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## New answers tagged fourier-transform

8

The definition of the normalized discrete Fourier transform (DFT) for any signal $x[n]$ is $$F(k)=\frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-j2\pi k n /N}$$ The DC component of the DFT is evaluated at $k=0$. Given $$e^{-j2\pi 0 n /N}=e^0=1$$ The above simplifies to $$F(0)=\frac{1}{N}\sum_{n=0}^{N-1}x[n]=\bar{x}$$

0

When you sample the signal in time it becomes periodic in frequency. The spectrum gets from -32k$\pi$rad/s to +32k$\pi$rad/s gets repeated with repetition rate of 64k$\pi$rad/s. Since $72 = 64+8$ the value at 72k$\pi$rad/s is the same as at 8k$\pi$rad/s (and at 136k$\pi$rad/s and -56k$\pi$rad/s etc)

1

So how come the magnitude of the Fourier transform is still the same? Because the Fourier Transform integrates from $-\infty$ to $+\infty$. There is no finite "observation window" and you always integrate over the entire signal. The time shift doesn't change that. Once you consider a finite interval, you are windowing the signal and that does ...

1

It looks like that signal is doing a pretty good job of passing the correct signal, but it's leaving in some of the fundamental at 470kHz. You probably only have a suppression ratio of 20dB or so, which isn't good, but it isn't hopeless. The critical thing that I see in that is that you can draw a line through it, and the desired signal consistently goes ...

1

First of all: yes, oscillator imperfections are one of the systemic sources of noise, self-interference, unintended interference from others, and general annoyance in receivers. What you should do to figure out whether this is a problem here is start by realizing what a mixer does: It takes your LO signal and the RF bandpass signal and multiplies them. The ...

2

Looking at the y-axis (vertical scale) on your plot I see the values are really small. I am assuming that your input signal is of magnitude order one. This suggests that all you are seeing is roundoff error. This is to be expected for a sine signal if you are only plotting the real part of the FFT. For a sine signal the Fourier transform is pure imaginary....

1

Time domain representation As Max have presented in their answer, acoustic pressure variations in two dimensions, that you are interested in, is given as a function of time $t$ and spatial (Cartesian in our case) coordinates $x$ and $y$, denoted by $p \left( x, y, t \right)$. Now, consider taking a snapshot of the function at some arbitrary time $t_{1}$. You ...

0

There is "time involved", as you put it. The measurement of the pressure field is taken over a span of time. Pressure as in sound pressure is an alternating quantity. The human ear is sensible to alternation of air pressure, not to pressure itself. The measurement gathers pressure as a function of time and two-dimensional space. $$p(t,x,y)$$ This ...

2

Is this some sort of correlation or convolution? No. For that you'd need to build products from the elements. You only do differences. It should be meaningful for calculating ifft(X), but I can't see how... I'm not immediately recognizing these terms, and the length of 5 seems very odd¹; so, this is probably an algorithmic step in a specific FFT algorithm ...

1

One way to interpret the DFT, that I personally find most useful, is by comparing it to the DTFT (Discrete Tome Fourier Transform, which has an infinite input domain) of a repeated function. That is, imagine tiling your image out to infinity, and applying the DTFT, to obtain a periodic, continuous Fourier domain that you then sample. Because of the ...

2

Fourier Transform: $$X(\omega) = \int_{t=-\infty}^\infty x(t)e^{-j\omega t}dt$$ Replace $\omega$ with $-\omega$: $$X(-\omega) = \int_{t=-\infty}^\infty x(t)e^{j\omega t}dt$$ Replace $t$ with $-t$: $$X(-\omega) = \int_{-t=\infty}^{-\infty} x(-t)e^{j\omega (-t)}d(-t)$$ $$= \int_{t=-\infty}^{\infty} x(-t)e^{-j\omega t}dt$$ To describe this property intuitively,...

3

I am assuming the original EEG data was real (which seems reasonable) and therefore can conclude that the result given represents just the positive frequencies; since the complete spectrum would be complex conjugate symmetric in the case of a real signal and due to the redundancy it is common to not show the negative frequencies. The comments on the window ...

0

As you've shown via differentiation, the Fourier transform of all piecewise constant signals that have a jump discontinuity of height $2$ at $t=0$ is $$X(j\omega)=\frac{2}{j\omega}+c\delta(\omega)\tag{1}$$ where $c$ is some real-valued constant. The easiest way to see that $c=0$ for $x(t)=\textrm{sign}(t)$ is to note that $x(t)$ is an odd function, and, ...

1

By taking the derivative you loose all information about the DC value of the original signal. Any signal $$x(t) = 2\cdot u(t)- a$$ has the same derivative, regardless of what $a$ is. So you do have to calculate the DC value by hand, which is simply the mean of the signal. $$X(0) = \lim_{\tau \to \infty} \int_{-\tau}^{+\tau} \operatorname{sgn}(t) \, \mathrm{d}... 1 Your method correctly produces the term$$\frac{1}{j\omega}\frac{\sin(\omega/2)}{\omega/2}$$of the given solution, but you forget that by taking the derivative you lose information about any constant terms in x(t). However, looking at the graph it's easy to figure out that the constant term in x(t) is \frac12. The Fourier transform of that constant ... 1 The result of zero padding is the same as sampling the DTFT which is a continuous (and periodic) function in the frequency domain. We see this when we compare the formula for the DFT (Discrete Fourier Transform) and the DTFT (Discrete Time Fourier Transform): DFT:$$X[k] = \sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}$$DTFT:$$X(\omega) = \sum_{n=\infty}^{\infty}x[n]...

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Perceval's theorem is satisfied with you apply scale factor of $1/\sqrt(N)$ for both the forward and inverse transform. I.e. If $$X(k) = \frac{1}{\sqrt{N}}\sum_{n=0}^{N-1} x(n) e^{-j2\pi\frac{kn}{N}}$$ then $$x(n) = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1} x(n) e^{j2\pi\frac{kn}{N}}$$ and $$\sum_{n=0}^{N-1}|x(n)|^2 = \sum_{k=0}^{N-1}|X(k)|^2$$

0

From the definition it is clear that computing a DFT of a sinusoid over an interval the length of which is an irrational multiple of the period will lead to a poor result. The most significant component in the frequency domain will yield parameters $A$, $F$ and $\theta$ such that $$A\cos(2\pi Fx + \theta)$$ is not a good approximation of the input sinusoid. ...

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