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# 1 a) If a composite signal has frequencies between 1.15 MHz and 850 KHz, what

ID: 3537538 • Letter: #

Question

# 1 a) If a composite signal has frequencies between 1.15 MHz and 850 KHz, what is its bandwidth? b) Given a period of 10 microseconds, calculate the corresponding frequency. # 2 What is the bit rate for each of the following signals? a) A signal in which 1 bit lasts 0.001 s b) A signal in which 10 bits last 20 ?s # 3 The amplification of a signal is +16 dB. What is the final signal power if it was originally 10 W? # 4 What is the length of a bit in a channel with a propagation speed of 2 x 108 m/s if the channel bandwidth is a. 10 Mbps? b. 100 Mbps? #5 The input stream to a 4B/5B block encoder is: 0100 0000 0000 0000 0000 0001 a. What is the output stream? b. What is the length of the longest consecutive sequence of 0s in the input? c. What is the length of the longest consecutive sequence of 0s in the output? #6 Using the two figures below: Problem 6a Problem 6b a). Determine the data stream of the eight bits in the left figure above if NRZ-L encoding us used. b) Determine the data stream of the four bits in the right figure above if Manchester encoding is used. #7 What is the significance of the twisting in twisted-pair cable? #8 Name the advantages of optical fiber over twisted-pair and coaxial cable #9 A light signal is travelling through a fiber. What is the delay in the signal if the length of the fiber-optic cable is a. 50 m? b. 2 Km? (Assume a propagation speed of 2 x 108 m)

Explanation / Answer

#1) a) bandwidth = 1150 KHz - 850 KHz = 300 KHz

b) f = 1/t = 100Hz


#2) a) bit rate = 1000 bits/s

b) bit rate = 500bits / s


#3) p2 = 10^(26/10)


#4) a) channel b/w is 10Mbps = 10^7 bits per second......

So 1 bit takes 1 /( 10)^7 second which is basically bit duration...
propogation speed = 2 x 10^8 m/s

So bit length = (2 x 10^8) * 1 / (10)^7 = 20 m


b) channel b/w is 100Mbps = 10^8 bits per second......

So 1 bit takes 1 / (10)^8 second which is basically bit duration...
propogation speed = 2 x 10^8 m/s

So bit length = (2 x 10^8) * 1 / (10)^8 = 2 m