CantorSet
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Hi everyone, this is not a homework question but from my reading of a signals processing paper.
This paper says if f(t) is the inverse Fourier transform of a function
f(\lambda) = e^{-2i\pi\lambda d}
then we can "easily see" that f(t) will have a peak d.
Part of the issue here is my shaky of the Fourier transform, which up til this point, I understand as a frequency decomposition of a signal. That is, let f(x) be a signal function supported on [-a,a]. Then,
f(\lambda) = \int_{-a}^{a}f(x)e^{-2i\pi x\lambda}dx
is the Fourier transform with the property that |f(\lambda)| quantifies the "amount" of frequency \lambda in the original signal function f(x).
But returning to my original problem, if we take the inverse Fourier transform of
f(\lambda) = e^{-2i\pi\lambda d}, then we have
f(t) = \frac{1}{2\pi}\int e^{-2i\pi\lambda d} e^{2i\pi\lambda t} d \lambda = \frac{1}{2\pi}\int e^{2i\pi\lambda (t-d)} d \lambda. But I can't see how |f(t)| is maximized at t=d, as it becomes the integral of 1.
Am I missing something?
This paper says if f(t) is the inverse Fourier transform of a function
f(\lambda) = e^{-2i\pi\lambda d}
then we can "easily see" that f(t) will have a peak d.
Part of the issue here is my shaky of the Fourier transform, which up til this point, I understand as a frequency decomposition of a signal. That is, let f(x) be a signal function supported on [-a,a]. Then,
f(\lambda) = \int_{-a}^{a}f(x)e^{-2i\pi x\lambda}dx
is the Fourier transform with the property that |f(\lambda)| quantifies the "amount" of frequency \lambda in the original signal function f(x).
But returning to my original problem, if we take the inverse Fourier transform of
f(\lambda) = e^{-2i\pi\lambda d}, then we have
f(t) = \frac{1}{2\pi}\int e^{-2i\pi\lambda d} e^{2i\pi\lambda t} d \lambda = \frac{1}{2\pi}\int e^{2i\pi\lambda (t-d)} d \lambda. But I can't see how |f(t)| is maximized at t=d, as it becomes the integral of 1.
Am I missing something?