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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, by ...

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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 Fourier ...

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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 ...

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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 ...

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. ...

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 $\tilde{g}(\omega)$: \...

6

Answer : When $x_2 = [-1024:1:1023]$, then $x_2[n]$ satisfies the condition $x_2[n] = x_2[(N-n)\mod N]$. That is why when $x_2 = [-1024:1:1023]$, then the FFT is real and hence the imaginary part is $0$. If you see the scale of the $y$-axis for imaginary part of $x_2$ plot, it is of the order of $10^{-17}$ which is almost $0$ in MATLAB. Detailed Explanation:...

6

I am thinking of two numbers. They add to the number 15. Tell me what the two numbers are.

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Note that each complex pole $s_{\infty}=\sigma + j\omega$ of the transfer function $H(s)$ contributes to the system's impulse response a complex exponential of the form $$e^{s_{\infty}t}=e^{\sigma t}e^{j\omega t}$$ The term $e^{\sigma t}$ is real-valued and represents exponential damping (assuming that the system is causal and stable, i.e. $\sigma<0$), ...

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 intuitive answer is that an impulse in time at t=0 contains all frequencies of equal magnitude, so applying an impulse to an LTI system is the same as applying all frequencies at once, thus the result is the response of the system to all frequencies, i.e., the frequency response. For a real world example, you can find the total frequency response of a ...

5

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: $$\left|\int_{-\infty}^{\infty}f(t)f(t-\tau) dt\right|^2 \leq \int_{... 5 The short answer is yes, if you have the Laplace or Z-transform of a function you do not need the Fourier transform. This is because the CFT is a special case of the Laplace transform and the DTFT is a special case of the Z transform. The Fourier transform is used to find the complex sinusoids that compose a function, whereas the Laplace transform finds ... 5 The DFT Matrix for Non Uniform Time Samples Series Problem Statement We have a signal  x \left( t \right)  defined on the interval  \left[ {T}_{1}, {T}_{2} \right] . Assume we have  N  samples of it given by  \left\{ x \left( {t}_{i} \right) \right\}_{i = 0}^{N - 1} . The samples time  {t}_{i}  is arbitrary and not necessarily uniform. We're ... 5 In the Fourier transform, the basis functions are complex exponentials. These functions are perfectly localized in the frequency domain, i.e., they exist at one frequency, but they have no time localization because of their infinite duration. The localization of a function depends on its spread in time and frequency. A complex exponential has zero spread in ... 5 Your first solution using the properties of the Fourier transform is correct. Your second solution is wrong, because you forgot to include the unit step function. Your function g(t) should be defined by$$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$$5 Well this goes to show that Fourier series is just approximation that gets more and more correct when you add more harmonics. Take a look at this: \dfrac{4\sin\theta}{\pi} is just first harmonic. Harmonics are integer multiple of base frequency as you can see: \sin3\theta, \sin5\theta etc. And the more of them you add to the first harmonic the more ... 5 In general, the characteristic function of a random variable is related to the fourier transform of the distribution as follows:$$\varphi_Z(-\omega) = \mathscr F \big\{ f_Z(z) \big\}$$Why? Because, fourier transform of a PDF f_Z(z) is:$$\mathscr F \big\{ f_Z(z) \big\} = \int^{\infty}_{-\infty} f_Z(z)\cdot e^{-j\omega z} \ \mathrm dz$$And ... 5 You are simply seeing the effect of the time delay due to being offset by a half a sample (a delay in time is a linear phase in frequency). If you have an odd number of samples then you can implement what would be a non-causal zero-delay signal since you can have the same number of samples for positive time as negative time. If a signal is symmetric in one ... 5 The Fourier series of the cycloid can be expressed in terms of the Bessel functions of the first kind:$$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}$$which results in a period of 2\pi and a maximum value of 2, the Fourier series of y(t) ... 4 Let's say that your signal is composed of two parts: even and odd:$$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) $$So now let's do the trick and add ... 4 JPEG is far simpler. It divides the image into 8x8 pixel blocks, and processes each using a Discrete Cosine Transform. The results are quantised and then encoded. The quality is fixed by the encoder. JPEG2000 uses a 2D wavelet function, the output of which is four "images", each a quarter the size of the original. One of those is actually an image, ... 4 If X(f) is the Fourier transform of x(t), then the Fourier transform of x^*(t) is given by X^*(-f). That's why for real signals X(f)=X^*(-f) holds. 4 It depends. Normalized autocorrelations have no units because the units are divided out as part of the normalization process. Non-normalized autocorrelations have the original data's units squared. 4 For power signals x(t) and y(t), the function$$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

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