ICT104 Arithmetic, Logical Operations And Signaling Homework Answer

Answer :

Answer to the questions – 

Q1.

a. Constructing a digital circuit where Z is the output and three inputs as A, B and C. Here is the diagram of digital circuit – 

The truth table of the derived digital circuit is here – 

b. The Boolean expression of the given digital circuit is here – 

(((¬ X v Y) ^ ¬ (Y v (X ^ ¬ Z))) v ^ ¬ (Y ^ Z)))

The truth table of the given digital circuit is here – 

Q2.

a. Here is a short report to justify the given statement – 

Introduction – 

In this short report we are going to justify the provided statement which is related to signal path limit with bandwidth of the channel. To justify the given statement, we can take help form Nyquist principles. Where he said that the number of independent pulses that can be put on a telegraph medium/ unit time is limited to the bandwidth of that channel. Based on this theory, we can explain the statement with some effective explanation.

Explanation – 

From the Nyquist principles, sampling is a way of converting analogue signals to digital signals. And in the terms of minimum sampling rate, the Nyquist principle says that it is beneficial of doubling the frequency rate of signal, can be seen at [2]. This approach will be helpful when and important while eliminating aliasing effects in the transmission of the signals. The original signals will have three sine curve features which are different for each amplitude and frequency range. To understand this more, we can take an example here of sampling the frequency. Here we can take a filter which is of an ideal low-pass filter that is used to filter high frequencies here. When it is set out, it will cut off the frequency when the frequency rate will be higher than set. Now, we leave all the components as it is and select low sampling frequency and aliasing will arise. This results that the generated signals will not match with the original signals. Now we try to limit the aliasing by filtering the frequency with the help of the filter. Here we can notice that when the frequency rate will go beyond the Nyquist rate, increasing the sampling frequency will not improve the quality of the resulted signals. From this demonstration, we can see that when sampling with higher frequency rate need fast conversion and much storage space, can be seen at [1]. So, sampling at higher rate will result better signal rate. As we know, sampling offer the discrete signals which are attained by analogue signals or called continuous signals. Here the amount of pulse will be double of the bandwidth rate and no more pulses can be put and sent via this signal path in this manner. It is also easy to see the effects of change in sampling frequency by deriving the transform plots. When the sampling frequency will be low, the signals will also low or decreasing. And when the sampling frequency will as low than the Nyquist rate, frequency will cause aliasing and crossover. 

Conclusion –

Here, we take help of Nyquist principle to understand and demonstrate the given statement which state that the maximum number of pulses that can be put through a signal path is limited to twice the bandwidth of the channel. To demonstrate and prove this we took an example of using a low-pass filter which is used here to filter out high frequency rate. When the sampling rate will become higher the frequency will also high and when we control the high frequency rate via the filter, sampling will be decreased with the decreased of frequency. By this demonstration and with the help of Nyquist principle, we can say and successfully explain that the rate of frequency will decide the carrying capacity of a pulse over a signal path in a channel.

1. Here are the performed operations, calculation can be seen at [3] – 
2. 00101111 + 00111110 = 

Addition of the both positive numbers is – 

00101111

00111110

11010001So, the answer is – 11010001

• 10101010 + 10101001 =

Addition of the both positive numbers is – 

10101010

10101001

01010111  So, the answer is - 101010111

• 11001010 – 10010011 =

We need to find out 2’s component of the negative number that will be – 01101100 and add 1 in it. So, the 2’s component of this negative number will be – 01101101. Now we can perform the calculation – 

11001010

01101101

10110111  So, the answer is - 10110111

• 01111111 + 11011010 =

Addition of the both positive numbers is – 

01111111

11011010

10100101  So, the answer is - 10100101

• 00100111 – 10110110 = 

We need to find out 2’s component of the negative number that will be – 01001001 and add 1 in it. So, the 2’s component of this negative number will be – 01101100. Now we can perform calculation on it – 

00100111

01101100

11011011 So, the answer is - 11011011

Q3. Research Case Study

1. UTP cable Cat6 – the unshielded twisted pair cable is mostly in used for connecting networking devices in a small to large network. This cable type is covered with a plastic cot and there is a group of twisted pair cables inside the plastic core. The number of twisted pair cables, is depended on the purpose as it comes with two pair (four wires) and four pair (eight wires) option. The Cat 6 cable is also of this type which contains twisted pair wires within a plastic coted jacket. The wires contained in it are made up of copper and the edges are connected with RJ-45 connector. This cable type is capable of carry data up to 10 Gbps for the distance of 100 meters at the frequency of 250 MHz. Cat 6 cable uses a thick plastic casing which is resistant from crosstalk effects. Due to its high capacity, it is mostly used as backbone cable, can be seen at [4]. 

2. Comparison of Cat 6 cable with Cat 7 cable – both the cable types is belonging to the UTP category but has different specifications and advantages to use in a network. The Cat7 cable is resistance from crosstalk and noise while the Cat 6 has such issues. Cat7 cable is backward compatible with Cat5 or Cat6 cables, which usesultraJAX connector. The advantages of Cat6 cable, as it provides speed of up to 10 Gbps at the frequency 250 MHz, four copper pairs and also backward compatible. While the disadvantages are it is expensive and full speed is not guaranteed. The advantages of Cat7 cable are higher bandwidth of 600 MHz, thicker, provide protection against interferences, durable and backward compatible. While the disadvantages are it is not still recognized by TIA.EIA, heavy, very expensive and prone to group loop issues, can be seen at [5] [6]. 

3. Criteria to choose a UTP cable – to choose a UTP cable, we will follow the below criteria can be seen at [7] – 

1. Bandwidth – the selected cable type should be capable to carry sufficient amount of data over a good bandwidth or speed. So that we do not need any repeater, switch, hub or signal enhancer.
2. Signal attenuation – the cable will be capable of transfer the signal and data to a particular distance without a disturbance. The cable should have great signal attenuation without the need of any signal booster. 
3. EMI interference – the cable should be resisted from EMI and other types of interference like crosstalk and also light weight, no grounding issues and not more expensive.
4. Capability of expansion – the cable is capable of expansion for the required distance. So that in case when we need to expend the network, it will be easy.
5. Cost – this factor is also very important while selecting a cable type. It should not so expensive. 

4. PC to PC connectivity – we can use both cable type as crossover and straight cable to connect a PC to other PC. Both these cables have different agreements of end wires which make them different and used for different purposes. In the crossover cable pins are connected as 1-3, 2-6, 3-1, 4-4, 5-5, 6-2, 7-7, 8-8 pin formats. It is simply used to directly connect two PCs, a router to switch or hub and connects two switches or hubs in a network. While the straight cable has the coding of a serial manner like 1 to 6 to 1 to 6 formats and it is used to connect a PC with switch or hub, to modem, a router and connect two switches or hubs at the uplink port. Both cable type uses RJ45 connect to connect to a port but different colour coding of wires, as discussed above, can be seen at [8]. 

5. Comparison of fibre optics connection with UTP – fibre optic cable is mostly used for large distance due to its high cost. It is majorly used in a network where high bandwidth is needed to work with network consumable applications. With compared to the UTP cables, fibre optic cable provides much higher speed as 10 Gbps and it resistant from EMI and crosstalk like interferences. While the UTP cable has issues of crosstalk. UTP cable is much susceptible to environmental effects while the fibre optic cable is not. Fibre optic cable did not carry electric signals so it will not radiate signals and it cannot be tapped, due to this reason is seems a more secure cable type than a UTP cable. While copper wires are used in an UTP cable that is prove to electricity, can be tapped and result the whole system to fail, can be seen at [9]. 

Q4 Reflective journal 

In these weeks, from week 1 to week 11, we have learnt so many techniques, technologies and skills related to arithmetic, logical operations and signalling mechanisms. We have learned about the digital circuits. Where we drive digital circuit diagrams, Boolean expression and deriving truth table for a digital circuit. We have learnt to draw a digital circuit from the given inputs and output in a syntax form. In the next week, we have learnt about the signal path, bandwidth channels, frequency and the combination of techniques which uses these all. There are so many techniques which uses these terms in different formats where different types and sort of forms are used of signals, frequency, bandwidth and more. Several principles are also learnt such as Nyquist principle which provide a standard to measure bandwidth and frequency rate in a transmitting channel. The 2’s components and their subtraction and additions are learnt with different methods. After this, we are able to perform operations on the two 2’s component values. In the next week, we have learnt various cable types, their features, bandwidth capacity, uses in different criteria and a detailed comparison between them that will help us in selecting an appropriate cable type for different network type. In this week, we majorly learnt about UTP cables and fibre optics cables. By learning all these techniques, methodologies, theories and practical aspects, we are able to discuss and demonstrate the digital circuits, Boolean expressions and truth table. We are also able to perform 2’s component operations like addition and subtraction of binary values (2’s components). We have now had enough understanding to different cable types with their uses in different network types in different conditions. We majorly gained knowledge about Cat 4, Cat 5, Cat 6 and Cat 7 cable types. A detailed understanding to unshielded twisted pair or UTP cables is learnt with enough understanding of straight and crossover cable and their uses to connect different networking devices and PCs. We can compare and discuss these UTP cable types within and with the fibre optics cable based on their different advantages and disadvantages.