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63

The Laplace and Fourier transforms are continuous (integral) transforms of continuous functions. The Laplace transform maps a function \$f(t)\$ to a function \$F(s)\$ of the complex variable s, where \$s = \sigma + j\omega\$. Since the derivative \$\dot f(t) = \frac{df(t)}{dt} \$ maps to \$sF(s)\$, the Laplace transform of a linear differential equation is ...


39

Devices using the Fourier Transform It is very difficult for an electronic device to decompose a signal in different frequencies. It's not. There's actually quite a few devices that do that, explicitly. First of all, you'll have to make a difference between the continuous Fourier transform (which you probably know as \$\mathcal F\left\{x(t)\right\}(f)=...


30

Sinusoidal waves don't have harmonics because it's exactly sine waves which combined can construct other waveforms. The fundamental wave is a sine, so you don't need to add anything to make it the sinusoidal signal. About the oscilloscope. Many signals have a large number of harmonics, some, like a square wave, in theory infinite. This is a partial ...


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This is a windowing artifact. The linked code pads out a 10,000 sample signal with zeroes so that the length is a power of two. %% Author :- Embedded Laboratory %%This Project shows how to apply FFT on a signal and its physical % significance. fSampling = 10000; %Sampling Frequency tSampling = 1/fSampling; %Sampling Time L = 10000; ...


23

Pentium100's answer is quite complete, but I'd like to give a much simpler (though less accurate) explanation. The reason because sinewaves have (ideally) only one harmonic is because the sine is the "smoothest" periodic signal that you can have, and it's therefore the "best" in term of continuity, derivability and so. For this reason is convenient to ...


23

The concept is good, but you will find it is not so simple in practice. Pitch is not simply the predominant tone, so there's problem number 1. The FFT frequency bins can't hit all (or even multiple) tones of the musical scale simultaneously. I would suggest playing with an audio program (for example, Audacity) that includes an FFT analyser and tone ...


22

a triangle wave is continuous Quote from here: - The triangle wave has no discontinuous jumps, but the slope changes discontinuously twice per cycle Having the slope change discontinuously also means an infinite range of sinusoidal components. For instance, if you time integrated a square wave you produce a triangle wave but, all the hamonics of the ...


22

instructor said that since the triangle wave is continuous it can be represented by a finite number of sine You either didn't get this right or the instructor misspoke. It's not sufficient for the signal itself to be continuous, but all derivatives must be continuous too. If there is any discontinuity in any derivative, then the repeating signal will have ...


20

As you know (since you mentioned the Fourier transform), a square wave can be represented (well, almost -- see below) as the sum of an infinite series of sine waves. But it would not be possible to send a true square wave through any real physical antenna: As you move along the infinite series, the frequencies get higher and higher, and eventually you'll ...


18

You can decompose any waveform into an infinite series of sine waves added together. This is called Fourier analysis (if the original waveform is repeating) or Fourier transform (for any waveform). In case of a repeating waveform (like a square wave), when you do Fourier analysis you find that all the sines that compose the waveform have frequencies that ...


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Laplace transforms may be considered to be a super-set for CTFT. You see, on a ROC if the roots of the transfer function lie on the imaginary axis, i.e. for s=σ+jω, σ = 0, as mentioned in previous comments, the problem of Laplace transforms gets reduced to Continuous Time Fourier Transform. To rewind back a little, it would be good to know why Laplace ...


12

Math proof: Take a function made up of the weighted sum of a finite series of sine/cosine components. Its derivative is also a weighted sum of a finite series of sine/cosine components. Same if you derivate any number of times. Since sine and cosine are continuous, the function and all its derivatives are continuous. Thus, a function having a ...


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You can't get much more "bare metal" and "hardware" than a set of vibrating reeds. http://www.stichtco.com/freq_met.htm So what hardware does a fourier transform, a bunch of resonant systems can do that


11

The Fourier series: \$ V_t = \dfrac{a_0}{2} + \displaystyle \sum_{i=1}^{\infty}[a_i sin(i \omega_0 t) + b_i cos(i \omega_0 t) ] \$ The term \$\dfrac{a_0}{2}\$ is a constant, that's the DC level. It could also have been written without dividing by two, but this is the convention. The terms of the infinite sum are the sum of a weighted sine and a ...


10

The 1/N scaling factor is almost arbitrary placed. An unscaled FFT followed by an unscaled IFFT using exactly the same complex exponential twiddle factors multiplies the input vector by scaler N. In order to get back the original waveform after an IFFT(FFT())round trip (thus making them inverse functions), some FFT/IFFT implementation pairs scale the FFT ...


10

Fourier transforms are for converting/representing a time-varying function in the frequency domain. A laplace transform are for converting/representing a time-varying function in the "integral domain" Z-transforms are very similar to laplace but are discrete time-interval conversions, closer for digital implementations. They all appear the same because ...


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I'm not sure what you're looking at, but you need to understand the exmples you link. None of them use the truth-value within the actual filtering. It's there so you have something to compare to with regard to the filter output. Here is the simple script: import random # intial parameters iteration_count = 500 actual_values = [-0.37727 + j * j * 0.00001 ...


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How long is the ground lead on your scope probe? your scope probe should be in the X10 position and be properly compensated. Remove the long ground lead that came with the scope probe and also remove the probe grabber hook from the front of the probe. Grab some #22 or #24 bare wire and wrap 2 or 3 turns around the exposed metal ring at the front of the ...


10

The Laplace variable \$s\$ relates to Fourier's \$j\omega\$ as follows: $$ s = \sigma + j\omega $$ Fourier transform can be seen as a Laplace transform when \$\sigma=0\$. The \$\sigma\$ allows the Laplace integral transformation to converge for signals that Fourier transform does not, e.g. a unitary step (Heaviside function). If you are working with real ...


9

The Fourier transform is linear, which means that if you sum two signals, then also their spectrum will be the sum. $$ \mathscr{F} [x(t) + y(t)] = \mathscr{F}[x(t)] + \mathscr{F}[y(t)] $$ If you consider the power, you can have two cases: The signals have different frequency Then their power just sums up, because: $$ P_{tot} = P_1 cos (\omega _1 t + \phi ...


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Good answers abound here, but it really depends on your interpretation of "can be represented by". One has to understand that a triangle wave is a theoretical mathematical construct that can not actually exist in reality. Mathematically speaking, in order to get a pure triangle wave you would need an infinite number of harmonic sine-waves, but to get a ...


8

Is the signal split by many narrow band pass filters in paralell? Are these filters RLC circuits? Are there chips that contains many such filters in paralell? No, these are typically collected as time signal, and then transformed digitally (by a DFT implemented by an FFT) to discrete Fourier domain for processing reasons. In the sense that yes, there's ...


7

In the specific example you gave, the square wave is much louder; mostly for physiological reasons. 1) Into the same load, a square wave will deliver twice the power as a sine wave of the same peak voltage. This is same as saying the square wave has an RMS voltage equal to its peak value; whereas a sinewave has an RMS value of 0.707 (actually sqrt(2)/2) ...


7

The derivative - rate of change - of a sinusoid is another sinusoid at the same frequency, but phase-shifted. Real components - wires, antennas, capacitors - can follow the changes (of voltage, current, field-strength, etc.) of the derivatives as well as they can follow the original signal. The rates of change of the signal, of the rate-of-change of the ...


7

Probability theory is very relevant to Electrical Engineering -- in fact several universities have courses just on this topic, such as University of Utah's ECE 3530/CS 3130 - Engineering Probability and Statistics and USC's EE503 Probability for Electrical and Computer Engineers. Take a look at the syllabi for these courses and you will get a feel for the ...


7

Trig theory to extract the phase & magnitude. If you only care about phase... Say you have a signal \$Acos(\omega t + \phi)\$ and you want to extract \$\phi\$. You can use an oscillator of the same frequency to extract this info BUT the issue is the phase. \$V_{sig} = Acos(\omega t + \phi)\$ \$V_{osc} = cos(\omega t)\$ \$V_{sig}V_{osc} = Acos(\...


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the individual parameters on an analog synthesizer could be controlled digitally. OK so far. instead of using mechanical potentiometers to adjust some quality of the waveform (eg the frequency of a LFO), use digital potentiometers. It seems you already know what you want the frequency, amplitude, and phase of each component to be in digital form. If you ...


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No, perfect mathematical square waves don't exist in real world because square wave is not a continuous function (it does not have a derivative at the step). Therefore you can only approximate a square wave and the approximation does have very high frequencies, and at some point the antenna would not be able to send these so it would be a low-pass filter.


6

The difference is that the digital Fourier transform (and FFT as well) gives a vector of size N (or M in some cases) that contains sums of N samples. So, basically, each point of the FFT transform is the result of a sum over a certain time interval of the time-based samples. That's why you divide by N. You can consider it this way: you take an interval of ...


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