15

If you want the shifted output of the IFFT to be real, the phase twist/rotation in the frequency domain has to be conjugate symmetric, as well as the data. This can be accomplished by adding an appropriate offset to your complex exp()'s exponent, for the given phase slope, so that the phase of the upper (or negative) half, modulo 2 Pi, mirrors the lower ...


12

Real-time low-latency partitioned convolution reverb with a long impulse response works by dividing the impulse response into unequally sized partitions. The shortest partitions (blocks) are at the beginning of the impulse response, and the partition length grows towards the end of the impulse response: Each partition length can be processed separately, ...


10

The resolution in the DFT is given by: $ \frac{{F}_{s}}{N} $. Hence you need 10e6 / 100 = 100,000 samples to get the resolution you want. You may bring the signal to the baseband (demodulation) and then you'll need a lower frequency of sampling to achieve what you want. Since the effective bandwidth of the signal is (Two Sided) 50 [KHz], it can be sampled at ...


9

The term Doppler Shift is actually a bit of a misnomer. The frequencies are not actually shifted but they are scaled (see http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html for definition of shifting vs. scaling). It's a relative change not an absolute one. Both time and frequency domains are scaled: when the source is moving towards you, the ...


8

Now, I would like to show what frequencies the speech has. However, I'm not sure what would be the best way to do that. It seems sometimes one calculates the absolute value of a Fourier transform, and sometimes power spectral density. If you want to attach physical meaning to your analysis, then go with the power spectral density, (PSD). This is ...


8

I did it for the DCT-I and DCT-II. At first I thought it is about circular convolution, but it is not. After desperate attempts to do it by myself I found the article: Convolution Using Discrete Sine and Cosine Transforms (Alternative Link). In this paper there is a derivation of circular convolution using DTT's, but it is constructed from sum of $C_1^{-1}$ ...


7

This is a nice question. I will try solving it using 2 approaches (Which are basically the same). The solution is the Least Squares Solution: $$ \hat{h} = \arg \min_{h} \frac{1}{2} \left\| h \ast x - y \right\|_{2}^{2} $$ We assume data is given in finite discrete form (As it is in practice). The convolution is done in valid mode (Like in MATLAB valid ...


7

Clipping is a non-linear operation. Consider a system that clips all input whose absolute value is larger than 1: $$ y(t) = \begin{cases}x(t)\text{ if }|x(t)|\leq 1 \\ 1 \text{ if } x(t) > 1 \\ -1 \text{ if } x(t) < -1 \end{cases}$$ While this system is linear for small inputs, it becomes non-linear as soon as the input becomes large enough. Non-...


7

Consider a liner discrete-time system. Assume we can define it in terms of an input-output relation as follows (you can assume a more general model but it is enough for our purpose): $$a_0y[n]+a_{1}y[n-1]+\cdots+a_{N}y[n-N]=b_0x[n]+b_{1}x[n-1]+\cdots+b_{M}x[n-M]\tag{1}$$ When the coefficients $\{a_i\}$ and $\{b_i\}$ are constant, we call it a finite-order ...


7

This is just "faking" the magnitude response of an IIR filter. The output's magnitude spectrum looks just like it has been filtered by the IIR filter with the given frequency response. Although it may somehow work, there are some limitations: Frequency-domain filtering is usually much more computationally demanding. It is not for real-time. The problem ...


6

The FFT of the FFT bascially gets you the original signal again, it's just scaled and time flipped, i.e. $$FFT\left \{ FFT \left \{ x(n) \right \} \right \} = N \cdot x(-n)$$ That's a simple consequence of the fact that the forward and inverse Fourier Transforms are almost the same. I have no idea what all the acronyms in the second question mean.


6

You get the exact same information you will receive if you analyse a 1-D signal using the Fourier analysis tools, for example. To illustrate this consider the following examples We perform Fourier Transform on it and obtain the following spectrum As you can see there are two symmetric dots representing the frequency that is present in the image and the ...


6

Regarding example 1: first of all, either the fft or the ifft needs to be normalised by the number of sampling points as you have done (actually you can normalise each by a factor of the square root of the number of points, it is just a meter of definition). However, in your case the ifft is half the length of the fft you performed. Hence your signal is ...


6

There can't be. One man's signal is another man's noise. In fact, a communication system making the absolute most of a bandwidth would be spectrally white, just like white noise, and hence be indistinguishable from noise to anyone but the receiver for that specific system.


6

As @Hilmar mentioned I think you get confused between Square wave and Rectangular function. In Wikipedia about Square Wave : A square wave is a non-sinusoidal periodic waveform in which the amplitude alternates at a steady frequency between fixed minimum and maximum values, with the same duration at minimum and maximum. Which its Fourier Transform is only ...


5

You need to use the Convolution Theorem. Namely, when you want to apply an LPF Kernel in one domain (Using convolution) you can use multiplication in the other domain. The tricky part is the dimensions of the signals (Since we're dealing with Discrete signals). Since the LPF Kernel in the convolution domain might be different then the size of the signal it ...


5

If you have image in the spatial domain, in order to calculate its DFT transform you should use fft. Once you have its DFT in order to get back to the spatial domain use the function ifft. Either way, the DFT transform gives you the data in the [0, 2pi] axis. Use fftshift to move it into the [-pi, pi] domain as seen in the paper. Good Luck. MATLAB Code ...


5

The frequency resolution $f_\Delta$ is $$ f_\Delta = \frac{f_\mathrm{s}}{N}, $$ where $f_\mathrm{s}$ is the sampling frequency and $N$ is the FFT size. So $$ N = \frac{f_\mathrm{s}}{f_\Delta} = \frac{10^7}{100} = 10^5. $$ This means your calculation is correct. Assuming that you don't zero-pad the FFT input vector, $N$ is the number of samples you should ...


5

Your calculation of the SNR is right and the DFT Bin Resolution is also OK. One thing you're missing is the effective resolution due to "Windowing" effect. The DFT of finite number of samples is interpolated by a Dirichlet Kernel (Like Sinc). It means you resolution is also limited by the main lobe width of the Sinc which is proportional to the inverse of ...


5

assuming there were no harmonics (frequency components other than the fundamental that are at integer multiples of the fundamental frequency) in the first place, that means the input to the clipping operation is a pure sine wave. a Fourier series analysis of that periodic sinusoid will show non-zero amplitude at only one frequency, the frequency of the ...


5

This sort of filtering is done all the time, but it doesn't have the effect you think it should. Suppose you have an IIR filter with an impulse response of $h[n]$ which is represented in the $z$ domain as: $$ H\left(z^{-1}\right) = \frac{h_n\left(z^{-1}\right)}{h_d\left(z^{-1}\right)} $$ where $h_n$ is the numerator polynomial in $z^{-1}$ of order $N$ and $...


5

Your parameters aren't correct for producing a whole number of cycles for each component. For each $i$ the value of $ \frac{\omega_{i}}{\omega_{s}} N $ has to be a multiple of $ 2 \pi $. Hope this helps. Ced Followup: Suppose your signal is $$ x[n] = A \cos( \alpha n + \phi ) $$ If $ \phi \neq 0 $ there is a possible solution which will give you a ...


5

There are a few different ways to interpret the math of the DFT. The one I find most suitable to explain this and other properties of the DFT assumes that the continuous Fourier domain was sampled, which causes the time domain to become periodic (just like the signal was sampled, causing the Fourier domain to become periodic). If your signal contains a sine ...


5

In simple terms, Image shown here speaks for itself. Before speaking about Fourier Transform black magic lets understand idea behind it. Work of the Mathematician Joseph Fourier demonstrated any arbitrary periodic (this is important) signal can be decomposed into bunch of sine-waves at with their corresponding amplitude and relative phases. Essentially, ...


5

Let's assume we have a sample rate of $f_s=10 kHz$ and FFT size of $N=1000$ Your bin spacing is $\delta f = 10 Hz$. It's simply the sample rate divided by the FFT size. That's all there is to it. Keep in mind that the FFT requires both the time domain and the frequency domain signal to be perodic. Most signals are not, so there needs to be some sort "...


4

Windowing artifacts can be considered as due to a loss of information about a larger data set (e.g.what the window cut out or zeroed). The "spread" of the Sync is related to the shortness of the data window (what's left after the information loss). Thus other information about the original data from outside the FFT results can be used to help "solve" this ...


4

Each DCT output bin is a (weighted) correlation against a cosine function of a certain frequency. A negative value would represent a negative correlation, e.g. something in the image data is of the opposite phase to that cosine (e.g. maybe dark when the cosine is 1, and light when the cosine is -1, instead of vice-versa for a positive correlation).


4

If I understand correctly, you are taking the entropy in the image from the distribution of black and white pixels. In image (a) you have 50% black, 50% white, so the entropy is 1 bit. In image (b) you have 85% black, 15% white, so the entropy is 0.6 bits. If you make another image (c) with black and white stripes equal in width to the black stripes in ...


4

Remember that as part of computing the gradients of an image, you are really just convolving your image, with (two) spatial filters. This then gives you two 'gradient' images, which you can then combine. For example, to compute the x-direction-gradient-image $G_x$, you would convolve your image with a filter $f_x$ that might look like this $\begin{bmatrix} ...


4

This can be explained by the Convolution theorem As you state, in the time domain you have the kernel (f) multiplied by a sinusoid (g). If we take the Fourier transform of f·g (i.e., F(f·g)), this is equivalent to the convolution between F(f) and F(g) (i.e., F(f)*F(g)). Note that the Fourier transform of the sinusoid is proportional to two delta functions ...


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