Satish
Junior Member
Posts: 30 
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Post by Satish on May 5, 2013 13:25:19 GMT -5
 Okay so I have done this question multiple times and I am not getting the correct answer the equation you are suppose to use is  . When I work out the integral I'm getting 32/35 and not 3/32. Can someone tell me if I'm going wrong somewhere... |
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Post by Hung Kieu on May 5, 2013 13:30:05 GMT -5
Just curious, but is this from one of your antenna's courses?
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Satish
Junior Member
Posts: 30
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Post by Satish on May 5, 2013 13:33:02 GMT -5
yeah it is actually
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Post by ZeroVash on May 5, 2013 13:49:49 GMT -5
I don't understand anything about antennas, but the first integer should give you 32/35  |
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Satish
Junior Member
Posts: 30
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Post by Satish on May 5, 2013 13:54:59 GMT -5
yeah I know my math is correct, but I doubt the question is wrong, it appeared in final exams two years in a row...so either something is missing from the question or the answer or the question itself is wrong...
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Post by justinec on May 5, 2013 14:11:49 GMT -5
Okay, just want to preface this by saying this isn't my subject at all lol. But according Wikipedia, the total radiated power would be given by \(P_{rad} = U(\frac{3 \pi ^2} {4})\). If you were to multiply that value with what's given to you as U, this would account for the 3/32, since 3 would be multiplied in the numerator, and 4x8 would be multiplied in the denominator. Also, the \(\pi ^2 \) would be cancelled in the denominator as well. Not sure how you would incorporate this information with this problem, I could only hope it helps in some way!  *Edit: Still trying to get the hang of LaTex LOL. Bear with me though. The equation I'm mentioning is found in the attached pic. |
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Satish
Junior Member
Posts: 30
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Post by Satish on May 5, 2013 14:26:52 GMT -5
That would be correct if i got sin^4, since i have (sin^3)^2, I'm getting sin^7 which would lead to the 32/35...I'm beginning to believe that the squared is a typo because I am getting the answer using your solution
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Post by justinec on May 5, 2013 14:51:26 GMT -5
Oh, right! Actually, I completely overlooked that your questions contained (sin^3)^2. I think a typo would be the most likely explanation!
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Post by Wirespeed on May 5, 2013 23:09:18 GMT -5
Okay, just want to preface this by saying this isn't my subject at all lol. But according Wikipedia, the total radiated power would be given by \(P_{rad} = U(\frac{3 \pi ^2} {4})\). If you were to multiply that value with what's given to you as U, this would account for the 3/32, since 3 would be multiplied in the numerator, and 4x8 would be multiplied in the denominator. Also, the \(\pi ^2 \) would be cancelled in the denominator as well. Not sure how you would incorporate this information with this problem, I could only hope it helps in some way! *Edit: Still trying to get the hang of LaTex LOL. Bear with me though. The equation I'm mentioning is found in the attached pic. try adding a \ right before the last close paren - that ought to do it. \(P_{rad} = U(\frac{3 \pi ^2} {4})\) and be careful about the apostrophes. I had to add those in there in my demonstration so that the script wouldn't trigger and get rid of them... |
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Post by justinec on May 5, 2013 23:14:26 GMT -5
\(P_{rad} = U(\frac{3 \pi ^2} {4})\)
I'm not sure what I'm doing wrong :/
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Post by Wirespeed on May 5, 2013 23:42:58 GMT -5
It shows up after I reload the page....
Strange.
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Post by justinec on May 5, 2013 23:54:38 GMT -5
Oh wow, it does for me too! After the reload! I thought I wasn't getting how to write it out at all!
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. When I work out the integral I'm getting 32/35 and not 3/32. Can someone tell me if I'm going wrong somewhere...
