Communication Systems 2021 Past Paper

Q1

Abstract

Frequency analysis is important for measuring the performance of telecommunication systems.

a)

Abstract

Explain the difference between discrete time and continuous time

Continuous time, as the name suggest is a flow of time, a “smooth” function over time. It is the analog version of time Discrete time is a set of points taken at a set(discrete) interval from each other, pieced together. It is the digital version of time

b)

Abstract

Sketch plots for time domain and frequency domain of a Sine Wave, Square Wave with a 50% duty cycle and Square Wave with a 25% duty cycle that are obtained from discrete sampling. Please explain why these plots look the way they do.

Comm20211b We can see for the transform of the 50% graoh that every other “harmonic” is surpressed, which is because want half of the total distribution

We can see for 25% that 1 in 4 harmonics is surpressed, which while being similar to 75% duty, the amplitude is much lower, as the initial amplitude is a lower power of the initial signal.

c)

Abstract

Explain the Nyquist Sampling Theorem and why this is important in the context of discrete sampling of continuous time signals.

Nyquist’s Sampling theorem of $f_s\ge 2\ast f_{max}$ essentially dictates for the sampling of any continuous signal, we must sample at a frequency of at least double the highest frequency present in the fourier decomposition of the signal, so as to avoid aliasing and get the full accuracy of the signal

Q2

Abstract

You are working in a company deploying a cable network between two new buildings separated by 200 metres and you are working with the client to determine the best technology to deploy. The connection requires less than 1 𝜇s latency between each building and will need at least 100 Gbps in speed

a)

Abstract

Explain the difference between the capacity and latency limits for single mode optical fibres, multimode optical fibres and cat-5 Ethernet cables. Based on these differences decide the type of cabled network to deploy, by providing appropriate approximations on capacity and latency determine the best technology for the client.

The general capacity limit is as follows:

• Cat 5 Ethernet Cables : 1Gb/s
• Single Mode Fibre: >100Gb/s
• Multi Mode Fibre: >100Gb/s

However it is important to note that the technology to achieve 100GBps in Single Mode in terms of QSFPs is relatively much more expensive and potentially more expensive, the added complexity could also potentially add latency to the Single Mode system, in general for high throughput systems over a distance of less than 500m, multimode is not just the cheaper, but also potentially faster option.

b)

Abstract

Outline the network architecture required to deploy your chosen communication system, including the technology required to interconnect users in the local area network the client is asking you to develop.

In general for this system the User will have Devices with their own MAC address that connect over a Local Area Network via Ethernet Cables in both buildings, each of these networks would connect into an ONT that would then using QSFPs transmit the data over the Multimode fiber cable that connects the two buildings.

c)

Abstract

The data received from high-order modulation formats is commonly shown on an IQ diagram, sketch one of these for the 8-PSK modulation format for both low and high SNR communication channels.

Transclude of Comm20212c.excalidraw

d)

Abstract

Explain how noise will affect the Bit Error Rate (BER) in an 8-PSK modulated system. In your answer, please state what limits the maximum achievable channel capacity of a noisy communication, what is the minimum theoretical required SNR for 8-PSK and describe approaches one can use to limit the impact of noise on the BER of a communication channel.

The formula that defines channel capacity is:

$$C=B{\log }_2(\frac{S}{N}+1)$$

So as we can see here the two main factors that limit the channel capacity of a noisy communication are $B$ , bandwidth of the system and $\frac{S}{N}$ or the Signal to Noise (Linear) of the system

For 8 psk we are transmitting ${\log }_{2}8=3$ bits per symbol, meaning we are aiming for $C=3$

so rearranging this we get:

$$\begin{array}{rl}3& ={\log }_2(\frac{S}{N}+1)\\ 2^3& =10^{\frac{SNR}{10}}+1\\ \frac{SNR}{10}& ={\log }_{10}7\\ SNR& =8.45dB\end{array}$$

Bit Error Rate is defined as

$$\begin{array}{rl}\text{erfc(z)}=& 1-\frac{2}{\sqrt{\pi }}{\int }_0^{z}e-x^{2}dx\\ ⟹\text{BER}& =0.5\text{erfc}(\text{SNR}\ast \frac{B}{v_s{\log }_{2}m})\end{array}$$

So we can see, as SNR decreases we can either increase the bandwidth, or decrease the baud/symbol rate, or decrease the modulation level of the system.

3

Abstract

As a new hobby you want to start building radio systems. You plan to start with a very simple radio communication system that has a range of up to 400 meters between two fixed locations.

a)

Abstract

What is a simple type of communications system you could make and how would you construct this system? In your answer explain what modulation format you would use, the components you need with appropriate diagrams for the sender and receiver.

For a simple hobbyist radio system i would use Amplitude Modulation

Transclude of Comm20213a.excalidraw

b)

Abstract

Choose an appropriate carrier frequency for your communication system and draw a plot of the signal that would be generated when modulated using your chosen format in frequency domain showing clearly the frequency components that will present in this signal.

460MHz is a free frequency for hobbyists:

Transclude of Comm20213b.excalidraw

c)

Abstract

Over this type of radio channel indicate what types of losses will occur during transmission, what effect they have on the channel and explain what causes these losses.

Free Space Path Loss, the largest source of loss, especially with non directional transmitters, this causes a massive loss in signal pwoer. This is simply based on the physical principle that because the signal must travel in every direction, the further it is from the transmitter the lower the signal power is.

Fading:

• Slow fading, this doesnt have much effect on the system beyond being extra power loss
• Rayleigh fading, this is caused by moving objects due to the doppler effect, and can shift the actual frequency of the signal to the point that it can exit the range of the input bandpass filter
• Fast fading, also caused by the doppler effect, it can cause individual bits to be “lost”

d)

Abstract

Your receiver has an electrical noise of 1 mV RMS and an antenna impedance of 50$\Omega$ . By assuming a sensible SNR for your chosen system calculate the minimum power in Watts for the transmitter required to receive a signal over a channel 400 meters in length, stating any assumption you make

,$$ \begin{align} \text{Let SNR = 16dB: SNR}{linear} &= 10^{\frac{16}{10}} = 39.81\ \text{Power at Recevier: } P{receiver}&= \frac{(V_{rms})^2}{R} = \frac{(V_{noise} * \sqrt{\text{SNR}{linear}})^2}{R} \ &=\frac{(6.31010^{-3})^2}{50}= 7.96210^{-7}\ \text{FPSL} &= (\frac{4\pi dv}{c})^2 \ &= (\frac{4\pi40046010^6}{310^8})^2 \ &= 59403613 \ \text{Power at Transmitter: }P{transmitter}&= \text{FPSL}* P_{receiver} \ &= 59403613 * (7.962*10^{-7}) \ &= 47.297W \end{align}

Assuming no additional sources of noise have a sizable effect on the power of the signal. Assuming an otherwise non built up area #### d) Firstly, the speed of the car you're driving will implicitly create doppler shift in your signal, so large enough changes in speed will shift the signal and affect the "beating" of the IF filter on a superheterodyne radio receiver Potentially if you're driving past other hobbyist radio systems contention and distortion can occur from overlapping signals. Driving in a car you are most likely going to drive away from the receiver at some point, which will lead to lower signal power.