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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


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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)\...


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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, ...


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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 ...


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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 ...


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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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