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I have searched this question but couldn't find the answer in this network. I know this is very confusing question for DSP beginners. Both DFT and Z-transform work for Discrete signal. I have read that "Z-transform is the general case of DFT, when we consider unit circle then, Z-transform becomes Discrete Fourier Transform (DFT)". What does this mean? Ok, I can understand the mathematical verification but what is the physical meaning of this and how this affect the analysis in DSP?

Bibek Subedi
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Actually, the Z transform is not really a proper transform, just a re-interpretation of the sequence of samples as coefficients of a formal Laurent series.

In some cases the formal Laurent series converges, if it does, it does so on an annular region in the complex plane. For useful signals (stable, summable, exponentially decaying) this annulus contains the unit circle, and the evaluation of the Laurent series on the unit circle corresponds to the Fourier series.

The interesting point of connecting a signal sequence to a periodic function on the unit circle is the inverse transformation, that many useful sequences are sequences of Fourier coefficients. And of course that convolution of signals corresponds to point-wise multiplication of the functions.

Lutz Lehmann
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Z transform is digital equivalent of laplace transform and it is used for steady state analysis of signals/systems,while DFT is digital analog of fourier transform,now can you describe difference between laplace transform and fourier transform?you can check this link

http://answers.yahoo.com/question/index?qid=20090819040222AAxskS9

illustration from DFT to z transform

http://fourier.eng.hmc.edu/e102/lectures/Z_Transform/node1.html

  • thank you for your answer. Yes I agree that Z transform is the digital equivalent of laplace transform. I have written the sentence above which is taken from the signal processing book. I just want the illustration of this. You can see mathematically that, for discrete case, when $$ z = e^{jw} $$, Z-transform actually becomes DFT. I think you can now understand what I am trying to say. – Bibek Subedi Feb 17 '14 at 05:22
  • because z transform is used for unit circle,you can now check what is magnitude of $z$ –  Feb 17 '14 at 05:23
  • Also, I have checked all the links that appears in first two pages of Google search using keyword "difference between z-transform and Fourier transform" but couldn't find the required answer. – Bibek Subedi Feb 17 '14 at 05:25
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    I already said in question that I can verify this mathematically. I just need the interpretation. – Bibek Subedi Feb 17 '14 at 05:25
  • please check second link –  Feb 17 '14 at 05:26
  • http://fourier.eng.hmc.edu/e102/lectures/Z_Transform/Z_Transform.html this one also –  Feb 17 '14 at 05:27
  • interpretation how?how to interpret it? –  Feb 17 '14 at 05:27
  • thank you for your second link. Can you state the important aspect of that link so that you can help others like me. – Bibek Subedi Feb 17 '14 at 05:27
  • let us continue this discussion in chat – Bibek Subedi Feb 17 '14 at 05:29
  • so what is idea behind the z transform?it is to convert given signal into z domain,physically it has interpretation to find transfer function from given impulse function,DFT,laplace transform ,fourier transform they are doing almost same things,but in different manner and different domain –  Feb 17 '14 at 05:31
  • exactly. Can you elaborate "Different manner" and "Different domain" term in answer. – Bibek Subedi Feb 17 '14 at 05:32
  • different manner ,for example let us consider quadratic equation like $(x^2-2*x+1)$ you can evaluate it directly inserting or $(x-1)^2$ by this way,but different domain introduce for example some additional parameter with additional propery,we know that palace transform is same as fourier transform multiplied by decaying exponential,it means that it introduce different paremeter for analyze signal –  Feb 17 '14 at 05:53