Difference between revisions of "Practice summation shifted geometric ECE438F11" - Rhea

Latest revision as of 08:23, 11 November 2013

Practice Question on "Digital Signal Processing"

Topic: Review of summations

Question

Simplify this summation:

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

Share your answers below

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!

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 1

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 2

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

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 3

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 4

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

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

Answer 5

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 6

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

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.

Back to ECE438 Fall 2011 Prof. Boutin

@@ Line 2: / Line 2: @@
 [[Category:ECE438Fall2011Boutin]]
 [[Category:problem solving]]
-= Simplify this summation=
+[[Category:geometric series]]
- <math>\sum_{n=-42}^5 3^{n+1} (1+j)^n </math>
+<center><font size= 4>
+'''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]'''
+</font size>
+Topic: Review of summations
+</center>
+----
+==Question==
+Simplify this summation:
+<math>\sum_{n=-42}^5 3^{n+1} (1+j)^n </math>
 ----
 ==Share your answers below==
 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===
+<span style="color:green"> TA's comments: Any complex number can be written as one single complex exponential. i.e.
-<span style="color:green"> TA's comments: Any complex number can be written as one single complex exponential. i.e. <math>a+jb=\sqrt{a^2+b^2}e^{j\theta}, where\   tan\theta = \frac{b}{a}</math> </span>
-===Answer 2===
+<math>a+jb=\sqrt{a^2+b^2}e^{j\theta}, where\   tan\theta = \frac{b}{a}</math> </span>
+----
+===Answer 1===
 Set <math>x=3+j3</math>.  Note that <math>|x|>1</math>.
@@ Line 19: / Line 32: @@
 :<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>
-===Answer 3===
+===Answer 2===
   <math>\sum_{n=-42}^5 3^{n+1} (1+j)^n = \sum_{n=-42}^5 3^{n+1} (\sqrt{2} e^{j\pi/4})^n</math>
 :By letting l = n+42,
@@ Line 28: / Line 41: @@
 :This is as far as I could go trying to type these long equations...
-===Answer 4===
+===Answer 3===
 use finite geometric series formula <math>S=\frac{a_1(1-r^n)}{1-r};a1=3,r=(3+3j)^(-42),n=47,then S=-4037-j2692</math>
+===Answer 4===
+<math>\sum_{k=0}^{n-1}ar^k=a\frac{1-r^n}{1-r}</math>
+<math>
+\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}
+</math>
+===Answer 5===
+By comparing with the formula:
+ <math>\sum_{k=0}^{n-1}ar^k=a\frac{1-r^n}{1-r}</math>
+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:
+<math> 3\frac{1-(3(1+j))^6}{1-3(1+j)}+3\frac{1-(3(1+j))^{-41}}{1-(3(1+j))^{-1}}-3 </math>
+===Answer 6===
+<math>\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}</math>
+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.
 ----
 [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]]