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31

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 ...


16

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 ...


14

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 ...


12

Welcome to windowing. Nothing to do with William G. The easiest cure which works by brute force burying the errors in noise by using averaging is to sample a large number of cycles so that the boundary conditions do not predominate. I have not looked at your numerical results, but: Look at your second and third graphs. The waveforms that you displayed ...


11

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

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 ...


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

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

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

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

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

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 $l=h-k\neq mN$): $$\sum_{n=0}^{N-1}e^{-j\frac{2\pi}{N}nl}=\frac{1-e^{-j\frac{2\pi}{N}Nl}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-e^{-j2\pi l}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-1}{1-e^{-j\frac{2\pi}{N}l}}=0,\quad l\neq mN$$


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

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 ...


4

The key ingredient is that the base functions of the Fourier transform $\exp(i \omega t)$ are eigenfunctions of LTI systems. That means the LTI system can be represented as a diagonal linear operator in the Fourier basis. Or in other words: To apply an LTI system in frequency domain, you just multiply their frequency responses. And applying an LTI system to ...


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

The signals that we measure in MRI are a combination of signals from all over the object being imaged. It so happens that any signal (even if you simply make one up and draw a squiggle) is composed of a series of sine waves, each with an individual frequency and amplitude. The Fourier transform allows us to work out what those frequencies and amplitudes are. ...


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

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

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 ...


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

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.


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