New answers tagged fourier
1
https://www.ques10.com/p/5768/state-and-prove-the-property-of-kernel-separatin-1/
It is possible due to the separability property of Fourier Transform
1
If $X(f)=\mathcal{F}\{x(t)\}$ you have, in analogy with Eq. $(7)$,
$$x(f)\Longleftrightarrow X(-t)\tag{1}$$
where we assume that functions with the independent variable $f$ are Fourier transforms of the corresponding functions with independent variable $t$.
Applying $(1)$ to
$$x(t-b)\Longleftrightarrow e^{-2\pi jbf} X(f)\tag{2}$$
gives
$$x(f-b)\...
4
Relating the DFT to the FT (CTFT) is a big issue. Let's start with the basic definitions without any domain specification using a $\frac{1}{N}$ normalization on the DFT.
$$ FT(x(t))(f) = \int x(t) e^{-i2\pi t f } dt $$
$$ DFT(x[n])(k) = \frac{1}{N} \sum x[n] e^{-i 2\pi \frac{n}{N} k } $$
They are obviously similar. Assuming they cover the same interval, ...
2
There are two factors involved here. One has been mentioned in a comment: your truncation error is larger for smaller values of $N$ because you truncate the time domain function at $NT$, where the sampling interval $T$ is constant.
The other factor - which is the more important one here - is that you divide the FFT result by $N$. That's why you see a ...
2
Most of the classical linear transformations (even filtering) may have three main types of scalings:
natural or none: sum or integral does not have an explicit scaling factor (but the inverse may need one)
in amplitude: because of linearity, at least one reference "unit" signal should have a unit amplitude in the transformed domain, for an easy ...
1
The autocorrelation of $x(t)$ is
$$r_x(t)=x(t)\star x(-t)\tag{1}$$
where $\star$ denotes convolution. Taking the Fourier transform of $(1)$ gives
$$S_x(\omega)=X(\omega)X^*(\omega)=|X(\omega)|^2\tag{2}$$
$S_x(\omega)$ is the energy density of $x(t)$, and according to $(2)$ it equals the squared magnitude of the Fourier transform of $x(t)$. So if $x(t)$ is ...
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