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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}) </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> |
===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 09:31, 3 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 $
Answer 3
write it here.
