1 Star2 Stars3 Stars4 Stars5 Stars (No Ratings Yet)
Print Print

Update : Correction to solution of GATE-2012 ECE Q38

by Krishna Sankar on November 9, 2012

Thanks to Mr. Raghava G D’s comments on the post discussing Question 38 on Communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper, realized that I had made an error in the solution. Have the updated the post with the right answer and additional explanations.

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) 9/10

(D) 1/10


I had earlier claimed that the right answer is 7/8 and given that this answer is not listed in the options provided, there might have been a typo in the question paper. However, after digesting the comments from Mr. Raghava GD, realized that 7/8 is the probability that the received symbol is not equal to the transmitted symbol, and not the probability of error for an optimum detector.

After applying Maximum A Posteriori Detector (MAP) rule to the problem, the right answer seems to be (D) 1/10.

Please refer to the Update section in the original post to see the relevant discussion. 


Thanks !



[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf


D id you like this article? Make sure that you do not miss a new article by subscribing to RSS feed OR subscribing to e-mail newsletter. Note: Subscribing via e-mail entitles you to download the free e-Book on BER of BPSK/QPSK/16QAM/16PSK in AWGN.

{ 0 comments… add one now }

Leave a Comment

Previous post:

Next post: