Talk:Linear canonical transformation: Difference between revisions

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Mathematics B‑class Mid‑priority

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I have corrected two simple details:

Use of i = sqrt(-1) is better than 'j'. i is used in Mathematics instead of j
In matrix for fourier transform the minus sign was wrong in the upper value -1. It's 1.

In the applications section there are so many i's already, and the figures use i as subscripts, so that they should be redone without the i. Without changing figure j is better. In any event,

i={\sqrt {-1}}

or

j={\sqrt {-1}}

must appear in the text. Either i or j is acceptable, but they must be defined in the text.--Rmoba (talk) 20:39, 1 August 2009 (UTC)[reply]

user cems2 says: I beleive the formula used for the spherical lens is wrong. specifically the x in the exponent numerator I think should be a constant, probably unity. as it stands the formula is not symmetric in x and y. —Preceding unsigned comment added by 192.12.184.2 (talk) 22:11, 21 April 2009 (UTC)[reply]

Correct. The x should be the k defined in the previous section. I have corrected this - jhealy 14:05 (GMT), May 20, 2009

I have two issues with the article as is, both related to the equation

$X_{(a,b,c,d)}(u)={\sqrt {-i}}\cdot e^{-i\pi {\frac {d}{b}}u^{2}}\int _{-\infty }^{\infty }e^{-i2\pi {\frac {1}{b}}ut}e^{i\pi {\frac {a}{b}}t^{2}}x(t)\;dt\,,$	when b ≠ 0,
$X_{(a,0,c,d)}(u)={\sqrt {d}}\cdot e^{-i\pi cdu^{2}}x(du)\,,$	when b = 0.

First, when I do a search of articles on the LCT, most use a kernel that does not have the factor of $2\pi$ that occurs in the kernel here. This difference is a bit like the choice of kernel for the Fourier transform, where some authors put the $2\pi$ in the exponent of the kernel, and some put it as a constant outside the exponential. (For my purposes, the form here is better! But the majority of the literature seems to have the other form.)

Second, I think that there is a factor of $1/{\sqrt {b}}$ missing from the kernel. It should be

X_{(a,b,c,d)}(u)={\sqrt {\frac {1}{ib}}}\cdot e^{-i\pi {\frac {d}{b}}u^{2}}\int _{-\infty }^{\infty }e^{-i2\pi {\frac {1}{b}}ut}e^{i\pi {\frac {a}{b}}t^{2}}x(t)\;dt\,,

when b ≠ 0,

If anyone is following this page, please check that this is correct --- I'm reluctant to make a change without confirmation. PiperArrow123 July 19, 2010

@@ Line 12: / Line 12: @@
 : Correct. The x should be the k defined in the previous section. I have corrected this - [[User:jhealy|jhealy]] 14:05 (GMT), May 20, 2009
+----
+I have two issues with the article as is, both related to the equation
+:{| border="0"
+|-
+| <math>X_{(a,b,c,d)}(u) = \sqrt{-i} \cdot e^{-i  \pi \frac{d}{b} u^{2}} \int_{-\infty}^\infty e^{-i 2 \pi \frac{1}{b} ut}e^{i  \pi \frac{a}{b} t^2} x(t) \; dt \, , </math> {{pad|3em}} || when ''b''&nbsp;≠&nbsp;0,
+|-
+| <math>X_{(a,0,c,d)}(u) = \sqrt{d} \cdot e^{-i  \pi cdu^{2}} x(du) \, ,</math> || when ''b''&nbsp;=&nbsp;0.
+|}
+First, when I do a search of articles on the LCT, most use a kernel that does not have the factor of <math>2\pi</math> that occurs in the kernel here. This difference is a bit like the choice of kernel for the Fourier transform, where some authors put the <math>2\pi</math> in the exponent of the kernel, and some put it as a constant outside the exponential. (For my purposes, the form here is better! But the majority of the literature seems to have the other form.)
+Second, I think that there is a factor of <math>1/\sqrt{b}</math> missing from the kernel. It should be
+:{| border="0"
+|-
+| <math>X_{(a,b,c,d)}(u) = \sqrt{\frac{1}{ib}} \cdot e^{-i  \pi \frac{d}{b} u^{2}} \int_{-\infty}^\infty e^{-i 2 \pi \frac{1}{b} ut}e^{i  \pi \frac{a}{b} t^2} x(t) \; dt \, , </math> {{pad|3em}} || when ''b''&nbsp;≠&nbsp;0,
+|}
+If anyone is following this page, please check that this is correct --- I'm reluctant to make a change without confirmation. [[User:PiperArrow123|PiperArrow123]] July 19, 2010