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## Hot answers tagged dtft

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It's a complex exponential that rotates forever on the complex plane unit circle: $$e^{-j\omega t} = \cos(\omega t) - j \sin(\omega t).$$ You can think of Fourier transform as calculating correlation between $f(t)$ and a complex exponential of each frequency, comparing how similar they are. Complex exponentials like that have the nice quality that they can ...

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Note that the sequence $$x[n]=\frac{u[n-1]}{n}\tag{1}$$ is in $\ell^2(\mathbb Z)$ because $$\sum_{n\in\mathbb{Z}}|x[n]|^2=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}<\infty\tag{2}$$ but it is not in $\ell^1(\mathbb Z)$ since $$\sum_{n\in\mathbb{Z}}|x[n]|=\sum_{n=1}^{\infty}\frac{1}{n}=\infty\tag{3}$$ We know that for sequences in $\ell^1(\... 7 The difference is pretty quickly explained: the CTFT is for continuous-time signals, i.e., for functions$x(t)$with a continuous variable$t\in\mathbb{R}$, whereas the DTFT is for discrete-time signals, i.e., for sequences$x[n]$with$n\in\mathbb{Z}$. That's why the CTFT is defined by an integral and the DTFT is defined by a sum: $$X(j\Omega)=\int_{-\... 6 HINT: If you have a sequence$$\hat{x}[n]=\begin{cases}x[n],&n\text{ even}\\0,&n\text{ odd}\end{cases}$$then the DTFT of x[2n] can be written as$$\sum_{n=-\infty}^{\infty}x[2n]e^{-jn\omega}=\sum_{n=-\infty}^{\infty}\hat{x}[n]e^{-jn\omega/2}=\hat{X}(e^{j\omega/2})\tag{1}$$You can obtain \hat{x}[n] as$$\hat{x}[n]=\frac12\left(x[n]+(-1)^nx[... 6 As mentioned in Batman's answer, the condition of the sequence being absolutely summable is only sufficient but not necessary. The Fourier transform can be extended to$\ell_2$sequences, i.e. sequences for which $$\sum_{n=-\infty}^{\infty}|f[n]|^2<\infty$$ is satisfied. A further generalization is possible if you allow distributions and their ... 6 If you don't like thinking about imaginary numbers, complex numbers and functions, you can alternatively think of the complex exponential in the FT as just shorthand for mashing together both a sinewave and a cosine wave (of the same frequency) into a single function that requires less chalk on the chalkboard to write. 5 It can also have a$+$sign, there's no difference. Write down a part of the sum (around index$l=0$) and try to see that in both cases you're summing the same terms, just in a different order. More formally, take one of the two sums, transform the index by changing its sign ($k=-l$) and see what you get. 5 Yes your understanding is basically correct. The 1st paragraph (2 lines) expresses the fundamental relation between the DFS and the DFT of a finite-length sequence$x[n]$while the 2nd paragraph tries to put down the relation between the DFT$X[k]$of a sequence and the DTFT$X(e^{j\omega})$of it (assuming it exists). However this 2nd paragraph shall better ... 5 do you see anywhere in your book where this "DTFT" is defined for your w.s.s. process,$x[n]$? the DTFT is normally defined as $$X(e^{j \omega}) \triangleq \sum\limits_{n=-\infty}^{+\infty} x[n] e^{-j \omega n}$$ but for this infinite sum to converge, we normally require that $$\sum\limits_{n=-\infty}^{+\infty} \Big|x[n]\Big| < +\infty$$ which is ... 5 5 The DFT is a sampled version of the DTFT only for finite length signals. Otherwise, there is no point in comparing the DTFT with the DFT because you can only compute the DFT for finite length (or periodic) signals. Your example$x[n]=\sin(\omega_0n)$is an infinitely long sequence. For specific choices of$\omega_0it is periodic, but nevertheless, you can'... 5 \begin{align}X(e^{j\omega})&=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\\&=\sum_{n=-\infty}^{\infty}\left[\frac{1}{2\pi}\int_{0}^{2\pi}X(e^{j\Omega})e^{jn\Omega}d\Omega\right]\;e^{-jn\omega}\\&=\int_{0}^{2\pi}X(e^{j\Omega})\left[\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}e^{jn(\Omega-\omega)}\right]d\Omega\\&=\int_{0}^{2\pi}X(e^{j\Omega})\delta(... 4 Whether it's the Fourier Transform or the Laplace Transform or the Z Transform, etc. the exponential is the eigenfunction of Linear and Time-invariant (LTI) operators. if an exponential function of "time" goes into an LTI, an exponential just like it (but scaled by the eigenvalue) comes out. what the F.T. does is break down a general function into a sum ... 4 The Fourier Transform:f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(t)e^{i\omega t} dt\\ F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-i\omega t} dt$$converts a function to an integral of harmonic functions. You can think of these as sins and cosines because e^{i\theta} = cos(\theta) + i \sin(\theta). The Fourier Transform as a continuous form of the ... 4 You defined the signal vector as x = [1 2 3 2 1]. Since the DFT is defined by$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}$$the command fft(x) computes the DFT of the signal$$x[n]=\delta[n]+2\delta[n-1]+3\delta[n-2]+2\delta[n-3]+\delta[n-4]$$This signal is not symmetrical with respect to n=0, and it is not equal to the signal you computed the DTFT of. ... 4 The explanation by @Maximilian Matthé is a standard and formal approach for this question. But I think it is not intuitive and easy to understand the reason. In the following, I will try to explain from another aspect. First of all, periodicity means infinity. For a signal in the temporal/spatial domain, the periodicity is somewhat easy for realization, ... 4 According to Oppenheim and Schafer's "Discrete Time Signal Processing", the Goertzel algorithm will be more efficient than the FFT in computing an N point DFT if less than 2 Log_2 N DFT coefficients are needed. So if you need to compute 15 frequency points, the Goertzel will be more efficient for total number of samples N > 65536. Anything below ... 4 Cedron Dawg posted an interesting initial point in this answer. It begins with these steps:$$ \begin{align} U(\omega) &= \sum\limits_{n=0}^{+\infty} e^{-j \omega n} \\ &= \lim_{ N \to \infty } \sum\limits_{n=0}^{N-1} e^{-j \omega n}\\ &= \lim_{ N \to \infty } \left[ \frac{ 1 - e^{-j \omega N} }{ 1 - e^{-j \omega } } \right] \\ &= \... 4 This is pretty straight forward using the definition of the Discrete Time Fourier Transform (DTFT). The definition of the DTFT: $$X \left( {e}^{j \omega} \right) = \sum_{m = -\infty}^{\infty} x \left[ m \right] {e}^{-j \omega m}$$ Differentiating with respect to\omega: \begin{align*} \frac{d}{d \omega} X \left( {e}^{j \omega} \right) & = \... 4 The DTFT relationshipsx_{even}[n]=\frac12\left(x[n]+x^*[-n]\right)\Longleftrightarrow\textrm{Re}\left\{X(e^{j\omega})\right\}$$and$$x_{odd}[n]=\frac12\left(x[n]-x^*[-n]\right)\Longleftrightarrow j\,\textrm{Im}\left\{X(e^{j\omega})\right\}$$hold for any sequence x[n] for which the DTFT exists. There is no assumption about x[n] being real-valued ... 4 We are analyzing a real signal with the DTFT. Since we are using a limited number of samples it's like we are transforming a finite signal. No. The DTFT takes an infinite discrete time signal as an input but the spectrum is continuous. I think you are confusing the DTFT and the DFT. They are different things. As I remember, the FT of a finite signal has ... 4 Assuming that future vistors won't take the time to read all the comments, I'd like to give a very simple and straightforward interpretation of the discrete Fourier transform (DFT) as an approximation of the continuous-time Fourier transform (CTFT). First, we need to assume that the signal to be transformed decays sufficiently for large value of |t|. (If ... 3 If you have a continuous-time signal x(t), then the two signals you're talking about are$$\begin{align} x_c(t) &=x(t)\cdot\sum_{n=-\infty}^{\infty}\delta(t-nT) \\ &=\sum_{n=-\infty}^{\infty}x(t)\delta(t-nT) \\ &=\sum_{n=-\infty}^{\infty}x(nT)\delta(t-nT) \\ \tag{1} \end{align}$$and you define$$x_d[n] \triangleq x(nT)\tag{2}$$The first ... 3 DTFT is the Z-transform at the unit circle. So if z=re^{j\omega} then for DTFT r = 1. i.e If you have the Z-transform of a signal then plug-in e^{j\omega} for every z 3 Consider the case \ f(t) = 2 \cos(\omega_0 t) = e^{+i \omega_0 t} + e^{-i \omega_0 t}.\ Then$$ F(\omega) = \int\limits_{-\infty}^{+\infty} e^{i (-\omega + \omega_0) t} \ dt + \int\limits_{-\infty}^{+\infty} e^{i (-\omega - \omega_0) t} \ dt\\ $$When |\omega| \ne |\omega_0|, both integrands oscillate around zero, and the integrals are effectively zero.... 3 Hint: Expand e^{j \omega} using Euler's formula. This will enable you to express H(\omega) = R(\omega) + j I(\omega) where R and I are the real and imaginary parts and then the magnitude of the frequency response is just \sqrt{ R(\omega)^2 + I(\omega)^2}. 3 The long road computes the modulus of the DTFT. This works in general, but can be tedious. When the formula possesses some symmetry, like here (the two coefficients are the same), you can more efficiently factor a term, so that you recover a known Euler formula for the sine or the cosine, in the shape of e^{j\nu}+e^{-j\nu} or e^{j\nu}-e^{-j\nu}. Here, ... 3 The result of a DTFT is periodic, because any discrete-time signal has a continuous spectrum. This can be e.g. explained by the following: Let x(t) be a time-continuous signal. Now, making it discrete corresponds to multiplying it with a Dirac-Train:$$ x[n] = x(t)\sum_{n\in\mathbb{Z}} \delta(t-nT) $$Considering the Convolution theorem of the (... 3 You should consider the DTFT pair:$$\delta[n] \xrightarrow{\text{DTFT}} 1$$and the time-shifting property of the Fourier transform$$x[n-n_0] \xrightarrow{\text{DTFT}} X(e^{j\omega})e^{-j\omega n_0}$$plus the linearity of the FT. That is,$$ax_1[n]+bx_2[n] \xrightarrow{\text{DTFT}} aX_1(e^{j\omega})+bX_2(e^{j\omega})\$ Now consider the ...

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