Hot answers tagged

30

It's because the simultaneous presence of two sinusoidal signals with the same frequency and different phases is actualy equivalent to a single sinusoidal at the same frequency, but, with a new phase and amplitude as follows: Let the two sinusodial components be summed like this : $$ x(t) = a \cos(\omega_0 t + \phi) + b \cos(\omega_0 t + \theta) $$ Then ...


15

Your work is OK except for the problem that the Fourier transform of $\cos(2\pi f_0 t)$ does not exist in the usual sense of a function of $f$, and we have to extend the notion to include what are called distributions, or impulses, or Dirac deltas, or (as we engineers are wont to do, much to the disgust of mathematicians) delta functions. Read about the ...


10

Let $h(t)$ denote the impulse response of an LTI system. Then, for any input $x(t)$, the output is $$y(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)\,\mathrm d\tau.$$ In particular, the response to the input $x(t) = \exp(j2\pi ft)$ is $$\begin{align} y(t) &= \int_{-\infty}^\infty h(\tau)\exp(j2\pi f(t-\tau))\,\mathrm d\tau\\ &= \exp(j2\pi ft)\int_{-\...


9

Time invariance plays a huge role in nature. Most systems (including your ear/brain) don't have an absolute time reference but treat all points in time equally. That results in a preference for the description of these systems with essentially time invariant basis functions, which is what (complex) sinusoids are. For linear time invariant systems, the ...


8

Complex exponentials (with decaying sinusoids being the real part) are the solutions to certain types of low-order linear differential equations. Modeling simple natural phenomena with these low-order linear differential equations turns out to be surprisingly useful. "Why did the real world turn out this way?" might be a good question for philosophers. ...


8

All real-life signals are finite energy. The universe contains a fixed (and finite) quantity of energy, which has been unchanged since it came into being. A signal's energy is given by $E =\int_{-\infty}^{\infty}|x(t)|^2dt$ Thus, the only way to make a signal's energy go to infinity is to allow it to continue for infinite time or reach an infinite peak ...


8

As was pointed out in a comment above, this is a simple consequence of the dynamics of a pendulum. There's nothing particularly signal-processing-related about this problem, just some simple physics and trigonometry. For a pendulum that is displaced from its angular equilibrium point by an angle of $\theta_0$ with an angular velocity of $\omega_0 = 0$ at $t ...


8

I'll use the non-unitary Fourier transform (but this is not important, it's just a preference): $$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-i\omega t}dt\tag{1}$$ $$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{i\omega t}d\omega\tag{2}$$ where (1) is the Fourier transform, and (2) is the inverse Fourier transform. Now if you formally take the ...


7

The FFT results actually do reveal everything about the original injected frequencies. But because the injected frequencies were not exactly periodic in the FFT aperture length, the frequencies have been convolved into Sinc waveforms due to this non-periodic-related windowing, and then resampled. To get the original frequencies back, you may need to ...


7

The cross pattern is typically a border effect, due to the periodicity induced by the standard implementation and hypotheses behind the Fast Fourier transform, when the image lacks periodicity from the right to the left, and the bottom to the top. In other words: if two opposite borders lacks continuity in values (when glued together), artifacts show. The ...


6

This is not a complete answer by any means, and I don't expect it to be accepted, but I also think there's significant educational value in this response. So I may infer, and I may be wrong(I hope I am), that if the number of samples in a complete waveform period are not a power of 2, the FFT of the same does not reveal anything without some kind of an ...


5

What might seem intuitive differs greatly between individuals. But let's start with a few basic things about a Fourier transform that people who have studied it might know about it. Intuitively or not. Basic concept 1: Something symmetric around t=0 in the time domain is strictly real in the frequency domain (as only cosine functions and DC are purely ...


5

The short answer is yes, if you have the Laplace or Z-transform of a function you do not need the Fourier transform. This is because the CFT is a special case of the Laplace transform and the DTFT is a special case of the Z transform. The Fourier transform is used to find the complex sinusoids that compose a function, whereas the Laplace transform finds ...


5

Your first solution using the properties of the Fourier transform is correct. Your second solution is wrong, because you forgot to include the unit step function. Your function $g(t)$ should be defined by $$g(t)=e^{-t}u(t)\tag{1}$$ which gives for $g(2t-1)$ $$g(2t-1)=e^{-(2t-1)}u(2t-1)=e^{-(2t-1)}u\left(t-\frac12\right)\tag{2}$$ Consequently, the Fourier ...


5

Disclaimer: I know this topic is older, but if one is looking for "fast accurate convolution high dynamic range" or similar this is one of the first of only a few decent results. I wanna share my insights I got on this topic so it might help somebody in the future. I apologize if I might use the wrong terms in my answer, but everything I found on this topic ...


5

some things I've read online show the Fourier transform should look more like a complex exponential. Don't believe everything you read online! A signal "contains all frequencies" is a vague description; it can apply to any finite-energy signal $x(t)$ whose Fourier transform $X(f)$ is nonzero for all values of $f$, $-\infty < f < \infty$. On the ...


5

Just use the formula for the geometric series (I use $l=h-k\neq mN$): $$\sum_{n=0}^{N-1}e^{-j\frac{2\pi}{N}nl}=\frac{1-e^{-j\frac{2\pi}{N}Nl}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-e^{-j2\pi l}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-1}{1-e^{-j\frac{2\pi}{N}l}}=0,\quad l\neq mN$$


4

Then just use a table of Fourier transform pairs to see that $\delta(t) \leftrightarrow 1$, and variable substitution ($f_1 = f+f_0$ and $f_2 = f-f_0$), to get what you need.


4

White noise implies no correlation between samples of the noise, even consecutive samples. Colored noise, therefore, implies that there is correlation of some sort between the noise samples, which in turn implies that we can take advantage of that correlation to get rid of some of the noise. Beyond that, there is not a lot that we can say about what it ...


4

Each frequency burst is equivalent to a rectangular window on an infinite sinusoid. A rectangular window in the time domain is the same as circular convolution with a Sinc function in the frequency domain. Thus, you end up with a spike with side humps (the spike for the sinusoid, and the humps due to the convolution with the transform of the rectangular ...


4

Let's say that your signal is composed of two parts: even and odd: $$s(t)=s_e(t)+s_o(t)$$ We also know following properties of this type of functions: Even: $f(-x)=f(x)$ Odd: $f(-x)=-f(x)$ Let's calculate the time inversion of your signal $s(-t)$ and apply above properties: $$s(-t)=s_e(-t)+s_o(-t)=s_e(t)-s_o(t) $$ So now let's do the trick and add ...


4

For power signals $x(t)$ and $y(t)$, the function $$R_{xy}(\tau)=\lim_{T\rightarrow\infty}\frac{1}{2T}\int_{-T}^{T}x(t)\bar{y}(t+\tau)dt\tag{1}$$ is the cross-correlation of $x(t)$ and $y(t)$. So the expression you're asking about is the cross-correlation of $x(t)$ and $y(t)$ evaluated at lag $\tau=0$: $$R_{xy}(0)=\lim_{T\rightarrow\infty}\frac{1}{2T}\...


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

The plot is of $$\mid X\left(i\omega\right) \mid = \sqrt{\left(\frac{1}{a+j\omega}\right)\left(\frac{1}{a-j\omega}\right)} = \frac{1}{\sqrt{a^2 + \omega^2}}$$ against $\omega$ In particular $\omega$ can be equal to $-a$. This checks out with Wolfram alpha


4

You're right, the first Fourier transform correspondence in your reference is wrong. It should be $$\mathcal{F}\{e^{-\alpha n}u[n]\}=\frac{1}{1-e^{-\alpha}e^{-j\omega}}\tag{1}$$ You just need to substitute $a=e^{-\alpha}$ and use the formulas you know.


4

The Cauchy Schwarz inequality states that: $$ \left|\int_{-\infty}^{\infty}g_1(t)g_2(t) dt\right|^2 \leq \int_{-\infty}^{\infty}|g_1(t)|^2 dt \int_{-\infty}^{\infty}|g_2(t)|^2 dt $$ I'm going to assume that $f(t)$ is real, just to make the math a little easier. From the above we can write: $$ \left|\int_{-\infty}^{\infty}f(t)f(t-\tau) dt\right|^2 \leq \int_{...


4

The answer to your last question is definitely 'no'. The point hotpaw2 makes in his answer is very relevant: the FFT is an efficient implementation of the DFT, and there are no equivalently efficient implementations for the numerical computation of the $\mathcal{Z}$-transform or the Laplace transform. But that's not the only reason. There are important ...


4

In the Fourier transform, the basis functions are complex exponentials. These functions are perfectly localized in the frequency domain, i.e., they exist at one frequency, but they have no time localization because of their infinite duration. The localization of a function depends on its spread in time and frequency. A complex exponential has zero spread in ...


4

Sparsity concept is extensively being used in computer vision and image processing. The Idea is that natural image can be pretty sparse when it is transformed to different bases. this bases can be predefined, e.g. FFT,DCT or can be learned from the image, e.g. sparse coding. Here are a few well know example of algorithms that uses the sparsity assumption on ...


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