Revision as of 08:57, 11 November 2011

Practice Problem: Simplify this summation

$ \sum_{n=-42}^5 3^{n+1} (1+j)^n  $

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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

$ \sum_{n=-42}^5 3^{n+1} (1+j)^n = \sum_{n=-42}^5 3^{n+1} (\sqrt{2} e^{j\pi/4})^n $

By letting l = n+42,

$ \sum_{l=0}^{47} 3^{l-41} (\sqrt{2} e^{j\pi/4})^{l-42} =        3^{-41}(\sqrt{2}e^{j\pi/4})^{-42}\sum_{l=0}^{47} (3\sqrt{2}e^{j\pi/4})^l =         \frac{1 - (3\sqrt{2}e^{j\pi/4})^{48}}{1 - 3\sqrt{2}e^{j\pi/4}}3^{-41}(\sqrt{2}e^{j\pi/4})^{-42}   $

This is as far as I could go trying to type these long equations...

Answer 4

use finite geometric series formula $ S=\frac{a_1(1-r^n)}{1-r};a1=3,r=(3+3j)^(-42),n=47,then S=-4037-j2692 $

Answer 5

$ \sum_{k=0}^{n-1}ar^k=a\frac{1-r^n}{1-r} $

$ \begin{align} \sum_{n=-42}^5 3^{n+1} (1+j)^n &= 3\sum_{n=-42}^5 (3(1+j))^n \\ &=3\frac{1-(3(1+j))^6}{1-3(1+j)}+3\frac{1-(3(1+j))^{-41}}{1-(3(1+j))^{-1}}-3 \end{align} $

Answer 6

By comparing with the formula:

$ \sum_{k=0}^{n-1}ar^k=a\frac{1-r^n}{1-r} $

We note:

       a = 3
       r = 3(1 + j)
       

Then we can break the sum into the part where n is negative, and the part where n is positive. By comparing with the formula again, we find the sum equal to:

$ 3\frac{1-(3(1+j))^6}{1-3(1+j)}+3\frac{1-(3(1+j))^{-41}}{1-(3(1+j))^{-1}}-3 $

Answer 7

$ \begin{align} \sum_{n=-42}^5 3^{n+1} (1+j)^n &= 3\sum_{-42}^{5} (3+3j)^n\\ &=3\frac{(3+3j)^{-42}-(3+3j)^5}{1-(3+3j)} \end{align} $

by finite geometric series

And if you want to simplify down farther be my guest, although I believe there is probably a quick trick I am missing atm.

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