# Difference between DFT and Z-Transform

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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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

illustration from DFT to z transform

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

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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 at 5:22
because z transform is used for unit circle,you can now check what is magnitude of $z$ –  dato datuashvili Feb 17 at 5:23
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 at 5:25
I already said in question that I can verify this mathematically. I just need the interpretation. –  Bibek Subedi Feb 17 at 5:25