New answers tagged dtft
1
I doubt that Hilmar will agree (maybe he will) but I will offer this answer as a counter to the one that has already been (prematurely) accepted by the questioner.
Hilmar's answer is wrong. It is factually wrong and analytically wrong.
1) Does the DTFT take only infinite input sequences?
No. There is nothing that prevents you from applying ...
3
We are analyzing a real signal with the DTFT. Since we are using a limited number of samples it's like we are transforming a finite signal.
No. The DTFT takes an infinite discrete time signal as an input but the spectrum is continuous. I think you are confusing the DTFT and the DFT. They are different things.
As I remember, the FT of a finite signal has ...
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1) Does the DTFT take only infinite input sequences?
No. There is nothing that prevents you from applying the formula to a finite sequence. However, a finite time sequence has an infinite spectrum, so you can't sample it without some amount of aliasing
2) If i apply the FT to a step function i get an infinite band spectrum.
Yes
If i sample the step ...
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$\displaystyle \hbox{ DTFT}(\alpha x_1 + \beta x_2) = \alpha\cdot \hbox{ DTFT}(x_1) + \beta \cdot\hbox{ DTFT}(x_2)$
Linearity property cannot applied here.
1
First of all, note that the discrete-time Fourier transform (DTFT) of a sequence is always periodic with period $2\pi$:
$$X(\omega)=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\tag{1}$$
This is obvious because $e^{-jn\omega}$ is $2\pi$-periodic in $\omega$.
Next, consider the inverse DTFT:
$$x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(\omega)e^{jn\omega}d\omega\...
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