Desolation Gargantua

Question No 1

A sinewave has a period (duration of one cycle) of 35 μs. What is its corresponding frequency, in kHz, expressed to 3 significant figures?

If the frequency of this sinewave is now reduced by a factor of 2.5, using the value calculated above to 3 significant figures, what will be the new period?

Express your answer in μs, to 3 significant figures.

Answer

Given,  

Time Period = 35

                                         = 35 * 10^-6  second

We know, frequency =

                                                =

                                                =

                                                = 28571.428 Hz

                                                F= 28.571 Khz

When Frequency is reduced by factor of 2.5

                                                F¹ =

                                                 =            

 

                                                = 11928.571 Hz

                                         F¹ = 11.429 khz

Question No 2

A mobile base station (BS) in an urban environment has a power measurement of 15 µW at 175 m. If the propagation follows an inverse cube-power law, what is a reasonable power value, in µW, to assume at a distance 0.7 km from the BS?

Give your answer in scientific notation to 2 decimal places

Answer

Given,

            Urban environment has a power measurement of 15  at 175m

Propagation follows inverse cube power low

Where K is Constant

As given, p= 15  d= 175m

Now, P at 0.71cm or 700m from Both sides will be

P = 0.234

This is the required reasonable power at a distance of 0.7 km from the both sides.

Question No 3

An analogue signal is sampled every 160 µs to convert it into a digital representation. What is the corresponding sampling rate expressed in kHz?

According to the Sampling Theorem, for this sampling rate, what will approximately be the highest frequency present in the signal, in kHz, assuming the lowest frequency is very close to zero?

If each sample is quantized into 128 levels, what will be the resulting bitrate in bps?

Give your answer in scientific notation to 2 decimal places.

Answer

Given,

            Sampling time ( t_s) = 160 μs

Now,

            Sampling rate ( f_s) =

                                              =

                                           = 6.25 khz

Question No. 4

The 7-bit ASCII code for the character ‘T’ is:

1010100

An even parity check bit is now appended to the end of the code so 8 bits are transmitted. What will be the transmitted code for the ASCII character ‘T’?

If even parity check bits are used, which of the following 8-bit received ASCII codes are invalid?

i.10001110

ii.11011111

iii.01110110

iv.11011011

Solution

The given 7-bit ASCII code for ‘T’ is 1010100.

Since the number of 1’s here is 3 which is odd, the even parity check bit would be 1.

Thus the 8-bit code for ‘T’ which would be transmitted would be 10101001

Now if we have used even-parity, the valid 8 bit codes must contain even no. of 1’s or we can say that performing xor operation on all the 8 bits should result in 0. Thus the codes with odd number of 1’s must be invalid.

i.) 10001110 : It is valid as it contains four 1’s, which is even.

ii.) 11011111 : It is invalid as it contains seven 1’s, which is odd.

iii.) 01110110 : It is invalid as it contains five 1’s, which is odd.

iv.) 11011011 : It is valid as it contains six 1’s, which is even.

Question No 5

Convert the following dotted decimal IP address into binary

60.231.159.202

Solution

The decimal number 60 in binary number is 01000100

60 : 01000100

The decimal number 231 in binary number is 11100111

231 : 11100111

The decimal number 159 in binary number is 10011111

159 : 10011111

The decimal number 202 in binary number is 11001010

202 : 11001010

Therefore, the converted binary address is :

01000100.11100111.10011111.11001010

Question No 6

Express the following 48-bit binary MAC address in hexadecimal.

1111 1100 – 01111110 – 10001011 – 11010110 – 00110101 – 10011010

Solution

Four bits are taken at a time to convert to corresponding hexadecimal representation.

So,

• 1111 = F

 1100 = C

 11111100 = FC

• 0111 = 7

1110 = E

01111110 = 7E

• 1000 = 8

 1011 = B

 10001011 = 8B

• 1101 = D

 0110 = 6

 11010110 = D6

• 0011 = 3

 0101 = 5

 00110101 = 35

• 1001 = 9

 1010 = A

 10011010 = 9A

So, the 48-bit binary MAC address in Hexadecimal is

 FC-7E-8B-D6-35-9A

Question No 7

In a high-quality coaxial cable, the power drops by a factor of 10 approximately every 5 km. If the original signal power is 0.25 W (=2.5 x 10-1), how far will a signal be transmitted before the power is attenuated to 25 μW?

As part of your answer, include a Table showing the signal power vs. distance in 5 km intervals (like Table 1.2 in Section 1.4.4).

If optical fiber is used instead of the coaxial cable, briefly explain how you would expect the above calculated distance value to change. You are not required to include another Table.

Solution

Initial Power = 2.5 * 10-1 W

Final Power = 25 μW = 2.5 * 10-6 W

So, the required factor by which the power would drop is

Initial Power/Final Power = (2.5 * 10-1)/ (2.5 * 10-6) = 105

Now, drop by a factor of 101 happens after 5 km

Hence, drop by a factor of 105 happens after 5 * 5 or 25 kms

Breakup of power drop is given in the below table,

 

Single Power (Watt)	Distance (KM)
2.5 * 10-2	5
2.5 * 10-3	10
2.5 * 10-4	15
2.5 * 10-5	20
2.5 * 10-6	25
2.5 * 10-2	5

 

1. b) The attenuation factor of an optical fiber depends on the wavelength of the light transmitted through the fiber. Typically, for a single mode optical fibre,0.5 db per km loss occurs for a 1310 nm light source.

Now, in fiber optics, Attentuation (db) = 10 log10(Input Power/Output Power)

So, Our required attenuation in db is = 10 log10(2.5 * 10-1 )/(2.5 * 10-6 ) = 10 log10 105 = 10 * 5 = 50 db

Therefore, amount of distance covered by the 1310 nm light source to reach that attenuated power

50 / 0.5 km or 100 km

It is obvious now that optical fiber has comparatively lesser attenuation than coaxial cables.

So, the distance value would increase 4 times when an optical fiber is used.