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Inverse Z-transforms (warning: ECE geekiness)
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Nov 16, 2004, 10:32 PM
 
Any Electrical or Computer Engineers out there who have taken a signals class recently?

I have a transfer function in the Z-domain:
H(Z) = cos( Z^-1 )

and I'm wondering how to find h(n) of that.

(Technically, you could use square brackets, i.e. H[Z] and h[n], I believe)

Anyone have a clue? If so, I've got a bugger of a homework problem that me and at least six other people cannot figure out
     
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Nov 16, 2004, 11:04 PM
 
I think your best bet would be to find someone who has Mathematica and ask it for the answer.

Chris
     
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Nov 16, 2004, 11:46 PM
 
My initial impression is you have to convert the cosine to its exponential form.
     
funkboy  (op)
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Nov 17, 2004, 07:37 AM
 
Correct. I ended up getting it when I realized to use the symbolic math toolbox in MATLAB, one must do like:
z = sym(z)
f = cos(z) % I did an inline, exponential expression of it
iztrans( f )

Thanks for the ideas though, I was desperate last night
     
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Nov 17, 2004, 10:06 AM
 
Originally posted by funkboy:
Correct. I ended up getting it when I realized to use the symbolic math toolbox in MATLAB, one must do like:
z = sym(z)
f = cos(z) % I did an inline, exponential expression of it
iztrans( f )

Thanks for the ideas though, I was desperate last night
Erm... you forgot to invert the z in that cos(z). Wasn't is cos(z^-1)
     
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Nov 18, 2004, 08:00 AM
 
this still gives me nightmares. i guess that's why i took a job in IT.
     
funkboy  (op)
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Nov 19, 2004, 10:29 PM
 
Originally posted by The Godfather:
Erm... you forgot to invert the z in that cos(z). Wasn't is cos(z^-1)
Thank you, good point. But I did do it properly, yes, and my homework was just handed back (very quickly, might I say!).

Today was the big test in the class, and I think I did fairly well. How many ECE majors do we have in here?

And I feel your pain, kikko... how are you doing in IT? I have no plans to deviate from my course of study, but it's always interesting to hear other people's opinions. I hope to do more software/hardware design - are you mostly doing technical support? Tech support is something I'll be glad to leave behind me someday.
     
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Nov 20, 2004, 07:53 AM
 
Not an ECE major, but it's the British equivalent - Electronic Engineering degree.

Technically, you could use square brackets, i.e. H[Z] and h[n], I believe)
I believe you should use square brackets. The Z-domain is discrete, isn't it?. Normal parenthesis for continuous, square ones for discrete (like an array in C, etc...).

Oh yeah - and I'd've went straight for Matlab too
Just who are Britain? What do they? Who is them? And why?

Formerly Black Book
     
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Nov 20, 2004, 12:07 PM
 
Originally posted by Black Book:
Not an ECE major, but it's the British equivalent - Electronic Engineering degree.



I believe you should use square brackets. The Z-domain is discrete, isn't it?. Normal parenthesis for continuous, square ones for discrete (like an array in C, etc...).

Oh yeah - and I'd've went straight for Matlab too
I thought that the Z domain was contiuous and the analized sequence h[n] was discrete.
     
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Nov 20, 2004, 01:51 PM
 
Originally posted by The Godfather:
I thought that the Z domain was contiuous and the analized sequence h[n] was discrete.
Hmm... You may be right, DSP is only a tiny part of my degree; and it's the part I'm least interested in.

However. When I've used Z-transforms, it's been in control too, where you have a controller in the S-domain, and see how you would implement it in a microprocessor. So digital, therefore discrete.
Also, isn't the stuff in the Z-domain all summations rather than integrals? Which would imply discrete rather than continuous...

I'd look it up, but I'm off out to the cinema
(Last edited by Black Book; Nov 23, 2004 at 05:40 AM. )
Just who are Britain? What do they? Who is them? And why?

Formerly Black Book
     
funkboy  (op)
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Nov 20, 2004, 08:30 PM
 
Originally posted by Black Book:
Also, isn't the stuff in the Z-domain all summations (?) rather than integrals? Which would imply discrete rather than continuous...
Very correct! I believe the Z transform requires a discrete input/output.

Enjoy the movie
     
   
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