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[[Category:ECE438Fall2011Boutin]] | [[Category:ECE438Fall2011Boutin]] | ||
[[Category:problem solving]] | [[Category:problem solving]] | ||
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| + | <center><font size= 4> | ||
| + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
| + | </font size> | ||
| + | |||
| + | Topic: Review of infinite summations | ||
| + | |||
| + | </center> | ||
| + | ---- | ||
| + | ==Question== | ||
A student in [[ECE301]] once wrote the following formula on his exam: | A student in [[ECE301]] once wrote the following formula on his exam: | ||
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===Answer 1=== | ===Answer 1=== | ||
First we know the summation of an infinity geometric series: | First we know the summation of an infinity geometric series: | ||
| + | |||
<math> \sum_{n=0}^{\infty} \alpha^n = \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1</math>; (eq1) | <math> \sum_{n=0}^{\infty} \alpha^n = \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1</math>; (eq1) | ||
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<math> \sum_{n=0}^{M-1} \alpha^n = \frac{\alpha - \alpha^M}{(1 - \alpha)} , \left| \alpha \right| < 1</math>; (eq1) | <math> \sum_{n=0}^{M-1} \alpha^n = \frac{\alpha - \alpha^M}{(1 - \alpha)} , \left| \alpha \right| < 1</math>; (eq1) | ||
| − | <math> \sum_{n=0}^{N} \alpha^n = \frac{\alpha - \alpha^{N-} | + | <math> \sum_{n=0}^{N} \alpha^n = \frac{\alpha - \alpha^{N-1}}{(1 - \alpha)} , \left| \alpha \right| < 1</math>; (eq2) |
| + | |||
| + | (eq2) - (eq1) = <math>\sum_{n = M}^N \alpha^n = \frac{\alpha^M - \alpha^{N-1}}{(1 - \alpha)}</math> | ||
| + | |||
| + | M and N also need to be positive integers for this argument | ||
===Answer 3=== | ===Answer 3=== | ||
| − | + | This fomular is wrong. To see that, let N=M+1 | |
| + | the correct fomular should be | ||
| + | |||
| + | <math>\sum_{n = M}^N \alpha^n = \frac{\alpha^M - \alpha^{N+1}}{(1 - \alpha)}</math> when a!=1 | ||
---- | ---- | ||
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | ||
Latest revision as of 08:35, 11 November 2013
Practice Question on "Digital Signal Processing"
Topic: Review of infinite summations
Question
A student in ECE301 once wrote the following formula on his exam:
$ \sum_{n = M}^N \alpha^n = \frac{\alpha^M - \alpha^{N-1}}{(1 - \alpha)} $
Is this formula correct? For what values of the parameters is the formula valid? Please comment.
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
First we know the summation of an infinity geometric series:
$ \sum_{n=0}^{\infty} \alpha^n = \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1 $; (eq1)
so we can compute
$ \sum_{n=M}^{\infty} \alpha^n = \left( \alpha \right)^M \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1 $; (eq2)
similarly,
$ \sum_{n=N+1}^{\infty} \alpha^n = \left( \alpha \right)^{N+1} \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1 $; (eq3)
then we can substract eq3 from eq2, if N+1> M
$ \sum_{n=M}^{\infty} \alpha^n - \sum_{n=N+1}^{\infty} \alpha^n = \frac{{\left( \alpha \right)^M } - {\left( \alpha \right)^{N+1}}}{(1 - \alpha)} , \left| \alpha \right| < 1 $;
for N larger or equal to M, $ \left| \alpha\right| < 1 $, the equation above holds.
//Did I make any mistake in the N+1 part?
- TA's comments: Looks good. How about $ |\alpha|>1\ or\ |\alpha|=1\ ? $
//ahhh, the series is finite, so the condition $ \left| \alpha\right| < 1 $ doesn't necessarily hold. but it requires an other approach with
$ \sum_{n=-\infty}^{0} \alpha^n = \frac{1}{(1 - \frac{1}{\alpha})} , \left| \alpha \right| > 1 $;
Anyone want to finish it for me? Thanks
- TA's comments: Actually, when the geometric series is finite length, we can write the sum in one closed form as long as the common ratio is not equal to 1. Can someone try to write down the equation?
Answer 2
$ \sum_{n=0}^{M-1} \alpha^n = \frac{\alpha - \alpha^M}{(1 - \alpha)} , \left| \alpha \right| < 1 $; (eq1)
$ \sum_{n=0}^{N} \alpha^n = \frac{\alpha - \alpha^{N-1}}{(1 - \alpha)} , \left| \alpha \right| < 1 $; (eq2)
(eq2) - (eq1) = $ \sum_{n = M}^N \alpha^n = \frac{\alpha^M - \alpha^{N-1}}{(1 - \alpha)} $
M and N also need to be positive integers for this argument
Answer 3
This fomular is wrong. To see that, let N=M+1 the correct fomular should be
$ \sum_{n = M}^N \alpha^n = \frac{\alpha^M - \alpha^{N+1}}{(1 - \alpha)} $ when a!=1
