Timeline for Noise added to a Random Process
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Nov 10, 2020 at 14:30 | comment | added | mark leeds | Matt: As I said, I'll look closer when I have more time ( I'm moving so it's chaos here so may not be this weekend ) but it's so interesting because if one squares your $h[n]$ term, it looks like one gets something like $ 0.2^{(2n)} u[n]^2 + 0.2^{(2n-2)}u[n-1]^2$ which looks quite similar to my term immediately after where I write "So it is". So, we're definitely doing something quite similar. I'll study it at some point and get back hopefully with what causes the difference. | |
| Nov 10, 2020 at 14:19 | comment | added | mark leeds | I was more interested in the approach so thanks. things like z-transforms are not my "thing" so I will look at what you did closely and learn something for sure. As far as the answers go, I could have made some algebra mistake also. I'll try to do it over this weekend and include the mean and see what I get. But, it's neat to know two different approaches to the problem. thanks again. | |
| Nov 10, 2020 at 10:09 | comment | added | Matt L. | @markleeds: You can use the Z-transform to compute the transfer function: $$H(z)=\frac{1+z^{-1}}{1-0.2z^{-1}}$$ The corresponding impulse response is $$h[n]=(0.2)^nu[n]+(0.2)^{n-1}u[n-1]$$ I think that the numbers in (3) are correct, yet they are different from yours. If you like you can check the results and I'll gladly correct my answer if it turns out there's an error somewhere. | |
| Nov 9, 2020 at 21:17 | comment | added | mark leeds | I think that my answer is correct ( assuming mean is zero ) but I'm interested in your approach so would you mind showing what you got for $h[k].$ It's for learning purposes because I'm not clear on how to handle the unit impulse since it happens twice. If it happened once, that would be easy because then it's just an AR(1). Thanks. | |
| Nov 9, 2020 at 8:06 | history | answered | Matt L. | CC BY-SA 4.0 |