Transmission Media Importance question And Answers

Transmission Media 

Q-1. On what parameters the quality of transmission depends in case of guided transmission media? 

Ans: It is mainly decided by the frequency of transmission and the characteristics of the transmission media.

 Q-2. Why wires are twisted in case of twisted pair of transmission medium?

 Ans: It minimizes electromagnetic interferences between the pairs of wires, which are bundled together, so that the cross talk is minimum.

 Q-3. Give a popular example where co-axial cables are used for broadband signaling. 

Ans: Use of co-axial cable for broadband signaling is cable TV (CATV) application.

 Q-4. What devices are used as source and detector in case of single mode of fiber? 

Ans: LASER is used as source and photodiode is used as detector in case of single mode of fiber.

Q-5. In what way multi-mode and single-mode fibers differ?

 Ans: The core diameter of single-mode fiber is much smaller than that of multi-mode fiber. For example, For multi-mode fiber: Core diameter lies in the range of 50-200μm Cladding diameter lies in the range of 125-400μm Repeater spacing is 2Km. For single-mode fiber: Core diameter lies in the range of 8-12μm Cladding diameter 125μm Repeater spacing is 20Km

Q-6. Why does single-mode fibres are used for large distance communications rather than multi-mode fibres? 

Ans: In a multi-mode fiber, the quality of signal-encoded light deteriorates more rapidly than single-mode fiber, because of interference of many light rays. As a consequence, single-mode fiber allows longer distances without repeater. For multi-mode fiber, the typical maximum length of the cable without a repeater is 2km, whereas for single-mode fiber it is 20km.

 Q-7. What is crosstalk? How is it minimized in case of twisted-pair of wire? 

Ans: (a) Crosstalk refers to the picking up of electromagnetic signals from other adjacent wires by electromagnetic induction. (b) When a pair of wires is twisted together, the electromagnetic signals generated by the two wires cancel each other as these are of opposite polarity. This helps to reduce the susceptibility of interference to the adjacent wires. 

 Q-8. What are the factors responsible for attenuation in case of terrestrial microwave communication? 

Ans: Attenuation due to distance is 10 log (4πd/λ) 2 . Factors responsible for attenuation are given below: i) Distance – Attenuation is more if distance increases. ii) Wavelength – Attenuation is less if wavelength is longer. (i.e high frequency components are attenuated more than the low frequency component) iii) Rainfall – Attenuation is less if there is no rain. 

 Q-9. What parameters decide the spacing of repeaters in case of terrestrial microwave communication? 

Ans: Parameters are the height of the antenna ‘h’ and adjustment factor ‘k’ based on the relation d = 7.14√kh, where d is the distance in Km between two the two antennas. 

 Q-10. Why two separate frequencies are used for uplink and downlink transmission in case of satellite communication? 

Ans: Two separate frequencies are used so that one cannot interfere with the other and full duplex communication is possible. And other reason is that the Power required to transmit a signal is proportional to the frequency of the signal. And more power requirement more would be the weight of the system. As there are constraints on the load that can be carried with the satellite, mainly down linking frequency is lower than the up linking one. 

 Q-11. Why uplink frequencies are higher than downlink frequencies in case of satellite communication? 

Ans: The satellite gets power from solar cell. So, the transmitter is not being of higher power. On the other hand the ground station can have much higher power. As we want less attenuation and better signal-to-noise ratio, lower frequency is more suitable for downlink and higher frequency is commonly used for uplink. 

To insert a function from the Function Library

To insert a function from the Function Library:

In our example below, we'll use a function to calculate the number of business days it took to receive items after they were ordered. In our example, we'll use the dates in columns B and C to calculate the delivery time in column D.

1. Select the cell that will contain the function. In our example, we'll select cell D3.
   
2. Click the Formulas tab on the Ribbon to access the Function Library.
3. From the Function Library group, select the desired function category. In our example, we'll choose Date & Time.
4. Select the desired function from the drop-down menu. In our example, we'll select the NETWORKDAYS function to count the number of business days between the ordered date and received date.
   
5. The Function Arguments dialog box will appear. From here, you'll be able to enter or select the cells that will make up the arguments in the function. In our example, we'll enter B3 in the Start_date: field and C3 in the End_date: field.
6. When you're satisfied with the arguments, click OK.
   
7. The function will be calculated, and the result will appear in the cell. In our example, the result shows that it took four business days to receive the order.
   

• Like formulas, functions can be copied to adjacent cells. Hover the mouse over the cell that contains the function, then click, hold, and drag the fill handle over the cells you want to fill. The function will be copied, and values for those cells will be calculated relative to their rows or columns.