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.

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 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

2

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.

3

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

4

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.

5

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.

6

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.

7

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

8

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?

9

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,

10

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

11

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)

15

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:

16

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:

17

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.

18

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?

19

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.

20

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:

21

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.

22

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:

23

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:

24

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:

25

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

26

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

27

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.

28