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(\...


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

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 ...


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

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

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

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

You should use the synthesis equation of an impulse train with period $T$ (which is easy to derive): $$x(t)=\sum_{k=-\infty}^{\infty}\delta(t-kT)=\sum_{k=-\infty}^{\infty}\frac{1}{T}e^{jk\frac{2\pi}{T} t}\tag{1}$$ That is: the Fourier coefficients for all terms is a constant ($\frac{1}{T}$). Now assume that there are two impulses with different ...


5

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 ...


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 ...


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

I think the main issue is you are jumping ahead of yourself. You probably remember or read somewhere that the Fourier transform of a $rect$ function is a $sinc$ function. This is true; however, no where in this section does he mention Fourier transform! In fact, what he is doing is not Fourier transform. What he does in this section is to represent any ...


4

Let $g(x) = f(ax)$. Then, using a change of variables $y = ax$, $$\begin{align*} G(u) &= \int_{-\infty}^{\infty} g(x)\exp(-j u x)\,\mathrm dx\\ &= \int_{-\infty}^{\infty} f(ax)\exp(-j u x)\,\mathrm dx\\ &= \frac{1}{|a|}\int_{-\infty}^{\infty} f(y)\exp(- j (u/a) y)\,\mathrm dy\\ &= \frac{1}{|a|}F\left(\frac{u}{a}\right) \end{align*}$$ not $\...


4

This doesn't require any complicated computations! Look up some tables of identities about the Fourier Series. Which operation would you have to apply to $f$ so that its Fourier series would become $\sum_{-\infty}^{\infty}(c_k)^2 e^{2\pi ikx}$? For example, we know that derivating in the time domain is equivalent to multiplying the Fourier series ...


4

The book "Dr. Euler's Fabulous Formula: Cures Many Mathematical Ills", by P. Nahin, Princeton University Press, leads up to and contains an explanation of the Gibbs phenomena which might be suitable for someone with a good undergraduate university level math background.


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

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

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 ...


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 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 ...


4

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 ...


4

HINT: Going from your last equation, $$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$ This can be simplified further down by considering the following: \begin{align} e^{j2\pi (f_1T-n)} &= e^{j\pi (f_1T-n)}\cdot e^{j\pi (f_1T-n)}\\ 1 &=e^{j\pi (f_1T-n)}\cdot e^{-j\pi (f_1T-...


4

If the size $N$ of the DFT is even, only one "extremal" point (after fftshift) is Nyquist. If $N$ is even, you cannot have an arbitary $c_{-5}$ and $c_5$. They must add to be whatever your $c_{-5}$ term is. If the input to the DFT is purely real and if $N$ is even, consider the Nyquist point, at $c_{N/2}$ to be split in half. One half is the negative-...


3

$cos(t)$ and $cos(\pi t)$ are both periodic with periods $2\pi$ and $2$ respectively. To find the period of the sum we need to find an integers $n,m$ such that $\pi n = 2 m$ or $\pi / 2 = m /n $, which is not possible since $\pi$ is irrational. I think the Fourier series uses specific frequencies to form the signal. You can't necessarily put a frequency ...


3

Dimension in mathematics is defined as the number of independent components of a structure (with some generalizations, like fractal dimensions). Now which number of dimensions you assign to e.g a discrete image (as in picture) depends on what aspect of the image you are describing. As a vector space where each pixel is a component, you get as many dimensions ...


3

One way to see the equivalence is to first factor out the $0.5$ in your initial transfer function, i.e. $$ H(f) = 0.5(1 + e^{-i2\pi f}). $$ Then you can rewrite the $1$ and $e^{-i2\pi f}$ as follows: $$ H(f) = 0.5(e^{i\pi f}e^{-i\pi f} + e^{-i\pi f}e^{-i\pi f}) = 0.5(e^{i\pi f} + e^{-i\pi f})e^{-i\pi f}. $$ Your result then follows as a trivial ...


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