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comparison main/signal/inst/bilinear.m @ 2710:2d64e669e231 octave-forge

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Fix docs so that the table will format nicely on the web.
author pkienzle
date Tue, 17 Oct 2006 02:02:03 +0000
parents 38e4c6572b8d
children 73fa4496fb07
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2709:c5d8c38f7efd 2710:2d64e669e231
30 ## s-plane positioned at 2/T tan(w*T/2) so that they will be positioned 30 ## s-plane positioned at 2/T tan(w*T/2) so that they will be positioned
31 ## at w after the bilinear transform is complete. 31 ## at w after the bilinear transform is complete.
32 ## 32 ##
33 ## The following table summarizes the transformation: 33 ## The following table summarizes the transformation:
34 ## 34 ##
35 ## Transform Zero at x Pole at x 35 ## +---------------+-----------------------+----------------------+
36 ## ---------------- ------------------------- ------------------------ 36 ## | Transform | Zero at x | Pole at x |
37 ## Bilinear zero: (2+xT)/(2-xT) pole: (2+xT)/(2-xT) 37 ## | H(S) | H(S) = S-x | H(S)=1/(S-x) |
38 ## 2 z-1 pole: -1 zero: -1 38 ## +---------------+-----------------------+----------------------+
39 ## S -> - --- gain: (2-xT)/T gain: (2-xT)/T 39 ## | 2 z-1 | zero: (2+xT)/(2-xT) | zero: -1 |
40 ## T z+1 40 ## | S -> - --- | pole: -1 | pole: (2+xT)/(2-xT) |
41 ## ---------------- ------------------------- ------------------------ 41 ## | T z+1 | gain: (2-xT)/T | gain: (2-xT)/T |
42 ## +---------------+-----------------------+----------------------+
42 ## 43 ##
43 ## With tedious algebra, you can derive the above formulae yourself by 44 ## With tedious algebra, you can derive the above formulae yourself by
44 ## substituting the transform for S into H(S)=S-x for a zero at x or 45 ## substituting the transform for S into H(S)=S-x for a zero at x or
45 ## H(S)=1/(S-x) for a pole at x, and converting the result into the 46 ## H(S)=1/(S-x) for a pole at x, and converting the result into the
46 ## form: 47 ## form: