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Old 2005-01-11, 22:04   #1
tinhnho
 
Jan 2005

13 Posts
Default Hard proplem for finding sine function

hi everyone

here is the question:

Find the sine function which satify these things below:
For t =< 0, f(t) = 0
For 0< t < 1/2, f(t) follows a sine curve with a period of T = 1 sec and peak distance = 1 m
For t >= 1/2, f(t)= 0

anyidea ? thanks everyone
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Old 2005-01-12, 01:31   #2
Guilherme
 
Nov 2004
Florianopolis - Brazil

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I think you could use DFT or FFT.
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Old 2005-01-12, 02:16   #3
tinhnho
 
Jan 2005

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sorry, i dont know what is DFT or FFT.
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Old 2005-01-12, 02:25   #4
Guilherme
 
Nov 2004
Florianopolis - Brazil

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Discrete Fourier Transform

Fast Fourier Transform
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Old 2005-01-12, 02:45   #5
tinhnho
 
Jan 2005

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Quote:
Originally Posted by Guilherme
Discrete Fourier Transform

Fast Fourier Transform
Here what i get:

sin(pi*t) ; 0< t <.5

is that correct ? it doesn't seem right
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Old 2005-01-14, 20:23   #6
ewmayer
2ω=0
 
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Sep 2002
República de California

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Quote:
Originally Posted by tinhnho
hi everyone

here is the question:

Find the sine function which satify these things below:
For t =< 0, f(t) = 0
For 0< t < 1/2, f(t) follows a sine curve with a period of T = 1 sec and peak distance = 1 m
For t >= 1/2, f(t)= 0

anyidea ? thanks everyone
First off, a simple sine function will never be zero outside some finite interval and nonzero inside it, so what you really want is the form of the sine function only in the interval [0, 0.5]. Period of 1 means sin(2*pi*t), so your desired function is a composite of 3 spliced-together pieces:

For t =< 0, f(t) = 0
For 0< t < 1/2, f(t) = sin(2*pi*t)
For t >= 1/2, f(t)= 0

Just plot sin(2*pi*t) between t = 0 and t = 0.5 on your graphing calculator and you'll see it hits zero at either end of the interval.

Last fiddled with by ewmayer on 2005-01-17 at 16:37
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Old 2005-01-17, 05:42   #7
tinhnho
 
Jan 2005

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thank you very much
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