FIT3165/FIT4165 Tutorial #9 TCP/IP Layered Architecture
Week 10 – Semester 1 – 2022 25 April 2022
Revision Status
Updated by Dr. and Rosanna F Alam, May 2022.
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©2022, Faculty of IT, Monash University
Instructions
1. Students work individually to solve this week’s exercise.
2. Each student must answer the following review questions.
PART 1- Review Questions
1. List the differences between guided media and unguided media?
With guided media, the electromagnetic waves are guided along an enclosed physical path whereas unguided media provide a means for transmitting electromagnetic waves but do not guide them.
2. What are the three major classes of guided media?
The three major categories of guided media are twisted-pair, coaxial, and fiber-optic cables.
3. Explain the following technical terms that are used in data communication.
a) Attenuation b) Modulation c) Noise
a) Attenuation: The loss of a signal’s energy due to the resistance of the medium.
b) Modulation: Modification of one or more characteristics of a carrier wave by an
information bearing signal.
c) Noise: Random electrical signals that can be picked up by the transmission
medium and cause degradation or distortion of the data.
4. A signal has passed through three cascaded amplifiers, each with a 4 dB gain. What is the total gain? How much is the signal amplified?
The total gain is 3 * 4 = 12dB. To find how much the signal is amplified, we can use the following formula:
12 = 10 log (P2 / P1) So, log10 (P2 / P1) = 1.2
Thus, P2 / P1 = 101.2 = 15.85 The signal is amplified almost 16 times.
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5. Define baseband transmission.
Baseband transmission means sending a digital or an analog signal without modulation using a low-pass channel.
6. How can we find the period of a sine wave when its frequency is given?
The period of a signal is the inverse of its frequency: T = 1/f.
7. Define channel capacity.
The rate at which data can be transmitted over a given communication path, or channel, under given conditions, is referred to as the channel capacity.
8. Define the theoretical maximum channel capacity bit rate of a noiseless channel using Nyquist theorem and Shannon’s theorem for noisy channel.
The Nyquist theorem defines the maximum bit rate of a noiseless channel. A very important consideration in data communications is how fast we can send data called capacity in bits per second, over a channel. Data rate depends on three factors:
• The bandwidth available
• The levels of the signals we use
• The quality of the channel (the level of noise)
Two theoretical formulas were developed to calculate the data Capacity:
1. Nyquist for a noiseless channel(Nyquist Equation)
C=2BLog2 L
2. Shannon for a noisy channel (Shannon’s Equation) C=BLog2 (1+SNR)
C = Capacity of the channel in bps
B = Bandwidth of the channel in Hz
L = Number of voltage levels in digital signal SNR = Signal to noise ratio
9. We measure the performance of a telephone line (4kHz of bandwidth). When the signal is 10V, the noise is 10 mV. What is the maximum data rate supported by this telephone line?
©2022, Faculty of IT, Monash University
SNR is the ratio of powers. The power is proportional to the voltage square (P = V2 / R). Therefore, we have
SNR = (102) / (10 X 10-3)2 = 106,
We then use the Shannon capacity to calculate the maximum data rate.
C = 4000 log2 (1+106) = 80 Kbps.
10. We have sampled a low-pass signal with a bandwidth of 200 kHz using 1024 levels of quantization.
a) Calculate the bit rate of the digitized signal.
b) Calculate the SNR for this signal.
c) Calculate the PCM bandwidth of this signal.
Solution: a.
f max = 0 + 200 = 200 kHz.
So, f s = 2 * 200,000 = 400,000 samples/s n b = log 2 1024 = 10 bits/sample.
So, n = 400,000 * 10 = 4 Mbps
SNRdB = 6.02 nb + 1.76 = 61.96
B PCM =( n b)* (B analog ) = 10 * 200 kHz = 2 MHz
11. A copper Shielded Twisted Pair (STP) cable has a loss of 1 dB per Km at 10 kHz. We want to have a link of 10km using this cable. What should the power of the signal be at the source if we want the signal to have 10mW power at the destination?
The loss of the cable for 10Km is 10dB = 10 log10 (P2/P1). This means, log10 (P2/P1) = – 1, or
(P2/P1) = 1/10, resulting in P1= 10 * P2 = 100 mW.
12. A digital signalling system is required to operate at 12000 bps. If a signal element encodes a 4-bit word, what is the minimum required bandwidth of the channel?
Using Nyquist’s equation: C = 2B log2M
We have C = 12000 bps
log2M = 4, because a signal element encodes a 4-bit word. Therefore, C = 12000 = 2B × 4, and B = 1500 Hz
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