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<math>\sum_{n=-42}^5 3^{n+1} (1+j)^n = 3\sum_{n=-42}^5 x^n = 3\sum_{n=-5}^{42}x^{-n} = 3\sum_{n=-5}^{42}(\frac{1}{x})^n </math> | <math>\sum_{n=-42}^5 3^{n+1} (1+j)^n = 3\sum_{n=-42}^5 x^n = 3\sum_{n=-5}^{42}x^{-n} = 3\sum_{n=-5}^{42}(\frac{1}{x})^n </math> | ||
<math> = 3(\sum_{n=-5}^{\infty}(\frac{1}{x})^n - \sum_{n=43}^{\infty}(\frac{1}{x})^n) = 3(\frac{(\frac{1}{x})^{-5}}{1-\frac{1}{x}} - \frac{(\frac{1}{x})^{43}}{1-\frac{1}{x}}) = 3(\frac{x^6-x^{-42}}{x-1}) = -4037-j2692 </math> | <math> = 3(\sum_{n=-5}^{\infty}(\frac{1}{x})^n - \sum_{n=43}^{\infty}(\frac{1}{x})^n) = 3(\frac{(\frac{1}{x})^{-5}}{1-\frac{1}{x}} - \frac{(\frac{1}{x})^{43}}{1-\frac{1}{x}}) = 3(\frac{x^6-x^{-42}}{x-1}) = -4037-j2692 </math> | ||
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| + | :<span style="color:blue">Instructor's comments: There is a much shorter solution using the finite geometric series formula. Note that, when the sum is finite, one does not have to worry about convergence. In particular, the formula works even if the norm of the argument is greater than one. -pm </span> | ||
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===Answer 3=== | ===Answer 3=== | ||
write it here. | write it here. | ||
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[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | ||
Revision as of 07:46, 4 September 2011
Contents
Simplify this summation
$ \sum_{n=-42}^5 3^{n+1} (1+j)^n $
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
TA's comments: Any complex number can be written as one single complex exponential. i.e. $ a+jb=\sqrt{a^2+b^2}e^{j\theta}, where\ tan\theta = \frac{b}{a} $
Answer 2
Set $ x=3+j3 $. Note that $ |x|>1 $.
$ \sum_{n=-42}^5 3^{n+1} (1+j)^n = 3\sum_{n=-42}^5 x^n = 3\sum_{n=-5}^{42}x^{-n} = 3\sum_{n=-5}^{42}(\frac{1}{x})^n $ $ = 3(\sum_{n=-5}^{\infty}(\frac{1}{x})^n - \sum_{n=43}^{\infty}(\frac{1}{x})^n) = 3(\frac{(\frac{1}{x})^{-5}}{1-\frac{1}{x}} - \frac{(\frac{1}{x})^{43}}{1-\frac{1}{x}}) = 3(\frac{x^6-x^{-42}}{x-1}) = -4037-j2692 $
- Instructor's comments: There is a much shorter solution using the finite geometric series formula. Note that, when the sum is finite, one does not have to worry about convergence. In particular, the formula works even if the norm of the argument is greater than one. -pm
Answer 3
write it here.
