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1

The significance is a constant time delay for all frequencies. Time delay is the derivative of phase with respect to frequency, so given a linear phase as shown, the time delay is constant. Notice that the abrupt steps in phase actually only occur when the magnitude goes through zero so does not effect the nature of it being a "linear phase" filter....


1

It must be added to the problem that $R(\omega)$ is a real-valued, possibly bipolar function. In that case, its inverse discrete-time Fourier transform must be even: $$r[n]=r[-n]\tag{1}$$ From the given relation between $H(e^{j\omega})$ and $R(\omega)$ it is clear that $$h[n]=r[n-25]\tag{2}$$ must hold. I'm sure that you'll manage to combine $(1)$ and $(2)$ ...


0

The short answer to "why do my DFT responses differ?" is that you have a chirp with a very small time-bandwidth product. Chirps that last longer in time, or sweep over a larger frequency range, become closer to an ideal chirp. (from https://en.wikipedia.org/wiki/Chirp_spectrum, though the page is also somewhat confusing overall) The shape of the ...


1

If you have a filter impulse response represented by H(z), replacing z with z^2 is equivalent to inserting a zero-valued sample in between each original impulse response sample. Such an operation compresses the frequency response of your original filter by a factor of two.


1

It is just like ordinary functions manipulation. You have to remember that z and the frequency are linked with $z = exp(j \omega \cdot T)$ $H(z^2)$ you are doing something equivalent to $H(2 \omega)$ on the continuous frequency domain, i.e. you are narrowing the filter band, and thus stretching its time response. On the second example, you have to remember ...


1

If it's not obvious from the other three (at this writing) answers: scale it to match he problem at hand. All of the three suggested scalings so far (energy = 1, DC gain = 1, maximum coefficient = full scale for your data type, then scale the output) are valid in different circumstances. And don't sweat over trying to find a universal "correct" ...


2

The gain is completely arbitrary and you can scale it as desired for the overall receiver or transmitter design. Where special attention must be paid is with fixed point design where the best practice is to let the filters grow the signal- do not scale the coefficients or the input as that only introduces more quantization noise and degrades SNR. Let the ...


1

I recommand to have the input and the output of the filter at same level (resampling included). This means a gain of 1 in linear or 0 dB. It is consistant and useful for reuse. The best way to preliminary verify for the gain on a low pass filter is to inject a DC constant signal. Usualy, I tune the taps level to compensate the fractional part of the gain. ...


1

My recommendation is to normalize the filter impulse response to have energy equal to 1. In continuous time, the digital communication system will transmit a symbol $a_i$ using the pulse $$s(t) = a_ip(t),$$ where $p(t)$ is an RRC pulse. The receiver will recover $a_i$ from $s(t)$ using a matched filter: \begin{align} a_i &= \int_{-\infty}^\infty a_i p(t)...


4

This is unfortunately a difficult problem. A microphone doesn't have a single frequency response. It has a different one for each direction of incidence. Especially for second order microphones (dipoles, cardioids, etc.), the frequency response also depends heavily on the distance and radiation impedance of the source. So first you need to decide: what ...


0

In the OFDM system, such as Wi-Fi, there are infinite paths exist, assume it is in free space, and only one path exist, Exactly. So if you use OFDM in a scenario where FSPL describes your transmission sufficiently, then you're doing something wrong, because you don't need OFDM, as you have a flat channel. We can still describe large scale fading in ...


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