#### Praat spectra analysis

Afia Nyarko <nyarkoafia2910@...>

Hi all,
Can anyone help me with this? I keep getting different answers.
What is the interval between the spectral components in an FFT spectrum if the sampling rate is 44100 Hz and the window is 12 ms long? If the window is 120 ms long? If the window contains 4096 samples?

Afia

Bogdan Rozborski

W dniu 23.03.2023 o 05:26, Afia Nyarko pisze:
Hi all,
Can anyone help me with this? I keep getting different answers.
What is the interval between the spectral components in an FFT spectrum if the sampling rate is 44100 Hz and the window is 12 ms long? If the window is 120 ms long? If the window contains 4096 samples?

Afia

If sampling frequency of your signal is 44100 Hz, then window length for 4096 samples of this signal is 0.09287979 s (1 / 44100  Hz * 4096 samples) So, it is neither 12 ms, nor 120 ms. Using FFT (option "Fast" checked) to calculate spectrum of 4096 samples produces bin width of 1.0009765625 Hz (44100 / 4096). If you take 12 ms of signal sampled @ 44100 Hz, you obtain 529,2 samples. For spectrum calculation with option "Fast" set (using FFT instead of DFT), you obtain spectrum calculated for 1024 samples (the closest higher power of two for 529 samples is 1024). Bin width for 1024 samples signal @ 44100 Hz sampling frequency is 43,06640625 Hz. By analogy, bin width for signal of 120 ms long (using FFT) will be 44100 Hz / 65536 samples which gives 0.6729125977 Hz.

Best, Bogdan.

Afia Nyarko <nyarkoafia2910@...>

Thank you so much Bogdan.

On Mon, Mar 27, 2023 at 03:49 Bogdan Rozborski <b.rozborski@...> wrote:
W dniu 23.03.2023 o 05:26, Afia Nyarko pisze:
Hi all,
Can anyone help me with this? I keep getting different answers.
What is the interval between the spectral components in an FFT spectrum if the sampling rate is 44100 Hz and the window is 12 ms long? If the window is 120 ms long? If the window contains 4096 samples?

Afia

If sampling frequency of your signal is 44100 Hz, then window length for 4096 samples of this signal is 0.09287979 s (1 / 44100  Hz * 4096 samples) So, it is neither 12 ms, nor 120 ms. Using FFT (option "Fast" checked) to calculate spectrum of 4096 samples produces bin width of 1.0009765625 Hz (44100 / 4096). If you take 12 ms of signal sampled @ 44100 Hz, you obtain 529,2 samples. For spectrum calculation with option "Fast" set (using FFT instead of DFT), you obtain spectrum calculated for 1024 samples (the closest higher power of two for 529 samples is 1024). Bin width for 1024 samples signal @ 44100 Hz sampling frequency is 43,06640625 Hz. By analogy, bin width for signal of 120 ms long (using FFT) will be 44100 Hz / 65536 samples which gives 0.6729125977 Hz.

Best, Bogdan.

Boersma Paul

The bin width for a window of 120 ms will be 44100 Hz / 8192 = 5.38330078125 Hz.

That's all done with FFT ("fast"), which is fast implementation of the DFT. With "fast" switched off, you'll get a slow implementation of the DFT, with a predictable bin width of 83.3648393194707 Hz and 8.333333333333334 Hz, for 12 and 120 ms, respectively. This is always close to 1 divided by the duration, independently of the sampling frequency. This works perfectly for 120 ms (1/0.120 = 8.333333333333334 Hz), but for 12 ms we have to round the duration of the window to 529 samples, so you get 1/0.012 * 529.2/529 Hz for the bin width.

On 27 Mar 2023, at 09:49, Bogdan Rozborski via groups.io <b.rozborski@...> wrote:

W dniu 23.03.2023 o 05:26, Afia Nyarko pisze:
Hi all,
Can anyone help me with this? I keep getting different answers.
What is the interval between the spectral components in an FFT spectrum if the sampling rate is 44100 Hz and the window is 12 ms long? If the window is 120 ms long? If the window contains 4096 samples?

Afia

If sampling frequency of your signal is 44100 Hz, then window length for 4096 samples of this signal is 0.09287979 s (1 / 44100  Hz * 4096 samples) So, it is neither 12 ms, nor 120 ms. Using FFT (option "Fast" checked) to calculate spectrum of 4096 samples produces bin width of 1.0009765625 Hz (44100 / 4096). If you take 12 ms of signal sampled @ 44100 Hz, you obtain 529,2 samples. For spectrum calculation with option "Fast" set (using FFT instead of DFT), you obtain spectrum calculated for 1024 samples (the closest higher power of two for 529 samples is 1024). Bin width for 1024 samples signal @ 44100 Hz sampling frequency is 43,06640625 Hz. By analogy, bin width for signal of 120 ms long (using FFT) will be 44100 Hz / 65536 samples which gives 0.6729125977 Hz.

Best, Bogdan.

_____

Paul Boersma
Professor of Phonetic Sciences
University of Amsterdam
Spuistraat 134, room 632
1012VB Amsterdam, The Netherlands
http://www.fon.hum.uva.nl/paul/

Bogdan Rozborski

W dniu 27.03.2023 o 19:27, Boersma Paul via groups.io pisze:
The bin width for a window of 120 ms will be 44100 Hz / 8192 = 5.38330078125 Hz.

That's all done with FFT ("fast"), which is fast implementation of the DFT. With "fast" switched off, you'll get a slow implementation of the DFT, with a predictable bin width of 83.3648393194707 Hz and 8.333333333333334 Hz, for 12 and 120 ms, respectively. This is always close to 1 divided by the duration, independently of the sampling frequency. This works perfectly for 120 ms (1/0.120 = 8.333333333333334 Hz), but for 12 ms we have to round the duration of the window to 529 samples, so you get 1/0.012 * 529.2/529 Hz for the bin width.

On 27 Mar 2023, at 09:49, Bogdan Rozborski via groups.io <b.rozborski@...> wrote:

W dniu 23.03.2023 o 05:26, Afia Nyarko pisze:
Hi all,
Can anyone help me with this? I keep getting different answers.
What is the interval between the spectral components in an FFT spectrum if the sampling rate is 44100 Hz and the window is 12 ms long? If the window is 120 ms long? If the window contains 4096 samples?

Afia

If sampling frequency of your signal is 44100 Hz, then window length for 4096 samples of this signal is 0.09287979 s (1 / 44100  Hz * 4096 samples) So, it is neither 12 ms, nor 120 ms. Using FFT (option "Fast" checked) to calculate spectrum of 4096 samples produces bin width of 1.0009765625 Hz (44100 / 4096). If you take 12 ms of signal sampled @ 44100 Hz, you obtain 529,2 samples. For spectrum calculation with option "Fast" set (using FFT instead of DFT), you obtain spectrum calculated for 1024 samples (the closest higher power of two for 529 samples is 1024). Bin width for 1024 samples signal @ 44100 Hz sampling frequency is 43,06640625 Hz. By analogy, bin width for signal of 120 ms long (using FFT) will be 44100 Hz / 65536 samples which gives 0.6729125977 Hz.

Best, Bogdan.

_____

Paul Boersma
Professor of Phonetic Sciences
University of Amsterdam
Spuistraat 134, room 632
1012VB Amsterdam, The Netherlands
http://www.fon.hum.uva.nl/paul/

Ups,  FFT length for 120 ms signal @ 44100 Hz sampling frequency is 8192 not 65536!

Afia Nyarko <nyarkoafia2910@...>

Thank you all but I am wondering why everyone is getting different answers?

On Mon, Mar 27, 2023 at 17:36 Bogdan Rozborski <b.rozborski@...> wrote:
W dniu 27.03.2023 o 19:27, Boersma Paul via groups.io pisze:
The bin width for a window of 120 ms will be 44100 Hz / 8192 = 5.38330078125 Hz.

That's all done with FFT ("fast"), which is fast implementation of the DFT. With "fast" switched off, you'll get a slow implementation of the DFT, with a predictable bin width of 83.3648393194707 Hz and 8.333333333333334 Hz, for 12 and 120 ms, respectively. This is always close to 1 divided by the duration, independently of the sampling frequency. This works perfectly for 120 ms (1/0.120 = 8.333333333333334 Hz), but for 12 ms we have to round the duration of the window to 529 samples, so you get 1/0.012 * 529.2/529 Hz for the bin width.

On 27 Mar 2023, at 09:49, Bogdan Rozborski via groups.io <b.rozborski@...> wrote:

W dniu 23.03.2023 o 05:26, Afia Nyarko pisze:
Hi all,
Can anyone help me with this? I keep getting different answers.
What is the interval between the spectral components in an FFT spectrum if the sampling rate is 44100 Hz and the window is 12 ms long? If the window is 120 ms long? If the window contains 4096 samples?

Afia

If sampling frequency of your signal is 44100 Hz, then window length for 4096 samples of this signal is 0.09287979 s (1 / 44100  Hz * 4096 samples) So, it is neither 12 ms, nor 120 ms. Using FFT (option "Fast" checked) to calculate spectrum of 4096 samples produces bin width of 1.0009765625 Hz (44100 / 4096). If you take 12 ms of signal sampled @ 44100 Hz, you obtain 529,2 samples. For spectrum calculation with option "Fast" set (using FFT instead of DFT), you obtain spectrum calculated for 1024 samples (the closest higher power of two for 529 samples is 1024). Bin width for 1024 samples signal @ 44100 Hz sampling frequency is 43,06640625 Hz. By analogy, bin width for signal of 120 ms long (using FFT) will be 44100 Hz / 65536 samples which gives 0.6729125977 Hz.

Best, Bogdan.

_____

Paul Boersma
Professor of Phonetic Sciences

Ups,  FFT length for 120 ms signal @ 44100 Hz sampling frequency is 8192 not 65536!

Henning Reetz

All gave the same answered – only Bogdan grabbed once a wrong value… Paul has the most detailed answer.

On 27 Mar 2023, at 23:31, Afia Nyarko <nyarkoafia2910@...> wrote:

Thank you all but I am wondering why everyone is getting different answers?

On Mon, Mar 27, 2023 at 17:36 Bogdan Rozborski <b.rozborski@...> wrote:
W dniu 27.03.2023 o 19:27, Boersma Paul via groups.io pisze:
The bin width for a window of 120 ms will be 44100 Hz / 8192 = 5.38330078125 Hz.

That's all done with FFT ("fast"), which is fast implementation of the DFT. With "fast" switched off, you'll get a slow implementation of the DFT, with a predictable bin width of 83.3648393194707 Hz and 8.333333333333334 Hz, for 12 and 120 ms, respectively. This is always close to 1 divided by the duration, independently of the sampling frequency. This works perfectly for 120 ms (1/0.120 = 8.333333333333334 Hz), but for 12 ms we have to round the duration of the window to 529 samples, so you get 1/0.012 * 529.2/529 Hz for the bin width.

On 27 Mar 2023, at 09:49, Bogdan Rozborski via groups.io <b.rozborski@...> wrote:

W dniu 23.03.2023 o 05:26, Afia Nyarko pisze:
Hi all,
Can anyone help me with this? I keep getting different answers.
What is the interval between the spectral components in an FFT spectrum if the sampling rate is 44100 Hz and the window is 12 ms long? If the window is 120 ms long? If the window contains 4096 samples?

Afia

If sampling frequency of your signal is 44100 Hz, then window length for 4096 samples of this signal is 0.09287979 s (1 / 44100  Hz * 4096 samples) So, it is neither 12 ms, nor 120 ms. Using FFT (option "Fast" checked) to calculate spectrum of 4096 samples produces bin width of 1.0009765625 Hz (44100 / 4096). If you take 12 ms of signal sampled @ 44100 Hz, you obtain 529,2 samples. For spectrum calculation with option "Fast" set (using FFT instead of DFT), you obtain spectrum calculated for 1024 samples (the closest higher power of two for 529 samples is 1024). Bin width for 1024 samples signal @ 44100 Hz sampling frequency is 43,06640625 Hz. By analogy, bin width for signal of 120 ms long (using FFT) will be 44100 Hz / 65536 samples which gives 0.6729125977 Hz.

Best, Bogdan.

_____

Paul Boersma
Professor of Phonetic Sciences

Ups,  FFT length for 120 ms signal @ 44100 Hz sampling frequency is 8192 not 65536!

Bogdan Rozborski

W dniu 28.03.2023 o 00:31, Afia Nyarko pisze:
Thank you all but I am wondering why everyone is getting different answers?

On Mon, Mar 27, 2023 at 17:36 Bogdan Rozborski <b.rozborski@...> wrote:
W dniu 27.03.2023 o 19:27, Boersma Paul via groups.io pisze:
The bin width for a window of 120 ms will be 44100 Hz / 8192 = 5.38330078125 Hz.

That's all done with FFT ("fast"), which is fast implementation of the DFT. With "fast" switched off, you'll get a slow implementation of the DFT, with a predictable bin width of 83.3648393194707 Hz and 8.333333333333334 Hz, for 12 and 120 ms, respectively. This is always close to 1 divided by the duration, independently of the sampling frequency. This works perfectly for 120 ms (1/0.120 = 8.333333333333334 Hz), but for 12 ms we have to round the duration of the window to 529 samples, so you get 1/0.012 * 529.2/529 Hz for the bin width.

On 27 Mar 2023, at 09:49, Bogdan Rozborski via groups.io <b.rozborski@...> wrote:

W dniu 23.03.2023 o 05:26, Afia Nyarko pisze:
Hi all,
Can anyone help me with this? I keep getting different answers.
What is the interval between the spectral components in an FFT spectrum if the sampling rate is 44100 Hz and the window is 12 ms long? If the window is 120 ms long? If the window contains 4096 samples?

Afia

If sampling frequency of your signal is 44100 Hz, then window length for 4096 samples of this signal is 0.09287979 s (1 / 44100  Hz * 4096 samples) So, it is neither 12 ms, nor 120 ms. Using FFT (option "Fast" checked) to calculate spectrum of 4096 samples produces bin width of 1.0009765625 Hz (44100 / 4096). If you take 12 ms of signal sampled @ 44100 Hz, you obtain 529,2 samples. For spectrum calculation with option "Fast" set (using FFT instead of DFT), you obtain spectrum calculated for 1024 samples (the closest higher power of two for 529 samples is 1024). Bin width for 1024 samples signal @ 44100 Hz sampling frequency is 43,06640625 Hz. By analogy, bin width for signal of 120 ms long (using FFT) will be 44100 Hz / 65536 samples which gives 0.6729125977 Hz.

Best, Bogdan.

_____

Paul Boersma
Professor of Phonetic Sciences

Ups,  FFT length for 120 ms signal @ 44100 Hz sampling frequency is 8192 not 65536!

It all depends whether you use FFT or DFT algorithm. I calculated bin width for "fast" option using an FFT algorithm. Paul also included calculations for DFT ("slower" option). \

Bogdan.

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