| Line 4: | Line 4: | ||
<math> x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(x-7k)^2} \ </math> | <math> x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(x-7k)^2} \ </math> | ||
| − | |||
| − | |||
should it be like this? | should it be like this? | ||
| Line 11: | Line 9: | ||
<math> x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(t-7k)^2} \ </math> | <math> x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(t-7k)^2} \ </math> | ||
| − | < | + | :<span style="color:red"> yes, it should be. The correction has been made. -pm</span> |
and I was trying to find out what the peak value is for this question but turns out to be very hard to calculate the sum | and I was trying to find out what the peak value is for this question but turns out to be very hard to calculate the sum | ||
| Line 17: | Line 15: | ||
<math> \sum_{t=-\infty}^\infty \frac{1}{1+t^2} \ </math> and wolfram said answer is '''π * coth(π)'''. is there any easier way to do that? | <math> \sum_{t=-\infty}^\infty \frac{1}{1+t^2} \ </math> and wolfram said answer is '''π * coth(π)'''. is there any easier way to do that? | ||
| + | :<span style="color:red"> You do not have to evaluate the sum. In particular, you do not need the peak value of that functions. Try to guess the period directly by looking at the sum. If you have no idea how to do this, read this [[Hw1periodicECE301f08profcomments| page]] first. -pm </span> | ||
Yimin. Jan 20 | Yimin. Jan 20 | ||
---- | ---- | ||
Revision as of 09:18, 20 January 2011
In question 2e
$ x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(x-7k)^2} \ $
should it be like this?
$ x(t)= \sum_{k=-\infty}^\infty \frac{1}{1+(t-7k)^2} \ $
- yes, it should be. The correction has been made. -pm
and I was trying to find out what the peak value is for this question but turns out to be very hard to calculate the sum
$ \sum_{t=-\infty}^\infty \frac{1}{1+t^2} \ $ and wolfram said answer is π * coth(π). is there any easier way to do that?
- You do not have to evaluate the sum. In particular, you do not need the peak value of that functions. Try to guess the period directly by looking at the sum. If you have no idea how to do this, read this page first. -pm
Yimin. Jan 20
