Solution for Question 6 : IES Conventional 1998 : Electronics & Telecommunication Engineering : Paper II

Q.6. (a): A signal is band limited to 3.6 KHz and three other signals are band limited to 1.2 KHz each. These signals are to be transmitted by means of Time-Division-Multiplexing.

 (i) If each signal is sampled at its Nyquist rate set up a scheme to achieve this multiplexing.

   (ii)   Specify the speed of commutator in samples per second.

  (iii) If the commutator output is quantized in 1024 levels with the result binary coded, what is the output bit rate?

   (iv)  Determine the minimum transmission bandwidth of the channel.  

Solution: (i) The Nyquisi rates at which the four signals are sampled are 7.2 kHz, 2.4 kHz, 2.4 kHz and 2.4 kHz respectively. As the sampling rate of m1(1) is thrice that of other three signals, there will be three samples of m1(1) and one each of other three signals in one rotation of the commutator of TDM scheme shown in figure below.

(ii) The samples per second of the four signals are 7200, 2400, 2400 and 2400 and thus there are 14400 samples in one second. Hence, the speed of commutator is 14400 samples per second.

(iii) The number of quantization levels M = 1024. Hence, the number of bits N is given by

 (2)^N = M = 1024

=>      N = 10

Hence, the output bit is 10 X 14400 bits/second or 144 Kb/second

(iv) The minimum channel bandwidth = 0.5 (7.2 + 2.4 + 2.4 + 2.4) X 10 = 72 KHz.




Q.6. (b): Draw refractive index profile for   
                                               (i)   step-index and
                                               (ii)  graded-index fibres.
Determine the cutoff wavelength for a step index fibre to exhibit single mode operation when the core refractive index and radius are 1.46 and 45 m respectively with the relative index difference being 0.25%. 

Solution: The refractive index profile for step-index and graded-index fibres are as shown in figure below - 

Now, Using equation (shown below) with Vc = 2.405 gives:

here, a = 4.5, n1 = 1.46, Δ = 0.25% = 0.0025
Hence, 

λc = 1.214 μm = 1214 nm

Hence, the fiber is single-moded to a wavelength of 1214 nm.