# Tag Info

9

This sum appears quite often in DSP. \begin{align} \sum_{n=0}^{N-1} \exp(-j\alpha n) &\stackrel{(a)}{=} \frac{1- \exp(-j\alpha N)}{1 - \exp(-j\alpha )}\\ &= \frac{e^{-j\alpha N/2}(e^{+j\alpha N/2}- e^{-j\alpha N/2})}{e^{-j\alpha /2}(e^{+j\alpha /2}- e^{-j\alpha /2})}\\ &\stackrel{(b)}{=} e^{-j\frac{\alpha}{2}(N-1)} \frac{\sin(N\alpha/2)}{\sin(\...

8

There are 4 versions of Fourier transforms that are all close cousins. It's all due to the basic property that "sampling in one domain corresponds to periodicity in the other domain". If a time signal is periodic, the frequency domain signal is discrete. If a frequency domain signal is periodic, the time domain signal is discrete. So you have four ...

7

We can figure out what's going on if we first understand a simple identity and then just compute the Fourier transform of the periodic function. A useful identity First let's prove that $$D(\omega - \omega') \equiv \int_{-\infty}^\infty dt \, e^{i (\omega-\omega') t} = 2\pi \, \delta(\omega - \omega')\,.$$ We just use a test function $\tilde{g}(\omega)$: \...

7

Consider a continuous-time periodic signal $x_1(t)$ whose fundamental period is $T_1$, fundamental radian frequency is $\omega_1 = \frac{2\pi}{T_1}$ and CTFS (continuous-time Fourier series) coefficients are: $$x_1 \longleftrightarrow c_k = \frac{1}{T_1} \int_{<T_1>} x_1(t) e^{-j\frac{2\pi k}{T_1}t} dt ~~ , ~~\text{for}~~ k=0,\pm 1, \pm 2...$$ We ...

7

Ok you are mixing the order of the transforms. When you decompose the argument $3t-6$ into $3\cdot(t-2)$ , then you should first scale $x(t)$ by $3$ and then shift the scaled result by $2$, that's where you have chosen the wrong order. Let's show the correct order: Let $x(t)$ be the original signal whose CTFT is $X(\omega)$, then define the following ...

7

I'm not understanding the comments. Of course you can do this. It is simply a matter of understanding what a DFT means, how to calculate DFT bin values, and how to interpret those bin values as continuous fourier series coefficients. First off, the plane you are looking at is the complex plane. Your points are a set of $N$ discrete samples. Each sample ...

6

In image processing, you are always dealing with discrete data, so the transform isn't really a Fourier transform, it is a discrete Fourier transform (DFT with a sum rather than an integral). The DFT can be understood as providing the coefficients of a Fourier series from which a periodic version of you're original image can be reconstructed. The "checker ...

6

I just finished writing an expository note on this topic recently. I hope you would find it useful. The link between Fourier transform (FT), Fourier series (FS), Discrete-Time Fourier Transform (DTFT) and Discrete Fourier Transform (DFT) is actually hidden in the Poisson summation formula $$\sum_{n=-\infty}^{\infty} f(x+nT) = \frac{1}{T} \sum_{n=-\infty}^{\... 6 alright, let's review a little bit of Euler before we get to the Fourier.$$ e^{j \theta} \ = \ \cos(\theta) \ + \ j \sin(\theta) $$from that you can get$$ \cos(\theta) = \frac{e^{j \theta} + e^{-j \theta}}{2} \quad\quad\quad \sin(\theta) = \frac{e^{j \theta} - e^{-j \theta}}{2j} $$so now let's look at Eq (1)$$ \begin{align} x(t) \ &= \ A_0 \ + \...

6

Well, when $m=k$ the integral is: $$\int_0^T e^{j(m-k)\Omega_0t} dt = \int_0^T e^{j \cdot 0 \cdot\Omega_0t} dt = \int_0^T dt = T$$ So as Juancho says in the comments, it's the same signal and so can't be orthogonal to itself.

5

This is related to Chirp Z-transform (CZT) (refer to the Bluestein's algorithm). Using this identity, the CZT can be expressed in terms of a convolution. Hence, it can be efficiently implemented using FFT.

5

Note that the value of the series $$\sum_{n=1}^{\infty}B_n\sin(nx)\tag{1}$$ for $x=0$ is always zero, regardless of the coefficients $B_n$ (assuming convergence). Furthermore, since the functions $\sin(nx)$ are odd, (1) can only represent odd functions. So you cannot represent general functions with the series (1). This is why you also need the cosine ...

5

@MattL suggested a nice, simple way to see the above result. But if you want to see the result in the normal analysis equations you mentioned, you can do like below. Say S(t) is a periodic train of impulses.So S(t) can be written as $$\ S(t)= \sum_{n=-\infty}^{\infty} \delta(t-nT)$$ Now if you take the fourier series of S(t),you can write S(t) as $$S(... 5 If you have two DFTs A[k] and B[k] (note the correct representation of a sinusoid at DFT bin number 1) A = [0,-j,0,0,j]; B = [1,1,1,1,1]; with the corresponding time-domain sequences a[n] and b[n] a = ifft(A); % [0, 0.38042, 0.23511, -0.23511, -0.38042]; b = ifft(B); % [1,0,0,0,0]; then the multiplication of the time-domain sequences c[n]... 5 The family of Fourier Transforms are specificaly developed for analysing frequency contents of the signals for which there is no definition of linearity or time invariance. Hence we can define the Fourier transform of any signal, as long as it's integrable (i.e. stable). On the other hand we can also define a Frequency Response H(\omega) for a particular ... 5 Yeah some of us can do it, you can speed up or slow down without affect the pitch, some guys call this applications of Time Stretch, there different ways to do it, you can do in frequency domain or time domain, you will need choose what is best for you, you will find some advantages and disadvantages of each. Time Domain: In Time Domain you can try some ... 5 A Hilbert transform is dispersive, as in any constant phase shift over frequency, which will distort a signal that has bandwidth. This is because a fixed DELAY, which would have no distortion, has linear phase versus frequency. If you fix the phase versus frequency to be constant (as we do in a Hilbert transform), the delay at each frequency will be ... 5 Yes, for time-limited functions it is possible to obtain the Fourier series coefficients by sampling the Fourier transform. This is the dual case of the more common form of the sampling theorem, stating that a band-limited function is fully characterized by its (time-domain) samples. The same is true in the discrete domain. Note that only for finite length ... 5 The Fourier series of the cycloid can be expressed in terms of the Bessel functions of the first kind:$$J_n(x)=\frac{1}{\pi}\int_0^{\pi}\cos(nt-x\sin t)dt,\qquad n\in\mathbb{Z}\tag{1}$$Using the cycloid parameterization$$y(t)=1-\cos t,\qquad x(t)=t-\sin t\tag{2}$$which results in a period of 2\pi and a maximum value of 2, the Fourier series of y(t) ... 5 If you did a continuous on off keying of a 10101010... pattern, then you would see sidebands as described since this is simply an up-conversion of the Fourier Transform of a 50% duty cycle square wave (moved to any carrier frequency). However if the data pattern for this case of a rectangular on-off keying was random, the resulting spectrum would be ... 4 The coefficients of the Fourier series that you have computed are, in effect, the spectrum of the periodic signal consisting of the sum of signals f(t) delayed in time or advanced in time by integer multiples of 1 second. Mathematically, the Fourier transform of a periodic function has impulses in it with the impulse amplitudes being the Fourier series ... 4 FFT and IFFT are algorithms that implement the (inverse) discrete Fourier transform (DFT). These transforms convert a signal into another representation, namely the frequency domain, and back. This other representation allows us to analyze certain properties of the signal and also to change these properites which would be hard to do in the original ... 4 I think part of the problem is an awkward and inconsistent naming convention. There are 4 flavors of Fourier Transforms depending on which domain is continuous or discrete (which maps to being aperiodic or perodic in the other domain). So we have Name Time Frequency Fourier Transform continous/aperiodic ... 4 The book "Blip, Ping & Buzz", by M. Denny (2007, John Hopkins University Press), explains signal processing, as used in sonar and radar, at the level of a supermarket science magazine. Some algebra required. More on the use of Fourier analysis than any theory. 4 Given  \left\{ x \left[ n \right] \right\}_{n \in M}  where  M  is the set of indices given for the samples of  x \left[ n \right] . The trivial solution (Which it would be great to have a faster more efficient solution is what I'm looking for) would be:$$ \arg \min_{y} \frac{1}{2} \left\| \hat{F}^{T} y - x \right\|_{2}^{2} $$Where  \hat{F}  is ... 4 I'm new to this exchange and I'm not sure how mathy you all get. I think the answer below is cool because it shows that in some sense the continuous-time Fourier transform is never periodic but that in another sense there are lots of ways to get periodic transforms. For the continuous-time Fourier transform on \mathbb{R}, both CMDoolittle's and Robert ... 4 You can't take the Fourier transform of a periodic signal, the integral diverges for all multiples of the period. This can be handled by the theory of distributions, but the Fourier series is a better fit. 4 You correctly figured out that the occurring integrals don't converge in the conventional sense. The easiest (and definitely non-rigorous) way to see the result is by noting the Fourier transform relation$$1\Longleftrightarrow 2\pi\delta(\Omega)$$By the shifting/modulation property we have$$e^{j\Omega_0t}\Longleftrightarrow 2\pi\delta(\Omega-\Omega_0)...

4

The common formulation of the forward DFT preserves energy (Parseval's theorem). This means that a longer constant magnitude sinewave input to a DFT, which has proportionally more energy, must be represented by a proportionally larger value in the DFT result, preserving that greater energy. Thus a factor of N, which scales with the length of the input. ...

4

Why are Fourier Analysis & Transform only applicable for LTI systems? That's simply not true. Won't Fourier analysis or Transform be possible? They are. The question is just whether they are meaningful. The point is that the continuous Fourier Transform (FT) is defined to integrate over all times from $t=-\infty$ to $t=\infty$; that works ...

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