e. In an optical fiber communication system, you have the choice of using system components of LED or semiconductor laser diode light source; multimode or single mode or graded index optical fiber; 850 nm or 1550 nm or 1310 nm operating wavelength. i. Which system components would you choose if you design a long distance backbone

telecommunication system? Semiconductor laser diode light source, single mode optical fiber, 1550 nm

operating wavelength.

ii. Which system components would you choose if you design a very high data rate backbone telecommunication system? Why?

Semiconductor laser diode light source, single mode optical fiber; 1310 nm operating wavelength.

iii. Explain the reasons of your choices in 5.e.i and 5.e.ii. above.

For choosing the laser, low spectral width and directionality.For choosing the single mode optical fiber, support of high data data needed in backbone.

1550 nm gives the minimum attenuation fiber, 1310 nm gives the minimum dispersion (minimum pulse spreading) fiber.

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iv. Repeat 5.e.i.above for SDH transmission.Same as 5.e.i above

v. Repeat 5.e.ii.above for SDH transmission.

Same as 5.e.ii above

48. Consider a telecommunication system using optical fiber.

a. Write the type of the light source, optical fiber, wavelength of operation you would prefer if it a long distance backbone application. Also explain the reasons of your preferences.

Semiconductor laser diode light source, single mode optical fiber, 1550 nm operating wavelength. For choosing the laser, low spectral width and directionality.For choosing the single mode optical fiber, support of high data data needed in backbone. 1550 nm gives the minimum attenuation in the fiber.

b. Write the type of the light source, optical fiber, wavelength of operation you would prefer if it a very high data rate backbone application. Also explain the reasons of your preferences.

Semiconductor laser diode light source, single mode optical fiber, 1310 nm operating wavelength. For choosing the laser, low spectral width and directionality.For choosing the single mode optical fiber, support of high data data needed in backbone. 1310 nm gives the minimum dispersion (minimum pulse spreading) in the fiber.

49. A simple schematic of an optical fiber link is given as follows:

Link specifications are: Laser output power = 1 dBm, loss per connector = 0.5dB, total loss in the optical fiber = 30 dB, receiver sensitivity threshold = -50 dBm

a. Find the optical power at the input of the optical receiver.

Optical power at the input of the optical receiver

= 1 dBm – 0.5 dB - 30 dB – 0.5 dB = - 30 dBm

b. Considering the result found in part a, can this link operate if only power budget is taken into account ? Explain. Assume that a power margin more than 15 dB is enough for reliable link operation.

Since the receiver sensitivity threshold = -50 dBm

Power margin = - 30 dBm - (-50 dBm) = 20 dB > 15 dB

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Thus there is enough power margin for reliable link operation.

c. Repeat part a if the link distance is doubled. Assume the other parameters are kept the same.

Doubling the link distance means increasing the total loss in the optical fiber by a factor of 2, i.e., total loss in the optical fiber = 2 x 30 = 60 dB

Then, optical power at the input of the optical receiver

= 1 dBm – 0.5 dB - 60 dB – 0.5 dB = - 60 dBm

d. Considering the result found in part c, can the link in part c operate ?

Since the receiver sensitivity threshold = -50 dBm

Power margin = - 60 dBm - (-50 dBm) = -10 dB < 15 dB,

Or, the received power level is even below the receiver sensitivity threshold,

Thus this link can not operate.

50. In the STM-1 frame of an SDH system, RSOH, AU Pointer and MSOH occupy their known columns of bytes and the Path Overhead (POH) occupies 3 columns from column 10 to column 12. Data packets of 576 bytes each are loaded in the remaining part (i.e., the payload area) of the STM-1 frame. 24 bytes of the data packets are overhead and the remaining 552 bytes contain the actual data. Assuming that the packets can be split between the two consecutive rows in the STM-1 payload area,

a. Maximum how many data packets can be transported in one STM-1 frame ? Note: Your answer should be an integer number)

In the STM-1 frame, first 9 columns of bytes are reserved for RSOH, AU Pointer and MSOH, leaving 270-9 = 261 columns of bytes

Three column of bytes are secured for the path overhead (POH) leaving 261 – 3 = 258 columns of bytes for the payload area.

One column is 9 bytes. Thus there are total space for 258 x 9 = 2322 bytes in the payload area.

Packet size is fixed and 576 bytes

Since packets can be split between the two consecutive rows in the payload area, the maximum number of data packets that can be transported in one STM-1 frame = 2322 / 576 = 4 data packets.

b. How many actual data bits exist within the total number of data packets found in part a ?

In one packet, there are 552 bytes of actual data, i.e., 552 payload bytes.

Thus, in one packet, there are 552 bytes x 8 bits / byte = 4416 bits of actual data in one packet, i.e. 4416 payload bits.

In part a, 4 packets are found in one STM-1 frame.

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İ.e., there are total of 4416 x 4 = 17,664 bits of actual data (carrying information) in one STM-1 frame

c. What is the actual data rate ?

One STM-1 frame is transmitted every 0.000125 seconds (1/8000th of a second)

In part b, it is found that the actual data in one STM-1 frame is 17,664 bits

Thus the rate of the actual data found in part b, is 17,664 bits in 0.000125 seconds. İ.e 17,664 bits x (1/0.000125 sec) = 141.312 Mbps

d. Find the percentage of the total STM-1 capacity which is occupied by none actual data.

Total STM-1 capacity is 9 rows X 270 columns = 2,430 bytes = 2,430 bytes x 8 bits / byte) = 19,440 bits

In part c, it is found that the actual data in one STM-1 frame is 17,664 bits,

So, none actual data in one STM-1 frame is 19,440 bits -17,664 bits = 1,776 bits

Thus 1,776 bits / 19,440 bits x 100 = 9.13 % of the total STM-1 capacity is occupied by none actual data.

e. Assuming that the packets can be split between the 2 consecutive frames in the STM-1 payload area, how many frames do you need to load 80 data packets of 53 bytes each.

To load 80 data packets of 53 bytes each, we need 80 x 53 bytes = 4,240 bytes in the payload area

There are total space for 258 x 9 = 2,322 bytes in the payload area in one STM-1 frame.

In two STM-1 frames, we have total of 2,322 bytes x 2 = 4,644 bytes of payload area which is more than 4,240 bytes (which is needed to load 80 data packets of 53 bytes each).

Thus we need 2 frames.

51. a. Write five points which you imagine the telecommunication networks would possess after 50 years from now. Extremely high rates in the backbone networks and very high rates in the access

networks, All optical networks, Almost all digital networks, Networks being able to follow multiprotocol, i.e., one huge network worldwide, Intelligent programmable network: From anywhere in the world, any service or feature

will be accessible irrespective of the connected network provider or network platform.

b. Write five of the most important elements in a telecommunication system.

Frequency of operation, Type of information to be transmitted, Rate of information to be transmitted, Transmission Medium,

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Cost.

52. a. In each of the below items (i, ii, iii, iv, v) 5 systems are named. For each item, write the name of the system which is unrelated to the other 4 systems.

i. Microwave, Radio Link, MMDS, Optical Fiber, LMDS Answer: Optical Fiber

ii. STM-1, E-1, T-3, PDH, 2 Mbps/8 MBps Multiplexer Answer: STM-1

iii. FSO, SDH, LMDS, Fiber, STM-16 Answer: LMDS

iv. PSTN, Circuit Switching, Packet Switching, PBX, E-1 Answer: Packet Switching

v. ISDN, ADSL, Twisted Pair Cable, Coaxial Cable, Local Loop Answer: Coaxial Cable

b. Compare Optical Fiber and FSO Links by indicating 5 points. Solution:

Both operate at optical carrier wavelengths, Both can handle very high information rates, Fiber Optics is used both in backbone and access networks, FSO is used mainly in the

access networks, Repeaterless link distance in Fiber Optics can be hundred (or even thousand) kilometers

where in FSO repeaterless link distance is maximum 4 – 5 km, Fiber Optics uses transmission medium of optical fiber cable whereas FSO uses

atmospheric transmission.

c. Compare Microwave and Satellite by indicating 5 points.

Both operate at microwave frequencies, Both are wireless systems, Repeaterless transmission distance in Microwave Systems is much shorter, Satellites can transmit to, and receive from, a large area (foot print or coverage), thus

advantageous in point-to-multipoint and broadcast applications, Propagation delay is quite important in satellite communications, not so important in

microwave. d. Compare the Twisted Pair and Coaxial Systems by indicating 5 points.

Both use cable transmission, Twisted Pair is not secure, coaxial is relatively more secure, Twisted Pair is usually for slow rate transmission and very high rate transmission can be

realized for very short distances. Coaxial is higher bandwidth for longer distances. Coaxial Systems are usually used mainly in CATV networks whereas Twisted Pairs are

employed in the classical analog local loop, xDSL and LAN applications,

Twisted Pair is usually less expensive than coax and easier to install and reconfigure.

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e. Considering Radio Link, FSO, Twisted Pair, Optical Fiber Communication and Satellite Systems, which can be the best fit for the below indicated applications: i. ISDN. Answer: Twisted Pair

ii. TRT TV broadcast to Holland. Answer: Satellite

iii. 30,000 channels of telephone transmission from Ankara to İstanbul. Answer: Optical Fiber Communication

iv. 2.5 Gbps access network in Kızılay, Ankara. Answer: FSO

v. 140 Mbps transmission from Trabzon to Erzurum. Answer: Microwave

53. Write the best match between the concepts / applications {indicated by item (i) through (v)} with the systems {indicated by items (A) through (E)}.

a. (i) Frequency Licence, (ii) SDH, (iii) Telephone Services, (iv) Cableless Cable TV, (v)

Fiberless Fiber and (A) MMDS, (B) FSO, (C) Microwave, (D) Optical Fiber, (E) WLL.

Answer: (i) and (C), (ii) and (D), (iii) and (E), (iv) and (A), (v) and (B). b. (i) Multi wavelength, (ii) Circuit Switching, (iii) Sampling, (iv) Add / Drop, (v) Transponder

and (A) Satellite, (B) DWDM, (C) PSTN, (D) Digital Communications, (E) SDH.

Answer: (i) and (B), (ii) and (C), (iii) and (D), (iv) and (E), (v) and (A).

54. a. Explain the basic difference between circuit switching and packet switching. Solution: In circuit switching, upon request, telecommunication network establishes a physical

circuit, either on temporary or permanent basis, connecting the caller and the called party. When the communication is over, the circuit is removed. In packet switching, the circuit is virtual, i.e., there is no physical circuit formed but the packets are switched based on certain control information (carried in the headers of the packets such as addressing) to choose the appropriate routes.

b. Explain SVC and PVC.

Solution: Both are virtual circuit types used in packet switched networks. In Switched Virtual Circuits (SVC), virtual circuits are formed by switching of the packets upon request on temporary basis. In Permanent Virtual Circuits (PVC), virtual circuits are formed by switching of the packets on permanent basis.

c. In an FSO telecommunication link design, name 3 factors that effect power budget analysis, and 2 other factors that should be considered in the design which are not in the power budget analysis.

Solution: Link length, laser power, detector’s sensitivity, and dispersion (pulse broadening) analysis, cost.

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d. In an optical fiber telecommunication link design, what type of light source, wavelength and fiber type would you prefer for a link

(i) With distance of 100 km and rate of 200 Mbps? (ii) With distance of 30 km and rate of 10 Gbps?

Solution: (i) For long distance transmission, we prefer semiconductor laser diode light source, 1550 nm wavelength and singlemode step index fiber. (ii) For high rate transmission, we prefer semiconductor laser diode light source, 1310 nm wavelength and singlemode step index fiber.

e. The bit sequence in an E1 frame carrying voice signal is given as: 1001100011010100111001110001000110001000010101010001101001101010010101010011010101101010011111001010110010011000001001100000100010011001001110101110100111000101110001101011110001101110010110011011001101100010010101010010111110111001110010011001110011010010

Write the voice signal sequence in the 27th channel, and the rate of voice signal in this channel.

Solution: 10001000, rate is 64 kbps

55. a. In each of the below items (i, ii, iii, iv, v) five of the telecommunication systems, and/or concepts are mentioned. For each item, write the one which is the most unrelated to the other 4.

i. STM-16, 622 Mbps, SDH, 34 Mbps, 155 Mbps Answer: 34 Mbps

ii. FSO, xDSL, Fiber Optics, STM-1, 10 Gbps Answer: xDSL

iii. Speed of light, frequency, information rate, wavelength, speed of electromagnetic wave. Answer: Information rate

iv. Fiber Optics, FSO, xDSL, LMDS, Radio Link Answer: Radio Link

v. Power budget, Atmospheric attenuation, Multiplexer, Power loss, 1550 nm wavelength Answer: Multiplexer

b. Compare Coaxial and FSO Links by indicating 5 points. Solution:

Coax operates at microwave, FSO at optical frequencies, FSO can handle much higher information rate than Coax, Coax is used both in backbone and LAN networks, FSO is used mainly in the access

networks, Repeaterless link distance for Coax is around 3 km, for FSO 2-5 km, Coax is cable technology, FSO is wireless.

c. Compare Fiber Optics and Satellite by indicating 5 points. Solution:

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FO operates at optical frequencies, Satellite at microwave frequencies, FO is cable system, Satellite is wireless, Repeaterless transmission distance in FO is around 50-100 km in practice, 20,000 km in

lab, for GEO satellites this can be 1000’s of km depending on the application. FO is mostly used in the core network in transmission of dense traffic, Satellites can

transmit to, and receive from, a large area (foot print or coverage), thus advantageous in point-to-multipoint and broadcast applications,

Information rate is much higher in FO than in Satellite. d. Compare the Twisted Pair and Radio Link Systems by indicating 5 points.

Solution: Twisted Pair is cable, Radio Link is wireless, Both are not very secure transmission, Twisted Pair is used usually for slow rate transmission, and very high rate transmission

can be realized only for very short distances. Coax supports much higher information rates for longer distances.

Twisted Pairs are employed in the classical analog local loop, xDSL and LAN applications, whereas Radio Links are used to carry long distance transmission,

Twisted Pair is usually less expensive than Radio Link and easier to install.

e. Which of the rate among 2 Mbps, 2.5 Gbps, 1 Tbps, 155 Mbps, 10 Gbps fits better to the telecommunication system among FSO, DWDM, Fiber optics, ADSL, Satellite?

Solution:2 Mbps, ADSL2.5 Gbps, FSO1 Tbps, DWDM155 Mbps, Satellite10 Gbps, Fiber optics

56. Consider an STM-1 frame in an SDH system. Nine columns of bytes are used for RSOH, AU Pointer and MSOH. Three columns of bytes are used for POH. In the remaining part of the frame (i.e., in the payload area) data packets are placed. Data packet size is 1024 total bytes, 28 of them being the control bytes (i.e., header) and the rest are the bytes carrying the actual information. Assuming the data packets can be split between the two consecutive rows in the STM-1 payload area,

a. What is the maximum integer number of data packets that an STM-1 frame can carry?

There are total space for 258 x 9 = 2322 bytes in the payload area.

Data packets size is fixed and 1024 bytes

Since data packet can be split between the two consecutive rows in the payload area, the maximum integer number of data packet that an STM-1 frame can carry = 2322 / 1024 = 2 data packets.

b. What is the number of bits carrying actual information in the payload area when the data packets are placed as found in Part a?

In one data packet, there are 1024-28= 996 actual information bytes.

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Thus, in one data packet, there are 996 bytes x 8 bits / byte = 7968 actual information bits.

In part a 2 data packets are found in one STM-1 frame.

i.e. there are 7968 x 2 = 15,936 bits of actual information in the payload area (carrying information) in one STM-1 frame

c. What is the rate of the actual information found in Part b?

One STM-1 frame is transmitted at every 0.000125 seconds (1/8000th of a second)

In Part b it is found that the information in one STM-1 frame is 15,936 bits

Thus rate of the information found in Part b is 15,936 bits in 0.000125 seconds, i.e., 15,936 bits x (1/0.000125 sec) = 127.488 Mbps

d. What is the rate of the data packet found in Part a?

In one STM-1 frame, 2 data packets are found in Part a.

2 data packets are

2 data packets x 1024 bytes/ data packet x 8 bits / byte =16,384 data packet bits in one STM-1 frame.

Thus rate of the data packet found in Part a is 16,384 bits in 0.000125 seconds, i.e., 16,384 bits x (1/0.000125 sec) = 131.072 Mbps

e. Find the percentage of actual information bits in the total number of bits in one STM-1 frame.

In one STM-1 frame there are total of 270 x 9 x 8 = 19,440 bits

It is found in part b that there are 15,936 bits of actual information

Thus, the percentage of actual information bits in the total number of bits in one STM-1 frame =15,936 bits /19,440 bits x 100 = % 81.975

57. Among Fiber Optics, FSO, Microwave, FDM, Satellite, Coax, Add / Drop, Twisted Pair, LMDS, MMDS, DWDM, FDD, TDMA, PSK, FDMA, Bluetooth, FSK, TDM, E-hierarchy, SDH, WLAN, Transponder, STM-hierarchy, xDSL, ISDN, CDMA, Sampling, PSTN, Modem, DS-SS, FH-SS, ASK, PCM, TDD,

a. Name 5 of them that are used in TV transmission

Solution: Fiber Optics, Microwave, Satellite, MMDS, Coax.

b. Name 5 of them that are used in the highest rate telecommunication networks

Solution: Fiber Optics, FSO, DWDM, SDH, STM- hierarchy

c. Name 5 of them that are used in the access networks

Solution: Fiber Optics, FSO, LMDS, xDSL, Twisted Pair

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d. Name 5 of them that are related to multiple access wireless telecommunication networks

Solution: TDMA, FDMA, CDMA, WLAN, DS-SS

e. List 10 of them which are related to wired transmission

Solution: Fiber Optics, FDM, Coax, Twisted Pair, DWDM, SDH, STM-hierarchy, xDSL, ISDN, PSTN.

58. Microwave links; M1 operates at L-Band, M2 operates at C-Band and M3 operates at milimeter wave. Transmitter power, atmospheric conditions, receiver sensitivity and all the other system parameters are the same for all these 3 links.

a. Which one of these links would you choose for transmitting the highest information bandwidth? Why?

Answer: Highest carrier frequency has the possibility to carry the highest information bandwidth. Thus M3 will transmit the highest information bandwidth.

b. Which one of these links would you choose for the longest possible link? Why? Answer: Longest possible link is achieved by the lowest frequency link which is M1.c. What is the Received Signal Level in dBm if 10 mile microwave link operating at 10 GHz is

used whose output power is 9 dBm? The antenna gains are 15 dBi each, cabling loss is 1.2 dB each for the transmitter and the receiver.

Answer: Free Space Loss FSL=96.6+20 log D+20 log F=96.6+20 log10+20 log10 = 96.6 + 20 + 20 = 136.6 dB Po - Lctx + Gatx - Lcrx + Gatx - FSL = RSL Received Signal Level, RSL=9 dBm–1.2 dB+15 dBi –136.6 dB –1.2 dB+15 dBi = - 100

dBm RSL = - 100 dBm d. If no fade margin is considered, find the maximum and minimum Receiver Sensitivity

Threshold levels that will operate the link given in part c. Answer: The maximum Receiver Sensitivity Threshold that will operate the link given in

part c is equal to RSL= - 100 dBm The minimum Receiver Sensitivity Threshold that will operate the link given in part c is

equal to RSL= dBm e. For the link given in part c, would you prefer to use a receiver with Receiver Sensitivity

Threshold of –110 dBm or –90 dBm? Why? What are the fade margins for each case? Answer: -110 dBm is preferred because –90 dBm will not operate. Fade margin for Receiver Sensitivity Threshold of –110 dBm is - 100 dBm – (–110 dBm) = 10 dB

No fade margin for Receiver Sensitivity Threshold of –90 dBm

59. a. In each of the below items (i, ii, iii, iv, v) 5 systems are named. For each item, write the name of the system which is unrelated to the other 4 systems.

i. FSO, SDH, ADSL, STM-1, Optical Fiber Communication Answer: ADSL

ii. PSTN, WLL, Circuit Switched Network, 8 Mbps / 34 MBps Multiplexer, Packet Switched Network

Answer: Packet Switched Network

iii. PSTN, 622 Mbps, T-2, 2 Mbps / 8 MBps Multiplexer, E-3

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Answer: 622 Mbps

iv. LMDS, MMDS, Satellite Communication System, DWDM, Microwave Systems Answer: DWDM

v. MMDS, Backbone Network, VDSL, Local Loop, Access Network Answer: Backbone Network

b. In a Fiber Optics Communication system, write

i. The carrier wavelength to be chosen for the longest possible link Answer: 1550 nm

ii. The carrier wavelength to be chosen for the highest rate link Answer: 1310 nm

iii. The types of the light sources to be used for the longest possible link and for the highest rate link

Answer: Semiconductor laser diode for both

iv. The range of the repeater distance in practical backbone network Answer: Around 50-75 km. v. The maximum rate-distance product (in Gbps-km) reached in laboratory work Answer: 1 Gbps x 20,000 km = 20,000 Gbps-km

c. Explain the reason why you have chosen your answers in subitems i, ii, iii of part b.

Answer: 1550 nm fiber yields minimum attenuation, 1310 nm fiber yields minimum dispersion, semiconductor laser diode provides low spectral width and good directionality

d. Write 2 similarities and 3 differences between FSO and Optical Fiber Systems. Answer: Both operate at light frequencies Both are used for high data rate transmission FSO uses atmosphere or free space where optical fiber systems use fibers for

transmission media FSO is used for link lengths of 1.5-5 km where optical fiber systems are used

for extremely long link lengths FSO is used in the access link where optical fiber systems are used mainly in

the backbone, also in the access links e. Write

i. The similarity between the step index multimode optical fiber and the graded index optical fiber

Answer: They both support multimode

ii. The difference between the step index multimode optical fiber and the graded index optical fiber

Answer: The refractive index profile of the core of the step index multimode optical fiber is constant whereas the refractive index profile of the core of the graded index optical fiber is the largest a the core center and it gradually decreases towards the cladding

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iii. The difference between the step index multimode optical fiber and the single mode optical fiber

Answer: The core diameter of the step index multimode optical fiber is much larger (50 m) than the core diameter of the single mode optical fiber (9 m)

iv. The similarity between the step index multimode optical fiber and the single mode optical fiber

Answer: Core refractive index of both the step index multimode optical fiber and the single mode optical fiber is constant

v. The similarity between the singlemode optical fiber and the graded index optical

fiber Answer: Both support small pulse broadening, i.e., low dispersion.

60. A simple schematic of an optical fiber link is given below:

Link specifications are: Laser output power = 10 dBm, loss per connector = 0.25dB, total loss in the optical fiber = 19.5 dB, receiver sensitivity threshold = -40 dBm. Assume that a power margin more than 20 dB is enough for a reliable link operation.

a. Find the optical power at the input of the optical receiver.

Answer: Optical power at the input of the optical receiver

= 10 dBm – 0.25 dB – 19.5 dB – 0.25 dB = - 10 dBm

b. Based on the result found in part a, can this link operate if only power budget is taken into account ? Explain.

Answer: Since the receiver sensitivity threshold = -40 dBm

Power margin = - 10 dBm - (-40 dBm) = 30 dB > 20 dB

Thus there is enough power margin for reliable link operation.

c. Repeat part a if the link distance is increased by three times. Assume the other parameters are kept the same.

Answer: Increasing the link distance by three times means increasing the total loss in the optical fiber by a factor of 3, i.e., total loss in the optical fiber = 3 x 19.5 = 58.5 dB

Then, the received optical power at the input of the optical receiver

= 10 dBm – 0.25 dB - 58.5 dB – 0.25 dB = - 49 dBm

d. Based on the result found in part c, can the link in part c operate ?

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Answer: Since the receiver sensitivity threshold = -40 dBm

The received power level is below the receiver sensitivity threshold,

Thus this link can not operate.

e. Is the analysis, you have made in parts a and b, enough to design the optical fiber link? Explain.

Answer: The analysis made in parts a and b covers only the power budget analysis.

However, to decide on the operation of the link, dispersion (pulse broadening) analysis of the link is also necessary. Dispersion analysis will provide indication about the rate requirements of the link.

61. ATM packets have 5 bytes header (i.e., control bits) and 48 bytes data (i.e., for information), i.e., total 53 bytes. In an STM-1 frame of an SDH system, ATM packets are placed in the payload area of STM-1. Assuming that the payload area of STM-1 is used for ATM transport only and knowing that the ATM packets can be split between the two consecutive rows in the STM-1 payload area;

a. Find the maximum integer number of ATM packets that an STM-1 frame can carry.

Answer: In the STM-1 frame, first 9 columns of bytes are reserved for RSOH, AU Pointer and MSOH, leaving 270-9 = 261 columns of bytes for the payload area.

One column is 9 bytes. Thus there are total space for 261 x 9 = 2349 bytes in the payload area.

ATM cell size is fixed and 53 bytes

Since ATM cells can be split between the two consecutive rows in the payload area, the maximum integer number of ATM cells that an STM-1 frame can carry = 2349 / 53 = 44 ATM cells.

b. Find the total number of ATM data bits (carrying information) in one STM-1 frame.

Answer: In one ATM packet, there are 48 bytes

Thus, in one ATM packet, there are 48 bytes x 8 bits / byte = 384 data bits

In part a, 44 ATM packets are found in one STM frame.

i..e. there are 384 x 44 = 16,896 ATM data bits in one STM-1 frame

c. What is the ratio in percentage of data bits found in part b to the total number of bits in STM-1?

Answer: STM-1 frame carries a total of (headers + data) 9 rows X 270 columns = 2,430 bytes = 2,430 bytes x 8 bits / byte) = 19,440 bits

Thus the ratio in percentage of data bits found in part b to the total number of bits in STM-1 is 16,896 bits / 19,440 bits x 100 = 86.914 %

d. Find the rate of the data found in part b.

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Answer: One STM-1 frame is transmitted every 0.000125 seconds (1/8000th of a second)

In part b, it is found that the data in one STM-1 frame is 16,896 bits

Thus rate of the information found in part b is 16,896 bits in 0.000125 seconds, i.e., 16,896 bits x (1/0.000125 sec) = 135.168 Mbps

e. Write 5 different types of signals that an SDH can transport?

Answer: E-hierarchy signal (2 Mbps, 8 Mbps, 34 Mbps, 140 Mbps, ....etc), T-hierarchy signals, J-hierarchy signals, ATM, Packet switching signals in general (frame relay, X. 25, IP)

62. a. i. Write the name of the system known as the cableless cable TV system. Answer: Multichannel Multipoint Distribution Services (MMDS).ii. Which system uses a lot of carrier frequencies? Answer: Frequency hopping (FH). iii. Write 2 types of duplexing used in LMDS. Answer: FDD, TDD. iv. Which signal differentiates the multi-user channels in CDMA. Answer: Chip sequence.

v. Write 4 types of frequency bands used in satellite communication. Answer: L-Band (390-1550 MHz), C-Band (Uplink around 6 GHz, Downlink around 4 GHz), Ku-Band (Uplink around 14 GHz, Downlink around 11 GHz), Ka-Band (Uplink around 30 GHz, Downlink around 20 GHz) b. Write 3 advantages and 2 disadvantages of optical fiber system when compared to FSO

systems. Answer: Advantages: Very high rate, very long distance, due to the use cable, more reliable Disadvantages: Installation is not easy, sometimes almost impossible

Cable is required which brings extra cost.

c. Write 5 basic differences between LMDS and Optical Fiber Communication systems.

Answer: LMDS uses microwave frequencies, fiber uses optical frequencies LMDS can be used up to 622 Mbps, fiber can be used up to 10 Gbps LMDS is used within 5 km distance, fiber is used within 50-100 km distances LMDS is wireless, fiber is cable system LMDS is an access system, fiber is a backbone system

d. Among Coaxial, Radio Link, Twisted Pair, Optical Fiber and Satellite Systems, which one is better to use for the following telecommunication applications:

i. TV Broadcast from İstanbul to Germany. Answer: Satellite ii. Video transmission within 40 km in a hilly terrain. Answer: Radio Link iii. ADSL Answer: Twisted Pair iv. Transmission of 120,000 telephone channels. Answer: Optical Fiber v. Cable TV distribution

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Answer: Coaxial

e. In designing a telecommunication system, write 10 factors to be considered.

Answer: Rate, distance, cost, frequency, modulation type, geographical conditions, amount of traffic, type of signal to be transmitted, noise level, bit error rate.

63. a. In each of the following items (i, ii, iii, iv, v) 5 different concepts are given. For each item, write the concept which is unrelated to the other 4 concepts.

i. Atmosphere, Fiber, FSO, Wireless, Laser. Answer: Fiberii. E1, SDH, J1, 622 Mbps, Twisted Pair Cable. Answer: Twisted Pair Cable iii. LMDS, TDD, Coaxial Cable, FDD, TDMA. Answer: Coaxial Cable iv. WLL, ADSL, Twisted Pair, Wireless, PSTN. Answer: Twisted Pair Cable

v. 155 Mbps, 2.5 Gbps, Satellite, Microwave, Ku-Band. Answer: 2.5 Gbps

b. Answer the following by “Yes” or “No”.

i. Can STM-1 carry E-5? Answer: Noii. Can E-7 be used in satellite communication? Answer: Noiii. Is TDD used to separate the multi-user channels? Answer: Noiv. Is WLL used as an access system for PSTN?

Answer: Yes v. Can you use FDD to separate the downlink and uplink communication?

Answer: Yes

c. In optical fiber communication systems:

i. Write the wavelength preferred for longest links. Answer: 1.55 μmii. Write the wavelength preferred for highest rate links. Answer: 1.31 μmiii. Write the type of fiber used for longest links. Answer: Single-mode fiberiv. Write the type of fiber used for highest rate links.

Answer: Single-mode fiber v. Write the type of optical source used for highest rate links. Answer: Semiconductor laser

d. Among FSO, optical fiber, satellite, twisted pair, coaxial and radio link communication systems, which one:

i. Supports the highest rate? Answer: Optical fiberii. Provides xDSL connection to end users? Answer: Twisted pair

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iii. Has the shortest repeater distance in the backbone? Answer: Coaxialiv. Has the highest carrier frequency?

Answer: FSO or optical fiber v. Is the best for wide area TV broadcast? Answer: Satellite

e. i. Write 3 types of multiple access systems. Answer: TDMA, FDMA, CDMAii. Write 4 types of SDH hierarchy. Answer: STM-1, STM-4, STM-16, STM-64iii. Write 1 advantage and 1 disadvantage of circuit switching. Answer: Advantage: Guaranteed transmission. Disadvantage: inefficient use of the network.iv. Write 1 advantage and 1 disadvantage of packet switching. Answer: Advantage: Efficient use of the network.

Disadvantage: Transmission is not guaranteed. v. Write 1 similarity and 1 difference between graded index and multimode step-index optical fiber. Answer: Similarity: They both support multimode. Difference: Refractive index profile of the core of multimode step-index is flat whereas

the refractive index profile of the core of graded index is graded.

64. In the STM-1 frame, first 9 columns of bytes are reserved for RSOH, AU Pointer and MSOH header. ATM packets have 5 bytes header (i.e., control bits) and 48 bytes data (i.e., for information), i.e., total 53 bytes. Ethernet packets have 26 bytes header (i.e., control bits) and 51 bytes data (i.e., for information), i.e., total 77 bytes. In an STM-1 frame of an SDH system, ATM and Ethernet packets are placed in turn next to each other in the payload area of STM-1. Assuming that the payload area of STM-1 is used for ATM and Ethernet transport only and knowing that the ATM and/or Ethernet packets can be split between the two consecutive rows in the STM-1 payload area;

a. Find the maximum integer number of ATM packets that an STM-1 frame can carry.

Answer: In the STM-1 frame, first 9 columns of bytes are reserved for RSOH, AU Pointer and MSOH, leaving 270-9 = 261 columns of bytes for the payload area.

One column is 9 bytes. Thus there are total space for 261 x 9 = 2349 bytes in the payload area.

ATM packet size is 53 bytes, Ethernet packet size is 77 bytes. Since ATM and Ethernet packets are placed in turn next to each other in the payload area of STM-1,

The maximum integer number of ATM plus Ethernet packets that an STM-1 frame can carry = 2349 / (53+77) = 18 packets

Thus, the maximum integer number of ATM packets that STM-1 frame can carry is 18.

b. Find the maximum integer number of Ethernet packets that an STM-1 frame can carry.

Answer: It is the same number of ATM packets found in part a, i.e.., 18 packets of Ethernet.

c. Find the total number of Ethernet data bits (carrying information) in one STM-1 frame.

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Answer: In one Ethernet packet, there are 51 data bytes

Thus, in one Ethernet packet, there are 51 bytes x 8 bits / byte = 408 data bits

In part b, 18 Ethernet packets are found in one STM frame.

i..e. there are 408 x 18 = 7,344 Ethernet data bits in one STM-1 frame

d. Find the rate of ATM carried in this STM-1 frame.

Answer: One STM-1 frame is transmitted every 0.000125 seconds (1/8000th of a second)

In part a, it is found that there are 18 ATM packets in one STM-1 frame which is 18 x 53 x 8 = 7,632 ATM bits

Thus the rate of ATM carried in this STM-1 frame is 7,632 bits in 0.000125 seconds, i.e., 7,632 bits x (1/0.000125 sec) = 61.056 Mbps

e. Find the rate of Ethernet data carried in this STM-1 frame. Answer: One STM-1 frame is transmitted every 0.000125 seconds (1/8000th of a second)

In part c, it is found that there are 7,344 Ethernet data bits in one STM-1 frame

Thus the rate of Ethernet data carried in this STM-1 frame is 7,344 bits in 0.000125 seconds, i.e., 7,344 bits x (1/0.000125 sec) = 58.752 Mbps

65. A simple optical fiber link schematic is shown as follows:

Optical power at the input of the optical receiver is -15 dBm. Each connector has a loss of 0.5 dB, Optical fiber has a total loss of 24 dB, receiver sensitivity threshold = -30 dBm.

a. Find the laser output power.

Answer: Optical power at the input of the optical receiver

= Laser output power – 0.5 dB – 24 dB – 0. 5 dB = -15 dBm

Laser output power = 10 dBm

b. Find the power margin left in this link for dependable operation.

Answer: Receiver sensitivity threshold = -30 dBm

Optical power at the input of the optical receiver is -15 dBm

Power margin = - 15 dBm - (-30 dBm) = 15 dB c. Find the minimum laser output power that will still operate this link.

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Answer: Minimum laser output power that will still operate this link should be at such a level that the optical power at the input of the optical receiver = receiver sensitivity threshold

Thus, Minimum laser output power – 0.5 dB – 24 dB – 0. 5 dB = -30 dBm Minimum laser output power = - 5 dBm

d. Do you prefer the design in part a or in part c ? Why?

Answer: Design in part a is preferrable because it has a power margin of 15 dB. However, the design in part c is just at the limit of operation since there is no power margin in the link.

e. The link distance in part a is doubled. If doubling the fiber length will double the fiber loss, what level of receiver sensitivity threshold is needed if the laser output power is 30 dBm and the power margin is 10 dB?

Answer: Optical power at the input of the optical receiver = Laser output power – 0.5 dB – 2x24 dB – 0. 5 dB = 30 dBm –0.5 dB – 48 dB –0.5 dB = -19 dBm

Power margin = Optical power at the input of the optical receiver - Receiver sensitivity threshold Thus 10 db = -19 dBm - Receiver sensitivity threshold i.e., Receiver sensitivity threshold = -29 dBm

66. Ethernet packets having 26 bytes header (i.e., control bits) and 74 bytes data (i.e., for information), i.e., total 100 bytes are placed in the payload of STM-1 frame of an SDH system. Assuming that the payload area of STM-1 is used for ethernet transport only and knowing that the ethernet packets can be split between the two consecutive rows in the STM-1 payload area;

a. Maximum how many integer number of ethernet packets can an STM-1 frame carry?

Answer: In the STM-1 frame, first 9 columns of bytes are reserved for RSOH, AU Pointer and MSOH, leaving 270-9 = 261 columns of bytes for the payload area.

One column is 9 bytes. Thus there are total space for 261 x 9 = 2349 bytes in the payload area.

Ethernet packet size is 100 bytes

Since ethernet packets can be split between the two consecutive rows in the payload area, the maximum integer number of ethernet packets that an STM-1 frame can carry = 2349 / 100, i.e., 23 ethernet packets.

b. How many total ethernet data bits (carrying information) are there in one STM-1 frame?

Answer: In one ethernet packet, there are 74 bytes

Thus, in one ethernet packet, there are 74 bytes x 8 bits / byte = 592 data bits

In part a, 23 ethernet packets are found in one STM-1 frame.

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So there are 592 x 23 = 13,616 ethernet data bits in one STM-1 frame

c. Find the ratio in percentage of data bits found in part b to the total number of bits in STM-1.

Answer: STM-1 frame carries a total of (headers + data) 9 rows X 270 columns = 2,430 bytes = 2,430 bytes x 8 bits / byte) = 19,440 bits

Thus the ratio in percentage of data bits found in part b to the total number of bits in STM-1 is 13,616 bits / 19,440 bits x 100 = 70.04%

d. What is rate of the data found in part b?

Answer: One STM-1 frame is transmitted every 0.000125 seconds (1/8000th of a second)

In part b, it is found that the data in one STM-1 frame is 13,616 bits

Thus the rate of the information found in part b is 13,616 bits in 0.000125 seconds, i.e., 13,616 bits x (1/0.000125 sec) = 108.928 Mbps

e. Which one of the hierarchies (E-1, E-2, E-5, E-6, STM-1) can not be transported by STM-4?

Answer: E-6

67. a. Write the name of the best transmission system which

i. broadcasts many TV channels from Türkiye to Kazakhistan. Answer: Satelliteii. connects the classical telephone subscriber to the telephone exchange. Answer: Twisted pairiii. supports the highest rate in telecommunication backbone systems. Answer: Optical fiber

iv. is widely used in cable TV. Answer: Coaxial

v. provides the highest rate without cable in telecommunication access systems. Answer: FSO

b. Write whether the following statements are correct or wrong: i. The information signal is directly modulated in CDMA. Answer: Wrong ii. 622 Mbps is a rate both in the SDH hierarchy and E- hierarchy. Answer: Wrong iii. In circuit switching, the network is used more efficiently as compared to packet switching. Answer: Wrong iv. MMDS is mainly used for high rate internet. Answer: Wrong

v. LMDS is used for TV transmission. Answer: Wrong

c. i. Which telecommunication system is known as the fiberless fiber system? Answer: Free Space Optics System (FSO).

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ii. In CDMA, for different users, is the same or different chip sequence used? Answer: Different. iii. Which telecommunication system uses more than one carrier frequency? Answer: Frequency hopping (FH).iv. Name the communication device which connects PC to mobile phone. Answer: Bluetooth.

v. Which telecommunication system can replace the twisted pair access telephone network in rural areas?

Answer: Wireless Local Loop (WLL).

d. Write one of the telecommunication system among (DWDM, VDSL, LMDS, FSO, Satellite) that can appropriately transmit the below rate:

i) 155 Mbps Answer: Satelliteii) 622 Mbps Answer: LMDS iii) 12 Tbps Answer: DWDM

iv) 52 Mbps Answer: VDSLv) 1 Gbps Answer: FSO

e. Write 2 similarities and 3 differences between FSO and Radio Link Systems.

FSO Radio Link

Similarity 1 Wireless Wireless

Similarity 2 Needs line of sight Needs line of sight

Difference 1 Optical carrier frequency Microwave carrier frequency

Difference 2 1-2-5 km link 30-50 km repeater distance

Difference 3 Up to 2.5 Gbps (without DWDM) can be transmitted Up to 622 Mbps can be transmitted

68. Free space loss (FSL) in a radio link is 116.6 dB and frequency of operation is 1 GHz.

a. What is the link distance?

Answer:

FSL = 96.6+20 log D+20 logF = 96.6+20 log D +20 log 1 = 96.6 +20 logD+0 =116.6 dB

20 log D +96.6 dB = 116.6 dB 20 log D = 20 dB → D = 10 miles

b. Transmitter output power is 16 dBm and the received signal level is – 71.6 dBm. Cable loss at the transmitter is 2 dB and the cable loss at the receiver is 1 dB. Transmitter and receiver antenna gains are equal. What is the antenna gain?

Answer:

RSL = 16 dBm – 2 dB + G dBi – 116.6 dB + G dBi – 1 dB = – 71.6 dBm 16 dBm + G dBi + G dBi = (– 71.6 +119.6) dBm = 48 dBm

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2G = 48 –16 → G = 16 dBi

c. If the fade margin used is 18.4 dB, what is the receiver sensitivity threshold? Answer: Fade Margin = Received Signal Level – Receiver Sensitivity Threshold

18.4 dB = – 71.6 dBm – Receiver Sensitivity Threshold

Receiver Sensitivity Threshold = – 71.6 dBm – 18.4 dB = – 90 dBm

d. The frequency of the link is made 10 GHz and all the other link parameters are kept the same. Find the new fade margin. Answer:

FSL = 96.6+20 log D+20 logF = 96.6+20 log 10 +20 log 10 = 136.6 dB RSL = 16 dBm – 2 dB + 16 dBi – 136.6 dB + 16 dBi – 1 dB = – 91.6 dBm Fade Margin = Received Signal Level – Receiver Sensitivity Threshold = – 91.6 dBm – (– 90 dBm) < 0 Since Received Signal Level < Receiver Sensitivity Threshold, there is no fade margin.

e. Can the link in part d operate? Why? Answer:

No because Received Signal Level < Receiver Sensitivity Threshold

69. a. In each of the following items (i, ii, iii, iv, v) 5 different concepts are given. For each item, write the concept which is unrelated to the other 4 concepts.

i. UTP, ADSL, STP, FSO, Twisted Pair. Answer: FSOii. FDMA, TDD, Optical Fiber Cable, FDD, TDMA. Answer: Optical Fiber Cable

iii. GEO, Ka-Band, 40 Gbps, 155 Mbps, Ku-Band. Answer: 40 Gbpsiv. Twisted Pair Cable, Radio Link, Local Loop, Fiber, FSO. Answer: Radio Link v. 2 Mbps, 34 Mbps, 140 Mbps, 565 Mbps, 622 Mbps.

Answer: 622 Mbps. b. Write the type of the optical fiber that can be most appropriately used

i. in the longest distance link. Answer: Singlemode optical fiber

ii. with laser diode. Answer: Singlemode optical fiber

iii. for transmission of 10 Gbps. Answer: Singlemode optical fiber

iv. in the highest rate transmission. Answer: Singlemode optical fiber

v. with LED.

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Answer: Multimode mode optical fiber

c. Which transmission system among (Twisted Pair, Radio Link, Optical Fiber, Satellite and FSO) is the best for the below applications?

i. xDSL Answer: Twisted Pairii. TV broadcast from Türkiye to Germany, Holland and Belgium. Answer: Satelliteiii. Transmission of 120000 telephone channels. Answer: Optical Fiberiv. Transmission of 10 Gbps from Europe to United States of America.

Answer: Optical Fiber v. GPS.

Answer: Satellite

d. In an optical fiber communication link, laser output power is 0 dBm, optical power at the input of the optical receiver is -20 dBm, optical fiber has a loss of 0.2 dB / km. If there are no other losses or gains in the link, find the link distance. If this link is designed for 10 Gbps transmission, write 2 points you need to know in order to decide whether this link will operate or not.

Answer: Loss in the fiber = 0 - (-20 dBm) = 20 dBOptical fiber has a loss of 0.2 dB / kmSo the link distance = 20 dB / 0.2 dB / km = 100 km.

2 points we need to know for operation: dispersion analysis, i.e., how much will be the pulse broadening, receiver sensitivity threshold.

e. For the below statements, write the one which describes TDD, FDD, TDMA, FDMA or CDMA.

i. Separates the uplink and downlink channels by using the full frequency bandwidth.

Answer: TDDii. Separates multi users by using the full frequency bandwidth and in a simultaneous manner. Answer: CDMA.iii. Separates multi users without using chip and by using the full frequency bandwidth. Answer: TDMA.iv. Separates the uplink and downlink channels in a simultaneous manner. Answer: FDD.

v. Separates multi users without using chip and and in a simultaneous manner. Answer: FDMA.

70. a. What is the best transmission system which is used for

i. cable TV connection in cities? Answer: Coaxialii. transmitting 40 Gbps for 6000 kilometers? Answer: Optical fiberiii. providing 2.5 Gbps in urban areas? Answer: FSO

iv. local loop connections of telephone subscribers? Answer: Twisted pair

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v. long distance wireless TV broadcast? Answer: Satellite

b. Would you prefer to use circuit switched network, packet switched network or both in terms of i. guaranteed transmission? Answer: Circuit switched networkii. efficient use of the network? Answer: Packet switched network iii. high rate transmission? Answer: Both iv. long distance transmission? Answer: Both

v. high priority transmission? Answer: Circuit switched network

c. Which telecommunication system

i. uses many carrier frequencies? Answer: Frequency hopping (FH)ii. can directly transmit signals at a distance of several ten thousand kilometers? Answer: Satellite iii. is effected by atmospheric turbulence? Answer: Free Space Optics (FSO)iv. can replace PSTN with a wireless extension? Answer: Wireless Local Loop (WLL)

v. uses chip code? Answer: Code Division Multiple Access (CDMA)

d. 5 different concepts are given in each of the following items (i, ii, iii, iv, v). For each item, one concept is unrelated to the other 4 concepts. Write the unrelated concept for each item.

i. Wireless, PSTN, WLL, Rural, Twisted Pair. Answer: Twisted Pair Cableii. Microwave, Ka-Band, X-band, 10 Gbps, Satellite. Answer: 10 Gbpsiii. Coaxial, Atmosphere, Wireless, FSO, Laser. Answer: Coaxialiv. FDMA, FDD, FSO, TDD, TDMA. Answer: FSOv. SDH, Fiber, STM-16, 155 Mbps, FDD.

Answer: FDD

e. Among the rates of 52 Mbps, 155 Mbps, 622 Mbps, 1 Gbps and 12 Tbps, which rate can the below telecommunication system transmit?

i) DWDM Answer: 12 Tbps ii) VDSL Answer: 52 Mbps iii) LMDS Answer: 622 Mbps

iv) FSO Answer: 1 Gbps v) Satellite Answer: 155 Mbps

71. a. i. Write the best type of the light source, optical fiber and the wavelength that should be employed in an optical fiber link for the transmission of 40 Gbps for 30 kilometers.

Answer: Laser, singlemode optical fiber, 1310 nm ii. Write the best type of the light source, optical fiber and the wavelength that should be

employed in an optical fiber link for the transmission of 1 Gbps for 300 kilometers.

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Answer: Laser, singlemode optical fiber, 1550 nm iii. What feature is requested in a light source to achieve high rate transmission? Answer: Narrow spectral width.

iv. What feature is requested in a light source in order to couple the maximum power to the core of optical fiber?

Answer: Narrow beamwidth or high directionalityv. Which type of optical fiber is the worst in terms of dispersion.

Answer: Multimode step index

b. Write whether the statement is correct or wrong. i. FDD separates the uplink and downlink channels by using the full frequency bandwidth.

Answer: Wrongii. TDMA separates multi users by using the full frequency bandwidth and chip code. Answer: Wrongiii. FDMA separates multi users without using chip and by using the full frequency bandwidth. Answer: Wrongiv. CDMA separates multi users by using the full frequency bandwidth where each user has a different chip code. Answer: Correct v. TDD separates the uplink and downlink channels by using different time slots.

Answer: Correct

c. Write 2 types of analyses needed in the design of a link and very briefly explain what you will obtain as the result of these analyses?

Answer: 1. Power budget analysis is made to find the power levels, losses and gains which will enable the link to operate within the requested power specifications.

2. Dispersion analysis is made to find the spreading of the pulse which will enable the link to operate within the requested data rate specifications.

d. Write 2 similarities and 3 differences between FSO and Fiber Optics Systems.

FSO Fiber Optics

Similarity 1 Extremely high rates Extremely high rates

Similarity 2 Optical carrier frequency Optical carrier frequency

Difference 1 Wireless Cable

Difference 2 1-2-5 km link > 50 km links

Difference 3 Easy to install Difficult to install

e. Two Wireless Local Area Networks (WLAN) operating in 2400 - 2483.5 MHz and 5470-

5725 MHz ranges, are assumed to have the same transmitter power, receiver sensitivity and all the other necessary system parameters. Which one would you prefer to use to have wider coverage? Why?

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Answer: Wider coverage is achieved by the lowest frequency range, i.e., in 2400 - 2483.5 MHz, because the other systems parameters, thus the lower frequency signals are attenuated less in the atmosphere.

72. Information packets, each having 26 bytes header (i.e., control bits) and 174 bytes data (i.e., for information), are replaced in STM-1 payload of SDH system. It is known that these packets can be split between the two consecutive rows in the STM-1 payload area and STM-1 payload carries only these same type information packets.

a. Find the maximum integer number of these information packets that an STM-1 frame can transport.

Answer: In the STM-1 frame, first 9 columns of bytes are reserved for RSOH, AU Pointer and MSOH, leaving 270-9 = 261 columns of bytes for the payload area.

One column is 9 bytes. Thus there are total space for 261 x 9 = 2349 bytes in the payload area.

Information packet size is 26 + 174 = 200 bytes

Since ethernet packets can be split between the two consecutive rows in the payload area, the maximum integer number of ethernet packets that an STM-1 frame can carry = 2349 / 200 = 11.7, i.e., 11 information packets.

b. Find the number of total data bits (that carry information) in 10 STM-1 frames.

Answer: In one information packet, there are 174 bytes

Thus, in one information packet, there are 174 bytes x 8 bits / byte = 1392 data bits

In part a, 11 information packets are found in one STM-1 frame.

So there are 1392 data bits / packet x 11 packets / STM-1 x 10 STM-1 = 153,120 information data bits in ten STM-1 frames

c. Find the rate of the data found in part b.

Answer: One STM-1 frame is transmitted every 0.000125 seconds (1/8000th of a second)

In part b, it is found that the data in 10 STM-1 frame is 153,120 bits, i.e., in one STM-1 frame is 15,312 bits

Thus the rate of the information found in part b is 15,312 bits in 0.000125 seconds, i.e., 15,312 bits x (1/0.000125 sec) = 122.496 Mbps

d. What percent of the total number of bits in one STM-1 is composed of data bits as found in part b?

Answer: STM-1 frame carries a total of (headers + data) 9 rows X 270 columns = 2,430 bytes = 2,430 bytes x 8 bits / byte) = 19,440 bitsIn part b, 153,120 information data bits are found in ten STM-1 frames, i.e., 15,312 information data bits in one STM-1 frame

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Thus the ratio in percentage of data bits found in part b to the total number of bits in STM-1 is 15,312 bits / 19,440 bits x 100 = 78.77%

e. Write 3 of E level hierarchies that an STM-1 can transport, and 2 of E level hierarchies that an STM-1 can not transport.

Answer: STM-1 can transport E-1, E-2, E-3, E-4 STM-1 can not transport E-5, E-6, E-7, E-8

73. A basic block diagram of a fiber optics communication system is given below:

The safety margin in the fiber link is 10 dB. Link distance is 100 km. Total loss in the optical fiber is 20 dB, receiver sensitivity threshold is – 30 dBm. Loss of the connector at the transmitter is 0.4 dB, loss of the connector at the receiver is 0.6 dB.

a. Find the minimum received signal level (RSL) that will operate the link. Answer: Safety Margin = RSL – receiver sensitivity threshold 10 dB = RSL – (– 30 dBm) RSL = – 30 dBm + 10 dB = – 20 dBm

b. Find the minimum output power of the laser that will operate the link. Answer: RSL = P0 – 0.4 dB – 20 dB – 0.6 dB = – 30 dBm P0 = 0.4 dB + 20 dB + 0.6 dB – 30 dBm = – 9 dBm

c. Keeping all the other parameters in the link the same, if we change the fiber which now has total loss of 40 dB, find whether the link will operate or not. Answer: RSL = – 9 dBm – 0.4 dB – 40 dB – 0.6 dB = – 50 dBm Receiver sensitivity threshold is – 30 dBm. Since RSL < Receiver sensitivity threshold the link will not operate.

d. Keeping all the other parameters in the link the same, if laser having an output of 1 dBm is used, find the maximum distance of the link so that the link will still operate. Note that the loss of the fiber used is 0.2 dB/km. Answer:

P0 – 0.4 dB – (0.2 dB/km) . x km – 0.6 dB = – 30 dBm 1 dBm– 0.4 dB – (0.2 dB/km) . x km – 0.6 dB = – 30 dBm – (0.2 dB/km) . x in km = – 30 dB 0.2 x = 30 x = 30 / 0.2 x = 30 / 0.2 =150 km

e. In a fiber optics link which does not operate, write 5 changes you need to make in the link parameters so that the link will operate.

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Answer: Use an optical fiber having smaller loss Decrease the link distance Increase the output power of the laser Use lower loss connectors Use a receiver having a smaller sensitivity threshold

74. In the optical fiber link given below:

The connector losses are 0.2 dB each. Laser output power is 0 dBm, receiver sensitivity threshold is -20 dBm. The loss in the fiber is 0.1 dB for each km and the length of the optical fiber is 96 km.

a. What is the optical power at the input of the optical receiver?

Answer: Optical power at the input of the optical receiver = Laser output power – 0.2 dB – 0.1 dB/km x 96 km – 0. 2 dB = 0 dBm – 0.4 db – 9.6 dB = -10 dBm

b. At which power margin, does this link operate?

Answer: Receiver sensitivity threshold = -20 dBm

Optical power at the input of the optical receiver is -10 dBm

Power margin = - 10 dBm - (-20 dBm) = 10 dB c. If you have no power margin, what is the minimum laser output power that will keep this link

operational?

Answer: Minimum laser output power that will keep this link operational should be at such a level that the optical power at the input of the optical receiver = receiver sensitivity threshold

Thus, Minimum laser output power – 0.2 dB – 9.6 dB – 0. 2 dB = -20 dBm Minimum laser output power = - 10 dBm

d. Does the link given in part a or given in part c operate better? Explain.

Answer: Design in part a is preferrable because it has a power margin of 10 dB. However, the link in part c is just at the limit of operation since there is no power margin in the link.

e. In the link given in part a, the laser output power is made 10 dBm and the length of the optical fiber is increased. If the other link parameters are kept the same, find the longest link distance you can have under these conditions.

Answer: Power margin of the link is 10 dB. Since the receiver sensitivity threshold is -20 dBm, the optical power at the input of the optical receiver should again be -10 dBm.

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Thus, the optical power at the input of the optical receiver = Laser output power – 0.2 dB – 0.1 dB/km x D – 0. 2 dB = 10 dBm – 0.4 db – 0.1 dB/km x D = -10 dBm – 0.1 D = -19.6 D = 196 km.

75. An SDH is used to transport signals with STM-1. This STM-1 carries ethernet of total 200 byte packets, out of which 26 bytes are header (i.e., control bits) and 174 bytes are data (i.e., for information) bits. Assuming that there are no POH bytes, the payload area of STM-1 is used for ethernet transport only and the ethernet packets can be split between the two consecutive rows in the STM-1 payload area,

a. Find the integer number of ethernet packets that is carried by this STM-1.

Answer: Since no POH bytes, in the STM-1 frame, first 9 columns of bytes are reserved for RSOH, AU Pointer and MSOH, leaving 270-9 = 261 columns of bytes for the payload area.

One column is 9 bytes. Thus there are total space for 261 x 9 = 2349 bytes in the payload area.

Ethernet packet size is 200 bytes

Since ethernet packets can be split between the two consecutive rows in the payload area, the maximum integer number of ethernet packets that an STM-1 frame can carry = 2349 / 200, i.e., 11 ethernet packets.

b. Find the number of total ethernet data bits (that carry information) in one STM-1 frame.

Answer: In one ethernet packet, there are 174 data bytes

Thus, in one ethernet packet, there are 174 bytes x 8 bits / byte = 1392 data bits

In part a, 11 ethernet packets are found in one STM-1 frame.

So there are 1392 x 11 = 15,312 ethernet data bits in one STM-1 frame

c. What is the percentage of data bits found in part b within the total number of bits carried in this STM-1?

Answer: STM-1 frame carries a total of (headers + data) 9 rows X 270 columns = 2,430 bytes = 2,430 bytes x 8 bits / byte) = 19,440 bits

Thus the ratio in percentage of data bits found in part b to the total number of bits in STM-1 is 15,312 bits / 19,440 bits x 100 = 78.77%

d. Find the rate of the data found in part b?

Answer: One STM-1 frame is transmitted every 0.000125 seconds (1/8000th of a second)

In part b, it is found that the data in one STM-1 frame is 15,312 bits

Thus the rate of the information found in part b is 13,616 bits in 0.000125 seconds, i.e., 15,312 bits x (1/0.000125 sec) = 122.496 Mbps

e. We have the choice of using STM-1, STM-4, STM-16 and STM-64. Which one would you choose if we need to transport

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i. E3 and 50 Mbps IP? Answer: STM-1ii. 300 Mbps IP and 100 Mbps ethernet? Answer: STM-4 iii. 622 Mbps ATM and 1 Gbps IP? Answer: STM-16 iv. E6 and 2.5 Gbps ATM? Answer: STM-64

v. Six different ethernets each at 10 Mbps? Answer: STM-1

76. a. Among the single mode optical fiber, step index multimode optical fiber and graded index optical fibers, which one

i. gives the lowest pulse broadening (i.e., lowest dispersion)?

Answer: Single mode optical fiberii. has the smallest core diameter? Answer: Single mode optical fiberiii. carries the total optical power with one mode? Answer: Single mode optical fiber. iv. has the same refractive index at the edge of the core and in the cladding? Answer: Graded index optical fiber.

v. has varying refractive index in the core? Answer: Graded index optical fiber.

b. You have gas laser, semiconductor laser, LED and a fluorescent lamp. Which one do you choose

i. to couple maximum optical power to the core of optical fiber?

Answer: Semiconductor laserii. for smallest spectral width? Answer: Semiconductor laseriii. for best directionality? Answer: Semiconductor laseriv. for highest data rate optical fiber communication? Answer: Semiconductor laser

v. for longest link optical fiber communication? Answer: Semiconductor laser

c. Write 2 differences and 3 similarities between LMDS and FSO systems.

Differences: 1. 2.Similarities: 1. 2. 3.

Answer: LMDS uses microwave frequencies, FSO uses optical frequencies. LMDS can be used up to 622 Mbps, FSO can be at 10 Gbps, even more. LMDS and FSO are used within 5 km distances. LMDS and FSO are wireless. LMDS and FSO are access systems.

d. For each of the following items, write whether the item is related to fiber optics, satellite, twisted pair, FSO, coaxial or radio link communication system.

i. GEO. Answer: satellite

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ii. 1 Tbps. Answer: fiber opticsiii. Wavelength of 1310 nm. Answer: fiber opticsiv. ADSL. Answer: twisted pair

v. Cable TV. Answer: coaxial

e. Among FDD, TDMA, CDMA, TDD and FDMA, which one is related to the following item?

i. Allocating different frequency bands to each user. Answer: FDMAii. Allocating different time slot to the base station. Answer: TDDiii. Multiplying the users signal by the chip code. Answer: CDMAiv. Allocating different time slot to each user. Answer: TDMA

v. Allocating different frequency band to the base station Answer: FDD

77. Write the relevant system for each of the following items:

a. i. System that provides PSTN facilities in a wireless environment (i.e., without twisted pair infrastructure). Answer: WLL

ii. Local Area Network that does not use cable connection. Answer: WLAN. iii. System that has multiple carrier frequencies. Answer: Frequency Hopping. iv. Digital communication using twisted pair access network. Answer: DSL.

v. Television broadcast system using wireless transmission at 2 frequencies, 30 GHz and 20 GHz. Answer: Satellite.

b. Write 5 communication systems that uses optics.

Answer: Fiber optics, FSO, DWDM, SDH, Infrared WLAN

c. Which transmission system is the best for the below telecommunication applications:

i. Transmission of video signals to 30 km distance in a mountainous geography. Answer: Radio Link ii. Transmission of STM-64. Answer: Optical Fiber iii. Telephone line modem. Answer: Twisted Pair iv. TV Broadcast from Türkiye to Holland. Answer: Satellite v. Wireless terrestrial TV distribution. Answer: Microwave

d. Write 5 wireless and 5 cable systems.

5 Wireless systems:

5 Cable systems:

Answer: Wireless systems: MMDS, Radio Link, Satellite, WLAN, LMDS, Cable Systems: Fiber optics, CATV, SDH, DSL, DWDM

e. Write 5 advantages of wireless and 5 advantages of cable systems. Advantages of wireless systems:

Advantages of cable systems:

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Answer: Advantages of wireless systems: No cable installation, easier coverage, easier broadcast, mobility, multi user network. Advantages of cable systems: Controlled environment, higher data rates, lower attenuation, no frequency allocation, better BER.

78. The first 12 columns of bytes in an STM-1 frame are reserved for RSOH, AU Pointer, MSOH and POH control bits. An IP packet composed of 576 bytes will be put in the payload of this STM-1. Assume that the payload of this STM-1 is used for IP transport only and the IP packets in the STM-1 payload area can be split between the two consecutive rows and there is no empty space in the payload, i.e., all the payload is filled with IP.

a. What is the maximum number of IP packets that can be put in one of this STM-1 frame?

Answer: In the STM-1 frame, first 12 columns of bytes are reserved for RSOH, AU Pointer, MSOH and POH control bits, leaving 270-12 = 258 columns of bytes for the payload area.

One column is 9 bytes. Thus there are total space for 258 x 9 = 2322 bytes in the payload area.

IP packet size is 576 bytes. The maximum integer number of IP packets that can be put in one of this STM-1 frame is 2322 / 576 = 4.03

b. What is the rate of IP carried in this STM-1 frame?

Answer: One STM-1 frame is transmitted every 0.000125 seconds (1/8000th of a second)

In part a, it is found that there are 2322 bytes of IP in one STM-1 frame= 2322 bytes x 8 bits / byte = 18,576 bits IP

Thus the rate of IP carried in this STM-1 frame is 18,576 bits in 0.000125 seconds, i.e., 18,576 bits x (1/0.000125 sec) = 148.608 Mbps

c. How many bits of IP can be placed in 10 STM-1 frames?

Answer: In one STM-1 frame, there are 2322 bytes x 8 bits / byte = 18,576 bits IPSo in 10 STM-1 frame, there are 18,576 bits x 10 = 185,760 bits IP

d. What is the rate of IP in part c?

Answer: The same as in part b, i.e., 148.608 Mbps

e. Maximum how many E-4 levels can be carried by STM-4?

Answer: Rate of STM-4 is 622 Mbps, rate of E-3 is 140 Mbps. Thus, 622 / 140 = 4 E-4 levels can be carried by STM-4.

79. In the below shown optical fiber link, laser output power = Po = 10 mWatt, the received power = Pr = 1 mWatt. The loss of each connector is 0.5 dB. The length of the optical fiber is 90 km. The margin used in the link design is 2 dB.

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a. Find Po and Pr in dBm.

Answer: Po = 10 log (10 mWatt / 1 mWatt) = 10 log (10) = 10 dBm Pr = 10 log (1 mWatt / 1 mWatt) = 10 log (1) = 0 dBm

b. What is the loss of the optical fiber ( Lf ) in dB?

Answer: Pr = Po - 0. 5 dB - Lf - 0.5 dB 0 dBm = 10 dBm - 1 dB - Lf Lf = 10 dBm - 1 dB = 9 dB

c. Find the maximum sensitivity threshold of this optical receiver so that this optical fiber link will operate.

Answer: Maximum sensitivity threshold = Pr – link margin = Pr – 2 dB = 0 dBm – 2 dB

= – 2 dBm

d. What will be the the loss of the optical fiber ( Lf ) if the length of the optical fiber is doubled, i.e., increased to 180 km?

Answer: In part b, Lf is found to be 9 dB which is for 90 km, the loss per km is 9 dB / 90 km = 0.1 dB/km.

If the length of the optical fiber is doubled, i.e., increased to 180 km, Lf = 0.1 dB/km x 180 km = 18 dB

e. A new optical fiber link will be designed with the fiber length given in part d and with a new laser. The other parameters of the link are kept the same. Find Po.

Answer: Pr = 0 dB = Po - 0. 5 dB - Lf - 0.5 dB = Po – 1 dB – 18 dB 0 dB = Po – 1 dB – 18 dB Po = 19 dBm

80. a. Among CDMA, TDMA, FDD, FDMA and TDD, which one is used to separate the i. uplink and downlink channels by utilizing the full time. Answer: FDD

ii. uplink and downlink channels by utilizing the full frequency bandwidth. Answer: TDD iii. multi users by utilizing the full frequency bandwidth and without using chip code. Answer: TDMA iv. multi users channels with the help of a chip code. Ans: CDMA

v. multi users by utilizing the full time and without using chip code. Answer: FDMA

b. In each of the following items (i, ii, iii, iv, v), 5 different concepts are given. For each

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item, write the concept which is unrelated to the other 4 concepts.

i. Satellite, Radio Link, TDD, Twisted Pair Cable, Optical Fiber. Answer: TDDii. 140 Mbps, STM-16, 155 Mbps, 622 Mbps, STM-64. Answer: 140 Mbps.

iii. Ka-Band, GEO, LEO, LMDS, Ku-Band. Answer: LMDS iv. FDMA, Optical Fiber, CDMA, Multiple Access. Answer: Optical Fiber

v. Twisted Pair, Optical Fiber, ADSL, VDSL, STP. Answer: Optical Fiber

c. Among the fiber and laser wavelength combinations of (singlemode fiber at 1550 nm), (graded index fiber at 1550 nm), (singlemode fiber at 1310 nm), (graded index fiber at 1310 nm) and (multimode step index fiber at 1310 nm), which one fits the best for the following applications?

i. The longest distance communication link. Answer: (singlemode fiber at 1550 nm) ii. The highest rate transmission. Answer: (singlemode fiber at 1310 nm)

iii. Minimum attenuation link. Answer: (singlemode fiber at 1550 nm) iv. Minimum dispersion link. Answer: (singlemode fiber at 1310 nm)

v. Transmission with the smallest core diameter and the smallest frequency. Answer: (singlemode fiber at 1550 nm)

d. For the following applications, will FSO, Radio Link, Satellite, Twisted Pair or Optical Fiber be preferred the most?

i. VSAT. Answer: Satellite ii. New installation for 2.5 Gbps data transmission in a populated urban area. Answer: FSOiii. 1 Tbps transmission. Answer: Optical Fiber

iv. VDSL. Answer: Twisted v. TV broadcast from Türkiye to Europe. Answer: Satellite

e. Each of the following items describes one system. Write the system which

i. uses multi carrier frequencies. Answer: Frequency hopping (FH).ii. multiplies the information signal by a code which is different for each multi user. Answer: CDMA.

iii. is used for telephone networks in the areas where there is no PSTN infrastructure. Answer: Wireless Local Loop (WLL). iv. is used at infrared frequencies for point to point communication.

Answer: Free Space Optics System (FSO). v. is used to broadcast many TV channels where there is no cable infrastructure. Answer: MMDS.

81. a. Among FSO, Microwave, FDM, Coax, Add / Drop, LMDS, MMDS, DWDM, FDD, TDMA, PSK, FSK, E-hierarchy, SDH, WLAN, STM-64, xDSL, ISDN, DS-SS, FH-SS, ASK, TDD, Fiber Optics, Satellite, Twisted Pair, CDMA, Sampling, TDM, FDMA, write 5 of them which are used in

i. TV transmission. Answer: Fiber Optics, Microwave, Satellite, MMDS, Coax.

ii. the highest rate telecommunication networks. Answer: Fiber Optics, FSO, DWDM, SDH, STM- 64

iii. the access networks. Solution: Fiber Optics, FSO, LMDS, xDSL, Twisted Pair

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iv. the multiple access wireless telecommunication networks.

Answer: TDMA, FDMA, CDMA, WLAN, DS-SS

v. the backbone network. Answer: Fiber Optics, DWDM, SDH, STM-64, Satellite

b. You have the telecommunication systems of Fiber Optics, ADSL, FSO, DWDM, Satellite. Which one of these systems will fit the best for the transmission of the

following data rates?

i. 10 Gbps. Ans: Fiber Opticsii. 8 Mbps. Ans: ADSLiii. 2.5 Gbps. Ans: FSOiv. 2 Tbps. Ans: DWDMv. 155 Mbps. Ans: Satellite

c. i. What is the most important advantage of circuit switched networks when compared to

packet switched networks? Answer: We are assured of the quality

ii. What is the most important disadvantage of circuit switched networks when compared

to packet switched networks? Answer: The network is used inefficiently

iii. Is IP connection oriented? Answer: No.

iv. Does the circuit switched network provide virtual circuits? Answer: No.

v. Can you send video by using packet switched network? Answer: Yes.

d. Write the type of satellite which

i. is mostly used in telecommunications. Answer: GEO ii. has equatorial orbit. Answer: GEO iii. has the longest distance from the Earth. Answer: GEO.

iv. always sees the same portion of the Earth. Answer: GEO.

v. is mostly used in remote sensing. Answer: LEO.

e. Write one similarity for each of the following three systems.

i. Fiber Optics, FSO, IR-WLAN. Answer: Optical systems ii. Radio Link, FSO, WLL. Answer: Wireless systems iii. Fiber Optics, Twisted Pair, Coaxial. Answer: Cable systems

iv. MMDS, LMDS, WLAN. Answer: Wireless systems

v. Satellite, Fiber Optics, Radio Link. Answer: Long distance systems

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82. a. Which system

i. utilizes the chip code? Answer: CDMA. ii. provides cable TV type application but without using cable? Answer: MMDS.

iii. uses optical wireless transmission? Answer: Free Space Optics System (FSO).

iv. employs many carrier frequencies? Answer: Frequency hopping (FH). v. extends the classical telephone infrastructure to rural areas with wireless connection? Answer: Wireless Local Loop (WLL).

b. Is FDD, FDMA, CDMA, TDMA or TDD used for the separation of the

i. uplink and downlink channels by allocating different time frames? Answer: TDD ii. multi users by allocating different bandwidths? Answer: FDMAiii. uplink and downlink channels by allocating different bandwidths? Ans: FDD iv. multi users by different time frames? Answer: TDMA

v. multi user channels without allocating different bandwidths and different time frames? Ans: CDMA

c. Will you choose to use Satellite, Twisted Pair Cable, FSO, Optical Fiber or Radio

Link for

i. TV broadcast from Türkiye to Kazakistan? Answer: Satelliteii. 40 Gbps transmission? Answer: Optical Fiber iii. 2.5 Gbps wireless connection? Answer: FSO

iv VSAT? Answer: Satellite v. ADSL? Answer: Twisted Pair Cable

d. We have graded index fiber at 1550 nm, singlemode fiber at 1550 nm, graded index fiber at 1310 nm, multimode step index fiber at 1310 nm and singlemode fiber at 1310 nm. Which one of these will you choose for

i. achieving minimum signal attenuation? Answer: singlemode fiber at 1550 nmii. achieving minimum pulse broadening? Answer: singlemode fiber at 1310 nm

iii. transmitting the largest data rate? Answer: singlemode fiber at 1310 nm iv. sending the signal to the longest distance? Ans: singlemode fiber at 1550 nm

v. sending the signal with the smallest frequency and the smallest core diameter? Answer: singlemode fiber at 1550 nm e. Five different concepts are given in each of the following items (i, ii, iii, iv, v). For each item, which concept is unrelated to the other 4 concepts?

i. FDD, GEO, Ka-Band, LEO, VSAT. Answer: FDDii. Satellite, Multiple Access, FDMA, CDMA, TDMA. Answer: Satellite

iii. GEO, UTP, STP, ADSL, VDSL. Answer: GEO iv. Optical Fiber, FSO, Satellite, WLL, Radio Link. Answer: Optical Fiber v. 34 Mbps, 155 Mbps, STM-64, STM-16, 622 Mbps. Answer: 34Mbps.

83. a. Which type of satellite

i. is located more than 36000 km away from the Earth? Answer: GEO. ii. can be in the atmosphere? Answer: LEO.

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iii. completes its one orbital rotation in one day? Answer: GEO

iv. used in remote sensing? Answer: LEO or MEO.

v. is used in TV broadcast? Answer: GEO

b. What is the similarity among

i. Radio Link, Satellite, Fiber Optics? Answer: Long distance systems ii. WLAN, LMDS, MMDS? Answer: Wireless systems iii. WLL, FSO, Radio Link? Answer: Wireless systems

iv. IR-WLAN, Fiber Optics, FSO? Answer: Optical systems

v. Fiber Optics, Coaxial, Twisted Pair? Answer: Cable systems

c. Among LMDS, CDMA, FSO, Fiber Optics and Satellite, which one fits the best for the application of

i. very high rate data transmission in the backbone network? Answer: Fiber Optics

ii. very high rate data transmission in the access network? Answer: FSO

iii. multiple access wireless telecommunication networks? Answer: CDMA

iv. TV broadcast? Answer: Satellite

v. microwave wireless access network? Answer: LMDS

d. For each of the following items (i, ii, iii, iv, v), write whether the indicated network is circuit swithed or packet switched?

i. Video transmission by using a continuous bit sequence. Answer: Circuit switched ii. Classical telephone voice (PSTN). Answer: Circuit switched iii. IP. Answer: Packet switched iv. Ethernet. Answer: Packet switched v. Video transmission by splitting the continuous bit sequence into smaller segments. Answer: Packet switched. e. Among the data rates of 8 Mbps, 155 Mbps, 2.5 Gbps, 10 Gbps, 2 Tbps, which one

is the best choice for the telecommunication system of

i. FSO? Ans: 2.5 Gbpsii. DWDM? Ans: 2 Tbps iii. Satellite? Ans: 155 Mbps iv. Fiber Optics? Ans: 10 Gbps v. ADSL? Ans: 8 Mbps

84. In the payload of an STM-1 frame, only IP packets of 680 bytes each are carried.

Assume that the IP packets in the STM-1 payload area can be split between the two

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consecutive rows and there is no empty space in the payload. The first 12 columns of bytes in this STM-1 frame involve RSOH, AU Pointer, MSOH and POH control bits.

a. Maximum how many IP packets can be placed in the payload of one such STM-1 frame?

Answer: In the STM-1 frame, first 12 columns of bytes are reserved for RSOH, AU Pointer, MSOH and POH control bits, leaving 270-12 = 258 columns of bytes for the payload area.

One column is 9 bytes. Thus there are total space for 258 x 9 = 2322 bytes in the payload area.

IP packet size is 680 bytes. The maximum number of IP packets that can be put in one of this STM-1 frame is 2322 / 680 = 3.4147

b. Find the rate of this IP carried in this STM-1 frame.

Answer: One STM-1 frame is transmitted every 0.000125 seconds (1/8000th of a second)

In part a, it is found that there are 2322 bytes of IP in one STM-1 frame= 2322 bytes x 8 bits / byte = 18576 bits IP

Thus the rate of IP carried in this STM-1 frame is 18,576 bits in 0.000125 seconds, i.e., 18,576 bits x (1/0.000125 sec) = 148.608 Mbps

c. Find the number of IP bits that can be carried in this STM-1 in 125 mseconds.

Answer: In part b, it is found that 18576 bits IP are carried in one STM-1 frame, i.e., in 125 sec. Thus in 125 mseconds, the number of IP bits that can be carried = (125 msec / 125 sec) x 18576 bits = 18576000 bits

d. How many STM-1 frames are needed to carry total (i.e., RSOH + AU Pointer + MSOH + POH control bits + STM-1 payload) 243000 bytes in STM-1?

Answer: In one STM-1 frame, there are total 270 x 9 = 2430 bytes. Thus to carry total 243000 bytes, 243000 / 2430 = 100 STM-1 frames.

e. Write four data rates used in STM hierarchies and indicate which one of these data rates is also used in E-hierarchy.

Answer: 155 Mbps, 622 Mbps, 2.5 Gbps, 10 Gbps. 2.5 Gbps or 10 Gbps rate is also used in E-hierarchy.

85. An optical fiber link is given below:

The length of the optical fiber is 96 km, the output power of the laser = Po = 0 dBm, the power received = Pr = -10 dBm, the margin used in the link design is 3 dB, the loss of

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each connector is 0.2 dB.

a. Find Po and Pr in mWatt.

Answer: Po = 10 log (x / 1 mWatt) = 0 dBm log (x / 1 mWatt) = 0 x / 1 mWatt = 1 x = 1 mWatt Pr = 10 log (y / 1 mWatt) = -10 dBm log (y / 1 mWatt) = -1 y / 1 mWatt = 0.1 y = 0.1 mWatt = 100 Watt

b. Find the optical fiber loss ( Lf ) in dB.

Answer: Pr = Po - 0. 2 dB - Lf - 0.2 dB -10 dBm = 0 dBm – 0.4 dB - Lf Lf = 10 dBm – 0.4 dB = 9.6 dB

c. If we want this link to operate, what should be the maximum sensitivity threshold of this optical detector in dBm?

Ans: Maximum sensitivity threshold = Pr – link margin = Pr – 3 dB = -10 dBm – 3 dB= – 13 dBm

d. Using the same fiber type and keeping the same link parameters, if only the fiber length is increased to 150 km, find the loss of the optical fiber, Lf.

Answer: In part b, Lf is found to be 9.6 dB which is for 96 km, i.e., the loss in the Fiber per km is 9.6 dB / 96 km = 0.1 dB/km.

If the length of the optical fiber becomes 150 km, Lf = 0.1 dB/km x 150 km = 15 dB

e. If the maximum sensitivity threshold of the optical detector is -14 dBm, find whether the optical fiber link given in part d will be operational or not.

Answer: From part d, Lf =15 dB, Pr = Po - 0. 2 dB - Lf - 0.2 dB = 0 dBm – 0.4 dB - L f = 0 dBm –0.4 dB –15 dB = -15.4 dBm Pr – link margin = -15.4 dBm – 3 dB = -18.4 dBmThis is smaller than the maximum sensitivity threshold of the optical detector which is -14 dBm. Since, Pr - maximum sensitivity threshold = 15.4 dBm -14 dBm =1.4 dB < 3 dB = MarginThe link will not be operational.

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