alt.math Google Group

Half Power Bandwidth of a 2nd Order Response

art...@sj.speakeasy.net (Artist) — Sun, 21 mar 2010 04:12:02 UT

I need to find the half power bandwidth of this Laplace Transform of a
secoond order response:
G = 1/(1 + ds/Wo + s^2/Wo^2)
where d is the damping constant and Wo the natural frequency.
So I set G = 1/( sqrt(2) ):
(1/2)^-.5 = 1/(1 + ds/Wo + s^2/Wo^2)
My expectations:
For small d the s^2/Wo^2 term will dominate and the bandwidth will be

alf Power Bandwidth of a 2nd Order Response

art...@sj.speakeasy.net (Artist) — Sun, 21 mar 2010 04:10:06 UT

Half Power Bandwidth of a 2nd Order Response

art...@sj.speakeasy.net (Artist) — Sun, 21 mar 2010 04:04:49 UT

I need to find the half power bandwidth of this Laplace Transform of a
secoond order response:
G = 1/(1 + ds/Wo + s2/Wo2)
where d is the damping constant and Wo the natural frequency.
So I set G = 1/( sqrt(2) ):
(1/2)^-.5 = 1/(1 + ds/Wo + s2/Wo2)
My expectations:
For small d the s2/Wo2 term will dominate and the bandwidth will be

Re: should be integer, clear fraction.** e.g. 19 23 25 30 31 32.

don.lo...@paradise.net.nz — Sun, 21 mar 2010 02:25:10 UT

(4 * ((219 411^2) - ((7 737 - (2 / 999))^2))) - 192 325 303 132 =
-0.0841064453 best.
> should be integer, clear fraction.** e.g. 19 23 25 30 31 32.
(4 * ((219 600^2) - (11 951.327^2))) - 192 325 303 132 = -0.243713379
(4 * ((219 617^2) - (12 259.727^2))) - 192 325 303 132 = -0.458129883
(4 * ((219 633^2) - (12 543.082^2))) - 192 325 303 132 = -0.234893799

should be integer, clear fraction.** e.g. 19 23 25 30 31 32.

don.lo...@paradise.net.nz — Sun, 21 mar 2010 02:10:56 UT

((4 * ((219 411^2) - ((7 737 - (2 / 999))^2)))
- 192 325 303 132) * (999^2) = -83 938.3165
should be integer, clear fraction.**
should be integer, clear fraction.** e.g. 19 23 25 30 31 32.
anyway, also pix 6 lucky nos ascending.
nz lotto 40/6 saturday 20/3/2010.. cheers,
e.g. 19 23 25 30 31 32.

Re: Q/. if 31 people shake hands with 30 people, how many handshakes.***??

b92...@yahoo.com (Bill) — Fri, 19 mar 2010 04:30:14 UT

OK Let's say that 31 women shake hands with 30 men. That
way we know that the set of 31 does not include members from the set
of 30. Then the answer is 930.
If the question was; "If 31 people shake hands with each other"
then the answer would be 465.
I think that my version is the correct one.

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esol...@gmail.com (Esolutions) — Thu, 18 mar 2010 00:51:09 UT

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dr...@teikyopost.edu (Dom) — Wed, 17 mar 2010 13:00:46 UT

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esol...@gmail.com (Esolutions) — Tue, 16 mar 2010 18:03:49 UT

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Re: Q/. if 31 people shake hands with 30 people, how many handshakes.***??

ldcol...@san.spamblocker.com (LDC) — Tue, 16 mar 2010 14:47:06 UT

On Mon, 15 Mar 2010 22:12:38 -0400, Ted Schuerzinger
Yes I do. Perhaps that is where the 0 went. :-)

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Re: Q/. if 31 people shake hands with 30 people, how many handshakes.***??

fe...@hughes.spam (Ted Schuerzinger) — Tue, 16 mar 2010 02:12:38 UT

You do realize you're wasting your time with a Don Lotto "puzzle"?

Re: Q/. if 31 people shake hands with 30 people, how many handshakes.***??

ldcol...@san.spamblocker.com (LDC) — Tue, 16 mar 2010 01:41:03 UT

On Mon, 15 Mar 2010 16:47:59 -0700 (PDT), don.lo...@paradise.net.nz
wrote:
How is 1 = 10? If we arbitrarily ignore digits, such "coincidences"
are endless (and meaningless).

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esol...@gmail.com (Esolutions) — Tue, 16 mar 2010 01:27:54 UT

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