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5

No, because this is a sufficient condition (for regularly sampled signals), and not a necessary one. This condition restricts the space of all possible continuous signals to a subspace of discrete sequences that contain the same information. Suppose that you can constrain the signal space, eg limited band-width, positivity, parametric models, sparsity, etc....


1

As a simple example, consider the discrete-time signal $$x[n]=a^nu[n],\qquad |a|<1\tag{1}$$ where $u[n]$ is the unit step function. The discrete-time Fourier transform (DTFT) of $x[n]$ is given by $$X(e^{j\omega})=\frac{1}{1-ae^{-j\omega t}}\tag{2}$$ Note that $\omega$ is a normalized angular frequency: $$\omega=2\pi f/f_s\tag{3}$$ where $f_s$ is ...


1

I think you consider that pulse in the spectrum as the noise. Then you should remove it by a lowpass filter of cutoff frequency $$ f_c = 5 ~\text{kHz.}$$ Slightly less than this should be used for guarding against the transition width of the lowpass filter. So you better select something like $4.8$ kHz for the lowpass filter cutoff frequency. Note that ...


0

Yes high frequency signals are harder to deal with due to their small wavelengths resulting in non-lumped circuit behaviour. But even circuits operating at 2.4 GHz (such as a wi-fi receiver-transmitter) are being analysed by properly designed oscilloscopes. 1 GHz is even within the range of terrestrial TV broadcast signals and ordinary oscilloscopes with ...


2

Assuming that the relevant portion of a continuous-time signal $x(t)$ is inside (or has been shifted to) the interval $[0,T]$, the DFT of a sampled version of the signal approximates the continuous-time Fourier transform (CTFT) in the following way: $$\begin{align}X(f)&=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt\\&\stackrel{\textrm{truncation}}{\...


2

Not sure to understand what you are asking, Microwaves wavelengths range between 300 MHz (1 m) and 300 GHz (1 mm) which is indeed of the same order of magnitude as the probe. 1GHz corresponds to a wavelengths of about 30 cm and is also in the Microwave range. Are you saying that it is possible to measure a frequency of 1GHz with a probe while it is ...


3

My first inclination is to say this is a meaningless question. The concept of "instantaneous" frequency really only pertains to a single pure tone with a slightly varying frequency. In this light, one may construct a definition saying "The instantaneous frequency at time $t$ is the same as that of a pure tone which matches the function (sum) in the ...


0

From an audio point of view, I would determine if the reciprocal of the least common multiple of a1,a2,a2 were within the range perceivable as pitch to a human; and, if so and stationary for about 6 periods or more, call that reciprocal the instantaneous pitch frequency.


0

A rectangular window of $N = 4$ samples, at a sampling rate of $32$ Hz would provide about $$ \Delta f = \frac{4\pi}{4} \frac{32}{2\pi} = 16 ~Hz$$ of frequency resolution... On the other hand, the apparent DFT bin frequencies, will be : $$ f_k = [0,8,16,24] $$ Hz. Interpret the last frequency as $24-32 = -8$ Hz due to the sampling theorem.


3

Eventhough Laurent has given a broader sense of the answer, let me put here the communications theory sense ot it. The concept of instantaneous freqency emerges when you consider Frequency Modulation or Phase Modulation systems, where the message is embedded into the change of the frequency or phase of a carrier signal. This carrier is typically a single ...


2

The notion of instantaneous frequency is (hopefully) consistent with the monochromatic wave model: $$x(t)=a \cos 2\pi \nu t\,,$$ where $a$ is the amplitude and $ \nu$ the frequency. It would be tempting to compute a similar formula for evolving amplitude and frequency cases, something like: $$x(t)=a(t) \cos \left(\phi(t)\right)\,.$$ However, this is not ...


1

Carry the units and it becomes understandable. T = 0.03125 seconds per sample Thus your sampling rate is 1/T = 32 samples per second Each bin corresponds to the cycles per frame. Your frame has four samples. Suppose the bin index is k. k (cycles per frame) * 32 (samples per second) / 4 (samples per frame) = 8k cycles per second = 8k Hz So the ...


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