GATE-2012 ECE Q47 (math)

Question 47 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper. Q47. Given that and , the value of is (A)  (B)  (C)  (D)  Solution To answer this question, we need to refer to Cayley Hamilton Theorem. This is discussed briefly in Pages 310-311 of Introduction to Linear Algebra, Glibert Strang (buy…

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GATE-2012 ECE Q36 (math)

Question 36 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper. Q36. A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is (A) 1/3 (B) 1/2 (C) 2/3 (D) 3/4 Solution Let us start by…

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Selection Diversity

This is the first post in the series discussing receiver diversity in a wireless link. Receiver diversity is a form of space diversity, where there are multiple antennas at the receiver. The presence of receiver diversity poses an interesting problem – how do we use ‘effectively‘ the information from all the antennas to demodulate the…

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GATE-2012 ECE Q39 (communication)

Question 39 on communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper. Q39. The signal  as shown is applied both to  a phase modulator (with  as the phase constant) and a frequency modulator (with as the frequency constant) having the same carrier frequency.  The ratio  for the same maximum phase deviation is,…

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Alamouti STBC

In the recent past, we have discussed three receive diversity schemes – Selection combining, Equal Gain Combining and Maximal Ratio Combining. All the three approaches used the antenna array at the receiver to improve the demodulation performance, albeit with different levels of complexity. Time to move on to a transmit diversity scheme where the information…

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GATE-2012 ECE Q38 (communication)

Question 38 on Communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper. Q38. A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X=0)=9/10, then the probability of error for an optimum receiver will be (A) 7/80 (B) 63/80 (C)…

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