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unscientific
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Homework Statement
Part(a)/(b)/(c): Fourier transform the following:
Part(d): Prove the convolution theorem
Part (e): Find total displacement
Homework Equations
The Attempt at a Solution
Part(a)
[tex]f = \frac {1}{sqrt{2\pi}} \int_{-\infty}{\infty} F e^{-iωt} dω[/tex]
[tex]\frac {1}{\sqrt{2\pi}} \int_{-T}{T}[/tex]
[tex]= \frac{1}{iω\sqrt{2\pi}} [e^{iωT} - e^{-iωT}] [/tex]
[tex] = \sqrt{\frac{2}{\pi}} \frac {sin (ωt)}{ω}[/tex]
Part (b)
[tex]F_2 = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{0}e^{-at}e^{iωt} dt + \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty}e^{-at}e^{iωt} dt [/tex]
[tex] = -\frac{2a}{\sqrt{2\pi}(a^2 + ω^2)} [/tex]
Part(c)
Using result from (b):
[tex] F_2 = -\frac{2a}{\sqrt{2\pi}(a^2 + ω^2)} [/tex]
Let a = 1,
[tex]e^{-|t|} = -\frac{1}{2\pi} \int_{-\infty}^{\infty} \frac {2}{1+ω^2} e^{-iωt} dω [/tex]
[tex] t→-t, [/tex]
[tex]e^{-|-t|} = \frac {1}{\pi} \int_{-\infty}^{\infty} \frac {1}{1+ω^2} e^{iωt} dω [/tex]
[tex] t→ω,[/tex]
[tex] e^{-|ω|} = \frac {1}{\pi} \int_{-\infty}^{\infty} \frac {1}{1+t^2} e^{iωt} dt [/tex]
Part(e)
[tex] V_{(ω)} = \sqrt{2\pi} F_{(ω)}G_{(ω)} = \frac {2}{\pi} \frac {a sin (ωT)}{a^2 + ω^2}[/tex]
[tex] sin (ωT) = \frac {1}{2i} (e^{iωT} - e^{-iωT}) ,[/tex]
Taking inverse Fourier transform of Vω:
[tex] V_{(t)} = \frac {1}{i\pi}(\frac{a}{\sqrt{2\pi}}) \int_{-\infty}^{\infty} \frac {1}{a^2 + ω^2} e^{iω(T-t)} dω + \frac {1}{i\pi}(\frac{a}{\sqrt{2\pi}}) \int_{-\infty}^{\infty} \frac {1}{a^2 + ω^2} e^{-iω(t+T)} dω [/tex]
My velocity is complex! Strange...
This is related to inverse Fourier transform of F2, f2 but as a function of (t-T) and (t+T):
[tex]i(\frac{1}{\sqrt{2\pi}}) f_2(t-T) - i(\frac{1}{\sqrt{2\pi}}) f_2(t+T) [/tex]
[tex]= i\frac{1}{\sqrt{2\pi}}[e^{-a|t-T|} - e^{-a|t+T|}] [/tex]
To find the displacement, simply integrate the above with respect to t from -∞ to ∞.
[tex] s = i\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-a|t-T|} dt - i\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-a|t+T|} dt [/tex]
[tex] = i\sqrt{\frac{2}{\pi a}}(e^{-aT} - e^{aT}) [/tex]
Again complex displacement.. (and negative)
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