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Inverse Z-transforms (warning: ECE geekiness)
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Professional Poster
Join Date: May 2001
Location: North Dakota, USA
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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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Addicted to MacNN
Join Date: Jun 1999
Location: Las Vegas, NV, USA
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I think your best bet would be to find someone who has Mathematica and ask it for the answer.
Chris
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Dedicated MacNNer
Join Date: Oct 2000
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My initial impression is you have to convert the cosine to its exponential form.
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Professional Poster
Join Date: May 2001
Location: North Dakota, USA
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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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Addicted to MacNN
Join Date: Dec 1999
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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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Senior User
Join Date: Nov 2002
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this still gives me nightmares. i guess that's why i took a job in IT.
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Professional Poster
Join Date: May 2001
Location: North Dakota, USA
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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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Grizzled Veteran
Join Date: Nov 2003
Location: Hebburn, UK
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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 
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Just who are Britain? What do they? Who is them? And why?
Formerly Black Book
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Addicted to MacNN
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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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Grizzled Veteran
Join Date: Nov 2003
Location: Hebburn, UK
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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.
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Just who are Britain? What do they? Who is them? And why?
Formerly Black Book
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Professional Poster
Join Date: May 2001
Location: North Dakota, USA
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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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