Hi Barry, Once again thanks a lot for these lectures. They've really helped me and others in our Dsp module. I might be mistaken but you know at 1.55, while marking the poles and zeros in the z plane. Should alpha be marked on the x axis instead of on the circle? Because z = alpha (so there's no imaginary component for it to be marked on the circle).
I need to ask one question.x[ n] = 7 ( 1/3)^n u[ n] - 6 ( 1/2)^n u[ n] .For the Z transform I found only one zero which is 3/2 but the answer says there are two zeros suc as 3/2 and 0.How is that possible?
+Anurag laumas Okay so checking the stability goes like this 1) if all the poles lye on the left half of the y-axis u say a system is stable 2)if any one of the pole is in right half of the y-axis u say a system is unstable 3)if pole(s) lye on y-axis then u say it is marginally stable i.e for some condition it will be stable and for some it is unstable hope this helps
At 01:24 you say there is a pole at z=0. But you haven't simplified the expression. The expression can be simplified to 1/(1-a*z^(-1)) and in this case there is no zero! So I think you are wrong. You must look at the most simplified expression when looking for zeros and poles.
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Thank you for explaining the z-transform in all these videos, it's helping me out!
Sorry for the slow response - alpha is a complex number in general, so the pole can be anywhere in the z-plane.
your the best sir . thanks it really helpedfor my exam :)
You are my hero... Thanks you a lot! :) Thanks to the development of technology, I can watch this awesome lecture in my room.
Thank you so much, sir.
thank you very much for this perfect explanation! :-)
Hello sir, i just wanted to ask , what is more efficient application of z transform ?
Thank you so much.
Hello sir, i just wanted to ask is this a WHOLE chapter about Z-transform? Is that all we need to know? Or is there more to it?
Hi Barry, Once again thanks a lot for these lectures. They've really helped me and others in our Dsp module.
I might be mistaken but you know at 1.55, while marking the poles and zeros in the z plane. Should alpha be marked on the x axis instead of on the circle? Because z = alpha (so there's no imaginary component for it to be marked on the circle).
,
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Thank u, I liked it
I need to ask one question.x[ n] = 7 ( 1/3)^n u[ n] - 6 ( 1/2)^n u[ n] .For the Z transform I found only one zero which is 3/2 but the answer says there are two zeros suc as 3/2 and 0.How is that possible?
hey man i really want to know how to check the stability of this..
please can u explain...
+Anurag laumas
Okay so checking the stability goes like this
1) if all the poles lye on the left half of the y-axis u say a system is stable
2)if any one of the pole is in right half of the y-axis u say a system is unstable
3)if pole(s) lye on y-axis then u say it is marginally stable i.e for some condition it will be stable and for some it is unstable
hope this helps
1:26 what does that mean? where does it come from??
thank you
How to implementation to diagrams
H(z)=1-z^(-1) Does it have pole at z=0 as we can write is as H(z)=(z-1)/z ?
If it has pole then y(n)=x(n)-x(n-1) Can not be FIR filter .
example 4 inverse z transform is worng
At 01:24 you say there is a pole at z=0. But you haven't simplified the expression. The expression can be simplified to 1/(1-a*z^(-1)) and in this case there is no zero! So I think you are wrong. You must look at the most simplified expression when looking for zeros and poles.
Ohm my gosh you should try this application! Bump into: 'Circuit Solver' by Phasor Systems on Google Play.
wow I learned nothing from this
your a fucking idiot then
I still dont know which region to shade after watching this video lmao
@@shawnkhoo1949 watch the video of region of convergence to get the idea.