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?
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2 Answers
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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.
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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
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$\begingroup$ 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. $\endgroup$ Commented Feb 17, 2014 at 5:22
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$\begingroup$ because z transform is used for unit circle,you can now check what is magnitude of $z$ $\endgroup$– user350Commented Feb 17, 2014 at 5:23
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$\begingroup$ 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. $\endgroup$ Commented Feb 17, 2014 at 5:25
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1$\begingroup$ I already said in question that I can verify this mathematically. I just need the interpretation. $\endgroup$ Commented Feb 17, 2014 at 5:25
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$\begingroup$ please check second link $\endgroup$– user350Commented Feb 17, 2014 at 5:26