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

1. The digital signal X(t) given below. X(t) 1 0 1 2 3 4 5 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. 1

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 25 levels, find the number of bits needed to transmit one sample. 25 levels mean that each sample is represented by 8 bits since 25 = 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 = 4 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 5 Kbps modem is used (assume full rate can be utilized)? 5 Kbps = 5000 bits / sec = ( 5000 bits / sec ) / (8 bits / character) = 7000 characters / sec No. of characters that can be downloaded in a minute = ( 7000 characters / sec ) x ( 0 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) 2

= 1.25 x 10 9 characters / sec No. of characters that can be downloaded in a minute For 15000 wavelengths, = (1.25 x 10 9 characters / sec ) x ( 0 sec / minute) = 7.5 x 10 10 characters / minute 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 / minute c. 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 bits / sec = (52 x 10 bits / sec) / (8 bits / character) =.5 x 10 characters / sec No. of characters that can be downloaded in a minute = (.5 x 10 characters / sec ) x ( 0 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. 5 Kbps modem is the slowest, VDSL is 3.9 x 10 8 / 4.2 x 10 5 905 times faster than 5 Kbps modem, DWDM in 3.b is 1.125 x 10 15 / 4.2 x 10 5 278 billion times faster than 5 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 = 9.+20 log D+20 logf = 9.+20 log10+20 log10 = 9.+20+20 = 13. db P o - L c t x + G a t x - L c r x + G a t x - FSL = RSL RSL = 10 dbm - 5 db + 28.3 dbi - 5 db + 28.3 dbi - 13. 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 3

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. FSL = 9.+20 log D+20 logf = 9.+20 log10+20 log1 = 9.+20+0 = 11. db P o - L c t x + G a t x - L c r x + G a t x - FSL = RSL RSL = 10 dbm - 5 db + 28.3 dbi - 5 db + 28.3 dbi - 11. db = - 0 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 = - 0 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. 25 levels is used to represent one sample. How many bits are required to transmit one sample value? 4

25 levels mean that each sample is represented by 8 bits since 25 = 2 8 The number of bits needed to transmit one sample = 8 bits 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 = 4 Kbps d. Would you allocate a 25 kbps channel to transmit this signal? Why? No because more than 4 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 4 kbps, the original signal will only be recovered with loss of information which is not desired.. 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 = 9.+20log1+20log10 = 9.+0+20 = 11. db For LINK-2: FSL = = 11. db = 9.+20logD+20log1 = 9.+20logD +0 11. db = 9.+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. dbm. P o - L c t x + G a t x - L c r x + G a t x - FSL = RSL RSL = 10. dbm - 2 db + 20 dbi - 2 db + 20 dbi - 11. db = - 70 dbm c. Find the output power of the transmitter in LINK-2 if the Received Signal Level in LINK-2 is -72. dbm. P o = RSL + L c t x - G a t x + L c r x - G a t x + FSL P o = -72. dbm + 2 db - 20 dbi + 2 db - 20 dbi + 11. db = 8 dbm d. The Receiver Sensitivity Threshold (R x ) for LINK-1 is - 90 dbm and the Receiver Sensitivity Threshold for LINK-2 is - 0 dbm. Can LINK-1 and LINK-2 operate? Why? For LINK-1, RSL = - 70 dbm > R x = - 90 dbm. i.e., LINK-1 can operate. For LINK-2, RSL = -72. dbm < R x = - 0 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? 5

For the given microwave transmitter and receiver system, to make the link operate, link distance should be reduced. 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 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. 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 7

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 5 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 5 Kbps modem is used (assuming full rate is utilized) = 3.2 Tbits/ 5 Kbps = 57.143.000 sec = 57.143.000 sec/ (35 days / year x 24 hrs/day x 0 min./hr x 0 sec / min ) = 57.143.000 sec/ (35 days / year x 8400 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. Rate of the download = 3.2 Tbits / (17.094 hours x 0 min./hr x 0 sec / min) = 52 Mbps The type of xdsl used is VDSL. 9. For a single TV channel, the bandwidth can be taken as 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 MHz band, there are ( x 10 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 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 25 levels? Sampling by twice the maximum frequency MHz x 2 = 12 M samples per second To represent the sample with 25 levels means 8 bits / sample Minimum Bit Rate= 12 M samples per second x 8 bits / sample = 9 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 4 Kbps In 9 Mbps, there are 9 M / 4 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) 139.24 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. 8

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=9.+20 log D+20 log F=9.+20 log1+20 log1 = 9. db P o - L c t x + G a t x - L c r x + G a t x - FSL = RSL Received Signal Level, RSL=1 dbm 0 db+20 dbi 0 db+20 dbi 9. db = - 55. dbm RSL = - 55. 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 µ sec. and the number of cycles=1 Thus,the carrier frequency = 1 cycle in 0.5 µ sec., i.e., 2 x 10 cycles/sec = 2 MHz. c. Find the rate of the information signal. Solution: One bit has duration of 0.5 µ sec. Thus, the rate of the signal = 2 x 10 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. 9

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? 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 4 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 ( x 10 x 2) samples / sec x 8 bits / sample = 9 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 5 kbps modem (assuming full rate can be utilized). Solution: Such modem will be able to download 10 15 bits / 5 kbps = 10 15 bits / 5 x 10 3 bps = (10 12 / 5) seconds = (10 12 / 5) seconds In one year there are 35 x 24 x 0 x 0 seconds Thus we need (10 12 / 5) / (35 x 24 x 0 x 0) years = 5.25 years c. Would you prefer to use the DWDM system as described in part a or the 5 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) 10

= 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, 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 5 kbps modem is used (assume full rate can be utilized). 5 kbps = 5000 bits / sec = ( 5000 bits / sec ) / (10 bits / character) = 500 characters / sec No. of characters that can be downloaded in a minute = (500 characters / sec ) x ( 0 sec / minute) = 3.3 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 bits / sec = (52 x 10 bits / sec) / (10 bits / character) = 5.2 x 10 characters / sec No. of characters that can be downloaded in a minute 11

= (5.2 x 10 characters / sec ) x ( 0 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 No. of characters that can be downloaded in a minute For 10000 wavelengths, = (1 x 10 9 characters / sec ) x ( 0 sec / minute) = x 10 10 characters / minute 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 = x 10 10 characters / minute x 10000 = x 10 14 characters / minute d. Compare and comment on the rates you have found in parts a, b and c. 5 Kbps modem in part a is the slowest, VDSL in part b is 3.12 x 10 8 / 3.3 x 10 5 928 times faster than 5 Kbps modem, DWDM in patc c is x 10 14 / 3.3 x 10 5 1.79 billion times faster than 5 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 x 10 14 characters / minute, so in one minute x 10 14 characters are downloaded With the DWDM given in part a, downloading rate is 3.3 x 10 5 characters / minute So, the number of years needed to download the same number of characters when the modem = x 10 14 characters / 3.3 x 10 5 characters / minute = 1.7857 x 10 9 minutes = 1.7857 x 10 9 / (0 x 24 x 35) years = 3397.5 years 1. 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. 12

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 1 2 3 4 5 7 8 t (microsec) a. If level 1 is presented by sin ( 10 πt) sin ( 4 10 πt) 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 sin ( 10 πt) cos ( 10 πt) 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 sin ( 10 πt), plot the Amplitude Shift Keying (ASK) Modulated signal. Clearly indicate the numbers on the time axis. 13

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) 1 0 1 2 3 4 5 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 - sec = 10 bits / sec = 1 Mbps 18. x ( t) = 2 3 + 0.7 cos( 3 10 πt) + 0.1sin ( 4 10 πt) 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 14

= 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 5 kbps modem is used, how many images can be downloaded in 20 hours? 5 kbps = 5000 bits/sec There are (0 minutes/hour) x (0 seconds/minute) = 300 seconds in an hour There are 300 x 20 = 72000 seconds in 20 hours So, 72000 seconds in 10 hours x 5000 bits/sec = 4032 x 10 bits in 10 hours No. of images that can be downloaded in 10 hours = 4032 x 10 bits / 10 bits = 4032 images b. If an ADSL with download rate of 1 Mbps is used, how many images can be downloaded in an hour? 1 Mbps = 10 bits/sec There are (0 minutes/hour) x (0 seconds/minute) = 300 seconds in an hour So, 300 seconds in 1 hour x 10 bits/sec = 300 x 10 bits in 1 hour No. of images that can be downloaded in 1 hour = 300 x 10 bits / 10 bits= 300 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? For one wavelength, 10 Gbps = 10 10 bits/sec = 10 7 bits/msec For 15000 wavelengths, 15000 x 10 7 bits/msec No. of images that can be downloaded in 1 msec = 15000 x 10 7 / 10 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. Rate of the system in part a is 5 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 15

= 1 Mbps / 5 Kbps = 17.8 The 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 / 5 Kbps = 2.8 billion e. Among the systems described in parts a, b and c, which one would you prefer to use? 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 7 8 t (nanosec) a. If level 0 is presented by cos ( 4 10 9 πt) and level 1 is presented by cos ( 2 10 9 πt), plot the Frequency Shift Keying (FSK) Modulated signal. Clearly indicate the numbers on the time axis. 9 b. If the carrier is cos( 4 10 πt ), 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. 1

9 c. If level 1 is represented by cos( 4 10 πt) 9 cos( 4 10 πt ) and level 0 is represented by, plot the Phase Shift Keying (PSK) Modulated signal. Clearly indicate the numbers on the time axis. d. What is the carrier frequency and the rate of the digital information signal f (t) in part b? 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? 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 13. db and frequency of operation is 10 GHz. a. Find the link distance. FSL = 9.+20 log D+20 logf = 9.+20 log D +20 log10=9. +20 logd+20 =13. db 20 log D +11. db = 13. db 20 log D = 20 db D = 10 miles 17

b. Cable loss of the transmitter cable is db, cable loss of the receiver cable is 4 db, transmitter output power is 20 db m and the received signal level is. dbm. Find the antenna gain if the transmitter and the receiver antenna gains are the same. RSL = 20 dbm db + G dbi 13. db + G dbi 4 db =. dbm 20 dbm + G dbi + G dbi = (. +14.) dbm = 80 dbm 2G = 80 20 G = 30 dbi c. If receiver sensitivity threshold is 9. dbm, find the fade margin. Fade Margin = Received Signal Level Receiver Sensitivity Threshold Fade Margin =. ( 9.) = 30 db d. Keeping all the other link parameters the same, what operation frequency should be chosen to have 50 db fade margin? Fade Margin = Received Signal Level Receiver Sensitivity Threshold 50 db = Received Signal Level ( 9.) db m Received Signal Level = 50 db + ( 9.) db m = 4. db m RSL = 20 dbm db + 30 dbi FSL db + 30 dbi 4 db = 4. dbm 70 db m FSL= 4. dbm FSL = 11. db FSL = 9.+20 log 10 +20 log F = 9. +20 +20 log F =11. 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. 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? Maximum frequency = 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. 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 2 10 = 1024 levels. c. Explain what happens to the rate of transmitted signal in multiplexing and in inverse multiplexing? In multiplexing rate increases but in inverse multiplexing rate decreases 18

d. Write five different points related to the E-hierarchy structure. In the existing infrastructure, the smallest hierarchy is E-1, the largest is E-7 Unit channel is 4 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. FSK ii. FM, PSK, Modulation, Statistical Multiplexing, Carrier Frequency. Statistical Multiplexing iii. Coaxial Cable, Microwave, FDM, Transmission Systems, Twisted Pair Cable. FDM iv. ADSL, Twisted Pair, VDSL, DWDM, Access Network. DWDM v. Bandwidth, Frequency Content, Rate, Virtual Reality Systems, Information Content. Virtual Reality Systems 23. a. How many times more voice channels can be transmitted in E- hierarchy as compared to E-2 hierarchy? 4 x 4 x 4 x 4 = 25 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. i. FSK, ASK, PSK, FDM, Modulation FDM ii. Information Signal, Carrier Frequency, Modulation, SDSL, FSK. SDSL iii. Fiber, Radio Link, Coaxial Cable, UTP, STP. Radio Link iv. Radio Link, Satellite, Fiber, FSO, Microwave. Fiber v. 3-D holography, Rate, Virtual Reality Systems, Virtual Museums, Multiplexing. Multiplexing c. Write 5 different multiplexer types. TDM, FDM, DWDM, Statistical multiplexing, WDM d. Write five telecommunication applications that require high rate. Virtual reality, holographic TV, Human Senses Added in Telecommunications, Machine-tomachine communication, Intelligent home and office utilities 19

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. Wrong ii. Carrier frequency is always larger than the maximum frequency involved in the information signal. Correct iii. Coaxial cable transmission can support higher frequencies as compared to twisted pair transmission. Correct iv. STP brings more security to the communication but interference is worse as compared to UTP. Wrong v. 40 Gbps transmission can only be done by using fiber optics. 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? Total bits in the video signal = 18 x 10 pictures x 28 x 10 bits / picture = 504 x 10 12 bits To find the no. of seconds to download this video signal 504 x 10 12 bits / 2 x 10 bits / sec = 252 x 10 sec To find the no. of hours to download this video signal 252 x 10 sec / 300 sec / hour = 7 x 10 4 hours b. How many hours do you need to upload this video signal using the same system as in part a? 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 10 4 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? Rate of DWDM = 100 wavelengths x 40 Gbps / wavelength = 10 2 x 40 x 10 9 bps = 40 x 10 11 bps To find the no. of seconds to download this video signal 504 x 10 12 bits / 40 x 10 11 bits/sec = 12 sec d. How many pictures of this video can you download in 500 seconds by using a 5 kbps modem? 20

With 5 kbps modem, in 500 seconds, total 5 x 10 3 bps x 500 sec = 28 x 10 bits are downloaded One picture is composed of 28 x 10 bits Thus only 1 picture of this video can be downloaded in 500 seconds by using a 5 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. None 25. A voice signal has a band of frequencies in between a minimum frequency of 50 Hz and a maximum frequency of 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. According to Nyquist, sampling rate is 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 2 9 = 512 levels b. Repeat part a, if the minimum frequency of this voice signal becomes 1 khz. 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 khz. In this case there is only one frequency in the signal which is khz which is also the maximum frequency so result found in part a will not change, i.e., 512 levels will be needed. 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. 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 Mhz Thus, the sampling rate required according to Nyquist = 7 Mhz x 2 = 14 x 10 samples / sec e. What is the rate of the digital signal obtained if 25 levels are used to digitize the signal in part c? 25 levels correspond to 8 bits / sample so the rate of digital signal is 000 x 2 samples / sec x 8 bits / sample = 9 bits / sec 2. Below modulated signal has 8 bits. 21

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. PSK and level 0 is represented by cos( 4 10 πt) cos( 4 10 πt), 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. 1 0 t (microsec) 1 2 3 4 5 7 8 c. What is the carrier frequency of the modulated signal? 2π f 4 10 = π so f = 2 10 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. ASK and level 0 is represented by cos( 4 10 πt) e. What is the rate of the information signal?, level 1 is represented by 0. 1 bit in 1 micro sec, thus the rate of the information signal is 10 bps or 1 Mbps. 22

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. Frequency Shift Keying (FSK). 11110001. b. Write the forms of the signals representing level 1 and level 0. Level 1 is sin ( 10 πt), level 0 is sin ( 4 10 πt) c. Find the bit rate of this sequence. 1 bit in 1 microsecond, i.e 10 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 cos( 10 πt ) e. What are the bit rates for the other 2 types of modulation? 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 11. db, receiver sensitivity threshold is 9. dbm. Cable loss of the transmitter cable is 4 db, cable loss of the receiver cable is 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. FSL = 9.+20 log D+20 logf = 9.+20 log 1 +20 logf=9. +0+20 logf =11. db 20 log F+9. db = 11. db 20 log F = 20 db F = 10 GHz b. Find the received signal level (RSL). Fade Margin = RSL receiver sensitivity threshold 20 db = RSL ( 9. dbm) RSL = 9. dbm + 20 db = 7. dbm c. Find the output power of the transmitter. RSL = P 0 4 db + 25 dbi 11. db + 25 dbi db = 7. dbm P 0 = 4 db 25 dbi + 11. db 25 dbi + db 7. dbm = 0 dbm 23

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. RSL = 0 dbm 4 db + 15 dbi 11. db + 10 dbi db = 101. dbm Receiver sensitivity threshold is 9. 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. 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 30 million pages and each page is expressed by 9 thousand bits. a. If the whole library is downloaded by using a 1 Mbps SDSL, find the number of days needed for this download. Total bits = 30 x 10 pages x 9 x 10 3 bits / page = 30 x 9 x 10 9 bits To find the no. of seconds to download the whole library 30 x 9 x 10 9 bits / 1 x 10 bits / sec = 30 x 9 x 10 3 sec To find the no. of hours to download the whole library 30 x 9 x 10 3 sec / 300 sec / hour = 9 x 10 2 hours To find the no. of days to download the whole library 9 x 10 2 hours / 24 hours / day = 4 x 10 2 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? 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 9 separate wavelengths and each wavelength is modulated at 10 Gbps, how many seconds will be needed to download this whole library? Rate of DWDM = 9 wavelengths x 10 Gbps / wavelength = 90 x 10 9 bps = 9 x 10 10 bps To find the no. of seconds to download this video signal 30 x 9 x 10 9 bits / 9 x 10 10 bits/sec = 3 sec d. If a 5 kbps modem is used, find the number of pages that can be downloaded in 9 seconds. 24

With 5 kbps modem, in 9 seconds, total 5 x 10 3 bps x 9 sec = 5 x 10 3 x 9 bits are downloaded. One page is composed of 9 x 10 3 bits Thus the number of pages that can be downloaded in 9 seconds by using a 5 kbps = 5 x 10 3 x 9 bits / 9 x 10 3 bits = 5 pages. e. Among the systems given in parts a, part c and d, which one would you prefer to use? Why? 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. Trunk ii. Wireless, Digital Modulation, ASK, FSK, PSK. Wireless iii. SDSL, HDSL, ISDN BRI, ADSL, ISDN PRI. ADSL iv. UHF, ADSL, VHF, EHF, Ka Band. ADSL v. Shielding, UTP, Coaxial Cable, STP, Radio Link. Radio Link b. If the number of required voice channels to be transmitted is 4 times the number of voice channels in E-2, what E hierarchy will you need? What is the rate of this E hierarchy? 4 x 4 x 4 = 4, i.e., E-5 will be needed. The rate is 55 Mbps. c. Write 5 different multiplexer types. 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? According to Nyquist, sampling rate is 3 MHz x 2 = million samples / sec Thus in a minute we require million samples / sec x 0 sec / minute = 30 million samples / minute. 25

e. If the digital sequence formed in part d has 0 million bits in two seconds, what is the number of levels to represent the sample? million samples / sec, i.e., 12 million samples in 2 seconds 0 million bits / 12 million samples = 5 bits / sample which means 2 5 = 32 levels 2