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## Hot answers tagged fourier

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:...

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

This is one of the best problems to demonstrate Fourier Series properties, and specifically the time derivative property: $$\frac{d}{dt}x(t) \overset{FS}\longleftrightarrow j2\pi kf_0 X_k$$ Instead of computing the integral $$X_k=\frac{1}{T_0}\int_{T_0}x(t)e^{-j2\pi kf_0t}dt$$ which is time consuming, you can take the first derivative of your signal and ...

4

Relating the DFT to the FT (CTFT) is a big issue. Let's start with the basic definitions without any domain specification using a $\frac{1}{N}$ normalization on the DFT. $$FT(x(t))(f) = \int x(t) e^{-i2\pi t f } dt$$ $$DFT(x[n])(k) = \frac{1}{N} \sum x[n] e^{-i 2\pi \frac{n}{N} k }$$ They are obviously similar. Assuming they cover the same interval, ...

3

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 ...

3

(Ideal) square waves are often drawn in a misleading way, because the vertical lines don't actually represent a signal value. The square wave actually jumps instantanously between two values, creating a discontinuity. In other words, a function that is defined like this: $$f(t) = \begin{cases} 0\,\,\,t<0\\1\,\,\,t>=0\end{cases}$$ is discontinuous. ...

3

Let me put a practical answer with the following Matlab / Octave Code : L = 2*1000; % signal sample count n = 0:L-1; % discrete-time index Fs = 44100; % sampling frequency am = [1, 0.5, 2, 0.5, 0.3, 0.6, 0.1, 0.2]; % magnitudes of 8 components fm = [882, 2646, 4410, 6615, 8820, 10000, 13230, 15876]; % frequencies % Time domain ...

3

Sketching the signal always helps in such cases. Let's consider a cosine of period $T_0$, that is, of fundamental frequency equal to $f_0=1/T_0$. That is the signal on the left. As you can see on the right, $|A\cos(2\pi f_0 t)|$ does not share the same period with $A\cos(2\pi f_0 t)$. The period of the signal you are looking for is $T_0^{\prime}=T_0/2$ and ...

3

If I understand correctly, you want to verify the energy calculation in the frequency domain by computing the energy as $$E_x=\int_{-\infty}^{\infty}|X(f)|^2df\tag{1}$$ with $$X(f)=\mathcal{F}\big\{x(t)\big\}=\frac{A}{A+i2\pi f}\tag{2}$$ From $(2)$ we get $$|X(f)|^2=\frac{A^2}{A^2+(2\pi f)^2}=\frac{1}{1+\left(\frac{2\pi f}{A}\right)^2}\tag{3}$$ With $(... 3 Not a full solution but a few hints: Note that your function has an odd symmetry. Hence the even Fourier coefficients should be zero. They don't seem to be in your case and I think that's a consequence of how you integrate: you integrate from 0 to$T/2$instead of$-T/2$to$T/2$. If you did both halves, they would turn out with opposing signs and then ... 3 Matlab has no notion of "negative time" for the FFT: it interprets the first sample of the time domain sequence to be at$t=0$As far as Matlab is concerned your vectors are delayed Gaussians, delayed by either$D = 1023$or$D = 1024$samples. A delay in the time domain corresponds to multiplication in the frequency domain by$e^{-i \cdot 2 \pi \cdot D \...

3

Duality in DFT would mean that if $x[n]$ has DFT coefficients as $X[k]$, then DFT of $X[n]$ would be $Nx[(N-k) \mod N]$ Proof: Given, $$X[k] = \sum^{N-1}_{n=0}x[n]e^{-j\frac{2\pi}{N}nk}, k=0,1,2,3,...,(N-1)$$ If we take DFT of the sequence $X[n]$, then what we get is the following : $$Y[k] = \sum^{N-1}_{n=0}X[n]e^{-j\frac{2\pi}{N}nk} = N \left(\frac{1}{N}\... 3 Initial comment: duality refers to the strong similarity of mathematical expressions and properties in two different domains, here time and frequency. This is not only decorative or made to annoy learners. Duality helps a lot: one can derive results much faster, interpret classes of transformations more easily. To make this serious, you can check the ... 3 We can! In your example, N=4, and the DFT is real-valued, so you get X[k]=X[4-k], and that's true:$$\begin{align}k=0:\;X[0]&=X[4]=0\\k=1:\;X[1]&=X[3]=0\\k=2:\;X[2]&=X[2]=-4\end{align}$$Note that by definition the DFT is N-periodic, so X[N]=X[0]. 2 DTFT is a periodic waveform;$$H(e^{j \omega}) = H(e^{j (\omega + 2 \pi k)}).$$Hence for every frequency interval [\omega_1, \omega_2] of length 2\pi it will repeat itself; i.e., beginning and ending at the same magnitude (and phase) on that interval. 2 Your problem is at this step. You're doing half of a shortcut, but not the other half:$$ x_k = \frac{2A}{T} \int_{0}^{T/2} \frac{2t-T}{T} e^{-i2 \pi k f_0â€™ t } $$Start with the Fourier series definition (with the notation tidied up):$$ x_k = \frac{A}{T_0} \int_{-T_0/2}^{T_0/2} \frac{2t-T_0}{T_0} e^{-i2 \pi \frac{k}{T_0} t } dt$$What you did with this ... 2 I doubt that Hilmar will agree (maybe he will) but I will offer this answer as a counter to the one that has already been (prematurely) accepted by the questioner. Hilmar's answer is wrong. It is factually wrong and analytically wrong. 1) Does the DTFT take only infinite input sequences? No. There is nothing that prevents you from applying ... 2 OK, so your signal is described in one period, T_0, as$$x_{T_0}(t) = \left\{\begin{array}{ll} 0, & -T_0/2 < t \leq 0 \\ 1, & 0 < t \leq T_0/2 \end{array}\right.$$Let's keep it that way (without any other shortcuts like \prod or \mathrm{rect(\cdot)} etc). One way to find the Fourier coefficients of the even and the odd part of a signal ... 2 If you already know the - quite well-known :) - FT pair$$e^{-at}u(t) \longleftrightarrow \frac{1}{a+j2\pi f}$$then by setting a = 1/T, you get$$e^{-\frac{t}{T}}u(t) \longleftrightarrow \frac{1}{\frac{1}{T} + j2\pi f}$$Time delay property$$x(t-t_0) \longleftrightarrow X(f)e^{-j2\pi ft_0}$$can be used now by setting t_0 = T, and then$$x(t-T) = e^{-\...

2

Simple: Euler's formula: $$e^{i\theta} = \cos(\theta) + i \sin(\theta)$$ and $$e^{-i\theta} = \cos(\theta) - i \sin(\theta)$$ Add them together: $$e^{i\theta} + e^{-i\theta} = 2 \cos(\theta)$$ You should see that in your equation. From there it is usually stated: $$\cos(\theta) = \frac{ e^{i\theta} + e^{-i\theta}}{ 2 }$$ See my article The ...

2

The definition of the Fourier Transform is $$F(\omega) = \int_{-\infty}^{+\infty} x(t) \cdot e^{- j \omega t} dt$$ Since you x(t) is only non-zero on $[0,1]$ you can simplify this to $$F(\omega) = \int_{0}^{1} x(t) \cdot e^{- j \omega t} dt$$ Pop in the definition of x(t) between 0 and 1, solve the integral and you are done.

2

There is a Fourier transform property called as "differenciation in frequency domain" which is as follows: If the Fourier transform of $x(t)$ is $X(\omega)$, then the Fourier transform of $tx(t)$ is as below: $$\mathcal{F}(tx(t)) = j\frac{d}{d\omega} (X(\omega))$$ Where $\omega = 2\pi f$, I think you would be able to easily derive the answer you ...

2

The short answer is that the sinusoidal components are only orthogonal if they are integer multiples of each other. Any other choice of frequency would not result in $<s1, s2> = 0$, which can be easily confirmed by comparing such cases (integer and non-integer related sinusoids). Thus the choice of basis signals is discrete in units of their frequency (...

2

Most of the classical linear transformations (even filtering) may have three main types of scalings: natural or none: sum or integral does not have an explicit scaling factor (but the inverse may need one) in amplitude: because of linearity, at least one reference "unit" signal should have a unit amplitude in the transformed domain, for an easy ...

2

There are two factors involved here. One has been mentioned in a comment: your truncation error is larger for smaller values of $N$ because you truncate the time domain function at $NT$, where the sampling interval $T$ is constant. The other factor - which is the more important one here - is that you divide the FFT result by $N$. That's why you see a ...

2

If $X(f)=\mathcal{F}\{x(t)\}$ you have, in analogy with Eq. $(7)$, $$x(f)\Longleftrightarrow X(-t)\tag{1}$$ where we assume that functions with the independent variable $f$ are Fourier transforms of the corresponding functions with independent variable $t$. Applying $(1)$ to $$x(t-b)\Longleftrightarrow e^{-2\pi jbf} X(f)\tag{2}$$ gives $$x(f-b)\... 2 The DFT is discrete in both time and frequency domains with the same number of samples in each domain, this means the input to the transform and the output of the transform are both discrete and both have N samples. This is defined in the formula for the DFT where we have N samples in time indexed as n = 0 to N-1 and N samples in frequency indexed ... 2 Why does the DFT have only N components in it ? As you pointed out, the DFT could have an infinity of components in it, that repeat every N entries. To some extent, this is what aliasing is all about -- you're putting those "missing" bits back in. There's two reasons (or perhaps one and a half) to limit the DFT to N components, though: One is ... 1 If I understand correctly, the OP is comparing cascading 5 time domain signals one after the other as in [f_1, f_2, f_3, f_4, f_5] and taking the DFT to summing longer samples of the signals and then taking the DFT. The following is what we would expect to see in this case. The DFT of the sum of functions is equal to the sum of their DFT's:$$\sum_{n=0}^{...

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