Timeline for Bandwidth visualization in frequency domain
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 7, 2022 at 15:52 | comment | added | Curious | ok, but in this case 15 bins will have bandwidth $2f_s/32=f_s/16$, isn't it? is it correct? | |
| May 7, 2022 at 15:27 | comment | added | MBaz | A two-sided analysis gives the same result: 32 bins, each of width $f_s/32$, gives a total bandwidth $f_s$. | |
| May 7, 2022 at 15:25 | comment | added | MBaz | He's assuming that the signal is real; 15 DFT samples are redundant, leaving 17 samples. These 17 samples define 17 bins. 15 of the bins have width $2f_s/32=f_s/16$ and the other two have width $f_s/32$. The total bandwidth is $15*2f_s/32 + 2*f_s/32 = 32f_s/32 = f_s$. | |
| May 7, 2022 at 15:06 | comment | added | Curious | but how could it be? in the text he considers "32-point signal", so the first dashed line corresponds to $1/N$, where $N=32$ otherwise the signal will have different sampling... | |
| May 7, 2022 at 14:55 | comment | added | MBaz | It's still $f_s$. He's only considering bins from 0 to $f_s/2$. | |
| May 7, 2022 at 12:35 | comment | added | Curious | that's clear now, thank you for explanation!) but here the author states that the bandwidth is $2f_s$, and I have no idea why... | |
| May 6, 2022 at 21:03 | comment | added | MBaz | Maybe this analogy will help. A signal is sampled 10 times, once per second, i.e. at times $0,1,2,3,\ldots,9$. Each sample can be thought of as "representative" of the signal during a time interval with duration one second, right?. In the DFT, each sample is "representative" of the actual spectrum, in a frequency interval with width $f_s/N$. | |
| May 6, 2022 at 20:22 | comment | added | Curious | I thought that $f_s/N$ is the center of the bin not a bandwidth; if $f_s/N$ is bandwidth, where is the center of the bin? $f_s/2N$? | |
| May 6, 2022 at 20:09 | comment | added | Curious |
bandwidth per sample - it is exactly what I needed! Why did I ask this question - because I'm trying to compare time domain signal amplitude and frequency domain signal amplitude, I use Python function np.fft.fft for calculating DFT and it calculates spectral density instead of sine/cosine amplitudes; dividing each sample (except first and last) by $N/2$ gives "correct" sine/cosine amplitudes for calculating inverse DFT.
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| May 6, 2022 at 13:36 | history | answered | MBaz | CC BY-SA 4.0 |