1 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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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.
2
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)
3
= 1.25 x 10 9 characters / sec 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 / minute For 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 / 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 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.
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?
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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. 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?
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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 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.
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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
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⇒ 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 107
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 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.
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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 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 µ sec. and the number of cycles=1
Thus,the carrier frequency = 1 cycle in 0.5 µ 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 µ 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.
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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 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)
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= 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
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
12
= (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 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 / minute For 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 / minute
d. 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.
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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 6 7 8 t (microsec)
a. If level “1” is presented by ( )6sin 6 10 tπ× and level “0” is presented by
( )6sin 4 10 tπ× , plot the Frequency Shift Keying (FSK) Modulated signal. Clearly
indicate the numbers on the time axis.
b. If level “1” is represented by ( )6sin 6 10 tπ× and level “0” is represented by
c. If the carrier is ( )6sin 6 10 tπ× , plot the Amplitude Shift Keying (ASK) Modulated
signal. Clearly indicate the numbers on the time axis.
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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 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. ( ) ( ) ( )6 62 3 0.7cos 3 10 0.1sin 4 10x t t tπ π = + × + × 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
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:
56 kbps = 56000 bits/sec
There are (60 minutes/hour) x (60 seconds/minute) = 3600 seconds in an hour
There are 3600 x 20 = 72000 seconds in 20 hours
So, 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 / 10
6 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 hour
So, 3600 seconds in 1 hour x 106 bits/sec = 3600 x 10
6 bits in 1 hour
No. of images that can be downloaded in 1 hour = 3600 x 106 bits / 10
6 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/msec
For 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
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= 1 Mbps / 56 Kbps = 17.86
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 / 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)
a. If level “0” is presented by ( )9cos 4 10 tπ× and level “1” is presented by ( )9cos 2 10 tπ× ,
plot the Frequency Shift Keying (FSK) Modulated signal. Clearly indicate the numbers on
the time axis.
Answer:
b. If the carrier is ( )9cos 4 10 tπ× , 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:
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c. If level “1” is represented by ( )9cos 4 10 tπ× and level “0” is represented by