what is magnitude of amplitude in real world?

in matlab

fs=1000;
fv=200
amplitude=0.1
t= 0 : 1/fs :10
y= amplitude * sin (2.*pi.*fv*t)
plot(t,y)

in this case

i convert mat file to wav file

so i need to know what is magnitude of amplitude in real world

in matlab, amplitude is 0.7,

In realworld, 0.7 means 0.7m or something?

[KSSV]

It depends on what context you are considering the amplitude and frequency.

[규호]

thank you

Answer 1

[Walter Roberson]

audio files do not store absolute units of anything. They store relative values, like 0.7 intended to mean 0.7 of the signal maximum.

The same audio file would be used for headphone output as would be used for speaker output, but headphone line voltage range is smaller than is used for speakers. Furthermore, the line levels used for professional equipment are higher than for consumer equipment.

If you want to indicate that your speaker cone should move forward to 70% of its maximum extension past neutral, and then back to 30% of its maximum towards the rear, behind neutral, then the audio file would have 0.7 -0.3 . The audio file would not contain "0.35 mm if you are using earbuds, 1.4mm if you connected to over-the-ear headphones, 4.2 mm if you are using the Den speakers with volume turned up to 6, 4.4 mm if you are using the Den speakers with the volume turned up to 7, 109mm if you are listening on the club loudspeaker" and so on. Absolute units in an audio file would need to be specific to one particular playback system with one particular volume setting.

Could a file contain absolute units like these? Yes: it is possible to calibrate a system with fixed settings. But you would use a data file for the purpose, not an audio file.

You might possibly be thinking about Sound Pressure Level.

The samples in an audio file do not refer to sound pressure. They refer to relative extensions of the speaker cone. If you extend a speaker cone as far as it goes and leave it there, then there is no sound during the time it is held extended! Sound is vibration which is due to changes. Sound pressure relates to relative changes in samples, not the samples themselves.

Sound pressure can be calculated by looking at the relative changes in samples, with respect to calibrated equipment. But the audio file doesn't store pressure levels, and the audio file doesn't know what equipment the file will be used with or how loud it will be turned up.

The loudest speakers in the world that I recall hearing about were made out of literal concrete. It takes a lot of energy to make a piece of concrete move back and forth quickly, but if you invest the energy then you can get them to move a lot of air. The audio file doesn't know anything about that.

Answer 2

[Sulaymon Eshkabilov]

Here in your exercise the amplitude is ~0.1 - see e.g.:

fs=1000; % 1kHz
fv=200;
amplitude=0.1;
t= 0 : 1/fs :10;
y= amplitude * sin (2.*pi.*fv*t);
Yamp = max(abs(y))

Yamp = 0.0951

Note that by icreasing the smapling rate (frequency), the amplitude will be almost 0.1 as assigned:

fs=10000;  % 10 kHz
fv=200;
amplitude=0.1;
t= 0 : 1/fs :10;
y= amplitude * sin (2.*pi.*fv*t);
Yamp = max(abs(y))

Yamp = 0.0998

[규호]

Thank you

[Sulaymon Eshkabilov]

Most welcome!