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Update : Correction to solution of GATE-2012 ECE Q38

Posted By Krishna Sankar On November 9, 2012 @ 8:58 pm In News | 2 Comments

Thanks to Mr. Raghava G D’s comments [1] 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 [2] to see the relevant discussion. 

http://www.dsplog.com/2012/11/03/gate-2012-ece-q38-communication/#update [2]

Thanks !

 

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf [3]

 


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URLs in this post:

[1] Mr. Raghava G D’s comments: http://www.dsplog.com/2012/11/03/gate-2012-ece-q38-communication/#comment-166627

[2] Update section in the original post: http://www.dsplog.com/2012/11/03/gate-2012-ece-q38-communication/#update

[3] http://gate.iitm.ac.in/gateqps/2012/ec.pdf: http://gate.iitm.ac.in/gateqps/2012/ec.pdf

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