a. In a broadcasting system an analog signal is sampled at a rate of 100KHz. The
ID: 2083836 • Letter: A
Question
a. In a broadcasting system an analog signal is sampled at a rate of 100KHz. The sampled signal is passed through a spectrum analyzer which gives the amplitude of the sampled signal for various frequency values. What will be the signal bandwidth for the following cases: i. The gap between the sample spectrum is 100Hz. ii. The gap between the sampled spectrum is 5KHz. If the sampled spectrum is passed through a low pass filter for reconstruction, which case mentioned in part(i) and (ii) requires less stringent requirement for the low pass filter? Justify your answer. b. The data rate of binary Amplitude Shift Keying (ASK) system is 4.8Mbps. Assuming Additive White Gaussian Noise (AWGN) channel having a bandwidth of 10MHz, the noise present in the channel has zero mean with power spectral density of 10^-15W/Hz. If the received signal amplitude is 1mV, determine the average probability of error for the coherent ASK detector.Explanation / Answer
Solution:
A) i. In this case The sampled signal gives bandwidths 200 KHz and 0 KHz
Because this gap is added to the sampled and it gives maximum and minimum frequency component
100 KHz + 100 KHz = 200 KHz
100 KHz - 100 KHz = 0 KHz
ii.
In this case The sampled signal gives bandwidths 200 KHz and 0 KHz
Because this gap is added to the sampled and it gives maximum and minimum frequency component
100 KHz + 5 KHz = 105 KHz
100 KHz - 5 KHz = 95 KHz
so if compared both the cases, For the second case requires less stringent low pass filter beacuse the the frequency components 105 KHz and 95 KHz are passed through it.
B) Data rate = 4.8 Mbps
BW = 10 MHz
PSD = 10-15 W / Hz
Pe = (1/2) erfc [sqrt (Ps T / 4N0)] = 0.5 erfc [sqrt ( 1 x 10-3 x 10-5 )/ 8(10-15 ) ] = 0.5 erfc( 12.5 x 105)
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