Revision as of 08:39, 12 September 2009 by Weim (Talk | contribs)


Discussion related to HW2 (ECE438BoutinFall09)



--Back to ECE438 (BoutinFall2009)


  • for Question 1, I obtained the Fourier transform in terms of the general zeros and poles, and then replaced, say z1 by r1.exp(jw1), where r1 and w1 are constants.

Now, for an approximate plot, do I fix r1 and w1, as the values that correspond to the approximate location of the zero/pole or do I let them remain general. for example for (a) the location of the upper zero would be something like (0.6)exp(j(pi/3))),

--Dlamba

  • For Question 1, is it ok to use MATLAB to compute magnitude of Fourier Transform? I am setting locations of zeroes and poles as a constant R1 * exp (j*phi), where R1 might be 0.33 and phi might be -pi/6 (as in 1a, second zero). And then I have a vector of omegas from 0 to 2*pi (all the way around unit circle), and then have "position" as 1*exp(j*omega). My vector of Fourier Transform magnitudes is then abs(position - z1) times the second zero, divided by the abs(position - p1), etc. But at any rate, are we supposed to be plotting magnitude of Fourier Transform over 0 to 2*pi radians? I think the directions are very unclear.
-- rscheidt
  • I'm doing a lot of heavy approximation for Q1. Basically using the fact that the magnitude of the the z-transform is K times the product of the absolute difference between any arbitrary point on the complex plane and the zeroes divided by the product of the absolute difference between that same point and the poles. So you basically choose your path to be the unit circle because we want the magnitude of the Fourier Transform. In this case, it would be the magnitude of the DTFT, which could be converted to the CTFT with more information. So you can start anywhere on the unit circle but you must go around once.
  • Since the zeroes are not complex conjugates of one another, this signal is imaginary. The magnitude of the DTFT from -pi to pi therefore is not symmetric. You can estimate which interval ([0,pi] vs [-pi,0]) grows faster although I can't tell if one side drops or not.

-- weim

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