Convolution property for InverseZtransform in Mathematica

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SUMMARY

The discussion centers on the convolution property of the Inverse Z-transform in Mathematica. When using the command ZTransform[Sum[f[k] g[k - n], {k, 0, n}], n, z] followed by InverseZTransform[%, z, n], the output does not utilize the convolution property for the Inverse Z-transform. Users are advised to define specific functions for f[] and g[] to test the behavior of the Inverse Z-transform and to be cautious about input formatting to avoid potential bugs.

PREREQUISITES
  • Familiarity with Mathematica syntax and functions
  • Understanding of Z-transforms and their properties
  • Knowledge of convolution in the context of signal processing
  • Basic experience with defining functions in Mathematica
NEXT STEPS
  • Define specific functions for f[] and g[] to test Inverse Z-transform behavior
  • Research the implementation of convolution properties in Mathematica's ZTransform and InverseZTransform
  • Explore potential bugs related to input formatting in Mathematica
  • Learn about the conditions under which the convolution property holds for Z-transforms
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Mathematica users, signal processing engineers, and anyone working with Z-transforms who seeks to understand the nuances of the Inverse Z-transform and its implementation in Mathematica.

Constantinos
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Hey!

So when I enter this in Mathematica

In[246] = ZTransform[Sum[f[k] g[k - n], {k, 0, n}], n, z]
InverseZTransform[%, z, n]

I get:
Out[246] = ZTransform[f[n], n, z] ZTransform[g[-n], n, z]
Out[247] = InverseZTransform[ZTransform[f[n], n, z] ZTransform[g[-n], n, z], z,
n]

Which means that although the software uses the convolution property of the Z-transform, it doesn't use it for the InverseZTransform. Any ideas how to make it use it?

Thanks!
 
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Perhaps it cannot determine what the inverse Z is. Did you define what f[] and g[] were or are you assuming this is true for all possible f and g? Are there special cases where it is not?

Try defining a really simple f[] and g[] where you know a simple Z exists and the simple inverse exists. See if that uses the inverse.

It also isn't clear to me what you input is. Not separating expressions with semicolons and using newlines instead may come back to bite you. There has been a history of bugs and surprises when doing that.
 

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