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The impulse response and frequency response are two attributes that are useful for characterizing linear time-invariant (LTI) systems. They provide two different ways of calculating what an LTI system's output will be for a given input signal. A continuous-time LTI system is usually illustrated like this: In general, the system $H$ maps its input signal $x(... 107 Negative frequency doesn't make much sense for sinusoids, but the Fourier transform doesn't break up a signal into sinusoids, it breaks it up into complex exponentials (also called "complex sinusoids" or "cisoids"): $$F(\omega) = \int_{-\infty}^{\infty} f(t) \color{Red}{e^{- j\omega t}}\,dt$$ These are actually spirals, spinning around in the complex plane:... 37 Let's say you had a spinning wheel. How would you describe how fast it is spinning? You'd probably say it's spinning at X revolutions per minute (rpm). Now how do you convey in what direction it's spinning with this number? It's the same X rpm if it's spinning clockwise or anti-clockwise. So you scratch your head and say oh well, here's a smart idea: I'll ... 29 FFT is actually not a great way of making a tuner. FFT has inherently a finite frequency resolution and it's not easy to detect very small frequency changes without making the time window extremely long which makes it unwieldy and sluggish. Better solutions can be based on phase-locked loops, delay-locked loops, auto correlation, zero crossing detection ... 15 Currently, my viewpoint (it is subject to change) is the following For sinusoidal repetition only positive frequencies makes sense. The physical interpretation is clear. For complex exponential repetition both positive and negative frequencies makes sense. It may be possible to attach a physical interpretation to negative frequency. That physical ... 15 Gabor filters are orientation-sensitive filters, used for texture analysis. The typically travel in packs, one for each direction. A gabor filter set with a given direction gives a strong response for locations of the target images that have structures in this given direction. For instance, if your target image is made of a periodic grating in a diagonal ... 15 The spectrum of a continuous tone is, as you said, of the form$\delta(f-f_0) + \delta(f+f_0)$: 2 impulses at frequencies$f_0$and$-f_0$. As a lowpass signal, this is said to have bandwidth$f_0$(the one-sided spectrum has components up to$f_0$). As a bandpass signal, it has zero bandwidth (there's nothing around the carrier frequency$f_0$). If you ... 14 Analog and digital First off, you need to understand what an "analog" signal is, and what a "digital" signal is, how they are different, and how they are similar. The term "analog" comes from the old distinction between "analog" and "digital" computers. A "digital computer", even a very primitive one of decades ago, has always been more or less what we ... 14 I would do a normalized autocorrelation to determine periodicity. If it is periodic with period$P$you should see peaks at every$P$samples in the result. A normalized result of "1" implies perfect periodicity, "0" implies no periodicity at all at that period, and values in between imply imperfect periodicity. Subtract the data sequence's mean from the ... 13 Bang on something sharply once and plot how it responds in the time domain (as with an oscilloscope or pen plotter). That will be close to the impulse response. Get a tone generator and vibrate something with different frequencies. Some resonant frequencies it will amplify. Others it may not respond at all. Plot the response size and phase versus the ... 13 One method that works if there's a relatively strong drum beat is to take the magnitude of the STFT of the waveform, and then auto-correlate it in only the time dimension. The peak of the auto-correlation function will be the beat, or a submultiple of it. This is equivalent to breaking up the signal into a lot of different frequency bands, finding the ... 13 Figure 1.(c) shows the Test image reconstructed from MAGNITUDE spectrum only. We can say that the intensity values of LOW frequency pixels are comparatively more than HIGH frequency pixels. Actually, this is not correct. The phase values determine the shift in the sinusoid components of the image. With zero phase, all the sinusoids are centred at the same ... 12 An FFT reports spectrum frequency peak or peaks (quantized by FFT bin size), which is different from musical pitch. It's possible for the perceived pitch frequency to be completely missing from an FFT spectrum. Some of the simplest guitar tuners just used low-pass or band-pass filtering and measured the time between zero-crossings. The reciprocal gives a ... 12 TL;DR? Google Scholar for harmonic partial separation. A good starting point would be sinusoidal modeling techniques that separate the signal into sines+noise (deterministic and stochastic) components. The deterministic component, made up of sines, can be resynthesized convincingly: http://mtg.upf.edu/files/projectsweb/sms-piano-original.wav http://mtg.... 12 The FFT can only be performed over a limited chunk of data. The basic math is based on the assumption that the time domain signal is periodic, i.e. your chunk of data is repeated in time. That typically results in a major discontinuity at the edges of the chunk. Let's look at a quick example: FFT size = 1000 points, Sample Rate = 1000 Hz, Frequency ... 11 suppose, Carrier signal frequency = 2800KHz message signal frequency = 3KHz Then you will get a signal that looks like this in the frequency plane. Obviously this is not to scale, but you get the idea. but what will happen if it is reversed ? i.e Message signal frequency = 2800KHz Carrie signal frequency = 3KHz please explain would happen here ? Then ... 11 Two remarks: I am assuming you are plotting the real (or imaginary) part of the Fourier transform. It is much more common to work with the magnitude or squared magnitude (power spectrum). The peak in the spectrum is a very poor measure of fundamental frequency (pitch). Take a piano note at 440 Hz, apply a notch filter to it to remove the 440 Hz component. ... 11 Posted for anyone who may find this useful... I created a picture that shows DFT frequency bin spacing for odd and even cases of N where N is the number of samples. FFTs usually operate on an even number of samples (the algorithm works by repeatedly breaking the problem into halves), so only the even case applies. The DC component (0*fs) is always part of ... 10 The impulse response is the response of a system to a single pulse of infinitely small duration and unit energy (a Dirac pulse). The frequency response shows how much each frequency is attenuated or amplified by the system. The frequency response of a system is the impulse response transformed to the frequency domain. If you have an impulse response, you ... 10 A Gabor filter is some parametrization of the idea of edges. This combines two somewhat contradictory ideas: an abrupt transition AND some fuzzy idea of where it is localized. It is mathematically a clever idea as it translates well in the Fourier domain: the Fourier transform of a Gabor is a Gaussian in Fourier space, and a Gaussian blob is the most ... 10 A beamformer is basically a spatial filter. It can be passive, just like a temporal filter. Instead of samples separated by time, they are separated by space. A passive temporal filter can be a bandpass that is "aimed" or "steered" at a particular frequency. For passive spatial filters (i.e. beamformers), the filter can be steered towards a particular ... 10 Well, first of all the Sound Level Pressure decreases by$6 \; \mathtt{dB}$when doubling the distance - this plays a big role. We do also have sound attenuation coming from our medium - air. Let's take a closer look onto sound absorption coefficient for different frequencies: Knowing that human speech is mostly concentrated at the range of$300\;\mathtt{Hz}...

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In many common applications negative frequencies have no direct physical meaning at all. Consider a case where there is an input and an output voltage in some electrical circuit with resistors, capacitors, and inductors. There is simply a real input voltage with one frequency and there is a single output voltage with the same frequency but different ...

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It is an edge detector. It just applies the Gabor Transform. The Gabor filter is basically a Gaussian (with variances sx and sy along x and y-axes respectively) modulated by a complex sinusoid (with centre frequencies U and V along x and y-axes respectively). See an example here.

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Make sure that your frequency doesn't reach values below 0 or above the half of your sample rate. Please post more information/code about how you generate your waveform! Chances are you are not doing it correctly. For example, if you want to generate a sine wave with a time-varying frequency $f(t)$ (for example to implement frequency modulation), ...

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You are right that the repetition is around 650 by how exactly do I compute that automatically? Seems like a peak-picking problem to me? Or is there some other methods that can be used? Yes, it's just peak-picking. Your period is the x value of the first strong peak: Your peaks are all similar in height, probably because you're doing the autocorrelation ...

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You can make a positive frequency spectrum quite simply (where fs is the sampling rate and NFFT is the number of fft bins). In the Matlab implementation of the FFT algorithm, the first element is always the DC component, hence why the array starts from zero. THis is true for odd and even values of NFFT. %//Calculate frequency axis df = fs/NFFT; fAxis = 0:df:...

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This comes from music terminology. The name "octave" comes from the fact that in the heptatonic musical scales (which are the prevalent scales in western music), the note with a 2:1 frequency ratio is the eighth note in the scale. For example, in the C major scale (C D E F G A B C) the eighth note is one octave above / has a 2:1 frequency ratio with the ...

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One trick, for even-length signals, is what to do with the "middle" sample. Here, I've split it half and half between each side of the FFT. The other trick is to ensure that you have the right amplitudes in the resampled signal. Here's it's a factor of 2. Try this in scilab: x = rand(1,100,'normal'); X = fft(x); XX = 2*[X(1:50) X(51)/2 zeros(1,99) X(51)/...

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Phase Noise and Frequency Noise are not two different noise sources, they are artifacts of the same noise, it is just a matter of what units you want to use. Frequency and Phase are directly related as frequency is phase changing with time, so if you have one you will always have the other; frequency and phase are related by derivatives and integrals: the ...

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