New answers tagged frequency-response
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It looks like the idea of that exercise is to compute the Fourier transform of $x_3(t)$, and then from that derive the Fourier transform of the original function $x_1(t)$.
Note that $x_3(t)$ is just a sum of scaled and shifted Dirac impulses, so computing its Fourier transform is trivial. The Fourier transform of $x_2(t)$ is then easily derived by ...
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Check this video for detail explanation-->
Frequency Response of Chirp Signal -Eye of Sauron(The Lord of the Rings)
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I agree with @user28715 answer. The best method is to apply a filter to your timeseries to get calibrated timeseries.
Filter
You did not specify which language you are using, but in Matlab I use the designfilt function. https://www.mathworks.com/help/signal/ref/designfilt.html
d = designfilt('arbmagfir',...);
a = 1;
b = d.Coefficients;
or you ...
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I am trying to do the same. My idea is to apply a digital filter based on the sensitivity/frequency curve, in order to calibrate my signal. Have you done this successfully?
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What is the link between them? Power should not be function of
frequency (it is the integral on all frequencies of power spectral
density)
As far as a Spectrum Analyzer is concerned, to display the power spectrum in dBm, what it does it to take the FFT and compute square of magnitude $P_n = 20 \times log_{10}(|X(f_n)|^2*1000/100)$ with $f_n$ being the ...
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In addition to the resolution bandwidth fact, i think power spectral density has something to do with stochastic signals. If your signal is predetermined, you have to work with power instead of power spectral density. One good example is the very sine wave you mentioned. The mean power for that signal in its main frequency can be defined using the integral ...
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These terms refer to the general duality properties of these two domains. It should be obvious that higher frequency components mean faster changes in time and lower rise/fall times. In most natural systems, some of these fundamental properties oppose each other:
Shorter rise time means more overshoot and higher settling times and vice versa; lower ...
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These are all frequency selective filters. Note the zeros on the unit circle, which correspond to zeros of the frequency response. The filter's passband is determined by range of the angles of the poles, and the stopband is determined by the range of the angles of the zeros.
As an example, let's consider PZ-diagram $\#1$: since the stopband is around DC (i....
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The key is to apply 'shift and scale'. So for $x(-2t + 4)$, you would first do $x_1(t) = x(t+4)$, then $x_2(t) = x_1(-2t) = x(-2t + 4)$.
So $X_1(f) = e^{-j2\pi f(-4)}X(f)$ = $e^{j8\pi f}X(f)$.
Then $X_2(f) = \frac{1}{|-2|}X_1(f/-2)$ = $\frac{1}{2}e^{j-4\pi f}X(f/-2)$
In your answer how did you assume it is symmetric? Also, in your third step, for $x(-t)$...
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