Sujet : Re: Omega
De : cd (at) *nospam* notformail.com (Cursitor Doom)
Groupes : sci.electronics.designDate : 06. Jul 2024, 00:31:25
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <3h0h8jpnnhusft850ht68t5es8op7cgkc7@4ax.com>
References : 1 2
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On Fri, 05 Jul 2024 00:33:31 +0100, JM <
sunaecoNoSpam@gmail.com>
wrote:
On Sun, 30 Jun 2024 08:44:34 +0100, Cursitor Doom <cd@notformail.com> wrote:
>
Gentlemen,
>
For more decades than I care to remember, I've been using formulae
such as Xc= 1/2pifL, Xl=2pifC, Fo=1/2pisqrtLC and such like without
even giving a thought as to how omega gets involved in so many aspects
of RF. BTW, that's a lower-case, small omega meaning
2*pi*the-frequency-of-interest rather than the large Omega which is
already reserved for Ohms. How does it keep cropping up? What's so
special about the constant 6.283 and from what is it derived?
Just curious...
>
What you are really asking (but may not realise) is why do mathematicians measure an angle in radians. If you draw two straught lines from the origin of a circle to its edge, then the angle (in radians) between the lines is the arc length (length of the part of the circle circumference between the two lines) divided by the radius. So if a single line rotates one complete revolution it rotates 2.PI radians.
>
One important thing to note is that if an angle x is measured in radians then the limit of sin(x)/x as x goes to 0 is 1. This leads to the derivative (wrt x) of sin(x) being equal to cos(x). *This is not true is x is not specified in radians*. (It also leads to simple series expansions of the trigonometric fuctions, to eulers formula etc.)
>
When performing calculus it is thus easiest to do so if angles are measured in radians. For a signal of frequency f it's corresponding phasor representation will rotate by 2.PI.f.t radians in an interval t. This is where the 2.PI comes from in your calculations.
>
For example for your inductor L the voltage across it and the current through it are related by e=L.di/dt. If the current i=I.sin(wt) (w in rad/s) then e=wLI.cos(wt). Thus the impedance e/i is wL cos(wt)/sin(wt) ie. it has a magnitude of wL (or 2.PI.f.L )and a phase angle of PI/2. And so forth.
Another piece of the jigsaw. Many thanks indeed for that contribution.