Timeline for partial fractions expansion inverse Z-transform, help
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| May 2, 2019 at 22:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 2, 2019 at 21:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 3, 2018 at 15:49 | history | edited | Fat32 | CC BY-SA 4.0 |
added 40 characters in body
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| Nov 21, 2018 at 21:17 | answer | added | Fat32 | timeline score: 1 | |
| Nov 21, 2018 at 18:13 | comment | added | Matt L. | This leaves you all options, and you should choose the quickest one. | |
| Nov 21, 2018 at 17:29 | comment | added | Late347 | both the plug-in style and the complicated style do indeed produce same output sequence y[k] as far as I understood it... so I would hope that in exam both would be accepted. The direct form the question by the teacher was as follows: "(c) Let the input to the system to be the unit step function. Define the corresponding output sequence." | |
| Nov 21, 2018 at 17:16 | comment | added | Matt L. | It's just a matter of taste how you express the solution. You can easily express the sum of shifted steps in terms of deltas for the first 3 samples. | |
| Nov 21, 2018 at 17:14 | comment | added | Late347 | yea, I suppose my teacher had that inverse-Z-transformed sequence as the correct answer... but I suppose both styles would in actual fact produce the same output sequence y[k]... It's just that the plug-in technique gives output as sums of unit step terms, where as the laborious technique has those delta terms and a unit step term... | |
| Nov 21, 2018 at 17:11 | comment | added | Matt L. | Yes, sure, as simple as that. | |
| Nov 21, 2018 at 17:07 | comment | added | Late347 | not required but what did u have in mind?, just plug-in u[n] into x[n]? i dont know what else could be done... | |
| Nov 21, 2018 at 16:50 | comment | added | Matt L. | Are you required to use partial fractions? Because I think the solution can be obtained much more easily. | |
| Nov 21, 2018 at 16:03 | history | asked | Late347 | CC BY-SA 4.0 |