- #1
katkatkat
- 4
- 0
Show G(k)=[itex]\sqrt{2π}[/itex]g1(k)g2(k)
Given that G(k) is the Fourier transform of F(x), g1(k) is Fourier trans of f1(x), g2(k) is Fourier trans of f2(X) and
F(x)=[itex]^{+∞}_{-∞}[/itex]∫dyf1(y)f2(x-y)
SO FAR
G(k)=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫F(x)e-ikxdx <-def'n of Fourier transform
=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫[itex]^{+∞}_{-∞}[/itex]∫dyf1(y)f2(x-y)e-ikxdx
mult by e-ikyeiky
=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫dyf1(y)e-iky[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)dx
=g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)dx
Problem starts here... I have arg of x-y in the remaining part, with dx...
Thought was that sub dx=dx-dy+dx=d(x-y)+dx <-Pretty sure I'm allowed to do this
=g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)d(x-y)+g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)d(y)
=[itex]\sqrt{2π}[/itex]g1(k)g2(k)+g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)d(y)
I'm definitely on the right track up to the problems part, do I say that ∫arg*dy is equal to zero because F(x) is a function of x... or a different reasoning... not sure... If I can make the right half of the equation = 0, I'll have the desired result
=[itex]\sqrt{2π}[/itex]g1(k)g2(k)
=G(k)
Given that G(k) is the Fourier transform of F(x), g1(k) is Fourier trans of f1(x), g2(k) is Fourier trans of f2(X) and
F(x)=[itex]^{+∞}_{-∞}[/itex]∫dyf1(y)f2(x-y)
SO FAR
G(k)=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫F(x)e-ikxdx <-def'n of Fourier transform
=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫[itex]^{+∞}_{-∞}[/itex]∫dyf1(y)f2(x-y)e-ikxdx
mult by e-ikyeiky
=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫dyf1(y)e-iky[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)dx
=g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)dx
Problem starts here... I have arg of x-y in the remaining part, with dx...
Thought was that sub dx=dx-dy+dx=d(x-y)+dx <-Pretty sure I'm allowed to do this
=g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)d(x-y)+g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)d(y)
=[itex]\sqrt{2π}[/itex]g1(k)g2(k)+g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)d(y)
I'm definitely on the right track up to the problems part, do I say that ∫arg*dy is equal to zero because F(x) is a function of x... or a different reasoning... not sure... If I can make the right half of the equation = 0, I'll have the desired result
=[itex]\sqrt{2π}[/itex]g1(k)g2(k)
=G(k)
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