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Rearranging Equations

  • Thread starter Kayne
  • Start date
  • #1
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Hi there,

In the attachment I have an equation, I have worked though to solve for C(z) which is is highlighted in green. The highlighted yellow equation is what is supposed to look like but I am unsure if I am correct or not. I have tried to used my answer to make a graph for a Proportional controller but it doesn't work out, which makes me believe i have rearranged it wrong

If someone can tell me if the answer i have found is correct or not that would be great.

Thanks for your time
 

Attachments

Last edited:

Answers and Replies

  • #2
33,647
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Your attached image is too small for me to read, which might be why no one has responded.
 
  • #3
24
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Thanks Mark44 for let me know, it seemed to work on my computer but I have now changed it to a work document so you should be able to see it and the answer I have come up with.
 

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  • #4
33,647
5,316
Everything looks fine down to your last unshaded equation for C(z). After that, the transition to c(k) is so abrupt, that I'm not able to decipher what's going on. I have no idea what the relationship is between c(k) and C(z).

In the last (shaded yellow) equation for C(z) it appears that you are replacing the terms involving R(z) with a finite power series in z-1, z-2, ..., z-m, and something similar with the terms involving C(z).

The complex analysis class I had was so long ago that I'm not able to follow what you're doing in the last two lines.

One thing that bothers me is the e-5x terms mixed in with powers of z, which I assume are complex numbers.
 
  • #5
24
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Hi Mark44,

This attachment is the question, it has to do with porportional controllers. In this attachment I have put in the tutorial that I have been following. maybe this will help with the changing from C(z) to C(k)
 

Attachments

  • #6
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It took me awhile, but it looks like you're working with the Z-transform. I don't have any experience with that transform, so I don't think I can be much help.

Here's a link to the Wikipedia article - http://en.wikipedia.org/wiki/Z_transform.

Here's another link to an introductory tutorial than might be more helpful - http://math.fullerton.edu/mathews/c2003/ZTransformIntroMod.html [Broken]
 
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  • #7
24
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Hi Mark44,

Thanks for the information it was a little better to understand than what i had in the text books. I was able to transform it in to the z domain

thank
 

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